% H = 100 - ( 52.14 + 34.73 )=13.13 %
<span>assume 100 g of this compound </span>
<span>mass H = 13.13 g </span>
<span>moles H = 13.13 g / 1.008 g/mol=13 </span>
<span>mass C = 52.14 g </span>
<span>moles C = 52.14 g/ / 12.011 g/mol=4 </span>
<span>mass O = 34.73 g </span>
<span>moles O = 34.73 g/ 15.999 g/mol=2 </span>
<span>the empirical formula is C4H13O2</span>
The answer cl atoms are in zn(cio3)2 is A:2 because I looked it up and it gave me 2.
True since ithave same number of electrons and protons
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
Answer:6.94
Explanation:
Molar mass of CaCO3=40+12+16×3
=40+12+48=100g/mol
Moles=mass of substance/molar mass
=97mg/100g=0.097/100=0.00097moles/L.
PH=-log[CaCo3]=-log(0.00097)=6.94
P.s it's log to base e
