Calcium hypochlorite (Ca(OCl)2, MW = 142.983 g/mol) is often used as the source of the hypochlorite ion (OCl–, MW = 51.452 g/mol) in solutions used for water treatment. A student must prepare 50.0 mL of a 55.0 ppm OCl– solution from solid Ca(OCl)2, which has a purity of 94.0%.

A) Calculate the mass of the impure reagent required to prepare the solution. Assume the density of the solution is 1.00 g/mL.

B) Which of the following methods would work best to accurately prepare 50.0 mL of the 55.0 ppm OCl– solution?

Use an analytical balance to weigh out an amount of reagent larger than what was determined in part A. In a volumetric flask, combine the reagent and water to create a solution with a concentration greater than 55.0 ppm. Use a pipet to transfer a portion of the concentrated solution to a separate 50.0 mL volumetric flask and dilute to the line with water to create a solution with a concentration of 55.0 ppm.

Use an analytical balance to weigh out an amount of reagent larger than what was determined in part A. In a volumetric flask, combine the reagent and water to create a solution with a concentration greater than 55.0 ppm. Use a graduated cylinder to transfer a portion of the more concentrated solution to a 50.0 mL volumetric flask and dilute to the line with water to create a solution with a concentration of 55.0 ppm.

Use an analytical balance to weigh out the amount of reagent determined in part A. Transfer the reagent to a 50.0 mL volumetric flask, add water to the line, and mix thoroughly.

Use an analytical balance to weigh out the amount of reagent determined in part A. Then use a graduated cylinder to measure out 50.0 mL of water. Combine the reagent and water in a 50.0 mL volumetric flask and mix it thoroughly.

Answer:

C

Explanation:

Pb + H2O + O2 => Pb(OH)2

Number of oxygen atom on left = 3

Number of oxygen atom on the right = 2

To balance number of oxygen

Change the coefficient of Pb and H2O to two each

2Pb + 2H2O + O2 => 2Pb(OH)2

The coefficient of Pb and H2O is 2,2

The correct coefficients for Pb and H₂O when balancing the equation Pb + H₂O + O₂ ⇒ Pb(OH)₂ are 1 and 2, respectively; therefore, the answer is A. 1,2. This ensures that the chemical equation is fully balanced with the correct number of atoms for each element on both sides.

When balancing the chemical equation Pb + H₂O + O₂ ⇒ Pb(OH)₂, the correct coefficients for Pb and H₂O, respectively, are 1 and 2. Balancing chemical equations requires that the same number of atoms for each element be present on both sides of the equation. Starting with lead (Pb), we see that one Pb atom is needed on both sides, so we keep the coefficient for Pb as 1.

Next, we balance the hydrogen atoms. There are two hydrogens in each hydroxide ion, and since there are two hydroxide ions in Pb(OH)₂, this means there are a total of four hydrogen atoms in the product. To balance this, we need 2 molecules of H₂O (since each molecule of water contains two hydrogen atoms, giving us the four we need), making the coefficient for H₂O equal to 2.

Lastly, the oxygen atoms will balance themselves once H and Pb are balanced. With 2 H₂O molecules, we have 2 oxygen atoms, and the Pb(OH)₂ molecule has 2 oxygen atoms as well, which means we don't need any additional O₂ for balance. The balanced equation is 1 Pb + 2 H₂O + O₂ ⇒ 1 Pb(OH)₂. Therefore, the correct answer is A. 1,2.

Answer:

The absolute error = 3.1m

The relative error = 0.31

Explanation:Please see attachment for explanation

Answer: voltage-gated channels

Explanation:

voltage-gated Ion channels: They are proteins of multi-subunit complexes that reacts with confrontations changes that occurs in the members potential which results in the opening and closing of the pore. They mainly help to generate electrical signals within the cell membrane. The locations of the voltage-gated ion channels includes in the cell body of the neurons, on the dendrites, on the axon hillock, on the node of Ranvier, etc. Majorly, when ion channels open, it is a form of response to stimuli.

Voltage-gated channels are the membrane ion channels that open in response to changes in membrane potential, cycling through closed, open, or inactivated states depending on the cell's electrical charge.

The membrane ion channels that open and close in response to changes in the membrane potential are known as voltage-gated channels. These channels can exist in one of three states - closed, open, or inactivated. Initially, voltage-gated ion channels are closed, but they open in response to a change in the electrical potential of the cell membrane. Specifically, when the inner membrane voltage becomes less negative in comparison to the outside, the channels sense the change through amino acids in their structure that are sensitive to charge, prompting the pore to open and allow ions to move across the membrane. Once activated, these channels will then become inactivated for a brief period, during which time they cannot open in response to a signal even if the membrane potential changes.

