Answer:
0.000273 M
Explanation:
Henry's states that at constant temperature the amount of a gas that dissolves in a liquid is directly proportional to the partial pressure in of that gas in equilibrium with that liquid.
Pressure of Oxygen = mole fraction of Oxygen × 1.00 atm
Mole fraction Oxygen = 21/100 × 1.00atm = 0.21 atm
Molar solubility of Oxygen = KH × PO2 = 0.0013 × 0.21 = 0.000273 M
The amount of a gas that dissolves in a liquid is proportional to the partial pressure of the gas above the liquid. Solubility of oxygen in water exposed to air at 1.00 atm is 0.000273 M.
Henry's Law:
It states that at constant temperature the amount of a gas that dissolves in a liquid is proportional to the partial pressure of the gas above the liquid.
C = k P
Where,
C = concentration of a dissolved gas
k = Henry's Law constant = 0.0013 M/atm.
P = partial pressure of the gas = [tex]\bold {\dfrac {21}{100} \times 1.00\ atm = 0.21\ atm}[/tex]
Put the values in the formula,
[tex]\bold {C = 0.0013 \times 0.21 = 0.000273\ M }[/tex]
Therefore, solubility of oxygen in water exposed to air at 1.00 atm is 0.000273 M.
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The pH of a 0.23M solution of acrylic acid HC3H3CO2 is measured to be 2.44. Calculate the acid dissociation constant Ka of acrylic acid. Round your answer to 2 significant digits. Step by step would be helpful...
Answer:
Kₐ = 5.7 x 10⁻⁵
Explanation:
The equilbrium for this acid is
HC₃H₃CO₂ + H2O ⇄ H₃O⁺ + C₃H₃CO₂ ⁻ ,
and the equilibrium constant for acrylic acid is given by the expression:
Kₐ = [ H₃O⁺][ C₃H₃CO₂⁻ ] / [ HC₃H₃CO₂ ]
Since the pH of the 0.23 M solution is known , we can calculate [ H₃O⁺].
The ][ C₃H₃CO₂⁻ ] is equal to [ H₃O⁺] from the above equilibria (1:1)
Finally [ HC₃H₃CO₂ ] is known.
pH = - log [ H₃O⁺]
taking antilog to both sides of the equation
10^-pH = [ H₃O⁺]
Substituting
10^-2.44 = [ H₃O⁺] = 3.6 x 10⁻³
[ C₃H₃CO₂⁻ ] = 3.6 x 10⁻³
Kₐ = ( 3.6 x 10⁻³ ) /0 .23 = 5.7 x 10⁻⁵
To find the Ka of acrylic acid, we use the measured pH to find [H+], set up an ICE table for the dissociation equilibrium, and solve for Ka, which is approximately 5.8 x 10^{-5} when rounded to two significant digits.
Explanation:To calculate the acid dissociation constant (Ka) of acrylic acid, we start by using the pH given. The pH is defined as the negative logarithm of the hydrogen ion concentration ([H+]), which can be represented as pH = -log[H+]. From the pH of 2.44, we find the concentration of hydrogen ions:
[H+] = 10^{-pH} = 10^{-2.44} ≈ 3.63 × 10^{-3} M
Given that acrylic acid (HC3H3CO2) is a weak acid and only partially dissociates in a solution, we can represent its dissociation as follows:
HC3H3CO2(aq) ⇌ H+(aq) + C3H3CO2⁻(aq)
The ICE table approach details the initial concentrations, the change in concentrations, and the final concentrations. For acrylic acid, we get:
Initial: [HC3H3CO2] = 0.23 M, [H+] = 0, [C3H3CO2⁻] = 0Change: [HC3H3CO2] decreases by x, [H+] increases by x, [C3H3CO2⁻] increases by xEquilibrium: [HC3H3CO2] = 0.23-x M, [H+] = x , [C3H3CO2⁻] = xSince the pH corresponds to the hydrogen ion concentration at equilibrium, we also know that x (the increase in [H+]) is equal to 3.63 × 10^{-3} M. Now we can calculate the Ka using the equilibrium expressions:
Ka = \frac{[H+][C3H3CO2⁻]}{[HC3H3CO2]}
Assuming x is small compared to the initial concentration, we can simplify [HC3H3CO2] to be approximately equal to 0.23 M:
Ka ≈ \frac{(3.63 × 10^{-3})^2}{0.23}
Thus, Ka ≈ 5.76 × 10^{-5}. Rounded to two significant digits, Ka ≈ 5.8 × 10^{-5}.
This alkene can be prepared via Wittig reaction. Draw structural formulas for the aldehyde and the Wittig reagent.
In a Wittig reaction, the structure of the desired alkene determines the structure of your starting aldehyde and Wittig reagent. The one (aldehyde or ketone) contains the carbonyl group that will form one half of the carbon-carbon double bond in the alkene, and the other (Wittig reagent) provides the rest of the alkene structure.
Explanation:The
Wittig reaction
is a method used in organic chemistry to create carbon-carbon double bonds (alkenes) from carbonyl compounds (such as aldehydes or ketones) and phosphonium ylides (a compound of a phosphorus cation and an organometallic compound). So, if you have a desired alkene you want to produce, you can actually reverse engineer what ingredients you need for this reaction. The carbon skeleton of the aldehyde and the Wittig reagent are directly related to the resulting alkene. The aldehyde or ketone compound will contain the carbonyl group (C=O) which will form one side of the carbon-carbon double bond in the alkene. The remainder of the alkene structure will be derived from the phosphonium ylide, the Wittig reagent. As an example, if you wanted to prepare 1-hexene, you would use hexanal as your aldehyde and methylenetriphenylphosphorane as your Wittig reagent.
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The Wittig Reaction is used to synthesize alkenes from aldehydes using a reagent known as a Wittig reagent. The Wittig reagent reacts with the aldehyde to create an intermediate molecule that eventually forms the alkene after the expulsion of a leaving group.
Explanation:The Wittig reaction is a chemical reaction used to synthesize alkenes from aldehydes or ketones using a triphenyl phosphonium ylide, commonly known as a Wittig reagent.
