53.5 g MgCl2 /95.21 g/mol = 0.5619 Moles MgCl2. Since the ratio is 3MgCl : 2Na3PO4 you multiply 0.5619 x 2/3 (since you're solving for NaPO, the 2 goes on top) which shows we have .3746 Moles of NaPO. Multiply times it's molar mass so 0.3746x 163.94g/mol= 61.4136 grams NaPO. I believe.
No i no idea why why did no one ever mean that i just got me a picture
Dmitri Mendeleev devised the first periodic table.
There are 2 elements in the equation that form a product
Answer:
ΔH0reaction = [ΔHf0 CO2(g)] - [ΔHf0 CO(g) + ΔHf0 O2(g)]
Explanation:
Chemical equation:
CO + O₂ → CO₂
Balanced chemical equation:
2CO + O₂ → 2CO₂
The standard enthalpy for the formation of CO = -110.5 kj/mol
The standard enthalpy for the formation of O₂ = 0 kj/mol
The standard enthalpy for the formation of CO₂ = -393.5 kj/mol
Now we will put the values in equation:
ΔH0reaction = [ΔHf0 CO2(g)] - [ΔHf0 CO(g) + ΔHf0 O2(g)]
ΔH0reaction = [-393.5 kj/mol] - [-110.5 kj/mol + 0]
ΔH0reaction = [-393.5 kj/mol] - [-110.5 kj/mol]
ΔH0reaction = -283 kj/mol