Calculate ΔHrxn for the following reaction: C(s) + H2O(g) --> CO(g) + H2(g) Use the following reactions and given ΔH values: C (s) + O2 (g) → CO2 (g), ΔH = -393.5 kJ 2 CO (g) + O2 (g) → 2 CO2 (g), ΔH= -566.0 kJ 2 H2 (g) + O2 (g) → 2 H2O (g), ΔH= -483.6 kJ

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Answer 1

Answer:

Explanation:

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What is the daughter nucleus produced when 63 Zn undergoes electron capture? Replace each question mark with the appropriate integer or symbol.

Answers

Answer: The daughter nuclei is $_(29)^(63)\textrm{Cu}$

Explanation:

Electron capture is defined as the process in which an electron is drawn to the nucleus where it combines with a proton to form a neutron and a neutrino particle.

$_Z^A\textrm{X}+e^-\rightarrow _(Z-1)^A\textrm{Y}+\gamma e$

The chemical equation for the reaction of electron capture of Zinc-63 nucleus follows:

$_(30)^(63)\textrm{Zn}+e^-\rightarrow _(29)^(63)\textrm{Cu}+\gamma e$

The parent nuclei in the above reaction is Zinc-63 and the daughter nuclei produced in the above reaction is copper-63 nucleus.

Hence, the daughter nuclei is $_(29)^(63)\textrm{Cu}$

Final answer:

When Zinc-63 undergoes electron capture, it results in the creation of a Copper-63 daughter nucleus. This is due to the atomic number decreasing by one (from 30 to 29) during electron capture, but the mass number remaining unchanged.

Explanation:

Electron capture is a process where a proton-rich nucleus absorbs an inner shell electron, which results in a conversion of a proton into a neutron, and the emission of an electron neutrino. In doing so, the atomic number decreases by one, while the mass number stays the same. Therefore, in the case of 63 Zn (zinc-63), the atomic number is 30 prior to electron capture. After electron capture, the atomic number will decrease by one to become 29, leading to the production of 63 Cu (copper-63).

Remember that the atomic number (bottom number), also known as the proton number, determines the element. Therefore, in our example, Zn changes to Cu. The fact that the mass number (top number) remains the same is due to the total number of protons and neutrons (nucleons) being conserved.

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Calculate the pH of the cathode compartment solution if the cell emf at 298 K is measured to be 0.660 V when (Zn^2+)=0.22 M and(P_H2)= 0.87atm.

Answers

Answer:

pH = 2.059

Explanation:

At the Cathode:

The reduction reaction is:

$2H^+ + 2e^- \to H_2 \ \ \ \mathbf{E^0_(red)= 0.00 \ V}$

At the anode:

At oxidation reaction is:

$Zn \to Zn^(2+) +2e^- \ \ \ \mathbf{E^0_(ox) = 0.76 \ V}$

The overall equation for the reaction is:

$\mathbf{Zn + 2H^+ \to Zn^(2+) + H_2}$

The overall cell potential is:

$\mathbf{E^0_(cell)= E^0_(ox) + E^0_(red)}$

$\mathbf{E^0_(cell)= 0.76 \ V +0.00 \ V}$

$\mathbf{E^0_(cell)= 0.76\ V}$

Using the formula for the Nernst equation:

$E = E^0 - ( (0.0591)/(n))log (Q)\n$

where;

E = 0.66

(Zn^2+)=0.22 M

Then

$0.66 =0.76- ( (0.0591)/(2))log \bigg ( ([Zn^(2+) ] PH_2)/([H^+]^2) \bigg )$

$0.66 =0.76- 0.02955 * log \bigg ( (0.22*0.87)/([H^+]^2) \bigg )$

3.4 = log ( 0.1914) - 2 log [H⁺]

3.4 = -0.7180 - 2 log [H⁺]

3.4 + 0.7180 = - 2 log [H⁺]

4.118 = - 2 log [H⁺]

pH = log [H⁺] = 4.118/2

pH = 2.059

The pH of the solution as described in the question is 2.7.

The equation of the reaction is;

Zn(s) + 2H^+(aq) ----> Zn^2+(aq) + H2(g)

The partial pressure of hydrogen can be converted to molarity using;

P= MRT

M = P/RT

M = 0.87atm/0.082 LatmK-1mol-1 × 298 K = 0.036 mol/L

We have to obtain the reaction quotient

Q = [Zn^2+] [H2]/[H^+]^2

Q = [0.22 ] [0.036]/[H^+]^2

Recall that, from Nernst equation;

E = E° - 0.0592/nlog Q

E° = 0.00V - (-0.76V) = 0.76V

0.660 = 0.76 - 0.0592/2logQ

0.660 - 0.76 = - 0.0592/2logQ

-0.1 = - 0.0592/2logQ

-0.1 × 2/ - 0.0592 = logQ

3.38 = log Q

Q = Antilog (3.38)

