<em>Answer:</em>
The equlibrium concentration sof Ca+2 ion willl be 4.9×10∧-3 M
<em>Data Given:</em>
Ksp of CaSO4 = 2.4 × 10∧-5
CaSO4 ⇔ Ca+2 + SO4∧-2
<em>Solution:</em>
Ksp = [Ca+2].[ SO4∧-2]
2.4 × 10∧-5 = [x].[x]= x²
x = 4.9×10∧-3 M
<em>Result:</em>
- The conc. of Ca+2 ion is 4.9×10∧-3 M
Your answer is D. Since there is little to no magnetic field to wire, if it is copper which most wires are, there will be no voltage in a wire.
Answer:
0.12693 mg/L
Explanation:
First we <u>calculate the concentration of compound X in the standard prior to dilution</u>:
- 10.751 mg / 100 mL = 0.10751 mg/mL
Then we <u>calculate the concentration of compound X in the standard after dilution</u>:
- 0.10751 mg/mL * 5 mL / 25 mL = 0.021502 mg/L
Now we calculate the<u> concentration of compound X in the sample</u>, using the <em>known concentration of standard and the given areas</em>:
- 2582 * 0.021502 mg/L ÷ 4374 = 0.012693 mg/L
Finally we <u>calculate the concentration of X in the sample prior to dilution</u>:
- 0.012693 mg/L * 50 mL / 5 mL = 0.12693 mg/L
The solution for this problem is:
1. number of mole / kg of vitamin K = (4.43 °C) / (40.0 °C/m) = 0.11075 m *molality freezing point depression constant for camphor is 37.7 C Kg/mo
2. (0.11075 mol/kg) x (0.0100 kg) = 0.0011075 mol
3. (0.500 g) / (0.0011075 mol) = 451 g/mol
