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5.11 g of MgSO₄ is placed into 100.0 mL of water. The water's temperature increases by 6.70°C. Calculate ∆H, in kJ/mol, for the

cupoosta [38]

Answer: Thus ∆H, in kJ/mol, for the dissolution of MgSO₄ is -66.7 kJ

Explanation:

To calculate the entalpy, we use the equation:

$q=mc\Delta T$

where,

q = heat absorbed by water = ?

m = mass of water = ${\text {volume of water}}\times {\text {density of water}}=100.0ml\times 1.00g/ml=100.0g$

c = heat capacity of water = 4.186 J/g°C

$\Delta T$ = change in temperature = $6.70^0C$

$q=100.0g\times 4.184J/g^0C\times 6.70^0C=2803.3J=2.8033kJ$

Sign convention of heat:

When heat is absorbed, the sign of heat is taken to be positive and when heat is released, the sign of heat is taken to be negative.

The heat absorbed by water will be equal to heat released by $MgSO_4$

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Given mass = 5.11 g

Molar mass = 120 g/mol

Putting values in above equation, we get:

$\text{Moles of }MgSO_4=\frac{5.11g}{120g/mol}=0.042mol$

0.042 moles of $MgSO_4$ releases = 2.8033 kJ

1 mole of $MgSO_4$ releases = $\frac{2.8033 kJ}{0.042}\times 1=66.7kJ$

Thus ∆H, in kJ/mol, for the dissolution of MgSO₄ is -66.7 kJ

3 0

3 years ago

A student has two samples of NaCl, each one from a different source. Assume that the only potential contaminant in each sample i

bija089 [108]

Answer:

The correct option is;

A. Which sample has the higher purity

Explanation:

The information given relate to the presence of two samples of NaCl, from different sources

The only potential contaminant in each of the sources = KCl

The content of the sample = NaCl

The molar mass of NaCl = 58.44 g/mol

The molar mass of KCl = 74.5513 g/mol

Let the number of moles of KCl in the sample = X

For a given mass of NaCl, KCl mixture, we have;

The molar mass of potassium = 39.0983 g/mol

The molar mass of chlorine = 35.453 g/mol

The molar mass of sodium ≈ 23 g/mol

Therefore;

Each mole of KCl, will yield 35.453 g/mol per 74.5513 g/mol of KCl

While each mole of NaCl will yield 35.453 g/mol per 58.44 g/mol of NaCl

Therefore, the pure sodium chloride sample will yield more chlorine per unit mass of sample.

As such if the two samples have the same mass, the sample with the contaminant of KCl will yield less mass of chlorine per unit mass of the sample, from which the student will be able to tell the purity of the solution.

The sample with the higher purity will yield a higher mass chlorine per unit mass of the sample.

6 0

3 years ago

What are two dangers associated with<br> nuclear fission?

Ostrovityanka [42]

Answer:

Nuclear energy produces radioactive waste

A major environmental concern related to nuclear power is the creation of radioactive wastes such as uranium mill tailings, spent (used) reactor fuel, and other radioactive wastes. These materials can remain radioactive and dangerous to human health for thousands of years

Explanation:

3 0

3 years ago

Select the pair of elements that is most likely to form an ionic bond. A. hydrogen and helium B. lithium and potassium C. potass

zhenek [66]

Ionic bonds are formed in the reaction of a metal and non-metal element. Therefore, the answer is C, potassium and iodine.

4 0

3 years ago

Read 2 more answers

A saturated aqueous solution of Ag2SO3 contains 3.2x10−5 M Ag+. What is the solubility-product equilibrium constant for Ag2SO3?

Alinara [238K]

Answer:

D) 1.6 x 10⁻¹⁴

Explanation:

The solubility-product equilibrium constant for Ag₂SO₃ is given by the expression

Ksp = [Ag⁺]² [SO₃²⁻]

where [Ag⁺] and [SO₃²⁻] are the concentration of the species dissolved in solution for the equlibrium

Ag₂SO₃ (s) ⇄ 2 Ag⁺ + SO₃²⁻

we are given the concentration of Ag⁺ and from the stoichiometry of the equilibrium, the concentration of SO₃²⁻ is half that value, so

[Ag⁺]² = 3.2 x 10⁻⁵ M

[SO₃²⁻] = 3.2 x 10⁻⁵ M / 2 = 1.6 x 10⁻⁵ M

plugging these values into the solubility product constant equation we have

Ksp = (3.2 x 10⁻⁵)² x (1.6 x 10⁻⁵) = 1.6 x 10¹⁴

Therefore D is the correct answer.

4 0

3 years ago