The balanced chemical reaction is:
2HCl + Ca = CaCl2 + H2
We are given the amount of the reactants to be used for the reaction. These values will be the starting point of our calculations.
100 g HCl ( 1 mol HCl / 36.46 g HCl ) = 2.74 mol HCl
100 g Ca ( 1 mol Ca / 40.08 g ) = 2.08 mol Ca
From the reaction, the mole ratio of the reactants is 2:1 where every 2 moles of hydrochloric acid, 1 mole of calcium is required. Therefore, the limiting reactant for this case is calcium.
Answer:
1. B
2. E
3. A
4. D
5. C
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Hope I helped!!!
GL :)
pH of the acetyl choline solution before incubation = 7.65
![[H_{3}O^{+}]=10^{-7.65}=2.24*10^{-8}M](https://tex.z-dn.net/?f=%5BH_%7B3%7DO%5E%7B%2B%7D%5D%3D10%5E%7B-7.65%7D%3D2.24%2A10%5E%7B-8%7DM)
pH of the solution after incubation = 6.87
![[H_{3}O^{+}]=10^{-6.87}=1.35*10^{-7}M](https://tex.z-dn.net/?f=%5BH_%7B3%7DO%5E%7B%2B%7D%5D%3D10%5E%7B-6.87%7D%3D1.35%2A10%5E%7B-7%7DM)
The difference in concentration of hydronium ion before and after incubation
=
-
=
This difference in hydronium ion concentration can be attributed to the increase in the concentration of acetic acid, which is formed when acetylcholine is hydrolyzed by acetycholinesterase. The mole ratio of acetylcholine to acetic acid is 1:1.
Therefore the moles of acetylcholine = 
Answer is: mass of unused sulfur is 5.87 grams.
Balanced chemical reaction: C + 2S → CS₂.
m(C) = 12.0 g; mass of carbon.
m(S) = 70.0 g; mass of sulfur.
n(C) = m(C) ÷ M(C).
n(C) = 12 g ÷ 12 g/mol.
n(C) = 1 mol; amount of substance.
n(S) = m(S) ÷ M(S).
n(S) = 70 g ÷ 32.065 g/mol.
n(S) = 2.183 mol.
From chemical reaction: n(C) : n₁(S) = 1 : 2.
n₁(S) = 1 mol · 2 = 2 mol.
Δn(S) = n(S) - n₁(S).
Δn(S) = 2.183 mol - 2 mol.
Δn(S) = 0.183 mol; amount of unused sulfur.
Δm(S) = 0.183 mol · 32.065 g/mol.
Δm(S) = 5.87 g.
Answer:
Convergent Plate Boundaries
Explanation:
I hope this helps :)
