Push the pushbutton to the first stop before inserting the tip into the solution, and then put the tip in the solution tube.
Answer:
(A) 4.616 * 10⁻⁶ M
(B) 0.576 mg CuSO₄·5H₂O
Explanation:
- The molar weight of CuSO₄·5H₂O is:
63.55 + 32 + 16*4 + 5*(2+16) = 249.55 g/mol
- The molarity of the first solution is:
(0.096 gCuSO₄·5H₂O ÷ 249.55 g/mol) / (0.5 L) = 3.847 * 10⁻⁴ M
The molarity of CuSO₄·5H₂O is the same as the molarity of just CuSO₄.
- Now we use the dilution factor in order to calculate the molarity in the second solution:
(A) 3.847 * 10⁻⁴ M * 6mL/500mL = 4.616 * 10⁻⁶ M
To answer (B), we can calculate the moles of CuSO₄·5H₂O contained in 500 mL of a solution with a concentration of 4.616 * 10⁻⁶ M:
- 4.616 * 10⁻⁶ M * 500 mL = 2.308 * 10⁻³ mmol CuSO₄·5H₂O
- 2.308 * 10⁻³ mmol CuSO₄·5H₂O * 249.55 mg/mmol = 0.576 mg CuSO₄·5H₂O
Answer:
Explanation:
half life = 4 h
initial concentration = 100
final concentration = 15
Time = t
No of half life x = t / 4
15 = 100 x ( 1/2 )ˣ
.15 = ( 1/2 )ˣ
ln .15 = - x ln2
x = - ln .15 / ln 2
= 1.897 / .693
x = 2.737
x = t / 4
t = 2.737 x 4 = 11 h approx .
To solve this, we simply equate the change in enthalpy for
the two substances since heat gained by water is equal to heat lost of aluminum.
We know that the heat capacity of aluminum is 0.089 J/g°C and that of water is
4.184 J/g°C. Therefore:
450.2 (95.2 - T) (0.089) = 60 (T – 10) (4.184)
3,814.45456 – 40.0678 T = 251.04 T – 2,510.4
291.1078 T = 6,324.85456
<span>T = 21.7°C</span>
