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zmey [24]
2 years ago
14

Which formula shows a correct representation of the combined gas law?(1 point)

Chemistry
1 answer:
weqwewe [10]2 years ago
8 0

Answer:

D

Explanation:

D. V1P1 / T1=V2P2 / T2 is correct

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Does anyone know how to do number 15?
vovikov84 [41]

Answer:

yes i do 980

Explanation:

i899 1/2

6 0
3 years ago
HELP PLEASE (Solve this problem using the appropriate law). what is the volume of 0.382 moles of hydrogen gas at 1.50 atmosphere
kaheart [24]

Answer:

V = 6.17 L

Explanation:

Given data:

Volume = ?

Number of moles = 0.382 mol

Pressure = 1.50 atm

Temperature = 295 k

R = 0.0821 L. atm. /mol. k

Solution:

According to ideal gas equation:

PV= nRT

V = nRT/P

V = 0.382 mol × 0.0821 L. atm. /mol. k ×295 k / 1.50 atm

V = 9.252  L. atm.  / 1.50 atm

V = 6.17 L

5 0
3 years ago
Which part of a chemical formula indicates the number of copies of molecules or compounds? A. plus signs B. subscripts C. coeffi
nadya68 [22]
I think it is C. coefficients but I don't know what subscripts are
3 0
4 years ago
Which of the following descriptions of gravitational force is NOT TRUE?
tia_tia [17]

Answer:

B

Explanation:

8 0
3 years ago
Calculate the time needed for a constant current of 0.961 a to deposit 0.500 g of co(ii) as
Vera_Pavlovna [14]

1.70 × 10³ seconds

<h3>Explanation </h3>

\text{Co}^{2+} + 2 e⁻ → \text{Co}

It takes two moles of electrons to reduce one mole of cobalt (II) ions and deposit one mole of cobalt.

Cobalt has an atomic mass of 58.933 g/mol. 0.500 grams of Co contains 0.500 / 58.933 = 8.484\times 10^{-3} \; \text{mol} of Co atoms. It would take 2 \times 8.484 \times 10^{-3} = 0.01697 \; \text{mol} of electrons to reduce cobalt (II) ions and produce the 8.484\times 10^{-3} \; \text{mol} of cobalt atoms.

Refer to the Faraday's constant, each mole of electrons has a charge of around 96 485 columbs. The 0.01697 mol of electrons will have a charge of 1.637 \times 10^{3} \; \text{C}. A current of 0.961 A delivers 0.961 C of charge in one single second. It will take 1.637 \times 10^{3} / 0.961 = 1.70 \times 10^{3} \; \text{s} to transfer all these charge and deposit 0.500 g of Co.

4 0
3 years ago
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