D.) There are 24 atoms in total in that compound.....
Answer:
The reaction quatient Q is 0.306
Explanation:
Step 1: data given
Volume = 2.50 L
Temperature = 1100 K
Number of moles C = 6.75 moles
Number of moles H2O = 13.5 moles
Number of moles CO = 3.10 moles
Number of moles H2 = 9.00 moles
Step 2: The balanced equation
C(s)+H2O(g)⇌CO(g)+H2(g)
Step 3: Calculate the concentration
Concentration = moles / volume
[C] = 6.75 moles / 2.5 L
[C] = 2.7 M
[H20] = 13.5 moles / 2.5 L
[H2O] = 5.4 M
[CO] = 3.10 moles / 2.5 L
[CO] = 1.24 M
[H2] = 9.00 moles / 2.5 L
[H2]= 3.6 M
Step 4: Calculate Q
Q = [CO][H2] / [C][H2O]
Q = (1.24 * 3.6)/(2.7*5.4)
Q = 0.306
The reaction quatient Q is 0.306
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.
Answer:An isotonic solution refers to two solutions having the same osmotic pressure across a semipermeable membrane. This state allows for the free movement of water across the membrane without changing the concentration of solutes on either side. i hope this helps :) .