I’m not 100% sure, but I think hydrogen gas is also produced :)
Answer:
A. N₂(g) + 3H₂(g) -----> 2NH₃ exothermic
B. S(g) + O₂(g) --------> SO₂(g) exothermic
C. 2H₂O(g) --------> 2H₂(g) + O₂(g) endothermic
D. 2F(g) ---------> F₂(g) exothermic
Explanation:
The question says predict not calculate. So you have to use your chemistry knowledge, experience and intuition.
A. N₂(g) + 3H₂(g) -----> 2NH₃ is exothermic because the Haber process gives out energy
B. S(g) + O₂(g) --------> SO₂(g) is exothermic because it is a combustion. The majority, if not all, combustion give out energy.
C. 2H₂O(g) --------> 2H₂(g) + O₂(g) is endothermic because it is the reverse reaction of the combustion of hydrogen. If the reverse reaction is exothermic then the forward reaction is endothermic
D. 2F(g) ---------> F₂(g) is exothermic because the backward reaction is endothermic. Atomisation is always an endothermic reaction so the forward reaction is exothermic
Answer:
8.34
Explanation:
1) how much moles of NH₃ are in the reaction;
2) how much moles of H₂ are in the reaction;
3) the required mass of the H₂.
all the details are in the attachment; the answer is marked with red colour.
Note1: M(NH₃) - molar mass of the NH₃, constant; M(H₂) - the molar mass of the H₂, constant; ν(NH₃) - quantity of NH₃; ν(H₂) - quantity of H₂.
Note2: the suggested solution is not the shortest one.
Mass 1 + %abundance of first isotope + Mass 2 + %abundance of second isotope
/ 100
This is RAM.
The mass of solute per 100 mL of solution is abbreviated as (m/v). Mass is not technically the same thing as weight, but the abbreviation (w/v) is also common. 262 grams of sucrose are needed to make 655 mL of a 40.0% (w/v) sucrose solution
<h3>Define Solute</h3>
A solute is a material that dissolves in a solution. The amount of solvent present in fluid solutions is greater than the amount of solute. The two most common examples of solutions in daily life are salt and water. Salt is the solute because it dissolves in water.
<h3>forms of ratios for product concentration or yield:-</h3>
- w/v:- Weight by volume or weight per volume are the terms used. Any solid compound's concentration in a liquid can be calculated using it. It is measurable in gm/ml.
- Weight by weight ratio is referred to as w/w.It is employed to determine the final yield of the compound obtained from the starting compound. as in —mg/—gm.
It provides the real yield of the substance or item.
- Volume/volume. It is used to specify a liquid's composition or percent in a liquid compound.
using w/v we can calculate the weight of sucrose:-
40.0% means 40 g sucrose/ 100 g solution
40.0g sucrose x (655/100)=grams of sucrose
262 grams of sucrose are needed to make 655 mL of a 40.0% (w/v) sucrose solution.
Learn more about Solute here:-
brainly.com/question/14397121
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