Answer:
2.53 L is the volume of H₂ needed
Explanation:
The reaction is: C₁₈H₃₀O₂ + 3H₂ → C₁₈H₃₆O₂
By the way we can say, that 1 mol of linolenic acid reacts with 3 moles of oxygen in order to produce, 1 mol of stearic acid.
By stoichiometry, ratio is 1:3
Let's convert the mass of the linolenic acid to moles:
10.5 g . 1 mol / 278.42 g = 0.0377 moles
We apply a rule of three:
1 mol of linolenic acid needs 3 moles of H₂ to react
Then, 0.0377 moles will react with (0.0377 . 3 )/1 = 0.113 moles of hydrogen
We apply the Ideal Gases Law to find out the volume (condition of measure are STP) → P . V = n . R . T → V = ( n . R .T ) / P
V = (0.113 mol . 0.082 L.atm/mol.K . 273.15K) 1 atm = 2.53 L
Answer:
The law of conservation of mass states that mass in an isolated system is neither created nor destroyed by chemical reactions or physical transformations. According to the law of conservation of mass, the mass of the products in a chemical reaction must equal the mass of the reactants.
Answer:
This experiment is uncontrolled because two different masses of substance A are used.
Explanation:
A controlled experiment is a structured experiment aimed at testing a particular observation or observations. The setup of a controlled experiment helps to determine the reason why a particular observation occurs and what must have led to it.
In the experiment highlighted above, different masses of a substance were used, they were heated to different temperatures. The set up does not show any correlation between the masses of substances heated and the temperatures. It is even difficult to try to predict the hypothesis for this kind of experimental set up. All the variables in play can best be assumed to be independent of one another.
Relative molecular mass or RMM is the answer.
Answer:
A. 145.2 g NH3
B. 0.76 g H2
C. 1.41 x 10^22 molecules NH3
Explanation:
A. 1 mol N2 -> 2 mol NH3
4.27 mol N2 -> x
x= (4.27 mol N2 * 2 mol NH3)/1 mol N2 x= 8.54 mol NH3
1 mol N2 -> 17 g
8.54 mol N2 -> x x= 145.2 g NH3
B.
2 g H2 -> 34 g NH3
x -> 13.01 g NH3
x= 0.76 g H2
C.
2 g H2 -> 34 g NH3
0.0235 g H2 -> x
x= 0.39 g NH3
0.39 g NH3 (1 mol NH3/17 g NH3)(6.023 x 10^23 molecules/1 mol NH3) =
1.41 x 10^22 molecules NH3
