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Oksanka [162]
2 years ago
14

Given the reaction _K(s) +_ Cl2(g) → _KCl(s) what is the amount of K, in grams, needed to completely react with 2 moles of Cl2(g

)?
Chemistry
1 answer:
damaskus [11]2 years ago
3 0

Answer:

156.4g K

Explanation:

I'm not sure if it is correct but I think it should be this

What do we know so far?: 2K + 1Cl2 -> 2KCl, 2 mol of Cl2

What are we looking for?: #g of K

What is the ratio of K to Cl2?: 2:1

Set up equation: 2molCl2 x \frac{2mol K}{1 mol Cl2}

Cancel unwanted units: 2 x \frac{2mol K}{1}

Answer we got: 2 x 2mol K = 4mol K

Converting moles to grams: 4 x 39.1 (molar mass of K) = 156.4g K

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What is the minimum pressure in kPa that must be applied at 25 °C to obtain pure water by reverse osmosis from water that is 0.1
Yuri [45]

Answer:

The minimum pressure should be 901.79 kPa

Explanation:

<u>Step 1: </u>Data given

Temperature = 25°C

Molarity of sodium chloride = 0.163 M

Molarity of magnesium sulfate = 0.019 M

<u>Step 2:</u> Calculate osmotic pressure

The formula for the osmotic pressure =

Π=MRT.

⇒ with M = the total molarity of all of the particles in the solution.

 ⇒ R = gas constant = 0.08206 L*atm/K*mol

⇒ T = the temperature = 25 °C = 298 K

NaCl→ Na+ + Cl-

MgSO4 → Mg^2+ + SO4^2-

M = 2(0.163) + 2(0.019 M)

M = 0.364 M

Π = (0.364 M)(0.08206 atm-L/mol-K)(25 + 273 K)

Π = 8.90 atm

(8.90 atm)(101.325 kPa/atm) = 901.79 kPa

The minimum pressure should be 901.79 kPa

6 0
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