Answer:
111 L
Explanation:
Calculation of moles of hydrogen gas:-
Mass of = 18.6 g
Molar mass of = 2.01588 g/mol
According to the given reaction:-
2 moles of hydrogen gas on reaction produces one mole of acetic acid gas.
So,
1 mole of hydrogen gas on reaction produces mole of acetic acid gas.
Also,
9.23 mole of hydrogen gas on reaction produces mole of acetic acid gas.
Moles of acetic acid gas = 4.615 moles
Given that:
Temperature = 35 °C
The conversion of T( °C) to T(K) is shown below:
T(K) = T( °C) + 273.15
So,
T = (35 + 273.15) K = 308.15 K
n = 4.615 moles
P = 1.05 atm
V = ?
Using ideal gas equation as:
where,
P is the pressure
V is the volume
n is the number of moles
T is the temperature
R is Gas constant having value = 0.0821 L atm/ K mol
Applying the equation as:
1.05 atm × V = 4.615 moles ×0.0821 L atm/ K mol × 308.15 K
<u>⇒V = 111 L</u>
Answer:
the correct answer is H2SO4 and HSO4-41
2Na(s) + 2 H2O (liquid) -> 2 NaOH (aq) + H2 (g)
1) 2 reactants are given, so we have to find which of them is an excess reactant and which of them will react completely.
Convert gram of Na and H2O into moles.
M(Na) =23.0 g/mol
10.0 g Na*1 mol Na/23.0 g Na≈0.4349 mol Na
M(H2O) = 2*1.0 +16.0 = 18.0 g/mol
50.0 g H2O*1 mol H2O/18.0 g H2O =2.778 mol H2O
2Na(s) + 2 H2O (liquid) -> 2 NaOH (aq) + H2 (g)
from equation 2 mol 2 mol
given 0.4349 mol 2.778 mol
We can see that H2O is an excess reactant, so we are going to find amount of H2 using Na.
2)
2Na(s) + 2 H2O (liquid) -> 2 NaOH (aq) + H2 (g)
from equation 2 mol 1 mol
given 0.4349 mol x mol
x=(0.4349*1)/2 =0.21745 mol H2
3) mass of H2
0.21745 mol H2 * 2.0 g H2/1 mol ≈ 0.435 g H2
Answer : 0.435 g H2.
The vapour pressure of a substance actually differs
relative to the temperature. So to know which compound has the vapour pressure
of 58 kilopascals at 65 degrees Celsius, we refer to standard tables of
temperature vs Vapour pressure.
From the tables or graph, the answer here is:
<span>2. ethanol</span>