In the most basic form of this reaction, an aldehyde is converted into an alkene through a series of steps. First, the Wittig reagent is prepared from a phosphonium salt, which when deprotonated by a base, forms a ylide or Wittig reagent. This reagent then reacts with the aldehyde in what's known as a [2+2] cycloaddition to form an oxaphosphetane intermediate, which then undergoes a reaction known as a retro-[2+2] cycloaddition to expel triphenylphosphine oxide, resulting in the formation of the alkene.
Without the exact structure of the desired alkene, it is impossible to provide the exact structures for the aldehyde and Wittig reagent, but hopefully, this general explanation of the reaction can assist you in figuring it out for your specific case.
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Based upon the intermolecular forces present, rank the following substances according to the expected boiling point for the substance.
Rank from highest to lowest boiling point. To rank items as equivalent, overlap them.
HCl, NaCl, N2, H2O
Boiling points of substances are dependent on the strength of their intermolecular forces. NaCl, with the strongest ionic bonds, has the highest boiling point, followed by H2O with hydrogen bonds, then HCl with dipole-dipole interactions, and finally N2 with the weakest van der Waals forces.
Explanation:The subject of this question is
intermolecular forces
and their impact on the boiling point of different substances. Boiling points are determined based on the strength of these forces. The stronger the intermolecular forces, the more energy is needed to break them, leading to a higher boiling point.
NaCl
, sodium chloride, has ionic bonds, which are the strongest of the intermolecular forces, so it has the highest boiling point.
H2O
, water, has hydrogen bonds, which are the next strongest, so it comes next.
HCl
, hydrogen chloride, has dipole-dipole interactions, weaker than ionic or hydrogen bonding, hence a lower boiling point than NaCl or H2O. Finally,
N2
, nitrogen, with its weakest van der Waals forces has the lowest boiling point. So the order from highest to lowest boiling point is: NaCl, H2O, HCl, N2.
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A balance measures mass to 0.001 g. If you determine the mass of an object that weighs about 30 g, would you record the mass as 30 g, 32 g, 32.1 g, or 32.075 g? Explain your choice by writing one or two complete sentences that describe your thinking.
Answer:
The mass is recorded as 32.075 g
Explanation:
"The first digit of uncertainty is taken as the last significant digit", this is the rule for significant figures in the analysis. The balance measures the mass up to three decimal places, so it makes the most sense to note the whole figure.
Answer:
30g
Explanation:
The true mass of the recorded will be 30g ±0.001g. This figure means that the actual mass recorded lies somewhere between either 29.999g or 30.001g. Both figures could be rounded to two significant figures giving 30g.
Hence the answer written is correct to two significant figures considering the uncertainty in the measurement of the mass.
If a sample of gas is located 10.2 cm from the injection point and the chart speed is 0.5 cm/min, what is the retention time?
Answer : The retention time is, 20 min
Explanation :
Retention time : It is defined as the amount of time a compound spends on the column after it has been injected.
Formula of retention time is:
[tex]\text{Retention time}=\frac{\text{Distance from injection point to center of peaks}}{\text{Chart recorded speed}}[/tex]
Given:
Distance from injection point to center of peaks = 10 cm
Chart recorded speed = 0.5 cm/min
Now put all the given values in the above formula, we get:
[tex]\text{Retention time}=\frac{10cm}{0.5cm/min}[/tex]
Retention time = 20 min
Thus, the retention time is, 20 min
Nitrogen and hydrogen combine to form ammonia in the Haber process. Calculate (in kJ) the standard enthalpy change ∆H° for the reaction written below, using the bond energies given. N2( g) + 3H2( g) → 2NH3( g)
Answer:
∆H° rxn = - 93 kJ
Explanation:
Recall that a change in standard in enthalpy, ∆H°, can be calculated from the inventory of the energies, H, of the bonds broken minus bonds formed (H according to Hess Law.
We need to find in an appropiate reference table the bond energies for all the species in the reactions and then compute the result.
N₂ (g) + 3H₂ (g) ⇒ 2NH₃ (g)
1 N≡N = 1(945 kJ/mol) 3 H-H = 3 (432 kJ/mol) 6 N-H = 6 ( 389 kJ/mol)
∆H° rxn = ∑ H bonds broken - ∑ H bonds formed
∆H° rxn = [ 1(945 kJ) + 3 (432 kJ) ] - [ 6 (389 k J]
∆H° rxn = 2,241 kJ -2334 kJ = -93 kJ
be careful when reading values from the reference table since you will find listed N-N bond energy (single bond), but we have instead a triple bond, N≡N, we have to use this one .
"Synthesis gas" is a mixture of carbon monoxide and water vapor. At high temperature synthesis gas will form carbon dioxide and hydrogen, and in fact this reaction is one of the ways hydrogen is made industrially.
A chemical engineer studying this reaction fills a 200.mL flask at 12.°C with 1.3atm of carbon monoxide gas and 3.6atm of water vapor. He then raises the temperature considerably, and when the mixture has come to equilibrium determines that it contains 0.82atm of carbon monoxide gas, 3.12atm of water vapor and 0.48atm of carbon dioxide.
The engineer then adds another 0.43atm of carbon monoxide, and allows the mixture to come to equilibrium again. Calculate the pressure of hydrogen after equilibrium is reached the second time. Round your answer to 2 significant digits.
Answer:
0.56 atm
Explanation:
The equilibrium occurs when, in a reversible reaction, the velocity of the formation of the products is equal to the velocity of the formation of the reactants. In this scenario, the partial pressures and the concentration of the components remains constant.
The equilibrium can be characterized by the equilibrium constant, which can be calculated by the concentration (Kc), or by the partial pressure (Kp). In the expression of Kc, solids and pure liquids are not put, and in the expression of Kp, only gases are considered.
The constant is calculated by the product of the concentration, or pressure, of the products, elevated by their coefficients, divided by the product of the concentration, or pressure, of the reactants, elevated by their coefficients. Its value only changes with the temperature.
