When calcium metal is placed in water, hydrogen gas bubbles out, leaving a highly alkaline (basic) solution. Write the balanced chemical equation for this reaction. Phases are optional. Do not write an ionic equation (i.e., the answer should not show any charges).

Answer:

[tex]Ag_2SO_4[/tex]

Explanation:

Formula for the calculation of no. of Mol is as follows:

[tex]mol=\frac{mass\ (g)}{molecular\ mass}[/tex]

Molecular mass of Ag = 107.87 g/mol

Amount of Ag = 5.723 g

[tex]mol\ of\ Ag=\frac{5.723\ g}{107.87\ g/mol} =0.05305\ mol[/tex]

Molecular mass of S = 32 g/mol

Amount of S = 0.852 g

[tex]mol\ of\ S=\frac{0.852\ g}{32\ g/mol} =0.02657\ mol[/tex]

Molecular mass of O = 16 g/mol

Amount of O = 1.695 g

[tex]mol\ of\ O=\frac{1.695\ g}{16\ g/mol} =0.10594\ mol[/tex]

In order to get integer value, divide mol by smallest no.

Therefore, divide by 0.02657

[tex]Ag, \frac{0.05305}{0.02657} \approx 2[/tex]

[tex]S, \frac{0.02657}{0.02657} \approx 1[/tex]

[tex]O, \frac{0.10594}{0.02657} \approx 4[/tex]

Therefore, empirical formula of the compound = [tex]Ag_2SO_4[/tex]

Answer:

The empirical formula is Ag2SO4

Explanation:

Step 1: Data given

A compound contains Ag, S and O

Mass of Ag = 5.723 grams

Mass of S = 0.852 grams

Mass of O = 1.695 grams

Molar mass of Ag = 107.87 g/mol

Molar mass of S = 32.065 g/mol

Molar mass of O = 16 g/mol

Total mass = Mass of Ag + mass of S + mass of O = 5.723g + 0.852g + 1.695g = 8.27 grams

Step 2: Calculate moles

Moles = mass / molar mass

Moles of Ag = 5.723 grams / 107.87 g/mol = 0.05305 moles

Moles S = 0.852 grams / 32.065 g/mol = 0.0266 moles

Moles O = 1.695 grams / 16.00 g/mol = 0.1059 moles

Step 3: Calculate mole ratio

We divide by the smallest amount of moles

Ag: 0.05305/0.0266 = 2

S: 0.0266/0.0266 = 1

O: 0.1059 / 0.0266 = 4

The empirical formula is Ag2SO4

Answer:

POH= 13

Explanation:

A student preparing for the experiments inadvertently adds an additional 400 mL of the same acid solution to the dissolution vessel. What will be the new pOH of this solution?

PH is the measure of the degree of acidity of a solution.

POH is the measure of the degree of alkalinity of a solution

Note that  pH + pOH = 14

if concentration remains the same, then volume changes will not affect pH.

The pH of the solution is given as

PH= -log[H+].

For this experiment, the dissolution vessel contains 0.1 M HCl, no matter the initial volume of the acid solution

For the molar concentration of the cation, we can propose that  a strong acid will  dissociate completely,

[H+] = 0.1 = 1 x 10-1 M

. substituting the concentration of the cation

PH=-log[1 x 10-1] = 1.

Note that  pH + pOH = 14 for any aqueous solution.

we say that

the pOH = 14 - pH.

pOH = 14-1 = 13.

Answer:

- Net Input: NAD⁺, coenzyme A, pyruvate

- Net Output: NADH, acetyl CoA, CO₂

- Not input or output: O₂, ADP, glucose and ATP

Explanation:

Hello,

In this case, it is important to recall that acetyl-CoA is produced either by oxidative decarboxylation of pyruvate derived from glycolysis, which is carried out into the mitochondrial matrix, by cause of the oxidation of high-order fatty acids, or by oxidative degradation of very specific amino acids. Acetyl-CoA then enters in the citric acid cycle where it is oxidized in the light of energy production.

In this manner, during such processes, there are some net inputs and outputs, therefore, they are sorted as show below, considering there some of them not classified neither as input nor output:

- Net Input: NAD⁺, coenzyme A, pyruvate

- Net Output: NADH, acetyl CoA, CO₂

- Not input or output: O₂, ADP, glucose and ATP

Best regards.

