Answer:
Therefore, 54.45 g of Ammonium sulphate is required
Explanation:
Molar mass of ammonium sulphate, (NH₄)₂SO₄ = 132 g/mol
Since density of solution is 1.10 g/ml, a 1 mL solution has mass = 1.10 g
275 mL solution will have a mass = 275 * 1.10 = 302.5 g
Since the solution is 18% ammonium sulphate by weight,
mass of ammonium sulphate present in 275 mL solution = 302.5 * 18/100
mass of ammonium sulphate present in 275 mL solution = 54.45 g
Therefore, 54.45 g of Ammonium sulphate is required
Answer:
Isotopes are atoms that have different number of neutrons, so they have a different mass number than the other atoms in an element.
Answer:
95.2 - 40.8 = 54.4 g of oxygen
number of moles = mass (g)/ Mr
no. of moles of carbon = 40.8/12 = 3.4
no. of moles of oxygen = 3.4
divide both by smallest value which is 3.4 and you’ll get 1 mole of carbon and 1 mole of oxygen therefore the empirical formula is CO
Explanation:
hope this helps :)
When aluminum metal is made to contact with chlorine gas (Cl₂), a highly exothermic reaction proceeds. This produces aluminum chloride (AlCl₃) powder. The balanced chemical equation for this reaction is shown below:
2Al(s) + 3Cl₂(g) → 2AlCl₃(s)
Since it was stated that aluminum is in excess, this means that the amount of AlCl₃ produced will only depend on the amount of Cl₂ gas available. The molar mass of Cl₂ is 70.906 g/mol. Using stoichiometry, we have the following equation:
(21.0 g Cl₂/ 70.906 g/mol Cl₂) x 2 mol AlCl₃/ 2 mol Cl₂ = 0.1974 mol AlCl₃
Thus, we have determined that 0.1974 <span>moles of aluminum chloride can be produced from 21.0 g of chlorine gas. </span>
