Percent yield of the equation at above condidtions will be 42.4%
Answer:
B) The pH of a buffer solution does not change significantly when a small amount of acid is added.
C) The pH of a buffer solution is determined by the ratio of the concentration of acid to the concentration of base.
Explanation:
Buffer solution is useful in maintaining the pH of a solution constant . It is made of a weak acid and its salt or a weak base and its salt . Ph of a buffer is given by the following relation .
pH = pKa + log [ A⁻ ] / [ HA ]
A- is a base and HA is a weak acid .
When we add acid , it reacts with A⁻ or base and gets neutralised .
When we add base , it reacts with acid HA and gets neutralised . In this way it maintains the pH of the solution .
Answer:
A. 2Cu(s) + S(s) → Cu₂S(s)
B. 2C₈H₁₈(g) + 25O₂(g) → 16CO₂(g) + 18H₂O(g)
C. 2SO₂(g) + O₂(g) → 2SO₃(g)
Explanation:
A. Let's think the reactants:
Cu and S
The product is: CuS
Equation: 2Cu(s) + S(s) → Cu₂S(s)
B. The reactants are:
O₂ and C₈H₁₈
The products are: water and CO₂
This is a combustion reaction: 2C₈H₁₈(g) + 25O₂(g) → 16CO₂(g) + 18H₂O(g)
C. Reactants: SO₂ and O₂
Products: SO₃
Equation: 2SO₂(g) + O₂(g) → 2SO₃(g)
This is a classical C1V1 = C2V2 question. We are given the initial concentration (2.0M), the final concentration (0.5M) and the final volume (200mL). We are asked how many mL of 2.0 M NaBr we require to make up the final concentration and volume, so we are calculating the initial volume (V1)
Therefore:
C1V1 = C2V2
2.0M(V1) = (0.5M)(200.0 mL)
V1 = 50 mL
So 50 mL of NaBr is needed to make a dilution to 0.5M in 200 mL.