So, for the reaction given:
CO(g) + H₂O(g) ⇄ CO₂(g) + H₂(g)
1.3 atm 3.6 atm 0 0 Initial
-x -x +x +x Reacts (stoichiometry is 1:1:1:1)
1.3-x 3.6-x x x Equilibrium
At equilibrium:
pCO = 1.3 - x = 0.82
x = 1.3 - 0.82 = 0.48 atm
pH₂O = 3.12 atm
pCO₂ = 0.48 atm
pH₂ = 0.48 atm
Thus,
Kp = pCO₂*pH₂/(pCO*pH₂O)
Kp = 0.48*0.48/(0.82*3.12)
Kp = 0.090
When more carbon monoxide (CO) is added, the equilibrium will shift to the right, and more products will be formed, is order to reestablish the equilibrium (Le Chatelier's pricniple), so:
CO(g) + H₂O(g) ⇄ CO₂(g) + H₂(g)
0.82 atm 3.12 atm 0.48 atm 0.48 atm 1st equilibrium
1.25 3.12 0.48 0.48 After the CO addition
-x -x +x +x Reacts
1.25-x 3.12-x 0.48 +x 0.48+x New equilibrium
Because the temperature is the same, Kp = 0.090
0.090 = (0.48+x)*(0.48+x)/[(1.25-x)*(3.12-x)]
0.090 = (0.2304 + 0.96x + x²)/(3.9 - 4.37x + x²)
0.2304 + 0.96x + x² = 0.351 - 0.3933x + 0.09x²
0.91x² + 1.3533x - 0.1206 = 0
Solving this 2nd grade equation at a graphic calculator, for x > 0 and x < 1.25
x = 0.084 atm
pH₂ = 0.48 + 0.084
pH₂ = 0.56 atm
Draw the major organic product in the reaction scheme below. Be sure to clearly show stereochemistry (if applicable).
Answer:
see explanation below
Explanation:
You are missing the reaction scheme, but in picture 1, I found a question very similar to this, and after look into some other pages, I found the same scheme reaction, so I'm gonna work on this one, to show you how to solve it. Hopefully it will be the one you are asking.
According to the reaction scheme, in the first step we have NaNH2/NH3(l). This reactant is used to substract the most acidic hydrogen in the alkine there. In this case, it will substract the hydrogen from the carbon in the triple bond leaving something like this:
R: cyclopentane
R - C ≡ C (-)
Now, in the second step, this new product will experiment a SN2 reaction, and will attack to the CH3 - I forming another alkine as follow:
R - C ≡ C - CH3
Finally in the last step, Na in NH3 are reactants to promvove the hydrogenation of alkines. In this case, it will undergo hydrogenation in the triple bond and will form an alkene:
R - CH = CH - CH3
In picture 2, you have the reaction and mechanism.
This is a question in Organic Chemistry that requires the student to draw the major product of an organic reaction scheme with clear display of stereochemistry if necessary. It entails simplifying larger molecules using line-angle structures and understanding the arrangement of reactants and products in a chemical equation.
Explanation:The subject of this question is in the discipline of
Chemistry
, specifically the area of
Organic Chemistry
. The task at hand is to draw the major product of an organic reaction scheme while demonstrating clear understanding of stereochemistry, if applicable. In Organic Chemistry, skeletal structures or line-angle structures are often employed to simplify the schematic representation of larger molecules. Here, carbon atoms are delineated by each end of a line or bend in a line. Hydrogen atoms, if attached to a carbon, are typically not drawn while other atoms are represented by their elemental symbols. Also, understanding the importance of the arrangement of reactants and products in a chemical equation is fundamental in reaction
stoichiometry
, which allows us to predict the direction and extent of a reaction. Regarding stereochemistry, it is the study of the three-dimensional structure of molecules. It is important when a molecule features chiral centers (carbon atoms connected to four different groups) because the orientation of these groups in space can lead to different products, known as stereoisomerism.
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his is the chemical formula for chromium(III) nitrate: . Calculate the mass percent of oxygen in chromium(III) nitrate. Round your answer to the nearest percentage.
Mass percent is a technique of expressing or defining a concentration or a component in a combination. The mass percent of oxygen in chromium(III) nitrate is approximately 60.5%.
To calculate the mass percent of oxygen in chromium(III) nitrate (Cr(NO₃)₃), we need to determine the molar mass of the compound and the molar mass of the oxygen atoms in the compound.
The molar mass of Cr(NO₃)₃ can be calculated by adding the molar masses of the individual atoms in the compound:
Molar mass of Cr(NO₃)₃ = (1 x molar mass of Cr) + (3 x molar mass of N) + (9 x molar mass of O)
Molar mass of Cr(NO₃)₃ = (1 x 52.00 g/mol) + (3 x 14.01 g/mol) + (9 x 16.00 g/mol)
Molar mass of Cr(NO₃)₃ = 238.03 g/mol
The molar mass of the oxygen atoms in Cr(NO₃)₃ can be calculated by multiplying the molar mass of oxygen by the number of oxygen atoms in the compound:
Molar mass of O in Cr(NO₃)₃ = 16.00 g/mol x 9
Molar mass of O in Cr(NO₃)₃ = 144.00 g/mol
To calculate the mass percent of oxygen in Cr(NO₃)₃, we can use the formula:
The mass percent of oxygen = (mass of O in compound / molar mass of compound) x 100%
Mass percent of oxygen = (144.00 g/mol / 238.03 g/mol) x 100%
The mass percent of oxygen ≈ 60.5%
Therefore, the mass percent of oxygen in chromium(III) nitrate is approximately 60.5%.
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To calculate the mass percent of oxygen in chromium(III) nitrate, we need to determine the molar mass of the compound and the molar mass of oxygen. The molar mass of Cr(NO3)3 is calculated to be 291.97 g/mol. The mass percent of oxygen in chromium(III) nitrate is approximately 55.24%.
Explanation:To calculate the mass percent of oxygen in chromium(III) nitrate, we need to determine the molar mass of the compound and the molar mass of oxygen. The formula for chromium(III) nitrate is Cr(NO3)3.
The molar mass of Cr(NO3)3 can be calculated as follows:
Molar mass of Cr = 51.996 g/mol
Molar mass of N = 14.007 g/mol
Molar mass of O = 15.999 g/mol
Molar mass of Cr(NO3)3 = (51.996 g/mol) + 3*(14.007 g/mol) + 9*(15.999 g/mol) = 291.97 g/mol
The molar mass of oxygen is 15.999 g/mol. The mass percent of oxygen in chromium(III) nitrate can be calculated as:
Mass percent of O = (mass of O / molar mass of Cr(NO3)3) * 100%
Mass percent of O = (9*(15.999 g/mol) / 291.97 g/mol) * 100% = 55.24%
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If 0.060 faraday is passed through an electrolytic cell containing a solution of In3+ ions, the maximum number of moles of In that could be deposited at the cathode isA) 0.010 mole
B) 0.020 mole
C) 0.030 mole
D) 0.060 mole
E) 0.18 mole
Answer: option B. 0.020 mole
Explanation:Please see attachment for explanation
The maximum number of moles of In that could be deposited at the cathode is 0.02 mole.
