CHEMISTRY COLLEGE

How much energy is released when 33.8 g of water freezes? The heat of fusion for water is 6.02 kJ/mol.

Answers

Answer 1

Answer:

Answer:

11.29Kj

Explanation:

1. find moles of 33.8g of water

Molar mass of H2O: 18.02g/Mol

33.8/18.02= 1.88mols

2. find energy

1.88 x 6.02= 11.29Kj

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A chemist titrates 80.0mL of a 0.3184M pyridine C5H5N solution with 0.5397M HBr solution at 25°C . Calculate the pH at equivalence. The pKb of pyridine is 8.77.

Answers

Answer:pH = 2.96

Explanation:

C5H5N + HBr --------------> C5H5N+ + Br-

millimoles of pyridine = 80 x 0.3184 =25.472mM

25.472 millimoles of HBr must be added to reach equivalence point.

25.472 = V x 0.5397

V =25.472/0.5397= 47.197 mL HBr

total volume = 80 + 47.197= 127.196 mL

Concentration of [C5H5N+] = no of moles / volume=

25.472/ 127.196= 0.20M

so,

pOH = 1/2 [pKw + pKa + log C]

pKb = 8.77

pOH = 1/2 [14 + 8.77 + log 0.20]

pOH = 11.0355

pH = 14 - 11.0355

pH = 2.96

Calculate the solubilities of the following compounds in a 0.02 M solution of barium nitrate using molar concentrations, first ignoring ionic strength and activities. a. silver iodate
b. barium sulfate
c. Repeat the above calculations using ionic strength and activities.

Answers

Answer:

a. 1.7 × 10⁻⁴ mol·L⁻¹; b. 5.5 × 10⁻⁹ mol·L⁻¹

c. 2.3 × 10⁻⁴ mol·L⁻¹; 5.5 × 10⁻⁸ mol·L⁻¹

Explanation:

a. Silver iodate

Let s = the molar solubility.

AgIO₃(s) ⇌ Ag⁺(aq) + IO₃⁻(aq); Ksp = 3.0 × 10⁻⁸

E/mol·L⁻¹: s s

$K_(sp) =\text{[Ag$^(+)$][IO$_(3)$$^(-)$]} = s* s = s^(2) = 3.0* 10^(-8)\ns = \sqrt{3.0* 10^(-8)} \text{ mol/L} = 1.7 * 10^(-4) \text{ mol/L}$

b. Barium sulfate

BaSO₄(s) ⇌ Ba²⁺(aq) + SO₄²⁻(aq); Ksp = 1.1 × 10⁻¹⁰

I/mol·L⁻¹: 0.02 0

C/mol·L⁻¹: +s +s

E/mol·L⁻¹: 0.02 + s s

$K_(sp) =\text{[Ba$^(2+)$][SO$_(4)$$^(2-)$]} = (0.02 + s) * s \approx 0.02s = 1.1* 10^(-10)\ns = (1.1* 10^(-10))/(0.02) \text{ mol/L} = 5.5 * 10^(-9) \text{ mol/L}$

c. Using ionic strength and activities

(i) Calculate the ionic strength of 0.02 mol·L⁻¹ Ba(NO₃)₂

The formula for ionic strength is

$\mu = (1)/(2) \sum_(i) {c_(i)z_(i)^(2)}\n\n\mu = (1)/(2) (\text{[Ba$^(2+)$]}\cdot (2+)^(2) + \text{[NO$_(3)$$^(-)$]}*(-1)^(2)) = (1)/(2) (\text{0.02}* 4 + \text{0.04}*1)= (1)/(2) (0.08 + 0.04)\n\n= (1)/(2) *0.12 = 0.06$

(ii) Silver iodate

a. Calculate the activity coefficients of the ions

$\log \gamma = -0.51z^(2)√(I) = -0.051(1)^(2)√(0.06) = -0.51* 0.24 = -0.12\n\gamma = 10^(-0.12) = 0.75$

b. Calculate the solubility

AgIO₃(s) ⇌ Ag⁺(aq) + IO₃⁻(aq)

$K_(sp) =\text{[Ag$^(+)$]$\gamma_(Ag^(+))$[IO$_(3)$$^(-)$]$\gamma_{IO_(3)^(-)}$} = s*0.75* s * 0.75 =0.56s^(2)= 3.0 * 10^(-8)\ns^(2) = (3.0 * 10^(-8))/(0.56) = 5.3 * 10^(-8)\n\ns =2.3 * 10^(-4)\text{ mol/L}$