Answer:

The answer will be that if there is a linear relationship exist between CO2 and the global temperature, then if one variable change, the other also will change correspondingly.

Answer:  The work energy lost by the system is -6078 Joules

Explanation:

According to first law of thermodynamics:

[tex]\Delta E=q+w[/tex]

[tex]\Delta E[/tex]=Change in internal energy

q = heat absorbed or released

w = work done or by the system

w = work done by the system=[tex]-P\Delta V[/tex]  {Work is done by the system as the final volume is greater than initial volume and is negative}

where P = pressure = 3 atm

[tex]\Delta V[/tex] = change in volume = (55-35) L = 20 L

w =[tex]-3atm\times (20)L=-60Latm=-6078Joules[/tex]  

{1Latm=101.3J}

Thus work energy lost by the system is -6078 Joules

Answer: 40731.8 grams of this gasoline would fill a 14.6gal tank

Explanation:

Density is defined as the mass contained per unit volume.

[tex]Density=\frac{mass}{Volume}[/tex]

Given : Mass of gasoline = ?

Density of the gasoline = [tex]0.737g/ml[/tex]

Volume of the gasoline = 14.6gal = 55267.01 ml           (1gal=3785.41ml)

Putting in the values we get:

[tex]0.737g/ml=\frac{mass}{55267.01ml}[/tex]

[tex]{\text {mass of gasoline}}=40731.8g[/tex]

Thus 40731.8 grams of this gasoline would fill a 14.6gal tank

Answer:

the final concentration is C=0.235 mg/L

Explanation:

doing a mass balance

ammonium mass outflow = ammonium inflow from river+ effluent discharge of ammonium  = 12000 L/s* 0.015 mg/L  + 1500 L /s * 2 mg/L = 3180 mg/s

outflow rate = 12000 L/s+ 1500 L /s  = 13500 L/s

then the final concentration is

C= ammonium mass outflow/outflow rate = 3180 mg/s/13500 L/s= 0.235 mg/L

Final answer:

To find the concentration of the mixture after becoming perfectly mixed downstream, we can use the principle of mass conservation.

Explanation:

To find the concentration of the mixture after becoming perfectly mixed downstream, we can use the principle of mass conservation. The total mass of ammonium-N in the effluent discharge is given by the product of its concentration and flow rate, while the total mass of ammonium-N in the river is given by the product of its concentration and flow rate. When the effluent discharge and river mix together, the total mass of ammonium-N remains constant.

We can use the equation:

(flow rate of effluent discharge * concentration of effluent discharge) + (flow rate of river * concentration of river) = (flow rate of mixture * concentration of mixture)

Plugging in the given values:

(1.5 m³/s * 2 mg/L) + (12 m³/s * 0.015 mg/L) = (flow rate of mixture * concentration of mixture)

Solving this equation will give us the concentration of the mixture after they have become perfectly mixed downstream.

Answer:

a. The balanced chemical equation for the reaction

[tex]NH_3(g)+HCl(g)\rightarrow NH_4Cl(g)[/tex]

b. HCl is the limiting reagent.

c. 7.3295 grams of ammonium chloride could form from the reaction mixture in part (b).

d. 0.6715 grams of ammonia is left over in the reaction mixture in part b.

Explanation:

a. The balanced chemical equation for the reaction

[tex]NH_3(g)+HCl(g)\rightarrow NH_4Cl(g)[/tex]

b.

Moles of ammonia =[tex]\frac{3.0 g}{17 g/mol}=0.1765 mol[/tex]

Moles of HCl = [tex]\frac{5.0 g}{36.5 g/mol}=0.1370 mol[/tex]

According to reaction, 1 mole of HCl reacts with 1 mole of ammonia . Then 0.1370 moles of HCl will react with :

[tex]\frac{1}{1}\times 0.1370 mol=0.1370 mol[/tex] of ammonia

Hence, HCl is the limiting reagent.

c.

Since, HCl is a limiting reagent amount of ammonium chloride will depend upon moles of HCl.

According to reaction, 1 mole of HCl gives with 1 mole of ammonium chloride . Then 0.1370 moles of HCl will give :

[tex]\frac{1}{1}\times 0.1370 mol=0.1370 mol[/tex] of ammonium chloride

Mass of 0.1370 moles of ammonium chloride :

53.5 g/mol × 0.1370 mol =  7.3295 g

7.3295 grams of ammonium chloride could form from the reaction mixture in part (b).

d.

Moles of ammonia =[tex]\frac{3.0 g}{17 g/mol}=0.1765 mol[/tex]

HCl is a limiting reagent and ammonia is an excessive reagent.