The given parameters;
number of ions, n = +3 ionscharge passed through the electrolytes, Q = 0.06 FThe maximum number of moles of In that could be deposited at the cathode is calculated as follows;
3 ions = 3 F
3 F ----------------------- 1 mole of the ion
0.06 F ---------------------- ? mole of the ion
[tex]= \frac{0.06\ F \times 1}{3 \ F } \\\\= 0.02 \ mole[/tex]
Thus, the maximum number of moles of In that could be deposited at the cathode is 0.02 mole.
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By titration, it is found that 65.5 mL of 0.117 M NaOH ( aq ) is needed to neutralize 25.0 mL of HCl ( aq ) . Calculate the concentration of the HCl solution.
Answer:
0.30 M
Explanation:
In case of titration , the following formula can be used -
M₁V₁ = M₂V₂
where ,
M₁ = concentration of acid ,
V₁ = volume of acid ,
M₂ = concentration of base,
V₂ = volume of base .
from , the question ,
M₁ = ? M
V₁ = 25.0 mL
M₂ = 0.117 M
V₂ = 65.5 mL
Using the above formula , the molarity of acid , can be calculated as ,
M₁V₁ = M₂V₂
Putting the respective values -
M₁ * 25.0mL = 0.117 M * 65.5 mL
M₁ = 0.30 M
In which of the following pure substances would hydrogen bonding be expected?
a. cyclobutane
b. propanoic acid
c. acetone
d. All of the Above
Hydrogen bonding typically involves a hydrogen atom attached to a highly electronegative atom. In the given choices, only b- propanoic acid, which contains a hydrogen bound to an oxygen atom in a carboxylic acid group, meets this criterion.
Explanation:In considering which of the following pure substances would hydrogen bonding be expected, you need to examine the molecular structure. Hydrogen bonding typically occurs when a hydrogen atom attached to a highly electronegative atom like nitrogen, oxygen or fluorine is in the vicinity of another electronegative atom.
cyclobutane does not have any such atoms, so it would not be expected to exhibit hydrogen bonding. Propanoic acid, on the other hand, contains a carboxylic acid group (COOH) which consists of a hydrogen atom bound to an oxygen atom, so it should display hydrogen bonding. Acetone, however, while it does contain oxygen, doesn't have hydrogen directly bonded to the oxygen, reducing its ability to form hydrogen bonds.
Therefore, out of cyclobutane, propanoic acid, and acetone, only propanoic acid would be expected to exhibit hydrogen bonding.
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For the data set: 0.09, 0.10, 0.11, 0.13, 0.09, 0.11, 0.10, 0.07 To obtain information of the precision of the data set the standard deviation would be:
a. 0.018
b. 0.022
c. 0.0166
d. 0.01
Answer:
Option A is correct (0.018)
S.D≅0.018
Explanation:
Option A is correct (0.018)
General Formula for Standard Deviation is:
[tex]Standard\ Deviation=\sqrt{\frac{\sum_{i=1}^{n}(x_{i}-\bar x)^2}{n-1}}[/tex]
Where:
[tex]x_{i}[/tex] is the data value
[tex]\bar x[/tex] is the mean/average of data
n is the total number of data elements
Calculating [tex]\sum_{i=1}^{n}(x_{i}-\bar x)^2}[/tex]
[tex]\bar x=\frac{0.09+0.10+ 0.11+ 0.13+ 0.09+ 0.11+ 0.10+0.07}{8} \\\bar x=0.1[/tex]
[tex]\sum_{i=1}^{n}(x_{i}-\bar x)^2}=(0.09-0.1)^2+(0.1-0.1)^2+(0.11-0.1)^2+(0.13-0.1)^2+(0.09-0.1)^2+(0.11-0.1)^2+(0.1-0.1)^2+(0.07-0.1)^2\\\sum_{i=1}^{n}(x_{i}-\bar x)^2}=2.2*10^{-3}[/tex]
Calculating n-1:
Total number of terms=8
n-1=8-1=7
Standard Deviation is:
[tex]S.D=\sqrt{\frac{2.2*10^{-3}}{7}}\\S.D=0.0177[/tex]
S.D≅0.018
Student mixed 25.0 mL of 0.100 M glucose, 15.0 mL of 0.500 M NaCl and 450. mL water. What are concentrations in his solution?
The question is incomplete, here is the complete question:
Student mixed 25.0 mL of 0.100 M glucose, 15.0 mL of 0.500 M NaCl and 450. mL water. What are concentrations in his solution?
A. 5.10 mM glucose, 15.3 mM NaCl
B. 5.56 mM glucose, 16.7 mM NaCl
C. 0.556 mM glucose, 0.167 mM NaCl
D. 0.222 mM glucose, 1.11 mM NaCl
E. 0.556 mM glucose, 0.0667 mM NaCl
Answer: The concentration of glucose and NaCl in the solution is 5.10 mM and 15.3 mM respectively.
