Answer:
Mg(NO4)2 is 180.3 g/mol
Explanation:
First find the substance formula.
Magnesium Nitrate.
Magnesium is a +2 charge.
Nitrate is a -1 charge.
So to balance the chemical formula,
We need 1 magnesium atom for every nitrate atom.
2(1) + 1(-2) = 0
So the substance formula is Mg(NO4)2.
Now find the molar mass of Mg(NO4)2.
Mg = 24.3 amu
N = 14.0 amu
O = 16.0 amu
They are three nitrogen and twelve oxygen atoms.
So you do this: 24.3 + 14.0(2) + 16.0(8) = 180.3 g/mol
So the molar is mass is 180.3 g/mol.
The final answer is Mg(NO4)2 is 180.3 g/mol
Hope it helped!
Gain .....
It loses electron when a positive charge is formed..ex Na+
Answer:
ΔH°rxn = - 433.1 KJ/mol
Explanation:
- CH4(g) + 4Cl2(g) → CCl4(g) + 4HCl(g)
⇒ ΔH°rxn = 4ΔH°HCl(g) + ΔH°CCl4(g) - 4ΔH°Cl2(g) - ΔH°CH4(g)
∴ ΔH°Cl2(g) = 0 KJ/mol.....pure element in its reference state
∴ ΔH°CCl4(g) = - 138.7 KJ/mol
∴ ΔH°HCl(g) = - 92.3 KJ/mol
∴ ΔH°CH4(g) = - 74.8 KJ/mol
⇒ ΔH°rxn = 4(- 92.3 KJ/mol) + (- 138.7 KJ/mol) - 4(0 KJ/mol) - (- 74.8 KJ/mol)
⇒ ΔH°rxn = - 369.2 KJ/mol - 138.7 KJ/mol - 0 KJ/mol + 74.8 KJ/mol
⇒ ΔH°rxn = - 433.1 KJ/mol
Answer:
800.0 mL.
Explanation:
- To solve this problem; we must mention the rule states the no. of millimoles of a substance before and after dilution is the same.
<em>(MV)before dilution of HCl = (MV)after dilution of HCl</em>
M before dilution = 12.0 M, V before dilution = 100.0 mL.
M after dilution = 1.5 M, V after dilution = ??? mL.
∵ (MV)before dilution of HCl = (MV)after dilution of HCl
∴ (12.0 M)(100.0 mL) = (1.5 M)(V after dilution of HCl)
<em>∴ V after dilution of HCl = (12.0 M)(100.0 mL)/(1`.5 M) = 800.0 mL.</em>
