Answer:
It contains 0.105 mole cu
Explanation:
1, 6, 2, 6
In the order you wrote them in
Answer:
Magnesium
0.003mole
Explanation:
The problem here entails we find the metal in the carbonate.
For group 2 member, let the metal = X;
The carbonate is XCO₃;
If we sum the atomic mass of the elements in the metal carbonate, we should arrive at 84g/mol
Atomic mass of C = 12g/mol
O = 16g/mol
Atomic mass of X + 12 + 3(16) = 84
Atomic mass of X = 84 - 60 = 24g/mol
The element with atomic mass of 24g is Magnesium
B.
Number of moles in 0.3g of CaCO₃:
Molar mass of CaCO₃ = 40 + 12 + 3(16) = 100g/mol
Number of moles = 
Number of moles = 
= 0.003mole
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!
It has 4 significant figures. If you see the Zero after the one decimal point you don’t count that and instead just started at 6