Answer:

The empirical formula is Li2CO3

Explanation:

Step 1: Data given

Suppose the mass of the compound = 100 grams

The compound contains :

18.7% Li = 18.7 grams of Li

16.3 % C = 16.3 grams of C

65.0% O = 65.0 grams of O

Total = 100%

Molar mass of Li = 6.94 g/mol

Molar mass of C = 12.01 g/mol

Molar mass of O = 16 g/mol

Step 2: Calculate moles

Moles = Mass / molar mass

Moles Li = 18.7 grams / 6.94 g/mol = 2.65 moles

Moles C = 16.3 grams / 12.01 = 1.36 moles

Moles O = 65.0 grams / 16.0 g/mol = 4.06 moles

Step 3: Calculate the mol ratio

We divide by the smallest number of moles

Li: 2.65/1.36 ≈ 2

C: 1.36/1.36 = 1

O: 4.06/1.36 ≈ 3

The empirical formula is Li2CO3

Answer: True the bicarbonate mixture can help save time and few routine.

Explanation:

For the purpose of making dialysate for hemodialysis patient therapies a bicarbonate mixing and delivering systems designed to prepare a liquid sodium bicarbonate formulation comes in handy.

Certain systems like the SDS unit also allow for the transfer and distribution of acid concentrate solutions. We also provide stand-alone acid concentrate delivery systems using a variety of holding tanks and delivery methods. 

A challenge for hemodialysis providers is to properly provide bicarbonate solution in a cost effective manner. Preparation and disinfection can be time-consuming and labor intensive.

Bicarbonate however can corrode certain metals and painted surfaces leaving your preparation area encrusted and grimy.

Furthermore, if not mixed properly, bicarbonate can negatively affect the dialysate solution.

The answer to the above is true the bicarbonate mixture can help save time and few routine.

Answer: 116 calories of energy

Explanation:

A calorie is a non-standard unit of energy.

On combustion,

1 gram Carbohydrates = 4 calories,

1 gram protein = 4 calories,

1gram fat = 9 calories.

Therefore,

15 grams of carbohydrates = (15*4) = 60

5 grams of protein = (5*4) = 20

4 grams of fats = (4*9) = 36

Then add up: 60 + 20 + 36 = 116 calories of energy

Answer:

The spatial separation between the two isotopes after they have traveled through a half cycle is [tex]1.65\times 10^{2}m[/tex]

Explanation:

The moving of the isotopes must excert centripetal force which is equals to the magnetic force on the ions due to the magnetic.

The centripetal force of the ions can be calculated by the following formula.

[tex]F_{c}=\frac{mv^{2}}{r}.............(1)[/tex]

Magnetic force on this ions calculated by the following formula.

[tex]F_{m}=qvB.............(2)[/tex]

Equate the equations (1) and (2)

[tex]F_{c}=F_{m}[/tex]

[tex]\frac{mv^{2}}{r}=qvB[/tex]

[tex]v=\frac{mv}{qB}[/tex]

Substitute the values of the both isotopes.

[tex]r_{12}=\frac{1.993\times 10^{-26}\times 6.13105}{1.6\times 10^{-19}\times 0.7700}=9.9041\times 10^{-2}m[/tex]

[tex]r_{13}=\frac{2.159\times 10^{-26}\times 6.13105}{1.6\times 10^{-19}\times 0.7700}=1.0729\times 10^{-1}m[/tex]

Now the distance traveled by both isotopes by half circle.

Therefore, the distance between the two isotopes is the diameter of the circle which is equal to the twice the radius.

[tex]Seperation=2\times(r_{13}-r_{12})[/tex]

[tex]Seperation=2\times(1.0729\times 10^{-1}-9.9401\times 10^{-2})=1.650\times 10^{-2}m[/tex]

Therefore, The spatial separation between the two isotopes after they have traveled through a half cycle is [tex]1.65\times 10^{2}m[/tex]

The question is incomplete, here is a complete question.

A sled slides along a horizontal surface for which the coefficient of kinetic friction is 0.25. Its velocity at point A is 8.0m/s and at point B is 5.0m/s.

Use the impulse-momentum theorem to find how long the sled takes to travel from A to B.

Answer : The sled takes time to travel from A to B is, 1.2 seconds.

