Answer:
[O2(g)][SO2(g)]^2/[SO3(g)]^2
Answer:
1.60x10⁶ billions of g of CO₂
Explanation:
Let's calculate the production of CO₂ by a single human in a day. The molar mass of glucose is 180.156 g/mol and CO₂ is 44.01 g/mol. By the stoichiometry of the reaction:
1 mol of C₆H₁₂O₆ -------------------------- 6 moles of CO₂
Transforming for mass multiplying the number of moles by the molar mass:
180.156 g of C₆H₁₂O₆ ----------------- 264.06 g of CO₂
4.59x10² g ---------------- x
By a simple direct three rule:
180.156x = 121203.54
x = 672.77 g of CO₂ per day per human
So, in a year, 6.50 billion of human produce:
672.77 * 365 * 6.50 billion = 1.60x10⁶ billions of g of CO₂
Answer:
N₂ + 3H₂ → 2NH₃ ΔH = - 92.2KJ
Explanation:
Let's write out the chemical equation between Nitrogen and Hydrogen to Form Ammonia.
Nitrogen + Hydrogen = Ammonia
N₂ + H₂ → NH₃
A Thermochemical Equation is a balanced stoichiometric chemical equation that includes the enthalpy change, ΔH.
The balanced stoichiometric chemical equation is given as;
N₂ + 3H₂ → 2NH₃
92.2 kJ of energy are evolved for each mole of N2(g) that reacts. And from the equation, 1 mole of N2 reacts.
The enthalpy change, ΔH = - 92.2KJ. The negative sign is because heat is being evolved.
The balanced thermochemical equation;
N₂ + 3H₂ → 2NH₃ ΔH = - 92.2KJ
Answer:
21.8 grams.
Explanation:
Molar mass data from a modern periodic table:
How many moles of MgO will be produced if Mg is the limiting reactant?
Number of moles of Mg:
.
The ratio between the coefficient of Mg and that of MgO is 2:2. Two moles of Mg will make two moles of MgO. 0.670644 moles of MgO will be produced if Mg is the limiting reactant.
How many moles of MgO will be produced if O₂ is the limiting reactant?
Number of moles of O₂:
.
The ratio between the coefficient of O₂ and that of MgO is 1:2. One mole of O₂ will make two moles of MgO. of MgO will be produced if O₂ is in excess.
How many moles of MgO will be produced?
0.541284 is smaller than 0.670644. Only 0.541284 moles of MgO will be produced since O₂ will run out before all 16.3 grams of Mg is consumed.
What's the mass of 0.541284 moles of MgO?
Formula mass of MgO:
.
Mass of 0.541284 moles of MgO:
.
Calcium carbonate has the formula: CaCO3
From the periodic table:
mass of calcium = 40 grams
mass of carbon = 12 grams
mass of oxygen = 16 grams
Therefore,
molar mass of CaCO3 = 40 + 12 + 3(16) = 100 grams
molar mass of carbonate = 12 + 3(16) = 60 grams
One mole of calcium carbonate contains one mole of carbonate. Therefore, 100 grams of CaCO3 contains 60 grams of CO3.
If the 0.5376 grams of the unknown substance is CaCO3, then the amount of carbonate will be:
amount of carbonate = (0.5376*60) / 100 = 0.32256 grams
Based on the above calculations, the sample is not CaCO3