1. Q = mcDT

= 2000 g( 4.19 J/[g^oC])(60.7-25.0)

= 299 166 J = 299.166 kJ

DH = -Q = -299.166 kJ

DH/n = -299.166 kJ/[(7.25-1.25)g/(58 g/mole)] = - 2890 kJ/mole = -2.9 X10³ kJ/mole

2. Q = mcDT

= 200 g( 4.19 J/[g^oC])(31.5-25.0)

= 5447 J = 5.447 kJ

DH = -Q = -5.447 kJ

DH/n = -5.447 kJ/ [4.0 / 56 g/mole] = -76 kJ/mole

3. Q = mcDT

= (200 + 200) g( 4.19 J/[g^oC])(6.6)

=11061.600 J = 11.061 kJ

DH = -Q = -11.061 kJ

n = CV = 0.50moles/L (0.200 L) = 0.100 moles

DH/n = -11.061 kJ/ 0.100 moles = -1 X 10² kJ/mole

4. 5.04 X 10^-1 kJ/mole = 504 J/mole

n = 4.0 / (55.8 g/mole) = 0.072 moles

Q = 504 J/mole (0.072 moles) = 36.288 J

Q = mcDT

36.288 J = 4c*(42-22)

c = 0.45 J/(g C)

5. m of alcohol = 30. mL ( 0.79 g/ml) = 23.7 g

m of water = ?

heat lost by alcohol = heat gained by water

- m_ac_aDT_a = m_wc_wDT_w

-2.37 (2.45)(15-22) = m(4.19)(15-11)

m = 24 g of water ( two sig figs if there was a decimal after 30 mL; should have been)

6. heat lost by metal = heat gained by water

- m_mc_mDT_m = m_wc_wDT_w

- 32.6(0.448)(x - 200) = 100(4.19)(x - 25.0)

x = 30.9 ^oC

7. Q = mcDT

= 2000 g( 4.19 J/[g^oC])(42.0-20.0)

= 184360 J = 184.36 kJ

DH = -Q = -184.36 kJ

DH/n = -184.36 kJ/ [(28.52-24.29)g / (25*12+52) g/mole] = -1.53 X 10⁴ kJ/mole, close enough to be pure ( again it should have read 2000. ml )

8. Q fondue = mcDT = 1.1 kg(8.9 kJ /(kg ^oC)* ( 98-22) = 744 kJ

744kJ/[639 kJ/(mole of CH₃OH)] = 1.16 moles of CH₃OH

= 1.16 moles *(12+3+16+1)g/mole = 37 g

9. Q = mcDT

= 100 g( 4.19 J/[g^oC])(28.0-22.0) = 2514 J = 2.514 kJ

DH = -Q = -2.514 kJ

DH/n = -2.514 kJ/( 4.01/(6.9 + 35.5) ) = -26.6 kJ/mole

-26.6 kJ/mole * 2.0 moles = -53 kJ

10. heat lost by metal = heat gained by water

- m_mc_mDT_m = m_wc_wDT_w

Note: kJ/kg = J/g; (k's cancel)

- 550g(0.448 J/[g^oC])(26.9-x) = 8500 g(4.19 J/[g^oC])(26.9 - 22.0)

x = 7.4 X 10^{2 oC}