Solid Magnesium is considered as active metal so it reacts with strong acids like HCl and H₂SO₄ liberating Hydrogen gas according to the following equations:
Mg(s) + 2 HCl(aq) → MgCl₂(aq) + H₂(g)
Mg(s) + H₂SO₄(aq) → MgSO₄(aq) + H₂(g)
so the amount of solid magnesium decrease by addition of strong acid to it.
Answer:
a)22.2°C after adding magnesium
b)17.3°C before adding magnesium
c) 4.9 is change
<span>In the field of science, usually, the product of an experiment is
computed ahead to understand if it reached a specific objective. It would reach
greater than 100% of percent yield if the factors include faster reaction rates;
proper handling of the reactants, no outside contaminants, and the procedure of
the experiment is followed smoothly. It would reach lesser than 100% percent yield
if the experiment is not followed, external factors such as contamination from
the environment (wind, moisture, etc). </span>
Answer:
0.74 grams of methane
Explanation:
The balanced equation of the combustion reaction of methane with oxygen is:
it is clear that 1 mol of CH₄ reacts with 2 mol of O₂.
firstly, we need to calculate the number of moles of both
for CH₄:
number of moles = mass / molar mass = (3.00 g) / (16.00 g/mol) = 0.1875 mol.
for O₂:
number of moles = mass / molar mass = (9.00 g) / (32.00 g/mol) = 0.2812 mol.
- it is clear that O₂ is the limiting reactant and methane will leftover.
using cross multiplication
1 mol of CH₄ needs → 2 mol of O₂
??? mol of CH₄ needs → 0.2812 mol of O₂
∴ the number of mol of CH₄ needed = (0.2812 * 1) / 2 = 0.1406 mol
so 0.14 mol will react and the remaining CH₄
mol of CH₄ left over = 0.1875 -0.1406 = 0.0469 mol
now we convert moles into grams
mass of CH₄ left over = no. of mol of CH₄ left over * molar mass
= 0.0469 mol * 16 g/mol = 0.7504 g
So, the right choice is 0.74 grams of methane
Answer:
Explanation:
1. Calculate the moles of copper(II) hydroxide
2. Calculate the molecules of copper(II) hydroxide