According to reaction, 1 mole of HCl reacts with 1 mole of ammonia . Then 0.1370 moles of HCl will react with :

[tex]\frac{1}{1}\times 0.1370 mol=0.1370 mol[/tex] of ammonia

Moles of Ammonia reacted = 0.1370 mol

Moles of ammonia left unreacted = 0.1765 mol - 0.1370 mol = 0.0395 mol

Mass of 0.0395 moles of ammonia :

0.0395 mol × 17 g/mol = 0.6715 g

0.6715 grams of ammonia is left over in the reaction mixture in part b.

Answer:

A) 0.20 cm³

B) 49.7 m²

C) 99.99%

D) 17.7 mg

Explanation:

A) The density of a material represents the mass that it occupies in a "piece" of volume. Thus, the density (d) is the mass (m) divided by the volume (v):

d =m/v

If the mass is 40.0 mg = 0.04 g, and the density is 0.20 g/cm³, the volume is:

0.20 = 0.04/v

v = 0.04/0.20

v = 0.20 cm³

B) The surface area (S) is the are that is presented in each gram of the material, so, it's the area (a) divided by the mass (m):

S = a/m

If the mass is 40.0 mg = 0.04 g, and the surface area is 1242 m²/g, so:

1242 = a/0.04

a = 49.7 m²

C) The percent of mercury removed is the mass removed divided by the initial mass, this multiplied by 100%. The mass removed is the initial mass (m0) less the final mass (m), so:

%removed = [(7.748 - 0.001)/7.748] *00%

%removed = 99.99%

D) The final mass of the spongy material is it mass (10 mg) plus the mass removed of the mercury (7.748 - 0.001 = 7.747 mg), so:

m = 10 + 7.747

m = 17.747 mg

m = 17.7 mg

Final answer:

This detailed answer explains how to calculate the volume, surface area, percentage of mercury removed, and final mass of a new material used to remove mercury from water.

Explanation:

A) To calculate the volume of a 40.0-mg sample of the material, you can use the formula: Volume = Mass / Density. Therefore, Volume = 40.0 mg / 0.20 g/cm³ = 200 cm³.

B) The surface area for a 40.0-mg sample can be calculated by multiplying the specific surface area by the mass: 1242 m²/g × 40.0 mg = 49680 m².

C) The percentage of mercury removed from the water is: ((Initial mercury mass - Final mercury mass) / Initial mercury mass) × 100 = ((7.748 mg - 0.001 mg) / 7.748 mg) × 100 = 99.9874%.

D) The final mass of the spongy material after exposure to mercury is the initial mass minus the mass used: 40.0 mg - 10.0 mg = 30.0 mg.

Answer: 40mL

Explanation: What is required for this question is to obtain the number of moles of NaOH solute in the required solution.

Number of moles of solute in solution = (Concentration of solution in mol/L) × (Volume of solution in L)

Number of moles of NaOH in the required solution = 0.2 × (300/1000) = 0.06 moles

This number of moles has to be in the volume of stock solution used to make the required solution.

Volume of stock solution needed = (Number of moles of NaOH in the stock solution that has to match that in the required solution in L)/(concentration of stock solution in mol/L)

Volume of stock solution required = 0.06/1.5 = 0.04L = 40mL

QED.

To make 300 mL of a 0.2000 M NaOH solution from a 1.5 M stock solution, you need 40 mL of the stock solution, as calculated using the dilution equation.

To find out how many milliliters of a 1.5 M stock solution is needed to make 300 mL of a 0.2000 M dilute NaOH solution, you can use the equation of dilution:
C1 times V1 = C2 times V2 where C1 and V1 are the concentration and volume of the stock solution, and C2 and V2 are the concentration and volume of the dilute solution, respectively.

The formula reorganized to solve for V1 is:
V1 = (C2  imes V2) / C1
Substituting the given values, we get:
V1 = (0.2000 M times 300 mL ) / 1.5 M
V1 = 40 mL

Therefore, you need 40 mL of the stock solution to prepare the required dilution.

Answer:

The reaction rate is increased.

Explanation:

The pressure is the force the gas molecules do when hitting each other and the walls of the container they're. The partial pressure is the pressure of a substance in a mixture would have if it was alone at the same conditions.

Thus, when the partial pressure increases, it means that the molecules collide more often. The reaction happens when the molecules collide in the right way, so, if the collisions are happening more often, the rate must be higher.

Answer: The Cu-63 isotope has the highest percent natural abundance

Explanation:

Average atomic mass is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.

Formula used to calculate average atomic mass follows:

[tex]\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i[/tex]

We are given:

Two isotopes of Copper, which are Cu-63 and Cu-65

Average atomic mass of copper = 63.54 amu

As, the average atomic mass of copper is closer to the mass of Cu-63 isotope. This means that the relative abundance of this isotope is the highest as compared to the other isotope.

