Answer:
The final volume of the gas is 36.1 L.
Explanation:
Given:
Initial pressure of the gas is, 
Final pressure of the gas is, 
Initial volume of the gas is, 
Final volume of the gas is, 
Here, we shall use Boyle's Law which states that for a process under constant temperature, the pressure of the gas changes inversely with the change in volume.
Here, the pressure is increased. So, the volume of the gas is decreased.
Therefore, as per Boyle's Law:
So, the final volume of the gas is 36.1 L.
Answer:
709 g
Step-by-step explanation:
a) Balanced equation
Normally, we would need a balanced chemical equation.
However, we can get by with a partial equation, as log as carbon atoms are balanced.
We know we will need an equation with masses and molar masses, so let’s <em>gather all the information</em> in one place.
M_r: 30.07 236.74
C₂H₆ + … ⟶ C₂Cl₆ + …
m/g: 90.0
(i) Calculate the moles of C₂H₆
n = 90.0 g C₂H₆ × (1 mol C₂H₆ /30.07 g C₂H₆)
= 2.993 mol C₂H₆
(ii) Calculate the moles of C₂Cl₆
The molar ratio is (1 mol C₂Cl₆/1 mol C₂H₆)
n = 2.993 mol C₂H₆ × (1 mol C₂Cl₆/1 mol C₂H₆)
= 2.993 mol C₂Cl₆
(iii) Calculate the mass of C₂Cl₆
m = 2.993 mol C₂Cl₆ × (236.74 g C₂Cl₆/1 mol C₂Cl₆)
m = 709 g C₂Cl₆
The reaction produces 709 g C₂Cl₆.
I think the chemical reaction is:<span>
N2H4 + 2 H2O2-> N2 + 4H2O
We are given the amount reactants allowed to react. This will be the starting point of the reaction. First, is to find the limiting reactant.
9.24 g H2O2 ( 1 mol / 34.02 g ) = 0.27 mol H2O2
6.56 g N2H4 ( 1mol / 32.06) = 0.20 mol N2H4
Since from the reaction we have 1:2 ratio of the reactants then the limiting reactant is hydrogen peroxide. We will use this to find the amount of N2 produced.
0.27 mol H2O2 ( 1 mol N2 / 2 mol H2O2 ) ( 14.01 g N2 / 1 mol N2) =1.89 g N2
</span>
Answer:
The answer to the question is;
The concentration of the Solution #1 in terms of molarity is
0.16704X moles/litre.
Explanation:
Let the concentration of the stock solution be X moles/liter
Therefore, 83.52 ml of the stock solution contains
83.52×(X/1000) moles
Dilution of 83.52 ml of X to 500 ml gives solution 1 with a concentration of
500 ml of solution 1 contains 83.52×(X/1000) moles
Therefore 1000 ml or 1 litre contains 2×83.52×(X/1000) moles = 0.16704X moles/litre
The molarity of solution 1 is 0.16704X moles/litre.
1 mole => 6.023 x 10^23 molecules
x => 8.5 x 10^25 molecules
x = 8.5x10^25/ 6.023 x 10^23 = 141.1
