Moles of H2SO4= 7.5x10^23/ 6.02x10^23 = 1.25 (3sf) moles of H2SO4
Mass of 1 mole of H2SO4= 98.1g
Therefore mass of 7.5x10^23 molecules of H2SO4= 122.63g
Molar mass of MgCO3 is 84.313 g/mol
You can calculate this from data on the periodic table:
Molar mass Mg = 24.305g/mol
molar mass C = 12.011g/mol
molar mass O = 15.999g/mol mass 3 mol = 47.997g
Total = 84.313g/mol
Mass to be used in 1.2L of 1.5M solution = 84.313g * 1.2L * 1.5mol /L = 151.763g
I have not taken significant figures into account
The balanced equation you provide is not necessary in this calculation
True, is the correct answer.
Answer:
ΔHr = -103,4 kcal/mol
Explanation:
<u>Using:</u>
<u>AH° (kcal/mol)
</u>
<u>Metano (CH)
</u>
<u>-17,9
</u>
<u>Cloro (CI)
</u>
<u>tetraclorometano (CCI)
</u>
<u>- 33,3
</u>
<u>Acido cloridrico (HCI)
</u>
<u>-22</u>
It is possible to obtain the ΔH of a reaction from ΔH's of formation for each compound, thus:
ΔHr = (ΔH products - ΔH reactants)
For the reaction:
CH₄(g) + Cl₂(g) → CCl₄(g) + HCl(g)
The balanced reaction is:
CH₄(g) + 4Cl₂(g) → CCl₄(g) + 4HCl(g)
The ΔH's of formation for these compounds are:
ΔH CH₄(g): -17,9 kcal/mol
ΔH Cl₂(g): 0 kcal/mol
ΔH CCl₄(g): -33,3 kcal/mol
ΔH HCl(g): -22 kcal/mol
The ΔHr is:
-33,3 kcal/mol × 1 mol + -22 kcal/mol× 4 mol - (-17,9 kcal/mol × 1 mol + 0kcal/mol × 4mol)
<em>ΔHr = -103,4 kcal/mol</em>
<em></em>
I hope it helps!
Gold is one of the softest metals. So YES