Percentage abundance of Cu-63 isotope = 69.2%

Percentage abundance of Cu-65 isotope = 30.8 %

Hence, the Cu-63 isotope has the highest percent natural abundance

Answer:

0,034 moles

Explanation:

Answer:

51:49 is the ratio of female participants to male participants.

Explanation:

Percentage of females in the race = 51 %

Percentage of males in the race = 100% - 51 % = 49%

Total participants in race = 400

Number of women participants = [tex]51% of 400=\frac{51}{100}\times 400=204[/tex]

Number of men participants  = [tex]49% of 400=\frac{49}{100}\times 400=196[/tex]

Ratio of female participants to male participants :

[tex]\frac{204}{196}=\frac{51}{49}=51:49[/tex]

Final answer:

To determine the ratio of female to male participants in a race with 400 participants and 51% females, we calculate 204 females and 196 males, resulting in a simplified ratio of 51:49.

Explanation:

To find the ratio of female to male participants in a local Thanksgiving 5K race with four hundred participants, where 51% are female, we first calculate the number of female and male participants. Since 51% of 400 participants are female, there are 204 female participants (0.51 × 400). The remaining participants are male, which accounts for 196 participants (400 - 204).

The ratio of female to male participants can be expressed as 204:196. To simplify this ratio, we divide both numbers by their greatest common divisor, which is 4. Therefore, the simplified ratio of female to male participants is 51:49.

Answer: A & C

Explanation:

Hexokinase is also one of the enzymes.

(A) Fructose 6-phosphate -> fructose 1,6-bisphosphate; phosphofructokinase-2

(C) Phosphoenolpyruvate -> pyruvate; pyruvate kinase

In gluconeogenesis, bypass reactions circumvent irreversible steps in glycolysis. The correct matching of the glycolytic reaction and the gluconeogenic enzyme is glucose to glucose 6-phosphate via glucose 6-phosphatase.

The process known as gluconeogenesis uses what are known as bypass reactions to circumvent three different steps in the glycolytic reaction process that are essentially irreversible. These bypass reactions occur via the use of different enzymes.

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Answer:  459 nm

Explanation:

The relation between energy and wavelength of light is given by Planck's equation, which is:

[tex]E=\frac{Nhc}{\lambda}[/tex]

where,

E = energy of the light  = [tex]261 kJ=261000J[/tex]   (1kJ=1000J)

N= avogadro's number  = [tex]6.023\times 10^{23}[/tex]

h = Planck's constant  = [tex]6.626\times 10^{-34}Js[/tex]

c = speed of light  = [tex]3\times 10^8m/s[/tex]

[tex]\lambda[/tex] = wavelength of light  = ?

Putting the values in the  equation:

[tex]261000J=\frac{6.023\times 10^{23}\times 6.626\times 10^{-34}Js\times 3\times 10^8m/s}{\lambda}[/tex]

[tex]\lambda=4.587\times 10^{-7}m=459nm[/tex]       [tex]1nm=10^{-9}m[/tex]

Thus the longest wavelength of light that can be used to eject electrons from the surface of this metal via the photoelectric effect is 459 nm

To calculate the pH of a solution containing a weak base and its conjugate weak acid, use the Henderson-Hasselbalch equation.

To calculate the pH of a solution that contains a weak base and its conjugate acid, we need to apply the Henderson-Hasselbalch equation.

The Henderson-Hasselbalch equation is given by: pH = pKa + log([A-]/[HA]), where pKa is the negative logarithm of the acid dissociation constant (Ka), [A-] is the concentration of the conjugate base, and [HA] is the concentration of the weak acid.

In this case, the weak base has a concentration of 0.047 M and the conjugate weak acid has a concentration of 0.057 M. The pKa value of the acid is 7.2 x 10^-8. Plugging these values into the Henderson-Hasselbalch equation, we can calculate the pH of the solution.

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The pH of a 75.0 mL solution that is 0.047 M in weak base and 0.057 M in the conjugate weak acid ( K a = 7.2 × 10⁻⁸) will be approximately 7.06.

To determine the pH of a solution that contains both a weak base and its conjugate acid, we can use the Henderson-Hasselbalch equation. The equation is:

pH = pKa + log([A⁻]/[HA])

Where:

Given that the Ka for the weak acid is 7.2 x 10⁻⁸, the pKa is:

pKa = -log(Ka) = -log(7.2 x 10⁻⁸)

The concentration of the weak base ([A⁻]) is given as 0.047 M, and the concentration of the conjugate weak acid ([HA]) is 0.057 M.

Substituting into the Henderson-Hasselbalch equation:

pH = pKa + log(0.047/0.057)

Performing the calculations:

pH = 7.14 + log(0.825)

pH 7.14 - 0.084 = 7.06

Therefore, the pH of the solution is approximately 7.06.

Answer:

Atomic mass of 35.5 g/mol is of chlorine.

