Answer:
0.0252mol/L
Explanation:
The following data were obtained obtained from the question:
Volume of solution = 284mL = 284/1000 = 0.284L
Mole of CuSO4 = 7.157 × 10^-3 mol
Molarity =?
Molarity = mole/Volume
Molarity = 7.157x10^-3 /0.284
Molarity = 0.0252mol/L
The concentration of the solution is 0.0252mol/L
Answer:
Explanation:
<u>1. Equilibrium equation</u>
<u>2. Equilibrium constant</u>
The liquid substances do not appear in the expression of the equilibrium constant.
![k_c=\dfrac{[HBr(g)]^2}{[H_2]}=4.8\times 10^8M](https://tex.z-dn.net/?f=k_c%3D%5Cdfrac%7B%5BHBr%28g%29%5D%5E2%7D%7B%5BH_2%5D%7D%3D4.8%5Ctimes%2010%5E8M)
<u>3. ICE table.</u>
Write the initial, change, equilibrium table:
Molar concentrations:
H₂(g) + Br₂(l) ⇄ 2HBr(g)
I 0.400 0
C - x +2x
E 0.400 - x 2x
<u>4. Substitute into the expression of the equilibrium constant</u>
<u>5. Solve the quadratic equation</u>
- 192,000,000 - 480,000,000x = 4x²
- x² + 120,000,000x - 48,000,000 = 0
Use the quadratic formula:
The only valid solution is x = 0.39999999851M
Thus, the final concentration of H₂(g) is 0.400 - 0.39999999851 ≈ 0.00000000149 ≈ 1.5 × 10⁻⁹M
Answer:
54.7°C is the new temperature
Explanation:
We combine the Ideal Gases Law equation to solve this.
P . V = n. R. T
As moles the balloon does not change and R is a constant, we can think this relation between the two situations:
P₁ . V₁ / T₁ = P₂ . V₂ / T₂
T° is absolute temperature (T°C + 273)
68.7°C + 273 = 341.7K
(0.987 atm . 564L) / 341.7K = (0.852 atm . 625L) / T₂
1.63 atm.L/K = 532.5 atm.L / T₂
T₂ = 532.5 atm.L / 1.63 K/atm.L → 326.7K
T° in C = T°K - 273 → 326.7K + 273 = 54.7°C
Answer:
0.0738 M
Explanation:
HNO3 +LiOH = LiNO3 + H2O
Number of moles HNO3 = number of moles LiOH
M(HNO3)*V(HNO3) = M(LiOH)*M(LiOH)
M(HNO3)*50.0mL = 0.100M*36.90 mL
M(HNO3) = 0.100*36.90/50.0 M = 0.0738 M
The electron configuration
1
s
2
2
s
2
2
p
6
3
s
2
3
p
2
is the element Silicon.
The key to deciphering this is to look at the last bit of information of the electron configuration
3
p
2
.
The '3' informs us that the element is in the 3rd Energy Level or row of the periodic table. The 'p' tells us that the element is found in the p-block which are all of the Groups to the right of the transition metals, columns 13-18. The superscript '2' tells us that the element is found in the 2nd column of the p-block Group 14.