(iii) Barium sulfate

a. Calculate the activity coefficients of the ions

$\log \gamma = -0.51z^(2)√(I) = -0.051(2)^(2)√(0.06) = -0.51*16* 0.24 = -0.50\n\gamma = 10^(-0.50) = 0.32$

b. Calculate the solubility

BaSO₄(s) ⇌ Ba²⁺(aq) + SO₄²⁻(aq

$K_(sp) =\text{[Ba$^(2+)$]$\gamma_( Ba^(2+))$[SO$_(4)$$^(2-)$]$\gamma_{ SO_(4)^(2-)}$} = (0.02 + s) * 0.32* s* 0.32 \approx 0.02*0.10s\n2.0* 10^(-3)s = 1.1 * 10^(-10)\ns = (1.1* 10^(-10))/(2.0 * 10^(-3)) \text{ mol/L} = 5.5 * 10^(-8) \text{ mol/L}$

When 1 mole of H2(g) reacts with F2(g) to form HF(g) according to the following equation, 542 kJ of energy are evolved H2(g) + F2(g2HF(g) Is this reaction endothermic or exothermic? What is the value of q?

Answers

Answer: The reaction is exothermic. The value of q is -542 kJ.

Explanation:

Endothermic reactions are defined as the reactions in which energy of the product is greater than the energy of the reactants. The total energy is absorbed in the form of heat and for the reaction comes out to be positive.

Exothermic reactions are defined as the reactions in which energy of the product is lesser than the energy of the reactants. The total energy is released in the form of heat and for the reaction comes out to be negative.

Thus $H_2(g)+F_2(g)\rightarrow 2HF(g)$ evolves heat , it is exothermic in nature. The value of q is -542kJ.

A piece of charcoal used for cooking is found at the remains of an ancient campsite. a 0.94 kg sample of carbon from the wood has an activity of 1580 decays per minute. find the age of the charcoal. living material has an activity of 15 decays/minute per gram of carbon present and the half-life of 14c is 5730 y. answer

Answers

Mass of sample of charcoal = 0.94 kg = 0.00094

∴, activity = decay rate / mass = 1580/0.00094
= 1.681 X 10^6 decays per min per gram

Using the half-life formula, we have:
activity of sample / activity of modern carbon = (1/2)^(age / half-life)
∴, Age = half-life x log (base 2) (modern activity / coal activity)
= 5730 x log(base 2)(1.681X10^6/ 15)
= 96115 years.

Answer: Age of the charcoal = 96115 years

Final answer:

Using the radiocarbon dating technique and applying the decay formula, it is calculated that the age of the charcoal from the an ancient campsite is approximately 9,500 years.

Explanation:

The age of the charcoal can be found using the technique of radiocarbon dating, which capitalizes on the process of radioactive decay. The isotope carbon-14 (¹4C) is used in this method as it has a known half-life of 5730 years. The number of decays per minute per gram of carbon in a live organism is known as its activity.

Initially, the activity was given as 15 decays per minute per gram. The present activity of the carbon in the charcoal is provided at 1580 decays per minute for a 0.94 kg or 940 gram sample. Thus, the current activity per gram is 1580/940 equals approximately 1.68 decays per minute per gram.

Given that the half-life of ¹4C is 5730 years, we can apply the formula for calculating the time passed using the rate of decay, which is given as T = (t1/2 / ln(2)) * ln(N0/N), where 'ln' is the natural logarithm, 'N0' is the initial quantity (15 decays/minute per gram), 'N' is the remaining quantity (1.68 decays/minute per gram).

Plugging in the given values, we get T = (5730 / ln(2)) * ln(15/1.68), which gives us approximately 9,500 years. Therefore, the age of the charcoal is around 9,500 years.

Learn more about Radiocarbon Dating here:

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Calculate the entropy change for the gas. (2b)Calculate the entropy change when 14 g of nitrogen expand from a volume of 10 L to a volume of 30 L at the same temperature. Assuming ideal behaviour for the nitrogen gas

Answers

Explanation:

because T is const so

deltaS=Q/T=nRLn(V2/V1)

=(14/28)x8.314xLn(30/10)=4,567 J/k

Object A has a molar heat of 31.2 J/mole∙°C and object B molar heat is 11.2 J/mole∙°C. Which object will heat up faster if they have the same mass and equal amount of heat is applied? Explain why.

Answers

Answer:

Substance B

Explanation:

Molar heat of A = 31.2J/mole.°C

Molar heat of B = 11.2 J/mole∙°C.

The molar heat of a substance is the amount of heat that must be added to a mole of a substance to raise the temperature by 1°C.

Substance B will heat up faster compared to A.
It has a smaller molar heat compared to A.
This suggests that it will require lesser heat to raise its temperature by 1°C.

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