Answer:
Tyre
In cold weather, you might have regularly kept a check on the pressure of the tyres of your car. Driving increases the temperature of the tyres, and, therefore, the air inside the tyre warms and expands. When you measure the pressure of the tyres at the time when you have just driven the car, it will be high. However, in cold weather, the pressure of the tyres will be low. So, it is recommended that you should always measure the pressure of the tyres.
Answer:
Average mass of 6 apple = 1 kg
Explanation:
Given:
Average mass of 5 dozen apple = 10 kg
Find:
Average mass of 6 apple
Computation:
Average mass of 6 apple = [Average mass of 1 apple] x [6 apple]
Average mass of 6 apple = [10/(12 x 5)] x [6]
Average mass of 6 apple = 1 kg
The answer is B...
because any of the others could have an opinion based answer but B will always be the same its a fact not opinion.
Answer:
3. 75.0%
Explanation:
2 ClO2(g) + F2(g) → 2 FClO2(g)
First order with respect to ClO2 and F2.
This means the rate equation is given as;
Rate = k [ClO2][F2]
When the initial concentrations of ClO2 and F2 are equal?
Let's assume an initial value of 1 for both reactants, so rate equation is given as;
Rate = k * 1 * 1 = k
The rate after 25% of the F2 has reacted is what percent of the initial rate?
The concentration left of F2 is 75% ( 100% - 25%) = 0.75
Concentration of ClO2 remains 1.
So rate equation is given as;
Rate = k * 1 * 0.75 = 0.75 k
Comparing 0.75k and k.
This means our answer is;
3. 75.0%
Answer:
ΔG° = -5.4 kJ/mol
ΔG = 873.2 J/mol = 0.873 kJ /mol
Explanation:
Step 1: Data given
ΔG (NO2) = 51.84 kJ/mol
ΔG (N2O4) = 98.28 kJ/mol
Step 2:
ΔG = ΔG° + RT ln Q
⇒with Q = the reaction quatient
⇒with T = the temperature = 298 K
⇒with R = 8.314 J / mol*K
⇒with ΔG° = ΔG° (N2O4) - 2*ΔG°(NO2
)
⇒ ΔG° = 98.28 kJ/mol - 2* 51.84 kJ/mol
⇒ ΔG° = -5.4 kJ/mol
Part B
ΔG = ΔG° =RT ln Q
⇒with G° = -5.4 kj/mol = -5400 j/mol
⇒
with R = 8.314 J/K*mol
⇒with T = 298 K
⇒with Q = p(N2O4)/ [ p(NO2) ]² = 1.63/0.36² = 12.577
ΔG = -5400 + 8.314 * 298 * ln(12.577)
ΔG = -5400 + 8.314 * 298 * 2.532
ΔG = 873.2 J/mol = 0.873 kJ/mol