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Oliga [24]
3 years ago
9

How many milliliters of a 3.0 M HCl solution are required to make 250.0 milliliters of 1.2 M HCl?

Chemistry
1 answer:
mixer [17]3 years ago
8 0
 formulae for finding is
(Mi)(Vi) = (Mf)(Vf) 
<span>(3.0M)(x) = (500.mL/1000)(1.0M)
</span>convert 500ml to liter

<span>x = (0.5 moles)/(3.0 moles/liter) </span>
<span>x = 0.17 Liters = 1.7 x 10^2 mL HCl</span>
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How much heat energy is required to convert 48.3 g of solid ethanol at -114.5 degree C to gasesous ethanol at 135.3 degree C? Th
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7.21 × 10⁴ J

Explanation:

Ethanol is solid below -114.5°c, liquid between -114.5°C and 78.4°C, and gaseous above 78.4°C.

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We need to calculate the heat required in different stages and then add them.

The moles of ethanol are:

48.3g.\frac{1mol}{46.07g} =1.05mol

Solid-liquid transition

Q₁ = ΔHfus . n = (4.60 kJ/mol) . 1.05 mol = 4.83 kJ = 4.83 × 10³ J

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ΔHfus: molar heat of fusion

n: moles

Liquid: from -114.5°C to 78.4°C

Q₂ = c(l) . m . ΔT = (2.45 J/g.°C) . 48.3g . [78.4°C-(-114.5°C)] = 2.28 × 10⁴ J

where,

c(l): specific heat capacity of the liquid

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Liquid-gas transition

Q₃ = ΔHvap . n = (38.56 kJ/mol) . 1.05 mol = 40.5 kJ = 40.5 × 10³ J

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ΔHvap: molar heat of vaporization

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Q₄ = c(g) . m . ΔT = (1.43 J/g.°C) . 48.3g . (135.3°C-78.4°C) = 3.93 × 10³ J

where

c(g): specific heat capacity of the gas

Total heat required

Q₁ + Q₂ + Q₃ + Q₄ = 4.83 × 10³ J + 2.28 × 10⁴ J + 40.5 × 10³ J + 3.93 × 10³ J = 7.21 × 10⁴ J

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