Atomic mass of 89.02 g/mol is of Yttrium.

Ytterium(III) chloride is the correct name for [tex]YCl_3[/tex].

1.835 grams of [tex]YCl_3[/tex]can be prepared.

Explanation:

[tex]2M+3X_2\rightarrow 2MX_3[/tex]

Moles of [tex]X_2[/tex] =n

Number of moleules of [tex]X_2=8.92\times 10^{20} molecules[/tex]

1 mole = [tex]6.022\times 10^{23} molecules[/tex]

[tex]n=\frac{8.92\times 10^{20} molecules}{6.022\times 10^{23} molecules}[/tex]

n = 0.001481 mole

Mass of [tex]X_2=0.105 g[/tex]

Molar mass of [tex]X_2=m[/tex]

[tex]n=\frac{Mass}{\text{Molar mass}}[/tex]

[tex]0.001481 mol=\frac{0.105 g}{m}[/tex]

m = 71 g/mol

Atomic mass of X = [tex]\frac{71 g/mol}{2}=35.5 g/mol[/tex]

Atomic mass of 35.5 g/mol is of chlorine.

The compound MX3 consists of 54.47% X by mass:

Molar mass of compound = M'

Percentage of element in compound :

[tex]=\frac{\text{number of atoms}\times text{Atomic mass}}{\text{molar mas of compound}}\times 100[/tex]

X:

[tex]54.47\%=\frac{3\times 35.5 g/mol}{M'}\times 100[/tex]

M' = 195.52 g/mol

Molar mass of compound = M'

M' = 1 × (atomic mass of M)+ 3 × (atomic mass of X)

195.52 g/mol = atomic mass of M + 3 × (35.5 g/mol)

Atomic mass of M = 89.02 g/mol

Atomic mass of 89.02 g/mol is of Yttrium.

Ytterium(III) chloride is the correct name for [tex]YCl_3[/tex].

[tex]2Y+3Cl_2\rightarrow 2YCl_3[/tex]

Moles of Yttrium = [tex]\frac{1g }{89.02 g/mol}=0.01123 mol[/tex]

Moles of chlorine gas= [tex]\frac{1 g}{71 g/mol}=0.01408 mol[/tex]

According to reaction, 3 moles of chlorine reacts with 2 moles of Y.

Then 0.01408 moles of chlorine gas will :

[tex]\frac{2}{3}\times 0.01408 mol=0.009387 mol[/tex] of Y.

This means that chlorine is in limiting amount., So, amount of yttrium (III) chloride will depend upon amount of chlorine gas.

According to reaction , 3 moles of chlorine gives 2 moles of [tex]YCl_3[/tex]

Then 0.01408 moles of chlorine will give :

[tex]\frac{2}{3}\times 0.01408 mol=0.009387 mol[/tex] of [tex]YCl_3[/tex]

Mass of 0.009387 moles of [tex]YCl_3[/tex]:

0.009387 mol × 195.52 g/mol = 1.835 g

1.835 grams of [tex]YCl_3[/tex]can be prepared.

To identify the elements in MX3, the calculations show that M is Iron (Fe) and X is Chlorine (Cl), forming Iron(III) Chloride (FeCl3). Given 1.00 g of each reactant, approximately 2.90 g of FeCl3 can be prepared. The limiting reagent in this process is Iron (Fe).

To determine the identities of M and X in the compound MX3 and the mass of MX3 that can be prepared, follow these steps:

Determine the molar mass of X2: The given data states that 0.105 g of X2 contains 8.92 × 1020 molecules. Using Avogadro's number (6.022 × 1023 molecules/mol), we can find the molar mass (MX2) of X2:

Calculate the atomic mass of X: Since X2 is diatomic, we divide the molar mass by 2:

Identify the element X - The atomic mass of 35.45 g/mol suggests that X is Chlorine (Cl).

Determine the molar mass and identity of M - Given that MX3 consists of 54.47% X by mass:

Solve for MM:

Identify the element M - The atomic mass of 55.85 g/mol suggests that M is Iron (Fe).

Name of MX3: Since M is Iron (Fe) and X is Chlorine (Cl), MX3 is Iron(III) Chloride (FeCl3).

Final Mass Calculation

Mole calculation for 1.00 g of M:

Limiting reagent: Each mole of X2 provides 2 moles of Cl, so: 0.0141 mol Cl2 × 2 = 0.0282 mol Cl (excess)

Iron (Fe) is the limiting reagent with 0.0179 mol producing 0.0179 mol of FeCl3

Calculate the mass of FeCl3:

Molar mass of FeCl3 = 55.85 g/mol + 3 × 35.45 g/mol ≈ 162.20 g/mol

Mass of FeCl3 = 0.0179 mol × 162.20 g/mol ≈ 2.90 g

Answer:

3.8 cm

Explanation:

Given data

Volume of a sphere: V = 4/3 × π × r³ = 4/3 × π × (2.5 cm)³ = 65 cm³

We can find the final volume using the combined gas law.

