C=2.41 g/L
m(NaCl)=0.291 g
c=m(NaCl)/v
v=m(NaCl)/c
v=0.291/2.41= 0.1207 L = 120.7 mL
Boyle's Law states: pV = constant.
24.43 x 1.895 = 46.29485
therefore, 15.6 x _____ = 46.29485
unknown = 2.968L
<u>Answer:</u> The concentration of hydrogen gas at equilibrium is 0.037 M
<u>Explanation:</u>
We are given:
Initial concentration of HI = 1.0 M
The given chemical equation follows:
<u>Initial:</u> 1.0
<u>At eqllm:</u> 1.0-2x x x
The expression of 
for above equation follows:
![K_c=\frac{[H_2][I_2]}{[HI]^2}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BH_2%5D%5BI_2%5D%7D%7B%5BHI%5D%5E2%7D)
We are given:
Putting values in above expression, we get:
Neglecting the negative value of 'x' because concentration cannot be negative
So, equilibrium concentration of hydrogen gas = x = 0.037 M
Hence, the concentration of hydrogen gas at equilibrium is 0.037 M
Your best bet is B or D. Most likely B
Answer:
Q₁- The concentration of HCl = 0.075 N = 0.075 M.
Q₂- The concentration of KOH = 7.675 mN = 7.675 mM.
Q₃- The concentration of H₂SO₄ = 0.2115 N = 0.105 M.
Q₄- The equivalence point is the point at which the added titrant is chemically equivalent completely to the analyte in the sample whereas the endpoint is the point where the indicator changes its color.
Explanation:
<u><em>Q₁:
</em></u>
- As acid neutralizes the base, the no. of gram equivalent of the acid is equal to that of the base.
- The normality of the NaOH and HCl = Their molarity.
∵ (NV)NaOH = (NV)HCl
∴ N of HCl = (NV)NaOH / (V)HCl = (0.15 N)(67 mL) / (134 mL) = 0.075 N.
∴ The concentration of HCl = 0.075 N = 0.075 M.
<em><u>Q₂:</u></em>
- As mentioned in Q1, the no. of gram equivalent of the acid is equal to that of the base at neutralization.
- The normality of H₂SO₄ = Molarity of H₂SO₄ x 2 = 0.050 M x 2 = 0.1 N.
∵ (NV)H₂SO₄ = (NV)KOH
∴ N of KOH = (NV)H₂SO₄ / (V)KOH = (0.1 N)(27.4 mL) / (357 mL) = 7.675 x 10⁻³ N = 7.675 mN.
∴ The concentration of KOH = 7.675 mN = 7.675 mM.
<em><u>Q₃:</u></em>
- As mentioned in Q1 and 2, the no. of gram equivalent of the acid is equal to that of the base at neutralization.
- The normality of NaOH = Molarity of NaOH = 0.5 N.
∵ (NV)H₂SO₄ = (NV)NaOH
∴ N of H₂SO₄ = (NV)NaOH / (V)H₂SO₄ = (0.5 N)(55 mL) / (130 mL) = 0.2115 N.
∴ The concentration of H₂SO₄ = 0.2115 N = 0.105 M.
<em><u>Q₄:</u></em>
- The equivalence point is the point at which the added titrant is chemically equivalent completely to the analyte in the sample whereas the endpoint is the point where the indicator changes its color.
- The equivalence point in a titration is the point at which the added titrant is chemically equivalent completely to the analyte in the sample. It comes before the end point. At the equivalence point, the millimoles of acid are chemically equivalent to the millimoles of base.
- End point is the point where the indicator changes its color. It is the point of completion of the reaction between two solutions.
- The effectiveness of the titration is measure by the close matching between equivalent point and the end point. pH of the indicator should match the pH at the equivalence to get the same equivalent point as the end point.
