Carbon is located in the fourth group, second period. It is also in the nonmetal section of the table as well
Answer : The partial pressure of and is, 216.5 mmHg and 649.5 mmHg
Explanation :
According to the Dalton's Law, the partial pressure exerted by component 'i' in a gas mixture is equal to the product of the mole fraction of the component and the total pressure.
Formula used :
So,
where,
= partial pressure of gas
= mole fraction of gas
= total pressure of gas
= moles of gas
= total moles of gas
The balanced decomposition of ammonia reaction will be:
Now we have to determine the partial pressure of and
Given:
and,
Given:
Thus, the partial pressure of and is, 216.5 mmHg and 649.5 mmHg
and .
Assuming complete decomposition of both samples,
First compound:
; of the first compound would contain
Oxygen and mercury atoms seemingly exist in the first compound at a ratio; thus the empirical formula for this compound would be where the subscript "1" is omitted.
Similarly, for the second compound
; of the first compound would contain
and therefore the empirical formula
.
"windweathered" because the use of wind weathering would bring the sand all around and it would be wind weathered
Answer:
Explanation:
<u>1) Rate law, at a given temperature:</u>
- Since all the data are obtained at the same temperature, the equilibrium constant is the same.
- Since only reactants A and B participate in the reaction, you assume that the form of the rate law is:
r = K [A]ᵃ [B]ᵇ
<u>2) Use the data from the table</u>
- Since the first and second set of data have the same concentration of the reactant A, you can use them to find the exponent b:
r₁ = (1.50)ᵃ (1.50)ᵇ = 2.50 × 10⁻¹ M/s
r₂ = (1.50)ᵃ (2.50)ᵇ = 2.50 × 10⁻¹ M/s
Divide r₂ by r₁: [ 2.50 / 1.50] ᵇ = 1 ⇒ b = 0
- Use the first and second set of data to find the exponent a:
r₁ = (1.50)ᵃ (1.50)ᵇ = 2.50 × 10⁻¹ M/s
r₃ = (3.00)ᵃ (1.50)ᵇ = 5.00 × 10⁻¹ M/s
Divide r₃ by r₂: [3.00 / 1.50]ᵃ = [5.00 / 2.50]
2ᵃ = 2 ⇒ a = 1
<u>3) Write the rate law</u>
This means, that the rate is independent of reactant B and is of first order respect reactant A.
<u>4) Use any set of data to find K</u>
With the first set of data
- r = K (1.50 M) = 2.50 × 10⁻¹ M/s ⇒ K = 0.250 M/s / 1.50 M = 0.167 s⁻¹
Result: the rate constant is K = 0.167 s⁻¹