Answer:75%
Explanation:
First, the balanced reaction equation must be written out clearly as a guide to solving the problem. The molar masses of H3PO4 and K3PO4 are then calculated as they will be consistently required in solving the problem. The theoretical yield is obtained from the amount of H3PO4 reacted. Since 1 mole of H3PO4 yields 1 mole of K3PO4, 0.05 moles of H3PO4 yields 0.05 moles of K3PO4. The mass of K3PO4 is produced is then the product of 0.05 and it molar mass hence the theoretical yield. The % yield is calculated as shown.
Answer:
smallest particle of an element which can take part in a chemical reaction
Explanation:
there are atoms of different elements is sodium atom
Answer:
ΔH°_rxn = -195.9 kJ·mol⁻¹
Explanation:
4NH₃(g) + 3O₂(g) ⟶ 2N₂(g) +6H₂O(g)
ΔH°_f/(kJ·mol⁻¹): -45.9 0 0 -241.8
The formula relating ΔH°_rxn and enthalpies of formation (ΔH°_f) is
ΔH°_rxn = ΣΔH°_f(products) – ΣΔH°_f(reactants)
ΣΔH°_f(products) = -6(241.8) = -1450.8 kJ
ΣΔH°_f(reactants) = -4(45.9) = -183.6 kJ
ΔH°_rxn = (-1450.8 + 183.6) kJ = -1267.2 kJ
Answer:
mass CaI2 = 23.424 Kg
Explanation:
From the periodic table we obtain for CaI2:
⇒ molecular mass CaI2: 40.078 + ((2)(126.90)) = 293.878 g/mol
∴ mol CaI2 = (4.80 E25 units )×(mol/6.022 E23 units) = 79.708 mol CaI2
⇒ mass CaI2 = (79.708 mol CaI2)×(293.878 g/mol) = 23424.43 g
⇒ mass CaI2 = 23.424 Kg
Assume it is 1 litre and weighs 1kg.
2 percent of 1 kg is 20g.
20g divided by molar mass of NaOH.
20g divide by 40 = 0.5 mole
0.5 mole in a litre would be 0.5M
That is the answer: 0.5M
