A second order reaction varies with the square of the concentration of the reactant. Therefore, halving the concentration will reduce the rate of reaction by a factor of 4.
The answer is E.
Answer:
Explanation:
The pressure, the volume and the temperature of an ideal gas are related to each other by the equation of state:
where
p is the pressure of the gas
V is the volume of the gas
n is the number of moles
R is the gas constant
T is the absolute temperature
For the gas in this problem:
n = 2.00 mol is the number of moles
V = 17.4 L is the gas volume
p = 3.00 atm is the gas pressure
is the absolute temperature
Solving for R, we find the gas constant:
Answer:
Explanation:
<u>According to Arrhenius concept of acid and base:</u>
"When a base in a solution, produces/yields OH- (Hydroxide) ions."
So, when a base is dissolved in a solution, it produces OH- ions.
<u>For example:</u>
NaOH ⇄ Na⁺ + OH⁻ (So, it is a base)
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Hope this helped!
<h3>~AH1807</h3>
Pb(NO₃)₂ ⇒limiting reactant
moles PbI₂ = 1.36 x 10⁻³
% yield = 87.72%
<h3>Further explanation</h3>
Given
Reaction(unbalanced)
Pb(NO₃)₂(s) + NaI(aq) → PbI₂(s) + NaNO₃(aq)
Required
- moles of PbI₂
- Limiting reactant
- % yield
Solution
Balanced equation :
Pb(NO₃)₂(s) + 2NaI(aq) → PbI₂(s) + 2NaNO₃(aq)
mol Pb(NO₃)₂ :
= 0.45 : 331 g/mol
= 1.36 x 10⁻³
mol NaI :
= 250 ml x 0.25 M
= 0.0625
Limiting reactant (mol : coefficient)
Pb(NO₃)₂ : 1.36 x 10⁻³ : 1 = 1.36 x 10⁻³
NaI : 0.0625 : 2 = 0.03125
Pb(NO₃)₂ ⇒limiting reactant(smaller ratio)
moles PbI₂ = moles Pb(NO₃)₂ = 1.36 x 10⁻³(mol ratio 1 : 1)
Mass of PbI₂ :
= mol x MW
= 1.36 x 10⁻³ x 461,01 g/mol
= 0.627 g
% yield = 0.55/0.627 x 100% = 87.72%
Answer:
4.5moles
Explanation:
First, let us balance the equation given from the question. This is illustrated below:
KClO3 —> KCl + O2
There are 2 atoms of O on the right side and 3 atoms on the left. It can be balance by putting 2 in front of KClO3 and 3 in of O2 as shown below
2KClO3 —> KCl + 3O2
Now, we have 2 atoms each of K and Cl on the left side and 1atom each of K and Cl on the right. It can be balance by putting 2 in front of KCl as shown below:
2KClO3 —> 2KCl + 3O2
Now the equation is balanced.
From the balanced equation,
2 moles of KClO3 produced 3 moles of O2.
Therefore, 3 moles of KClO3 will produce = (3 x 3) /2 = 4.5moles of O2.
Therefore 3 moles of KClO3 will produce 4.5 moles of O2