Explanation:
To calculate the number of moles for given molarity, we use the equation:
[tex]\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}[/tex] .....(1)
For glucose:Molarity of glucose solution = 0.100 M
Volume of solution = 25.0 mL
Putting values in equation 1, we get:
[tex]0.100M=\frac{\text{Moles of glucose}\times 1000}{25.0}\\\\\text{Moles of glucose}=\frac{(0.100\times 25.0)}{1000}=0.0025mol[/tex]
For NaCl:Molarity of NaCl solution = 0.500 M
Volume of solution = 15.0 mL
Putting values in equation 1, we get:
[tex]0.500M=\frac{\text{Moles of NaCl}\times 1000}{15.0}\\\\\text{Moles of NaCl}=\frac{(0.500\times 15.0)}{1000}=0.0075mol[/tex]
Total volume of solution = 25.0 + 15.0 + 450. = 490. mL
Now, calculating the concentration of glucose and NaCl in the solution by using equation 1:
Using conversion factor: 1 M = 1000 mM
For glucose:Moles of glucose = 0.0025 moles
Volume of solution = 490. mL
Putting values in equation 1, we get:
[tex]\text{Concentration of glucose}=\frac{0.0025\times 1000}{490}\\\\\text{Concentration of glucose}=0.0051M=5.10mM[/tex]
For NaCl:Moles of NaCl = 0.0075 moles
Volume of solution = 490. mL
Putting values in equation 1, we get:
[tex]\text{Concentration of NaCl}=\frac{0.0075\times 1000}{490}\\\\\text{Concentration of NaCl}=0.0153M=15.3mM[/tex]
Hence, the concentration of glucose and NaCl in the solution is 5.10 mM and 15.3 mM respectively.
The average radius of a molecule of lysozyme, an enzyme in tears, is 1430. pm. What is its radius in centimeters (cm)? (Enter your answer in scientific notation.)
Answer:
1.43 ×10⁻⁷ cm
Explanation:
Scientific notation is the way to express the large value in short form.
The number in scientific notation have two parts.
The digits (decimal point will place after first digit)
× 10 ( the power which put the decimal point where it should be)
for example the number 6324.4 in scientific notation will be written as = 6.3244 × 10³
Radius of molecule of lysozyme:
1430 pm
In centimeter:
1430/10¹⁰
1.43 ×10⁻⁷ cm
The expanded notation is standard notation of writing the numerical values which is normal way. The numbers are written as they are, without the power of 10 such as, 1430 pm.
The weight of H2SO4 (molecular weight 98.1) in 50.0 milliliters of a 6.00-molar solution is....
Answer:
The solution has 29.43 g of H2SO4
Explanation:
A.molar solution is in molarity concentration. It means moles of solute in 1L of solution.
This solution which is 6 M has 6 moles of sulfuric in 1L of water.
Let's determine moles of solute qith a formula:
Volume (L) . molarity = mol
0.050 L . 6 mol/L = 0.3 mol
Now we can convert the moles to mass (mol . molar mass)
0.3 mol . 98.1 g/mol = 29.43 g
A 50.0 milliliters of a 6.00-molar solution of H₂SO₄ contains 29.43 grams of the acid. This is calculated by determining the number of moles and then converting to mass using the molecular weight.
Calculating the Weight of H₂SO₄ in a 50.0 mL of a 6.00 M Solution:
To determine the weight of H₂SO₄ in a 50.0 mL solution with a molarity (M) of 6.00, follow these steps:
Calculate the number of moles of H₂SO₄ . Use the formula:Therefore, the weight of H₂SO₄ in 50.0 milliliters of a 6.00-molar solution is 29.43 grams.
Calculate the volume in liters of a 0.360 mol/l barium acetate solution that contains 100.g of barium acetate (Ba(C2H3O2)2) . Round your answer to 3 significant digits.
Answer:
1.08 liters is the volume of 100 g of Ba(CH₃COO)₂
Explanation:
This data is a sort of concentration (Molarity) → 0.360 M [Ba(CH₃COO)₂]
It means that 0.36 moles of solute are contained in 1L of solution.
We should determine the moles of salt, that corresponds to 100 g of solute.
Mass / Molar mass
Ba(CH₃COO)₂ molar mass = 255.3 g/m
100 g / 255.3 g/m = 0.391 moles
So, we can apply a rule of three to calculate the volume of salt.
0.360 moles of solute are contained in 1L of solution
0.391 would be contained in (0.391 .1) / 0.360 = 1.08 liters
A sample of nitrogen gas occupies 1.55 L at 27.0°C and 1.00 atm. What will the volume be at −100.0°C and the same pressure?
Answer:
0.89L ( 2 decimal place)
Explanation:
Using Combined Gas law.
( P₁V₁)/T₁ = (P₂V₂) / T₂
P₁ = 1.00 atm
V₁ = 1.55L
T₁ = 27.0 °C = converting to kelvin ( 27 + 273k) = 300k
P₂ = same as P₁ = 1.00 atm
V₂ = ?
T₂ = -100°C = converting to kelvin ( -100 + 273k) = 173k
(1 x 1.55)/300 = (1 x V)/173
1.55/300 = V/173
V / 173 = 0.00516666666
V = 173 x 0.00516666666 = 0.89383333333 ≈ 0.89L ( 2 decimal place)
Answer:
The volume at -100 °C is 0.894 L
Explanation:
Step 1: Data given
The initial volume = 1.55L
The initial temperature = 27.0 °C = 300 K
The pressure is 1.00 atm and stays constant
The temperature lowers to -100 °C = 173 K
Step 2: Calculate the volume
V1 / T1 = V2 / T2
⇒ with V1 = the initial volume = 1.55L
⇒ with T1 = the initial temperature = 300 K
⇒ with V2 = The new volume = TO BE DETERMINED
⇒ with T2 = the final temperature = 173 K
1.55 / 300 = V2 /173
V2 = 0.894 L
The volume at -100 °C is 0.894 L
From the list of the instructions and tips below, select the statement that would NOT lead to a successful recrystallization O Dry the crystals by pulling air through after vacuum filtration (to help them dry). O Keep the crystals open to the air for quite a while during the drying step. O Precipitate the crystals as fast as possible. O Rinse the crystals on the filter with a small amount of the ice-cold solvent. O Use a very small amount of ice-cold solvent and a spatula to remove any crystals left in the flas
Answer:
C. Precipitate the crystals as fast as possible
Explanation:
C. Precipitate the crystals as fast as possible
Precipitation if done fast , can lead to the formation of impure crystals , as impurities also get stuck inside the crystals. So it must be done slowly to obtain the pure crystals . Slower the crystals form , purer they are .
All the other options can lead to a successful recrystallization.
Final answer:
The statement 'Precipitate the crystals as fast as possible' is incorrect for successful recrystallization since it contradicts the ideal slow cooling process required for forming pure crystals.