Explanation :

Impulse-momentum theorem:

Impulse = Change in momentum

[tex]\Delta P=F\times \Delta t\\\\m\nu_f-m\nu_i=F\times \Delta t\\\\m(\nu_f-\nu_i)=\mu \times m\times g\times \Delta t\\\\(\nu_f-\nu_i)=\mu \times g\times \Delta t[/tex]

where,

m = mass

g = acceleration due to gravity = [tex]9.8m/s^2[/tex]

[tex]\mu[/tex] = coefficient of kinetic friction = 0.25

[tex]\nu_f[/tex] = final velocity = 8.0 m/s

[tex]\nu_i[/tex] = initial velocity = 5.0 m/s

t = time

Now put all the given values in the above formula, we get:

[tex](\nu_f-\nu_i)=\mu \times g\times \Delta t[/tex]

[tex](8.0-5.0)=0.25\times 9.8\times \Delta t[/tex]

[tex]\Delta t=1.2s[/tex]

Therefore, the sled takes time to travel from A to B is, 1.2 seconds.

The sled is decelerating due to kinetic friction, and we can calculate this deceleration using the equations of motion and the given coefficient of kinetic friction. The normal force on the sled can also be found by dividing the frictional force by the coefficient of kinetic friction.

This problem pertains to the concept of kinetic friction and motion in Physics. We know that the sled's velocity decreases from 8.0m/s to 5.0m/s over some distance, which implies that it's decelerating due to kinetic friction. We can make use of the equation of motion v^2 = u^2 - 2*a*d where v is final velocity, u is initial velocity, a is deceleration (friction) and d is distance.

First, let's find the deceleration caused by the friction force using this equation. As we know that the frictional force can be expressed as F_kinetic = μk*m*g, where m is the mass of the sled (not given) and g is the acceleration due to gravity (9.8 m/s^2). From here, we can also find the normal force on the sled by knowing that the normal force is equal to kinetic friction divided by the coefficient of kinetic friction.

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Answer:

4. More energy and a different charge.

Electrons in the second shell of a beryllium atom possess more energy due to being further from the nucleus but have the same negative charge as those in the first shell.

When comparing the energy and charge of electrons in different shells of a beryllium (Be) atom, we find that electrons in the first shell (1s) are closer to the nucleus and thus have lower energy than electrons in the second shell (2s and 2p). The second shell is at a higher energy level due to being further away from the nucleus and its electrons experiencing less attractive force because of the shielding effect of the inner-shell electrons. Electron-electron repulsion also impacts the energies of electrons in the atom. Despite the differences in energy levels, the charge on electrons in any shell of an atom remains the same, negative in nature.

Therefore, compared to the electrons in the first shell of a Be atom, the electrons in the second shell have more energy but the same charge. This leads us to the correct answer to the student's question: (3) more energy and the same charge.

Answer:

See explanation below

Explanation:

The question is incomplete, however, I found a question very similar to this, and I'm assuming this is the question you are asking to answer. If it's not, please tell me which one it is. Here's a tip for you to get an idea of how to solve.

Picture 1, would be the original question. Picture 2 is the answer of it.

Now, This is a E1 reaction where this type of reactions are taking place in two steps. The first step is the formation of the carbon cation, this step is always slow. The secon step is the addition of a nucleophyle, or, in this case, formation of a pi bond, and we get a alkene.

Hope this can help you

The minor alkene product in the reaction of 2-bromobutane with methanol in the presence of HBr is trans-2-butene. The major product is cis-2-butene. The stepwise mechanism for the formation of the major product is shown above.

The reaction of 2-bromobutane with methanol in the presence of HBr is a Markovnikov addition reaction. This means that the bromide ion will add to the carbon of the 2-bromobutane molecule that is more substituted with alkyl groups. In this case, the carbon that is more substituted with alkyl groups is the secondary carbon.

The carbocation intermediate that is formed in step 1 of the mechanism is a secondary carbocation. Secondary carbocations are more stable than primary carbocations, so they are more likely to form.

In step 2 of the mechanism, the methanol molecule can attack the carbocation intermediate from either side. However, the methanol molecule is more likely to attack the carbocation intermediate from the side that is less hindered by the other alkyl groups. In this case, the side of the carbocation intermediate that is less hindered by the other alkyl groups is the side that is facing the hydrogen atom of the methanol molecule.

In step 3 of the mechanism, the proton from the HBr molecule is transferred to the oxygen atom of the methyl ether intermediate. This is a simple acid-base reaction.

