Given the balanced equation:
( Reaction type : double replacement)
CaF2 + H2SO4 → CaSO4 + 2HFI
We can determine the number of grams prepared from the quantity of 75.0 H2SO4, and 63.0g of CaF2 by converting these grams to moles per substance.
This can be done by evaluating the atomic mass of each element of the substance, and totaling it to find the molecular mass.
For H2SO4 or hydrogen sulfate it's molecular mass is the sum of the quantity of atomic mass per element. H×2 + S×1 + O×4 = ≈1.01×2 + ≈32.06×1 + ≈16×4 = 2.02 + 32.06 + 64 = 98.08 u (Dalton's or Da) or g / mol.
For CaF2 or calcium fluoride, it's molecular mass adds 1 atomic mass of calcium and 2 atomic masses of fluoride due to the number of atoms.
Ca×1 + F×2 = ≈40.07×1 + ≈19×2 = 40.08 + 38 = 78.07 u (Da or Dalton's) or g / mol.
Answer:
The combustion of 59.7 grams of methane releases 3320.81 kilojoules of energy
Explanation:
Given;
CH₄ + 2O₂ → CO₂ + 2H₂O, ΔH = -890 kJ/mol
From the combustion reaction above, it can be observed that;
1 mole of methane (CH₄) released 890 kilojoules of energy.
Now, we convert 59.7 grams of methane to moles
CH₄ = 12 + (1x4) = 16 g/mol
59.7 g of CH₄ 
1 mole of methane (CH₄) released 890 kilojoules of energy
3.73125 moles of methane (CH₄) will release ?
= 3.73125 moles x -890 kJ/mol
= -3320.81 kJ
Therefore, the combustion of 59.7 grams of methane releases 3320.81 kilojoules of energy
If you change the subscripts it would change the reactants or products and then you would be solving a different formula, you would change what the chemical is
Answer is 3
s -1
p - 3
d - 5
f - 7
Mass is the property of a physical body and the resistance to acceleration when a net force is applied on the body.
The atomic mass of sodium (Na) is = 22.98
The atomic mass of nitrate (N) is = 14.00
The atomic mass of oxygen (O) is = 15.99
The sodium nitrate (NaNO3) consists of the atomic masses of Na+N+(O)3 = 85 grams
Therefore, the mass of 6.5 mol of sodium nitrate is = 6.5 * 1 mol of NaNO3
= 6.5 * (85)
= 552.50 grams
