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3 years ago

Help me please? I need to pass this tomorrow

Chemistry

1 answer:

kramer3 years ago

6 0

Here is what I have come up with, and I have used Mariam Webster for the definitions.

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Calculate ℰ° values for the galvanic cells described below. (a) cr3+(aq) + cl2(g) equilibrium reaction arrow cr2o72-(aq) + cl -(

Ad libitum [116K]

Answer:

$\boxed{\text{(a) 0.00 V; (b) 0.424 V}}$

Explanation:

We must look up the standard reduction potentials for the half-reactions.

ℰ°

Cr₂O₇²⁻ + 14H⁺ + 6e⁻ ⇌ 2Cr³⁺ + 7H₂O 1.36

Cl₂ + 2e⁻ ⇌ 2Cl⁻ 1.35827

2IO₃⁻ + 12H⁺ + 10e⁻ ⇌ I₂ + 6H₂O 1.195

Fe³⁺ + e⁻ ⇌ Fe²⁺ 0.771

(a) Cr³⁺/Cl₂

We reverse the sign of ℰ° for the oxidation half-reaction. Then we add the two half-reactions and their ℰ° values.

ℰ°/V

2Cr³⁺ + 7H₂O ⇌ Cr₂O₇²⁻ + 14H⁺ + 6e⁻ -1.36

Cl₂ + 2e⁻ ⇌ 2Cl⁻ 1.358 27

2Cr³⁺ + 3Cl₂ + 7H₂O ⇌ Cr₂O₇²⁻ + 6Cl⁻ + 14H⁺ 0.00

(b) Fe²⁺/IO₃⁻

ℰ°/V

Fe²⁺ ⇌ Fe³⁺ + e⁻ -0.771

2IO₃⁻ + 12H⁺ + 10e⁻ ⇌ I₂ + 6H₂O 1.195

10Fe²⁺ + 2IO₃⁻ + 12H⁺ ⇌ 10Fe³⁺ + I₂ + 6H₂O 0.424

The ℰ° values for the cells are $\boxed{\textbf{(a) 0.00 V; (b) 0.424 V}}$

4 0

4 years ago

PLEASE

sesenic [268]

Answer:

How many grams of H2 can be produced from the reaction of 11.5 grams of sodium with an excess of water? Hint: 2Na + 2H2O ---> 2NaOH + H2. Ans: 0.505g .

Explanation:

7 0

2 years ago

You are given an unknown mixture containing NaCl and NaHCO3. When you carry out the heating exactly as described in part A, only

SOVA2 [1]

Answer:

1.52g NaHCO3 were in the original mixture.

Mass percent: 64.1%

Explanation:

When NaHCO3 heats it descomposition occurs as follows:

2 NaHCO3(s) → Na2CO3(s) + CO2(g) + H2O(g).

The loss in mass is because of the evaporization of CO2 and H2O. As both are in the same porportion, its molar mass is the sum of both compounds (44g/mol + 18g/mol = 62g/mol)

Loss in mass: 74.80g - 74.24g = 0.56g.

In moles:

0.56g * (1mol / 62g) = 0.00903 moles of gas.

As 1 mole of the gases comes from 2 moles of NaHCO3:

Moles NaHCO3:

0.00903 moles of gas * (2 moles NaHCO3 / 1 mole gas) = 0.018 moles NaHCO3.

In grams (Molar mass NaHCO3: 84g/mol):

0.018 moles NaHCO3 * (84g / mol) = 1.52g NaHCO3 were in the original mixture.

The mass of the mixture was:

74.80g - 72.428g = 2.372g

That means mass percent of NaHCO3 is:

(1.52g / 2.372g) * 100 = 64.1%

7 0

4 years ago

What is the total number of moles of solute contained in 0.5 liter of 3.0 m

myrzilka [38]

Based on science 4) 1) 0.25M

2) 0.66M

3) 1.5M

4) 4.0M ––> Therefore, this is the most concentrated

5) 30g NaOH(1 mol NaOH / 40gNaOH) = 0.75 mol NaOH / 0.500L = 1.5M NaOH (3)

6) 1) 1.0 mol / L(1 L) = 1 mol H2SO4

2) 1 mol / L (2 L) = 2 mol H2SO4

3) 0.50 mol / L ( 1.0L) = 0.50 mol H2SO4 ––>

7) 0.200 mol / L (1 L)(74.6g / mol) = 14.92g (2)

8) 0.25 mol / 0.250 L = 1 mol / L = 1M (1)

This is the answer : 1m

4 0

3 years ago

How would you measure the specific latent heat of vaporisation of a liquid?

son4ous [18]

Answer:

Ramsey and Marshall method.

Explanation:

The specific latent heat of vapourization of a liquid is measured by a modification of the method of Ramsey and Marshall in the year 1896.

8 0

3 years ago

Help me please? I need to pass this tomorrow ​

Help me please? I need to pass this tomorrow