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zhannawk [14.2K]
2 years ago
13

What effect does temperature have on reaction rate?

Chemistry
1 answer:
barxatty [35]2 years ago
7 0
The answer for this multiple choice question is C



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A 6.000L tank at 19.2°C is filled with 18.0g of carbon monoxide gas and 10.6g of chlorine pentafluoride gas. You can assume both
Jobisdone [24]

Answer:

Total pressure: 2.89 atm

Mole fraction CO: 0.88

Partial pressure CO: 2.56 atm

Mole fraction ClF₅: 0.12

Partial pressure ClF₅: 0.33 atm

Explanation:

We should apply the Ideal Gases Law to solve this:

P . V = n . R . T

We need n, which is the total moles for the mixture

Total moles = Moles of CO + Moles of ClF₅

Moles of CO = mass of CO / molar mass CO → 18 g/28 g/mol = 0.643 mol

Moles of ClF₅ = mass of ClF₅ / molar mass ClF₅ → 10.6g/ 130.45 g/m = 0.0812 mol

0.643 mol + 0.0812 mol → 0.724 moles in the mixture

So we have the total moles so with the formula we would know the total pressure.

P . 6L = 0.724 mol . 0.082L.atm/mol.K . 292.2K

P = ( 0.724 mol . 0.082L.atm/mol.K . 292.2K) / 6L

P = 2.89 atm

Mole fraction is defined as the quotient between the moles of gas over total moles, and it is equal to partial pressure of that gas over total pressure

Moles of gas X /Total moles = Partial pressure of gas X/Total pressure

(Moles of gas X / Total moles) . Total pressure = Partial pressure of gas X

Mole fraction CO = 0.643 / 0.724 = 0.88

Partial pressure CO = 0.88 . 2.89 atm → 2.56 atm

Mole fraction ClF₅ = 0.0812 / 0.724 = 0.12

Partial pressure ClF₅ = 0.12 . 2.89 atm → 0.33 atm

5 0
3 years ago
Different isotopes of the same element must have a different
Naddika [18.5K]
Different isotopes of the same element must have a different mass number
5 0
3 years ago
Simply parmanent tissue bibliography please help it's for my project ​
daser333 [38]

Answer:

Do your project by yourself

6 0
2 years ago
A cell was prepared by dipping a Cu wire and a saturated calomel electrode into 0.10 M CuSO4 solution. The Cu wire was attached
lara [203]

Answer:

a)  cu2+ + 1Hg (l) 1Cl- equilibrium cu (s) + Hg2Cl2 (s)

b)  0.068 V.

Explanation:

A) Cu2+ + 2e- euilibrium cu (s)

 Hg2Cl2 + 2e- equilibrium 2Hg (l) + 1cl-

Cell Reaction: cu2+ + 1Hg (l) 1Cl- equilibrium cu (s) + Hg2Cl2 (s)

B) To calculate the cell voltage

E = E_o Cu2+/Cu - (0.05916 V / 2) log 1/Cu2+

putting values we get

 = 0.339V + (90.05916V/2)log(0.100) = 0.309V

 E_cell = E Cu2+/Cu - E SCE = 0.309 V - 0.241 V = 0.068V.

6 0
3 years ago
Current is applied to a molten mixture of AgF , FeCl2 , and AlBr3 . What is produced at each electrode? STRATEGY Rank the cation
ratelena [41]

Answer:

Cathode: Ag

Anode: Br₂

Explanation:

In the cathode must occur a reduction, so it's more likely to a metal atom be in the cathode. For the metals given the reduction reactions and the potential of reduction are:

Ag⁺ + e⁻ ⇒ Ag⁰ E° = + 0.80 V

Fe⁺² + 2e⁻ ⇒ Fe⁰ E° = - 0.44 V

Al⁺³ + 3e⁻ ⇒ Al⁰ E° = -1.66 V

As the potential for Ag is the higher, the reduction will occur for it first, so in the cathode will produce Ag.

For the anode an oxidation must occurs, so the reactions for the nonmetals are:

F₂ + 2e⁻ ⇒ 2F⁻ E° = +2.87 V

Cl₂ + 2e⁻ ⇒ 2Cl⁻ E° = +1.36 V

Br₂ + 2e⁻ ⇒ 2Br⁻ E° = +1.07 V

For oxidation, the less the E°, the faster the reaction will occur, so Br₂ will be formed in the anode.

5 0
3 years ago
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