Q= 2.39 × 10^3

Now;

2.39 × 10^3 = [0.22 ] [0.036]/[H^+]^2

2.39 × 10^3 = 7.92 × 10^-3/[H^+]^2

[H^+]^2 = 7.92 × 10^-3/2.39 × 10^3

[H^+] = 1.82 × 10^-3

pH = -log[H^+]

pH = -log[ 1.82 × 10^-3]

pH = 2.7

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FILL IN THE BLANKS (IMAGES ATTACHED)

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Answer:

1chemical properties

2.chrmistry

3.precipitate

4.endothermic reaction

5.matter

6.physical property

7.chemical change

8.exothermic reaction

9.physical change

22. What is the mass in grams of each of the following?a. 3.011 x 1023 atoms F
b. 1.50 x 1023 atoms Mg
c. 4.50 x 1012 atoms Cl
d. 8.42 x 1018 atoms Br
e. 25 atoms W
f. 1 atom Au

Answers

The mass in grams of 3.011 x 10²³ atoms of F is 9.5 g.

The mass in grams of 1.50 x 10²³ atoms of Mg is 5.98 g.

The mass in grams of 4.50 x 10¹² atoms of Cl is 2.65 x 10⁻¹⁰ g.

The mass in grams of 8.42 x 10¹⁸ atoms of Br is 1.12 x 10⁻³ g.

The mass in grams of 25 atoms of W is 3.1 x 10⁻²¹ g.

The mass in grams of 1 atom of Au is 3.27 x 10⁻²² g.

What is the mass in grams of 3.011 x 10²³ atoms F?

The mass in grams of 3.011 x 10²³ atoms of F is calculated as follows;

6.023 x 10²³ atoms = 19 g of F

3.011 x 10²³ atoms F = ?

= (3.011 x 10²³ x 19 g)/(6.023 x 10²³)

= 9.5 g

The mass in grams of 1.50 x 10²³ atoms of Mg is calculated as follows;

6.023 x 10²³ atoms = 24g of Mg

1.5 x 10²³ atoms F = ?

= (1.5 x 10²³ x 24 g)/(6.023 x 10²³)

= 5.98 g

The mass in grams of 4.50 x 10¹² atoms of Cl is calculated as follows;

6.023 x 10²³ atoms = 35.5 g of Cl

4.5 x 10²³ atoms Cl = ?

= (4.5 x 10¹² x 35.5 g)/(6.023 x 10²³)

= 2.65 x 10⁻¹⁰ g

The mass in grams of 8.42 x 10¹⁸ atoms of Br is calculated as follows;

6.023 x 10²³ atoms = 80 g of Br

8.42 x 10¹⁸ atoms Br = ?

= (8.42 x 10¹⁸ x 80 g)/(6.023 x 10²³)

= 1.12 x 10⁻³ g

The mass in grams of 25 atoms of W is calculated as follows;

6.023 x 10²³ atoms = 74 g of W

25 atoms W = ?

= (25 x 74 g)/(6.023 x 10²³)

= 3.1 x 10⁻²¹ g

The mass in grams of 1 atom of Au is calculated as follows;

6.023 x 10²³ atoms = 197 g of Au

1 atom of Au = ?

= (1 x 197 g)/(6.023 x 10²³)

= 3.27 x 10⁻²² g

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Final answer:

This solution provides the calculations necessary to convert the number of atoms of various elements (smallest particle of an element) to grams. It does so by using the molar mass of each element and Avogadro's number.

Explanation:

The mass of atoms can be determined by using Avogadro's number (6.022 x 1023 atoms/mol) and the molar mass of the specific element (g/mol). We use these to create a conversion factor and multiply by the number of atoms given.

For F (fluorine), which has a molar mass of about 18.9984 g/mol, 3.011 x 1023 atoms F is 9.00 g F.
For Mg (magnesium), with molar mass of about 24.3050 g/mol, 1.5 x 1023 atoms Mg is 6.07 g Mg.
For Cl (chlorine), with molar mass of about 35.453 g/mol, 4.50 x 1012 atoms Cl is 2.67 x 10-10 g Cl.
For Br (bromine), with molar mass about 79.904 g/mol, 8.42 x 1018 atoms Br is 0.12 g Br.
For W (tungsten), with molar mass about 183.84 g/mol, 25 atoms W is 7.65 x 10-22 g W.
For Au (gold), with molar mass about 197.0 g/mol, 1 atom Au is 3.28 x 10-22 g Au.

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An aqueous CsCl solution is 8.00 wt% CsCl and has a density of 1.0643 g/mL at 20°C. What is the boiling point of this solution? Kb = 0.51°C/m for water. Enter your answer with 2 decimal places and no units.