Answer:
The answer is
<h2>2.00 %</h2>
Explanation:
The percentage error of a certain measurement can be found by using the formula
From the question
actual measurement = 46.37 g
error = 47.25 - 46.37 = 0.88
The percentage error of the measurement is
We have the final answer as
<h3>2.00 %</h3>
Hope this helps you
Answer is: <span>the temperature of the reaction is 546</span>°C.
k₁ = 1,1·10⁻⁴ 1/s.
T₁ = 470 °C = 743,15 K.
Ea = 264 kJ/mol = 264000 J/mol.
k₂ = 4,36·10⁻³ 1/s.
R = 8,314 J/K·mol.
T₂ = ?
Natural logarithm<span> of Arrhenius' equation:
</span>lnk₁ = lnA - Ea/RT₁.
lnk₂ = lnA - Ea/RT₂.
ln(k₂/k₁) = (Ea/R) · (1/T₁ - 1/T₂).
ln( 4,36·10⁻³ 1/s / 1,1·10⁻⁴ 1/s) = (264000J/mol ÷ 8,314 J/K·mol) · ·(1/743,15K - 1/T₂).
3,68 = 31753,66 K · (0,00134 1/K - 1/T₂).
3,68 = 42,728 - 31753,66 · (1/T₂).
1/T₂ = 0,00122.
T₂ = 819 K = 546 °C.
Answer:
The correct answer is 199.66 grams per mole.
Explanation:
Based on law of effusion given by Graham, a gas rate of effusion is contrariwise proportionate to the square root of molecular mass, that is, rate of effusion of gas is inversely proportional to the square root of mass. Therefore,
R1/R2 = √ M2/√ M1
Here rate is the rate of effusion of the gas expressed in terms of number of mole per uni time or volume, and M is the molecular mass of the gas.
Rate Q/Rate N2 = √M of N2/ √M of Q
The molecular mass of N2 or nitrogen gas is 28 grams per mole and M of Q is molecular mass of Q and based on the question Q needs 2.67 times more to effuse in comparison to nitrogen gas, therefore, rate of Q = rate of N2/2.67
Now putting the values we get,
rate of N2/2.67/rate of N2 = √28/ √M of Q
√M of Q = √ 28 × 2.67
M of Q = (√ 28 × 2.67)²
M of Q = 199.66 grams per mole
I’m so sorry if it’s wrong but I think it’s this.....
Answer:
2NaHCO3(s) = Na2CO3(s) + CO2(g) + H2O(g).
And I found it on Quora.com
By the way NaHCO3 is Sodium bicarbonate but a.k.a Baking soda
Hoped this helped :)