[tex]\frac{P_{1}\times V_{1}}{T_{1}} =\frac{P_{2}\times V_{2}}{T_{2}}\\\frac{3.2atm\times 65cm^{3} }{278.6K} =\frac{1.0atm\times V_{2}}{298.2K}\\V_{2}=223cm^{3}[/tex]

The final radius of the bubble is:

V = 4/3 × π × r³

223 cm³ = 4/3 × π × r³

r = 3.8 cm

Final answer:

The balanced half-reaction for the oxidation of gaseous nitrogen dioxide to nitrate ion in acidic aqueous solution is 4NO2(g) + O2(g) + 4H+(aq) -> 4NO3-(aq).

Explanation:

The balanced half-reaction for the oxidation of gaseous nitrogen dioxide to nitrate ion in acidic aqueous solution is:

4NO2(g) + O2(g) + 4H+(aq) → 4NO3-(aq)

In this reaction, nitrogen dioxide (NO2) is oxidized to nitrate ion (NO3-), and oxygen gas (O2) is reduced to water (H2O). The reaction takes place in acidic surroundings, indicated by the presence of hydrogen ions (H+).

Answer: The p-function of [tex]Zn^{2+}[/tex] and [tex]NO_3^{-}[/tex] ions are 2.51 and 2.14 respectively.

Explanation:

p-function is defined as the negative logarithm of any concentration.

We are given:

Millimolar concentration of zinc nitrate = 3.1 mM

Millimolar concentration of calcium nitrate = 4.2 mM

Converting this into molar concentration, we use the conversion factor:

1 M = 1000 mM

1 mole of zinc nitrate produces 1 mole of zinc ions and 2 moles of nitrate ions

Concentration of zinc ions = 0.0031 M

Concentration of nitrate ions in zinc nitrate, [tex]M_1=(2\times 0.0031)=0.0062M[/tex]

1 mole of calcium nitrate produces 1 mole of calcium ions and 2 moles of nitrate ions

Concentration of calcium ions = 0.0042 M

Concentration of nitrate ions in calcium nitrate, [tex]M_2=(2\times 0.0042)=0.0084M[/tex]

To calculate the concentration of nitrate ions in the solution, we use the equation:

[tex]M=\frac{M_1V_1+M_2V_2}{V_1+V_2}[/tex]

Putting values in above equation, we get:

[tex]M=\frac{(0.0062\times 1)+(0.0084\times 1)}{1+1}\\\\M=0.0073M[/tex]

Calculating the p-function of zinc ions and nitrate ions in the solution:

[tex]\text{p-function of }Zn^{2+}\text{ ions}=-\log[Zn^{2+}][/tex]

[tex]\text{p-function of }Zn^{2+}\text{ ions}=-\log(0.0031)\\\\\text{p-function of }Zn^{2+}\text{ ions}=2.51[/tex]

[tex]\text{p-function of }NO_3^{-}\text{ ions}=-\log[NO_3^{-}][/tex]

[tex]\text{p-function of }NO_3^{-}\text{ ions}=-\log(0.0073)\\\\\text{p-function of }NO_3^{-}\text{ ions}=2.14[/tex]

Hence, the p-function of [tex]Zn^{2+}[/tex] and [tex]NO_3^{-}[/tex] ions are 2.51 and 2.14 respectively.

The p-function values for Zn2+ and NO3- are 9.6 and 15.5, respectively.

The p-function for Zn2+ is 9.6, and the p-function for NO3- is 15.5.

Answer:

The answer to your question is pH = 2.28

Explanation:

Chemical reaction

                    HCOOH(ac)  +  H₂O  ⇔   H₃O⁺   +  HCOO⁻

I                        0.15                 --              0               0

C                       - x                   --             +x              +x

F                     0.15 - x             --               x                x

Write the equation of equilibrium

                             [tex]ka = \frac{[H3O][HCOO]}{[HCOOH]}[/tex]

Substitution

                             [tex]1.8 x 10^{-4} = \frac{[x][x]}{0.15 - x}[/tex]

But     0.15 - x ≈   0.15

                             [tex]1.8 x 10^{-4} = \frac{x^{2} }{0.15}[/tex]

Solve for x

                             x² = (0.15)(1.8 x 10⁻⁴)

Simplification

                             x² = 0.000027

Result

                             x = 0.0052

                          pH = -log [H₃O⁺]

Substitution

                          pH = - log [0.0052]

Simplification and result

                         pH = 2.28

The pH of a 0.15M solution of HCOOH with a Ka of 1.8×10⁻´ is calculated using an ICE table, which leads to an equilibrium pH of approximately 2.78.

