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Aloiza [94]
4 years ago
8

The answer ............................:

Chemistry
1 answer:
Oduvanchick [21]4 years ago
7 0
3.22 jk sorry idk hehehe
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How much water should be added to 4.3 moles of LiBr to prepare a 2.05 m solution?
arsen [322]

Answer:

2.1 kg of water

Explanation:

Step 1: Given data

  • Moles of lithium bromide (solute): 4.3 moles
  • Molality of the solution (m): 2.05 m (2.05 mol/kg)
  • Mass of water (solvent): ?

Step 2: Calculate the mass of water required

Molality is equal to the moles of solute divided by the kilograms of solvent.

m = moles of solute/kilograms of solvent

kilograms of solvent = moles of solute/m

kilograms of solvent = 4.3 mol /(2.05 mol/kg) = 2.1 kg

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3 years ago
A sample of ideal gas is in a sealed container. The pressure of the gas is 485 torr , and the temperature is 40 ∘C . If the temp
Fantom [35]

Answer:

537.68 torr.

Explanation:

  • We can use the general law of ideal gas:<em> PV = nRT.</em>

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

<em>(P₁T₂) = (P₂T₁).</em>

P₁ = 485 torr, T₁ = 40°C + 273 = 313 K,

P₂ = ??? torr, ​T₂ = 74°C + 273 = 347 K.

∴ P₂ = (P₁T₂)/(P₁) = (485 torr)(347 K)/(313 K) = 537.68 torr.

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