Explanation:
The statement that would NOT lead to a successful recrystallization is "Precipitate the crystals as fast as possible." This practice is contrary to the ideal method of recrystallization where slow cooling of the solution is necessary to form pure crystals. Rapid precipitation can trap impurities within the crystal lattice and lead to the formation of small, impure crystals. The other statements regarding drying the crystals, keeping them open to air, rinsing with cold solvent, and using a small amount of ice-cold solvent to remove remaining crystals from the flask are in line with successful recrystallization techniques.
What mass (in g) of KIO3 is needed to prepare 50.0 mL of 0.20 M KIO3? b) What volume (in mL) of 0.15 m H2SO4 is needed to prepare 50.0 mL of 0.080 M H2SO4?
Final answer:
To prepare a 0.20 M KIO3 solution, 2.14 g of KIO3 is required for 50.0 mL. For the 0.080 M H2SO4 solution, 26.7 mL of 0.15 M H2SO4 is needed to dilute to 50.0 mL.
Explanation:
To calculate the mass (in g) of KIO3 needed to prepare 50.0 mL of a 0.20 M solution, use the formula:
mass = molarity × volume × molar mass
We begin by calculating the moles of KIO3 needed:
Moles of KIO3 = Molarity × Volume (in L)
= 0.20 mol/L × 0.0500 L = 0.010 moles of KIO3
The molar mass of KIO3 is K (39.10 g/mol) + I (126.90 g/mol) + 3×O (3× 16.00 g/mol) = 214.00 g/mol.
Now calculate the mass:
Mass of KIO3 = Moles × Molar Mass
= 0.010 moles × 214.00 g/mol = 2.14 g
For the second part, to find the volume (in mL) of 0.15 M H2SO4 needed to prepare 50.0 mL of a 0.080 M solution, we use the dilution formula:
M1V1 = M2V2, where M is molarity and V is volume.
The initial molarity (M1) is 0.15 M, the final molarity (M2) is 0.080 M, and the final volume (V2) is 50.0 mL. We solve for V1:
V1 = (M2V2) / M1
= (0.080 M × 50.0 mL) / 0.15 M = 26.7 mL
Accordingly, you would need 26.7 mL of the 0.15 M H2SO4.
The recommended daily allowance (RDA) of the trace metal magnesium is 410 mg/daymg/day for males. Express this quantity in μg/dayμg/day
Answer:
[tex]410mg/day *\frac{1000 ug}{1 mg}[/tex]
Recommended daily Amount (RDA) of magnesium is 410,000 μg/day.
Explanation:
There are 1000μg in 1 mg and 1000 mg in 1g
1 mg=1000μg
The Recommended daily Amount (RDA) is 410 mg/day of magnesium. Converting 410 mg/day into μg/day
[tex]410mg/day *\frac{1000 ug}{1 mg}[/tex]
It will become 410,000 μg/day
So Recommended daily Amount (RDA) of magnesium is 410,000 μg/day.
Consider the Williamson ether synthesis between 2-naphthol and 1-bromobutane in strong base. A reaction was performed in which 0.51 g of 2-naphthol was reacted with a slight excess of 1-bromobutane to make 0.29 g of 2-butoxynaphthalene. Calculate the theoretical yield and percent yield for this reaction.Figure:Chemical bonds
Answer:
0.70 g
41 %
Explanation:
We can write the Williamson ether synthesis in a general form as:
R-OH + R´-Br ⇒ R-O-R´
where R-OH is an alcohol and R´-Br is an alkyl bromide.
We then see that the reaction occurs in a 1:1 mole ratio to produce 1 mol product.
Therefore what we need to calculate the theoretical yield and percent yield is to compute the theoretical number of moles of 2-butoxynaphthalene produced from 0.51 g 2-naphthol, and from there we can calculate the percent yield.
molar mass 2-naphthol = 144.17 g/mol
moles 2-naphthol = 0.51 g / 144.17 g/mol = 0.0035 mol 2-naphthol
The number of moles of produced:
= 0.0035 mol 2-naphthol x ( 1 mol 2-butoxynaphthalene /mol 2-naphthol )
= 0.0035 mol 2-butoxynaphthalene
The theoretical yield will be
= 0.0035 mol 2-butoxynaphthalene x molar mass 2-butoxynaphthalene
= 0.0035 mol x 200.28 g/ mol = 0.70 g
percent yield= ( 0.29 g / 0.70 ) g x 100 = 41 %
The theoretical yield of 2-butoxynaphthalene in the reaction between 2-naphthol and 1-bromobutane can be calculated by comparing the number of moles of the limiting reagent to the number of moles of the product. The percent yield of the reaction can be determined by dividing the actual yield by the theoretical yield and multiplying by 100.
Explanation:In the Williamson ether synthesis reaction between 2-naphthol and 1-bromobutane in a strong base, 0.51 g of 2-naphthol reacted with a slight excess of 1-bromobutane to produce 0.29 g of 2-butoxynaphthalene. To calculate the theoretical yield, we need to compare the number of moles of the limiting reagent, which is 2-naphthol, to the number of moles of the product. The molar masses of 2-naphthol and 2-butoxynaphthalene are calculated and the theoretical yield is determined to be 0.348 g.
The percent yield of the reaction can be calculated by dividing the actual yield (0.29 g) by the theoretical yield (0.348 g) and multiplying by 100. The percent yield for this reaction is approximately 83.3%.
water has a heat of vaporization and boils at 100. what is the boiling point of water at the top of mount everest where the atmospheric pressure is only 34 as stong as the pressure at sea level
Answer: 71°C
Explanation: Liquids tends to boil when their vapour pressure equals atmospheric pressure. As we go higher, the atmospheric pressure reduces. Since Everest is at high altitude, water will boils at a much lower value.
What the sample will look like after the temperature is increased from 300 K to 450 KK at constant pressure?
Answer:
The sample will look expanded and occupy more space.
Explanation:
Since, the pressure is constant here, but the temperature is changed. Therefore, according to Charles' Law Volume is directly proportional to Temperature, provided the pressure is kept constant. Mathematically:
V1/T1 = V2/T2
V1 = (T1/T2)(V2)
V1 = (300 k/450 k)(V2)
V1 = (0.67)V2
The equation indicates that The fina volume of the gas V2 will be greater than the initial volume V1. Thus, sample will look expanded and occupy more space than the previous state.
