20 points for 2 chemistry questions!

1.) consider the specific heats.
h2o(s) = 2.09 j/g · ◦c
h2o (ℓ) = 4.18 j/g · ◦c
h2o(g) = 2.03 j/g · ◦c.
the heat of fusion for water is 334 j/g and its heat of vaporization is 2260 j/g. calculate the amount of heat required to convert 90 g of ice at −15◦c completely to liquid water at 56◦c. answer in units of kj.

2.) how much energy is required to turn 12 g ice at 227 k into steam at 382 k?
specific heat of ice = 2.077 j/g·k
specific heat of water (ℓ) = 4.180 j/g·k
specific heat of steam = 2.042 j/g·k
h2o heat of fusion = 6.00 kj/mol
h2o heat of vaporization = 40.6 kj/mol
answer in units of j.

The answer to your question is:

1.- Q = 53.95 kJ

2.- Q = 37450.2 J  

Explanation:

1.- Data

H2O(s) = 2.09 J/g · ◦C

H2O (ℓ) = 4.18 J/g · ◦C

H2O(g) = 2.03 J/g · ◦C

heat of fusion = 334 J/g

heat of vaporization = 2260 J/g

Heat = ? kJ

T1 = -15°C

T2 = 56°C

Process

Calculate    heat from -15°C to 0°C

                   heat of fusion

                   heat from 0°C to 56°C

a) heat from -15°C to 0°C

                  Q = mCp(T2 - T1)

                  Q = (90)(2.09) (0 + 15)

                  Q = 2821.5 J

b) heat of fusion

                  Q = (334)(90)

                  Q = 30060 J

c) heat from 0° to 56°C

                  Q = (90)(4.18)(56 - 0)

                  Q = 21067.2 J

Total heat = Q = 2821.5 + 30060 + 21067.2

                    Q = 53948.7 J

                    Q = 53.95 kJ

2.-

Data

Energy = ?

mass = 12 g

T1 = 227°K

T2 = 382 °K

Specific heat of ice = 2.077 J/g·K

Specific heat of water (ℓ) = 4.180 J/g·K

Specific heat of steam = 2.042 J/g·K

H2O heat of fusion = 6.00 kJ/mol  = 6000 J/ mol = 333.3 J/g

H2O heat of vaporization = 40.6 kJ/mol = 40600 J/ mol = 2255.6 J/g

Process

Calculate heat from 227 to 273°K

                 fusion heat

                heat from 273 to 373°K

                vaporization heat

                 heat from 373 to 382°K

heat from 227 to 273°K

                                      Q = (12)(2.077)(273 - 227)

                                      Q = 1146.5 J

fusion heat

                                      Q = (12)(333.3)

                                      Q = 4000 J

heat from 273 to 373 °K

                                       Q = (12)(4.18)(373 - 273)

                                       Q = 5016 J

vaporization heat

                                       Q = (12)(2255.6)

                                       Q = 27067.2 J

heat from 373 to 382

                                       Q = (12)(2.042)(382 - 373)

                                       Q = 220.5 J

Total heat

                Q = 1146.5 + 4000 + 5016 + 27067.2 + 220.5

                Q = 37450.2 J                                  

answeridentity;

percent abundance for cu-63 can be shown as:

63 x + 65 (1-x) = 62.9296

65 - 62.9296 = 2 x

x = 1.035

thus, percent abundance for cu-63 is 1.035

percent abundance for cu-65 can be shown as:

63 x + 65 (1-x) = 64.9278

65 - 64.9278 = 2 x

x = 0.0361

thus, percent abundance for cu-63 is 0.0361

the mass spectrum of cu-63 showing the intense peak as, this isotope is found in nature more abundantly around 92.38 % in comparison to cu-65 which is found 62.58 % naturally