Answer:
New volume V2 = 92.7 Liter (Approx)
Explanation:
Given:
V1 = 106 l
T1 = 45 + 273.15 = 318.15 K
P1 = 740 mm
T2 = 20 + 273.15 = 293.15 K
P2 = 780 mm
Find:
New volume V2
Computation:
P1V1 / T1 = P2V2 / T2
(740)(106) / (318.15) = (780)(V2) / (293.15)
New volume V2 = 92.7 Liter (Approx)
The volume of water vapour gas that would be produced will be 2,581.73 L
<h3>Stoichiometric calculations</h3>
From the equation of the reaction, the mole ratio of propane to oxygen is 1:5.
Mole of 535.55 grams of propane = 535.55/44.1 = 12.12 moles
Mole of 2665.06 grams of oxygen = 2665.06/32 = 83.28 moles
Going by the mole ratio, it appears propane is limiting while oxygen is in excess.
From the equation, 1 mole of propane produces 4 moles of water vapour. Thus, the equivalent mole of water vapour will be:
12.12 x 4 = 48.48 moles.
Using the ideal gas equation: pv = nRt
v = nRT/p = 48.48 x 0.08206 x 623/0.96 = 2,581.73 L
More on ideal gas equations can be found here: brainly.com/question/4147359
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Answer:- Formula of the hydrate is and it's name is Iron(III)sulfate pentahydrate.
Solution:- As per the given information, there is 18.4% water in the hydrate. If we assume the mass of the hydrate as 100 grams then there would be 18.4 grams of water and 81.6 grams of Iron(III)sulfate present in the hydrate.
Molar mass for Iron(III)sulfate is 399.88 gram per mol and the molar mass for water is 18.02 gram per mol.
We will calculate the moles of Iron(III)sulfate and water present in the compound on dividing their grams by their molar masses as:
=
=
Now, the next step is to calculate the mol ratio and for this we divide the moles of each by the least one of them means whose moles are less. Here, the moles of Iron(III)sulfate are less than moles of water. So, we divide the moles of each by 0.204.
= 1
= 5
There is 1:5 mol ratio between Iron(III)sulfate and water. So, the formula of the hydrate is and the name of the hydrate is Iron(III)sulfate pentahydrate.
Answer:
Explanation:
<u>1. Chemical equilibrium equation:</u>
<u>2. Species</u>
In an equilibrium constant expression you do not include the solid substances; only gases and dissolved substances.
The symbol means solid, thus Cu(s) and Ag(s) shall not appear in your equilibrium constant expression.
The symbol means in aqueous solution, thus the both and must appear in the equilibrium constant expression.
<u>3. Equilibrium constant expression.</u>
It is the quotient of the product of the concentrations of the species on the right hand side of the equilibrium equation, each raised to its corresponding coefficient, and the product of the concentrations of the species on the left hand side, each raised to its coresponding coefficient.