The critical temperature at which steel trusses can fail from exposure to heat alone without any flames is
Answer: 1000°F
Explanation:
As heat starts to attack the steel trusses, the steel will fail without flame at 1000°F
Sulfuric acid is essential to dozens of important industries from steelmaking to plastics and pharmaceuticals. More sulfuric acid is made than any other industrial chemical, and world production exceeds per year. The first step in the synthesis of sulfuric acid is usually burning solid sulfur to make sulfur dioxide gas. Suppose an engineer studying this reaction introduces of solid sulfur and of oxygen gas at into an evacuated tank. The engineer believes for the reaction at this temperature. Calculate the mass of solid sulfur he expects to be consumed when the reaction reaches equilibrium. Round your answer to significant digits.
The question is incomplete, here is the complete question:
Sulfuric acid is essential to dozens of important industries from steel making to plastics and pharmaceuticals. More sulfuric acid is made than any other industrial chemical, and world production exceeds [tex]2.0\times 10^{11}[/tex] per year.
The first step in the synthesis of sulfuric acid is usually burning solid sulfur to make sulfur dioxide gas. Suppose an engineer studying this reaction introduces 4.4 kg of solid sulfur and 6.90 atm of oxygen gas at 950°C into an evacuated 50.0 L tank. The engineer believes [tex]K_p=0.71[/tex] for the reaction at this temperature.
Calculate the mass of solid sulfur he expects to be consumed when the reaction reaches equilibrium. Round your answer to 2 significant digits.
Answer: The mass of sulfur that is expected to be consumed is 0.046 kg
Explanation:
We are given:
Initial partial pressure of oxygen gas = 6.90 atm
The chemical equation for the formation of sulfur dioxide follows:
[tex]S(s)+O_2(g)\rightleftharpoons SO_2(g)[/tex]
Initial: 6.90
At eqllm: 6.90-x x
The expression of [tex]K_p[/tex] for above equation follows:
[tex]K_p=\frac{p_{SO_2}}{p_{O_2}}[/tex]
We are given:
[tex]K_p=0.71[/tex]
Putting values in above equation, we get:
[tex]0.71=\frac{x}{(6.9-x)}\\\\x=2.9[/tex]
So, equilibrium partial pressure of sulfur dioxide = x = 2.9 atm
To calculate the number of moles, we use the equation given by ideal gas, which follows:
PV = nRT
where,
P = pressure of sulfur dioxide gas = 2.9 atm
V = volume of the container = 50.0 L
n = number of moles of sulfur dioxide gas = ?
R = [tex]0.0821\text{ L. atm }mol^{-1}K^{-1}[/tex]
T = temperature of the container = [tex]950^oC=[950+273]K=1223K[/tex]
Putting values in above equation, we get:
[tex]2.9\times 50.0=n\times 0.0821\times 1223\\\\n=\frac{2.9\times 50.0}{0.0821\times 1223}=1.44mol[/tex]
Moles of sulfur dioxide = 1.44 moles
By Stoichiometry of the reaction:
1 mole of sulfur dioxide is produced from 1 mole of sulfur
So, 1.44 moles of sulfur dioxide will be produced from [tex]\frac{1}{1}\times 1.44=1.44mol[/tex] of sulfur
To calculate the mass of a substance, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]
Moles of sulfur = 1.44 moles
Molar mass of sulfur = 32 g/mol
Putting values in above equation, we get:
[tex]1.44mol=\frac{\text{Mass of sulfur}}{32g/mol}\\\\\text{Mass of sulfur}=(1.44mol\times 32g/mol)=46.08g=0.046kg[/tex]
Hence, the mass of sulfur that is expected to be consumed is 0.046 kg
The molecules of a gas at constant volume are somehow given a lower average kinetic energy.
What two measurable properties of the gas will change and in what direction?
Answer:
Kinetic energy is proportional to temperature. An increase in the average kinetic energy, decrease the temperature and vice versa. pressure and temperature have a direct relationship between them, if temperature decreases, pressure also decreases and vice versa.
Explanation:
This concept is explained by taking into consideration the gas laws; GayLussac's law of combining volume and charles's law . considering the molecules of a gas at constant volume, there is a direct relationship between the pressure and temperature at constant volume, this is what gay lussac's law entails i.e gases who combine at constant volume they do so in fixed ratio and to the volume of their container.
As for charles's law, there exist a Volume - Temperature relationship at constant pressure, because volume and temperature have a direct relationship, when the average kinetic energy of gas molecules is decreased, this in turn decreases the temperature of the gas molecules.
And when the temperature decreases as a result of decrease in the average kinetic energy, this automatically affect the pressure thereby causing a reduction in the pressure of the gas container.
However, if the average kinetic energy of the gas molecules is increased again, then there will be an increase in the temperature and pressure and vice versa. Kinetic energy and temperature also have a direct relationship.
Final answer:
When the average kinetic energy of a gas at constant volume is decreased, the temperature and pressure of the gas will also decrease due to fewer and less forceful collisions of molecules with the container walls.
Explanation:
If the molecules of a gas at constant volume are given a lower average kinetic energy (KEavg), two measurable properties of the gas that will change are the pressure and temperature of the gas. A decrease in the average kinetic energy of gas molecules, which is directly proportional to the gas's absolute temperature, leads to a lower temperature. As the kinetic energy decreases, molecules move slower, resulting in fewer collisions with the walls of the container. This decrease in collision frequency and force leads to a decrease in gas pressure. The direction of these changes is downward; both temperature and pressure will decrease.
Below is the Lewis structure of the hydrogen peroxide (H202) molecule. H: O:O:H Count the number of bonding pairs and the number of lone pairs around the right oxygen atom. bonding pairs:
The two bonding pairs and two lone pairs on each oxygen atom in the Lewis structure of hydrogen peroxide (H₂O₂) molecule.
What is the lewis electron dot diagram?A lewis structure can be used to represent the number of chemical bonds, the bonding atoms, and the lone pairs reaming on the atoms in a given molecule.
Lines are used to showing the bonds between atoms that are bonded to one another and lone pairs are depicted as dot pairs and are placed next to the respective atoms. As the valence electrons of each oxygen atom are equal to six from the electronic configuration of the oxygen atom.
From the lewis structure of hydrogen peroxide, we can see that each oxygen atom form one bond with a hydrogen atom and another bond with an oxygen atom. The two lone pairs are still present on each oxygen atom in the molecule.
