Chemistry College

A solution is made by dissolving 38.81 grams of nickel (II) sulfate, NiSO4, in enough water to make 0.467
liters of solution. Calculate the molarity of this solution.

Answers

Answer 1

The molarity of the NiSO₄ solution made by dissolving 38.81 grams of nickel (ii) sulfate, NiSO₄, in enough water to make 0.467 liters of solution is 0.535 M

How do i determine the molarity of the solution?

First, we shall obtain the mole of 38.81 grams of nickel (ii) sulfate, NiSO₄. Details below:

Mass of NiSO₄ = 38.81 grams Molar mass of NiSO₄ = 154.75 g/molMole of NiSO₄ = ?

Mole of NiSO₄ = mass / molar mass

= 38.81 / 154.75

= 0.25 mole

Now, we shall determine the molarity of the solution. Details below:

Mole of NiSO₄ = 0.25 moleVolume of solution = 10.467 LMolarity of solution = ?

Molarity of solution = mole / volume

= 0.25 / 0.467

= 0.535 M

Thus, the molarity of the solution is 0.535 M

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Related Questions

What is the coefficient for sodium chloride when this equation is balanced?

Answers

Answer:

To resolve this, we need to place the coefficient “2” in front of the sodium in the reactant, to give the equation shown below. 2 Na (s) + Cl 2 (g) → 2 NaCl (s) In this equation, there are two sodiums in the reactants, two sodiums in the products, two chlorines in the reactants and two chlorines in the products; the equation is now balanced.

Explanation:

how do you balance this equation
2h2s+3o2+so2

Answers

The balanced equation is: 4 [tex]H_2S[/tex]+ 3 [tex]O_2[/tex]→ 4 [tex]SO_2[/tex]+ 8 [tex]H_2O[/tex]

The given chemical equation is unbalanced. To balance it, we need to adjust the coefficients in front of each chemical species until the number of atoms on both sides of the equation is equal.

The unbalanced equation is:

2 [tex]H_2S[/tex]+ 3 [tex]O_2[/tex]→ [tex]SO_2[/tex]

Let's start by balancing the sulfur (S) atoms. We have two sulfur atoms on the left side and one sulfur atom on the right side. To balance the sulfur, we can place a coefficient of 2 in front of the [tex]SO_2[/tex]:

2 [tex]H_2S[/tex]+ 3 [tex]O_2[/tex]→ 2 [tex]SO_2[/tex]

Now, let's balance the hydrogen (H) atoms. We have four hydrogen atoms on the left side (2 from each [tex]H_2S[/tex]) and none on the right side. To balance the hydrogen, we can place a coefficient of 4 in front of the water (H2O) on the right side:

2 [tex]H_2S[/tex]+ 3 [tex]O_2[/tex]→ 2 [tex]SO_2[/tex]+ 4 [tex]H_2O[/tex]

Finally, let's balance the oxygen (O) atoms. We have six oxygen atoms on the right side (3 from [tex]O_2[/tex] and 3 from 2 [tex]SO_2[/tex]) and three on the left side (2 from [tex]H_2S[/tex]). To balance the oxygen, we can place a coefficient of 3/2 in front of the O2:

2 [tex]H_2S[/tex]+ (3/2) [tex]O_2[/tex]→ 2 [tex]SO_2[/tex]+ 4 [tex]H_2O[/tex]

To remove the fractional coefficient, we can multiply all coefficients by 2:

4 [tex]H_2S[/tex] + 3 [tex]O_2[/tex]→ 4 [tex]SO_2[/tex]+ 8 [tex]H_2O[/tex]

Now the equation is balanced, with an equal number of atoms on both sides. The balanced equation is:

4 [tex]H_2S[/tex]+ 3 [tex]O_2[/tex]→ 4 [tex]SO_2[/tex]+ 8 [tex]H_2O[/tex]

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which of the following choice fits in the blank to complete the equation

Answers

To complete the equation, the blank should be filled with "ΔH°- ".

Option B is correct.

How do we explain?

The equation represents the standard Gibbs free energy change (ΔG°) in a chemical reaction in terms of the enthalpy change (ΔH°) and the entropy change (ΔS°). The equation is:

ΔG° = ΔH° - TΔS°

ΔG° represents the change in Gibbs free energy,

ΔH° represents the change in enthalpy,

T represents the temperature in Kelvin, and ΔS° represents the change in entropy.

The minus sign indicates that the change in Gibbs free energy is determined by the difference between the enthalpy and the product of temperature and entropy.

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which of the following choices fits in the blank to complete the equation

Answers

The equation that can be able to complete the reaction as it has been shown is option C

What is the free energy?

Free energy takes into account both the enthalpy (heat content) and entropy (degree of disorder) of a system. It provides a measure of the system's ability to do useful work or drive chemical reactions.

The formula for calculating free energy is:

G = H - TS

where:

G is the Gibbs free energy

H is the enthalpy (heat content) of the system

T is the temperature in Kelvin

S is the entropy (degree of disorder) of the system

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Use your data, the equation to the right, and the specific heat of water (4.184 J/g C) to compute the specific heat values of each metal. Use a calculator and round to the nearest hundredth place.

Answers

The heat capacity for the metals are;

Aluminum -  0.89

Copper - 0.11

Iron - 0.44

Lead - 0.12

What is the specific heat?

The specific heat of a substance is denoted by the symbol "C" and is typically measured in units of J/g·°C (joules per gram per degree Celsius) or cal/g·°C (calories per gram per degree Celsius).

The specific h

We have that;

For Aluminum;

c = 4.184 * 39.85 * 4.7/11.98 * 72.9

= 783.6/873.3

= 0.89

For Copper;

c =  4.184 * 12.14 * 1.9/12.14 * 75.4

= 96.5/915.3

= 0.11

For Iron

c =  4.184 * 40.24 * 2.4/12.31 * 75.1

= 404.1/924.5

= 0.44

For Lead

c = 4.184 * 39.65 * 0.7/12.46 * 76.7

c = 116.1/955.68

= 0.12

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The table describes a gas stored in four different containers. Properties of Stored Gas Container Properties 1 · Low number of collisions with container walls · Medium average kinetic energy · Large number of particles 2 · Large number of collisions with container walls · Medium average kinetic energy · Small number of particles with little spaces between them 3 · Large number of collisions with container walls · High average kinetic energy · Large number of particles with large spaces between them 4 · Few collisions with container walls · Low average kinetic energy · Small number of particles Which container has gas stored at the highest temperature? 1 2 3 4

Answers

Container 3 has the gas stored at the highest temperature.

Temperature is a measure of the average kinetic energy of the particles in a substance. In the given table, it is stated that container 3 has a large number of collisions with container walls, high average kinetic energy, and large number of particles with large spaces between them.

These properties indicate that the gas in container 3 has higher kinetic energy and more vigorous movement compared to the other containers.

Container 1 has a low number of collisions with container walls and a medium average kinetic energy. This suggests that the gas in container 1 has lower energy and less movement than the gas in container 3.

Container 2 has a large number of collisions with container walls, but it also has a small number of particles with little spaces between them. While the collisions may be frequent, the limited number of particles and the lack of space between them may result in lower overall kinetic energy compared to container 3.

Container 4 has few collisions with container walls, low average kinetic energy, and a small number of particles. These properties indicate that the gas in container 4 has the lowest energy and least movement among all the containers.

Container 3

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