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SVEN [57.7K]
2 years ago
11

1. Suppose you had 58.44 grams of NaCl and you dissolved it in exactly 2.00 L of solution.

Chemistry
1 answer:
Margarita [4]2 years ago
3 0
.500M or .500 mol/liters
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What was the weight percent of water in the hydrate before heating?
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Data:

weight of water before heating = 0.349

weight of hydrate before heateing = 2.107

Formula:

Weight percent of water = [ (weight of water) / (weight of the hydrate) ] * 100

Solution:

Weight percent of water = [ 0.349 / 2.107] * 100 ≈ 16.6 %

Answer: 16.6%
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A balloon that had a volume of 3.50 L at 25.0°C is placed in a hot room at 40.0°C. If the pressure remains constant at 1.00 atm,
inna [77]

Answer:

3.676 L.

Explanation:

  • We can use the general law of ideal gas: PV = nRT.

where, P is the pressure of the gas in atm.

V is the volume of the gas in L.

n is the no. of moles of the gas in mol.

R  is the general gas constant,

T is the temperature of the gas in K.

  • If n and P are constant, and have different values of V and T:

(V₁T₂) = (V₂T₁)

  • Knowing that:

V₁ = 3.5 L, T₁ = 25°C + 273 = 298 K,

V₂ = ??? L, T₂ = 40°C + 273 = 313 K,

  • Applying in the above equation

(V₁T₂) = (V₂T₁)

∴ V₂ = (V₁T₂)/(T₁) = (3.5 L)(313 K)/(298 K) = 3.676 L.

7 0
3 years ago
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