Use the References to access Important values If needed for this question The fo

Question

Use the References to access Important values If needed for this question The following information is given for manganese at 1atm: Hvap(2095 °C)-225 kJ/mol s(1244 °C)- 14.6 kJ/mol boiling point-2095 melting point-1244 specific heat solid-0.477 JgoC specific heat liquid-0.837 J/g What is Alin kJ for the process of freezing a 32.1 g sample of liquid manganese at its normal melting point of 1244°C Submit Answer Retry Entire Group 9 more group attempts remaining Cengage Learning I Cengage Technical Support MacBoo 0 F8 10 FA FS F6

Explanation / Answer

Molar mass of Mn = 54.94 g/mol


mass(Mn)= 32.1 g

use:
number of mol of Mn,
n = mass of Mn/molar mass of Mn
=(32.1 g)/(54.94 g/mol)
= 0.5843 mol

for freezing, heat is released
So, delta H will be negative
delta H = -delta Hfus * number of mol
= (14.6 KJ/mol)*( 0.5843 mol)
= 8.53 KJ
Answer: 8.53 KJ

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