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An air stream enters a variable area channel at a velocity of 30 m/s with a pressure of 120 kPa and a temperature of 10°C. At a certain point in the channel, the velocity is found to be 250 m/s. Using Bernoulli’s equation (i.e., p + rhoV2 /2 = constant), which assumes incompressible flow, find the pressure at this point. In this calculation, use the density evaluated at the inlet conditions. If the temperature of the air is assumed to remain constant, evaluate the air density at the point in the flow where the velocity is 250 m/s. Compare this density with the density at the inlet to the channel. Based on this comparison, do you think that the use of Bernoulli’s equation is justified
Answer:
P₂= 74 kPa under constant density and ρ₂ = 1.06 kg/m³ (-38.6% of error compared with incompresible assumption) . Thus Bernoulli’s equation should not be applied
Explanation:
Assuming ideal gas behaviour of air , then
P*V= n*R*T = m / M * R *T
since
ρ= m/V = P*M /( R *T)
where
n= moles , V= volume , m= mass
ρ= density
P= pressure = 120 kPa= 120000 Pa
M= molecular weight of air = 0.21*32+0.79*28= 28.24 gr/mol = 0.02824 kg/mol
T= absolute temperature = 10°C + 273 = 283 K
R= ideal gas constant = 8.314 J/mol K
solving for ρ
ρ= P*M /( R *T) = 120000 Pa*0.02824 kg/mol/(8.314 J/mol K*283 K) = 1.47 kg/m³
then from Bernoulli's equation
P₁ + ρ*v₁²/2 = P₂ + ρ*v₂²/2
where 1 denotes inlet and 2 denotes other point , p = pressure and v= velocity . Then solving for p₂
P₁ + ρ*v₁²/2 = P₂ +ρ*P₂²/2
P₂= P₁ +ρ*v₁²/2 - ρ*v₂²/2 = P₁ +ρ/2*(v₁² - v₂²)
replacing values
P₂= P₁ +ρ/2*(v₁² - v₂²) = 120000 Pa + 1.47 kg/m³/2*[(30 m/s)²-(250 m/s)²] = 74724 Pa = 74 kPa
P₂= 74 kPa
then if the temperature remains constant
ρ₁= P₁*M /( R *T) and ρ₂= P₂*M /( R *T)
dividing both equations
ρ₂/ρ₁ = P₂/ P₁
ρ₂ = (P₂/ P₁)*ρ₁
then from Bernoulli's equation
P₁ + ρ₁*v₁²/2 = P₂ + ρ₂*v₂²/2
P₂ = P₁ + ρ₁*v₁²/2 - ρ₂*v₂²/2
therefore
ρ₂ = (P₂/ P₁)*ρ₁ = (P₁ + ρ₁*v₁²/2 - ρ₂*v₂²/2 ) /P₁ *ρ₁
P₁ * ρ₂ = P₁ *ρ₁ + ρ₁²*v₁²/2 - ρ₂*ρ₁ * v₂²/2
P₁ * ρ₂ + ρ₂*ρ₁ * v₂²/2 = P₁ *ρ₁ + ρ₁²*v₁²/2
ρ₂* (P₁ + ρ₁ * v₂²/2) = P₁ *ρ₁ + ρ₁²*v₁²/2
ρ₂ = (P₁ *ρ₁ + ρ₁²*v₁²/2)/(P₁ + ρ₁ * v₂²/2) = (P₁ + ρ₁*v₁²/2)/(P₁/ρ₁ + v₂²/2)
replacing values
ρ₂ = ( 120000 Pa + 1.47 kg/m³/2*(30 m/s)²)/(120000 Pa/1.47 kg/m³+1/2*(250 m/s)²)
ρ₂ = 1.06 kg/m³
the error of assuming constant ρ would be
e = (ρ₂ - ρ)/ρ₂= 1- ρ/ρ₂= 1- 1.47 kg/m³/1.06 kg/m³ = -0.386 (-38.6%)
thus Bernoulli’s equation should not be applied
Which of the following would require the largest volume of 0.100 M sodium hydroxide solution for neutralization?:
(A) 20.0 mL of 0.0500 M nitric acid
(B) 15.0 mL of 0.0500 M hydrobromic acid
(C) 5.0 mL of 0.0100 M sulfuric acid
(D) 10.0 mL of 0.00500 M phosphoric acid
Answer:10.0 mL of 0.00500 M phosphoric acid
Explanation:
If we look at the Ka values of the acids, we will realize that phosphoric acid has a Ka of 7.1 * 10-3. It is the only acid in the list having acid dissociation constant less than 1. This means that it does not ionize easily in solution and a very large volume of base must be added to ensure that it reacts completely. Acids with Ka >1 are generally regarded as strong acids. All the acids listed have Ka>1 except phosphoric acid.
20.0 mL of 0.0500 M nitric acid would require the largest volume of 0.100 M sodium hydroxide solution for neutralization based on calculations using molarity, volume, and the number of protons the acid can donate.
Explanation:An acid-base neutralization reaction takes place when an acid reacts with a base to produce a salt and water. The volume of base needed to neutralize an acid depends on the molarity and volume of the acid, as well as the number of protons (H+) each acid molecule can donate.
For each option, we can calculate the number of moles of H+ using the formula: moles = Molarity * Volume(L), and then using the 1:1 stoichiometry with sodium hydroxide (NaOH) for one-proton acids, or 1:2 stoichiometry for two-proton acids.
(A) 20.0 mL of 0.0500 M nitric acid (HNO₃): 0.0500 mol/L * 0.020 L = 0.00100 mol of H+(B) 15.0 mL of 0.0500 M hydrobromic acid (HBr): 0.0500 mol/L * 0.015 L = 0.00075 mol of H+(C) 5.0 mL of 0.0100 M sulfuric acid (H₂SO₄): 2(0.0100 mol/L * 0.005 L) = 0.00010 mol of H+(D) 10.0 mL of 0.00500 M phosphoric acid (H₃PO₄): 0.00500 mol/L * 0.010 L = 0.00005 mol of H+So, option (A) with 20.0 mL of 0.0500 M nitric acid would require the largest volume of 0.100 M sodium hydroxide solution for neutralization.
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