Take This 10th Grade Chemistry Quiz

Explanation
The relative formula mass of a compound is the sum of the atomic masses of all the atoms in its formula. In copper sulfate (CuSO4), we have one copper atom (Cu) with an atomic mass of 63.55 amu, one sulfur atom (S) with an atomic mass of 32.07 amu, and four oxygen atoms (O) with an atomic mass of 16.00 amu each. Adding these up, we get a total of 63.55 + 32.07 + (16.00 x 4) = 159.55 amu. Rounding it off, the relative formula mass of copper sulfate is 160 amu.

Explanation
The relative formula mass of MgF2 can be calculated by adding the atomic masses of magnesium (Mg) and two fluorine (F) atoms. The atomic mass of Mg is 24 amu and the atomic mass of F is 19 amu. Adding these together gives a total of 62 amu, which is the relative formula mass of MgF2.

Explanation
When 9g of aluminium reacts with 35.5g of chlorine, the ratio of their masses can be used to determine the empirical formula of the compound formed. By dividing the mass of each element by their respective atomic masses (27g/mol for aluminium and 35.5g/mol for chlorine), we find that the ratio is approximately 1:3. This indicates that the empirical formula of the compound is AlCl3, where one aluminium atom combines with three chlorine atoms.

Explanation
The percent yield is calculated by dividing the actual yield (140g) by the theoretical yield (200g) and multiplying by 100. In this case, the calculation would be (140g/200g) * 100 = 70%. This means that the reaction only produced 70% of the expected amount of product.

Explanation
The empirical formula of a compound represents the simplest whole number ratio of the elements present in the compound. To determine the empirical formula, we need to calculate the moles of each element.

Given:
Mass of calcium (Ca) = 13.5g
Mass of oxygen (O) = 10.8g
Mass of hydrogen (H) = 0.675g

To find the moles, we divide the mass of each element by its molar mass:
Moles of Ca = 13.5g / 40.08g/mol = 0.336 mol
Moles of O = 10.8g / 16.00g/mol = 0.675 mol
Moles of H = 0.675g / 1.01g/mol = 0.669 mol

Now, we need to find the simplest whole number ratio of these moles. Dividing each mole value by the smallest value (0.336 mol) gives us:
Moles of Ca = 0.336 mol / 0.336 mol = 1
Moles of O = 0.675 mol / 0.336 mol ≈ 2
Moles of H = 0.669 mol / 0.336 mol ≈ 2

Therefore, the empirical formula of the compound is Ca(OH)2.

Explanation
In the given reaction, the stoichiometric ratio between sodium hydroxide (NaOH) and chlorine (Cl2) is 2:1. This means that for every 2 moles of NaOH, we need 1 mole of Cl2. To find the amount of Cl2 needed, we first need to convert the given mass of NaOH (100g) to moles. The molar mass of NaOH is 40g/mol, so 100g of NaOH is equal to 2.5 moles. Since the ratio is 2:1, we need half the number of moles of Cl2, which is 1.25 moles. The molar mass of Cl2 is 71g/mol, so the mass of Cl2 needed is 1.25 moles multiplied by 71g/mol, which equals 88.75g. Therefore, the correct answer is 88.75g.

Explanation
Methane (CH4) consists of one carbon atom and four hydrogen atoms. To calculate the percent mass of hydrogen in methane, we need to determine the molar mass of hydrogen and methane. The molar mass of hydrogen is 1 g/mol, and the molar mass of methane is 16 g/mol (12 g/mol for carbon + 4 g/mol for hydrogen). The percent mass of hydrogen in methane can be calculated by dividing the molar mass of hydrogen by the molar mass of methane and multiplying by 100. Thus, (4 g/mol / 16 g/mol) * 100 = 25%. Therefore, the correct answer is 25%.

Explanation
In the given reaction, 2 moles of calcium react with 1 mole of oxygen to form 2 moles of calcium oxide. The molar mass of calcium is 40 g/mol and the molar mass of oxygen is 16 g/mol. Therefore, 2 moles of calcium will react with 32 g of oxygen. To find the mass of oxygen that will react with 60 g of calcium, we can set up a proportion: (32 g of oxygen) / (2 moles of calcium) = (x g of oxygen) / (60 g of calcium). Solving for x, we find that x = 24 g of oxygen.

Explanation
The correct answer is 60% because it is asking for the percentage of magnesium in magnesium oxide. Since magnesium oxide consists of one magnesium atom and one oxygen atom, and the atomic mass of magnesium is 24.31 g/mol while the atomic mass of oxygen is 16.00 g/mol, the molar mass of magnesium oxide is 40.31 g/mol. Therefore, the mass of magnesium in magnesium oxide is 24.31 g/mol, which is 60% of the total mass of magnesium oxide.