To calculate the pH of a 0.15M solution of HCOOH with a Ka of 1.8×10⁻´, we can set up an ICE (Initial, Change, Equilibrium) table to find the equilibrium concentrations of the species in solution:

Applying the expression for Ka, we have:

Ka = 1.8×10⁻´ = (x)(x) / (0.15 - x)

Assuming x is small and can be ignored in the denominator, we simplify to:

Ka = x² / 0.15

Solving for x (which represents [H⁺]), we find that x is approximately equal to the square root of (Ka × [HCOOH]), thus:

[H⁺] = √(1.8×10⁻´ × 0.15) ≈ 1.65×10⁻²

To find the pH, we take the negative logarithm of [H⁺]:

pH = -log(1.65×10⁻²) ≈ 2.78

The pH of the solution at equilibrium is approximately 2.78.

Answer:

Hi

Williamson's ether reactions imply that an alkoxide reacts with a primary haloalkane. Alkoxides consisting of the conjugate base of an alcohol and are formed by a group R attached to an oxygen atom. They are often written as RO–, where R is the organic substituent (Step 1).

Sn2 reactions are characterized by the reversal of stereochemistry at the site of the leaving group. Williamson's synthesis is no exception and the reaction is initiated by the subsequent attack of the nucleophile. This requires that the nucleophile and electrophile be in anti-configuration (Step 2).

As an example (figure 3).

In the attached file are each of the steps of Williamson's synthesis.

Explanation:

The Williamson ether synthesis involves the reaction of an alkyl halide and an alcohol to form an ether. In the specific case of methyl t-butyl ether, methanol and t-butyl bromide are used. The reaction proceeds via the formation of an alkoxide ion which attacks the alkyl halide.

The Williamson ether synthesis is a method used to synthesize ethers. In this reaction, an alkyl halide (in this case, an alkyl bromide) reacts with an alcohol in the presence of a strong base to form an ether. The general mechanism involves the formation of an alkoxide ion, which then attacks the alkyl halide to form the desired ether.

In the specific case of methyl t-butyl ether, the alkyl bromide used is t-butyl bromide and the alcohol used is methanol. The reaction is as follows:

CH3OH + (CH3)3CBr → [(CH3)3CO]– + CH3Br
CH3Br + [(CH3)3CO]– → (CH3)3COCH3 + Br–

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Answer:

MnO- Manganese Oxide

Explanation:

Empirical formula: This is the formula that shows the ratio of elements

present in a  

compound.

How to determine Empirical formula

1. First arrange the symbols of the elements present in the compound

alphabetically to  determine the real empirical formula. Although, there

are exceptions to this rule, E.g H2So4

2. Divide the percentage composition by the mass number.

3. Then divide through by the smallest number.

4. The resulting answer is the ratio attached to the elements present in

a compound.

                                                                              Mn                         O    

% composition                                                      72.1                      27.9    

Divide by mass number                                       54.94                     16  

                                                                               1.31                      1.74    

Divide by the smallest number                         1.31                      1.31                          

                                                                               1                    1.3

The resulting ratio is 1:1

Hence the Empirical formula is MnO, Manganese oxide

Answer:

Xenon

Explanation:

Avogadro’s number represent the number of the constituent particles which are present in one mole of the substance. It is named after scientist Amedeo Avogadro and is denoted by [tex]N_0[/tex].

Avogadro constant:-

[tex]N_a=6.023\times 10^{23}[/tex]

Let the molar mass of the element is x g/mol

So,

[tex]6.023\times 10^{23}[/tex] atoms have a mass of x g

Also,

[tex]3.89\times 10^{24}[/tex] atoms have a mass of [tex]\frac{x}{6.023\times 10^{23}}\times 3.89\times 10^{24}[/tex] g

This mass is equal to 848 g

So,

[tex]\frac{x}{6.023\times 10^{23}}\times 3.89\times 10^{24}=848[/tex]

x= 131.3 g/mol

This mass correspond to xenon.

Answer:

When the neuron is at rest, what is primarily responsible for moving potassium ions OUT of the cell?​

The answer is "a concentration gradient"

Explanation:

A neuron is the main component of nervous tissue and it transmits information by electro-chemical signalling. For the nervous system to function, neurons must be able to send and receive signals.

A neuron at rest is negatively charged. The negative charge within the cell is created by the cell membrane being more permeable to potassium ion movement than sodium ion movement. At rest, there is a high concentration of potassium ions (K+) inside the cell compared to the extracellular fluid due to a net movement with the concentration gradient.

A concentration gradient acts on K+ (potassium ions). High number of potassium ions reside inside the cell, a chemical gradient occurs and pushes potassium out of the cell. The neuron membrane is more permeable to potassium ions than to other ions allowing it to selectively move out of the cell taking a positive charge with it down its concentration gradient.

