From another answer, it is most likely E. Horizon. Hope this helps!
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:
If this is a True or False question the answer is False
Answer:
9.28 g/L
Explanation:
We will be using the ideal gas law to solve this problem:
PV = nRT where P is the pressure (atm)
V is the volume (L)
R is the gas constant 0.08205 Latm/Kmol
T is the temperature (K)
n is the number of moles
The number of moles is the mass divided by the molecular weight, and from here we can solve for the density. (Note here we use the atomic weight of radon since its is a monoatomic noble gas)
PV = m/AW RT ⇒ P = (m/V ) RT/AW ⇒ P AW /RT =D
0.950 atm x 222.0 g/mol / [( 0.08205 Latm/Kmol ) x 277 K ] = D
9.28 g/L = D
Answer:
33 %
Explanation:
Step 1: Given data
Initial moles of the acid (nHA(0)): 6 mol
Moles of the conjugate base at equilibrium (nA⁻(eq))
Step 2: Write the balanced generic acid dissociation reaction
HA(aq) ⇄ A⁻(aq) + H⁺(aq)
Step 3: Calculate the percent ionization
We will use the following expression.