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2 years ago

How much energy (kJ) is required to change 0.18 mole of ice (s) at 0 C to water (l) at 0 C?

Chemistry

1 answer:

Dmitriy789 [7]2 years ago

6 0

Answer:25,06 kJ of energy must be added to a 75 g block of ice.

ΔHfusion(H₂O) = 6,01 kJ/mol.

T(H₂O) = 0°C.

m(H₂O) = 75 g.

n(H₂O) = m(H₂O) ÷ M(H₂O).

n(H₂O) = 75 g ÷ 18 g/mol.

n(H₂O) = 4,17 mol.

Q = ΔHfusion(H₂O) · n(H₂O)

Q = 6,01 kJ/mol · 4,17 mol

Q = 25,06 kJ.

Explanation:

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Answer:

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Explanation:

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In this case, since at 60 °C, 108 grams of ammonium bromide are completely dissolved in 100 grams of water for a saturated solution, once it is cooled to 30 °C, wherein only 83.2 grams are completely dissolved in 100 grams of water, the following mass will precipitate:

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Answer:

$m_{B}^{theoretical}=0.365gB$

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Explanation:

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In this case, since the reaction (A->B) have an initial amount of pure 4-aminobenzoic acid, the first step to compute the theoretical yield is to solve the following stoichiometric setup:

$m_{B}^{theoretical}=0.303gA*\frac{1molA}{137.14gA}*\frac{1molB}{1molA}*\frac{165.19 gB}{1molB}\\\\ m_{B}^{theoretical}=0.365gB$

Whereas A stands for 4-aminobenzoic acid and B for the benzocaine. Moreover, we compute the percent yield by dividing the actual yield (0.318 g) by the theoretical one (0.365 g):

$Y=\frac{0.318g}{0.365g} *100\%\\\\Y=87.1\%$

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