Answer:

The unknown gas is H2

Explanation:

Step 1: Data given

A mixture contains:

5.4 grams of Ar

5.4 grams of Ne

5.4 grams of X2

Molar mass of Ar = 39.95 g/mol

Molar mass of Ne = 20.18 g/mol

Step 2: Calculate moles of gases

.Moles of Ne = 4.5grams /20.18 g/mol = 0.223 moles

Moles of Ar= 4.5 grams /39.95 g/mol = 0.113 moles

Step 3: Calculate volume of gases

Volume of Ne =22.4 * 0.223 = 5

Volume of Ar =22.4 * 0.1525 = 2.53 L

Volume of unknown gas = 69.03 - 5 - 2.53 = 61.5 L

Step 4: Calculate moles of unknown gas

Moles of unkown gas =61.5/22.4 = 2.75 moles

Step 5: Calculate molar mass of unknown gas

Molar mass = mass / moles

Molar mass = 5.4 grams /2.75 moles ≈ 2 g/mol

The unknown gas is H2

The amount of hydrogen chloride that can be made is 1064 g

Why?

The two reactions are:

2H₂O → 2H₂ + O₂ 75.3 % yield

H₂ + Cl₂ → 2HCl 69.8% yield

We have to apply a big conversion factor to go from grams of water (The limiting reactant), to grams of HCl, the final product. We have to be very careful with the coefficients and percentage yields!

[tex]500.0gH_2O*\frac{1moleH_2O}{18.01 gH_2O}*\frac{2 moles H_2}{2 moles H_2O}*\frac{2.015g H_2}{1 mole H_2}*\frac{75.3 actual g}{100 theoretical g}=42.12 g H_2[/tex]

[tex]42.12H_2*\frac{1 mole H_2}{2.015gH_2}*\frac{2 moles HCl}{1 mole H_2}*\frac{36.46g}{1 mole HCl}*\frac{69.8 actualg}{100 theoreticalg} =1064gHCl[/tex]

Have a nice day!

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The amount of hydrogen chloride yield in the second reaction is 1065.7 g.

The two given reactions

The amount of hydrogen gas yield in the first reaction is calculated as follows;

[tex]\frac{500 \ g\ H_2O}{18 \ g \ H_2O} \times (2 \ mol\ H_2) \times 0.753= 41.83 \ g \ H_2[/tex]

The amount of hydrogen chloride yield in the second reaction is calculated as follows;

[tex]41.83 \ g \times (36.5 \ HCl) \times 0.698 = 1065.7 \ g \ HCl[/tex]

Thus, the amount of hydrogen chloride yield in the second reaction is 1065.7 g.

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Answer: 6.63

Explanation:Please see attachment for explanation

Final answer:

The pKa value for the weak acid HA can be calculated using the equation pKa = -log(Ka). Given the equilibrium concentrations, we can use the equation Ka = ([H+][A-])/[HA] to solve for Ka. After finding Ka, we can calculate pKa using the formula pKa = -log(Ka).

Explanation:

The pKa value for the weak acid HA can be calculated using the equation:

pKa = -log(Ka)

Given the equilibrium concentrations of [HA] = 0.170 M, [H+] = 2.00 × 10^-4 M, and [A-] = 2.00 × 10^-4 M, we can use the equation:

Ka = ([H+][A-])/[HA]

Plugging in the values gives us:

Ka = (2.00 × 10^-4)(2.00 × 10^-4)/(0.170)

After solving for Ka, we can calculate pKa using the formula pKa = -log(Ka).

Answer: sulfur pentafluoride

Explanation:

The rules for naming of binary molecular compound :

In the given formula, the lower group number element is written first in the name and keep its element name and the higher group number is written second.

First element i.e. sulphur in the formula is named first and keep its element name.

1) Gets a prefix if there is a subscript on it such as mono for 1, di for 2, tri for 3 and so on.

Second element i.e. fluorine is named second.

1) Use the root of the element name, if it is an anion then use suffix (-ide).

2) Always use a prefix on the second element such as mono for 1, di for 2, tri for 3 and so on.

Therefore, the chemical name of compound [tex]SF_5[/tex] is sulfur pentafluoride

The name of the molecular compound SF₅ is sulfur pentafluoride. Therefore, option A is correct.

A molecular compound is a compound composed of two or more nonmetallic elements. In molecular compounds, atoms are joined together by covalent bonds, which involve the sharing of electrons between atoms.

Examples of molecular compounds include water (H₂O), carbon dioxide (CO₂), methane (CH₄), and ammonia (NH₃). Molecular compounds often have specific naming conventions based on the elements present and their respective ratios, such as using prefixes to indicate the number of atoms for each element.

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