The overall mechanism of the reaction is shown below:

2-bromobutane + methanol + HBr → cis-2-butene + methyl bromide + water

For more such information on: alkene

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Answer:

C. All electron carriers are mobile and hydrophobic

Explanation:

Hello,

In this case, it is widely known that the electron carriers move inside the inner mitochondrial membrane and consequently move electrons from one to another. In such a way, they are mobile, therefore they are largely hydrophobic as long as they are inside the membrane.

For instance, the cytochrome c is a water-soluble protein in a large range, therefore, the answer is: C. All electron carriers are mobile and hydrophobic.

Best regards.

Answer:

A) False

B) True

Explanation:

The question is incomplete. The complete question is as follows.

There are two statements. Identify which statement is true or false.

A) For any substance, solid, liquid or gas, mass increases as volume increases.  

B) Density represents mass per volume.

Statement A

Since the amount of matter is fixed, an increase in volume doesn't cause an increase in mass. For example, a test tube, with a volume of 10ml, contains liquid fragrance. When heated, the fragrance vaporizes and fill the entire room, until you can smell it in every corner. Note that previously the fragrance of same mass was occupying a 10ml space in the test-tube. After heating, the same mass of fragrance occupied the entire room (volume increased but mass remained constant).

It happened because as the fragrance is heated, the space between molecules increases causing them to occupy more space BUT in the whole scenarios, the number of molecules stayed unaffected. Hence statement A is false.

Statement B

Density measures the compactness of the substance. It's defined as the amount of mass per unit volume. It increases as the atoms/molecules in a substance comes closer. Density of solids are greatest as the atoms/molecules are tightly packed together. It is least in gases as the molecules are far apart from each other. Hence the statement is true.

Answer:

The answer to your question is below

Explanation:

There are 4 types of chemical reactions:

- Synthesis is when two elements or compounds form only one compound.

- Decomposition is when 1 compound is broken into 2 or more products.

- Single replacement is when one element is replaced by another element.

- Double replacement is when the cations of two compounds are interchanged.

6.- Synthesis                            Al   +   3Cl   ⇒    AlCl₃

7.- Double replacement       2NaOH  +   H₂SO₄   ⇒   Na₂SO₄  +   2H₂O

8.- Decomposition                   Ni(ClO₃)₂   ⇒   NiCl₂   +   3O₂

9.- Combustion                      2C₄H₁₀  + 13O₂   ⇒   8CO₂   +   10H₂O

10.- Single replacement         Zn  +  2HCl  ⇒   ZnCl₂   +  H₂

(a) True - Dispersion forces strengthen with increased molecular polarizability.

(b) False - Dispersion forces in noble gases increase down the periodic table.

(c) False - Dispersion forces can be significant compared to dipole–dipole interactions.

(d) False - Dispersion forces depend on molecular size and electron cloud, not just molecular shape.

**(a) True:**

For molecules with similar molecular weights, dispersion forces (Van der Waals forces) become stronger as the molecules become more polarizable. Polarizability is a measure of how easily the electron cloud in a molecule can be distorted, and larger, more complex molecules exhibit greater polarizability, leading to stronger dispersion forces.

**(b) False:**

For noble gases, dispersion forces increase as you go down the column in the periodic table, not decrease. This is because larger atoms have more electrons and a greater electron cloud, resulting in stronger dispersion forces. As a consequence, boiling points generally increase down the noble gas column.

**(c) False:**

In terms of total attractive forces, dipole–dipole interactions are not always greater than dispersion forces. Dispersion forces can be significant, especially in large, nonpolar molecules where temporary dipoles can induce other molecules to develop temporary dipoles.

**(d) False:**

Dispersion forces between linear molecules are not always greater than those between nearly spherical molecules. The shape of the molecule influences the surface area of contact, affecting dispersion forces. Spherical molecules can have significant dispersion forces, especially if they are large and have a high number of electrons.

Answer:

The answer to your question is 0.33 liters

Explanation:

Data

Volume = ?

Molarity = 1.5 M

number of moles = 0.5

Formula

[tex]Molarity = \frac{number of moles }{volume}[/tex]

Solve for V

[tex]Volume = \frac{number of moles}{molarity}[/tex]

Substitution

[tex]Volume = \frac{0.5}{1.5}[/tex]

Simplification and result

       Volume = 0.33 l