Question 2Tips1 ptsHow many moles of OH− are there in 2.286 g of Mg(OH)2?Give your answer in moles and to three decimal places.(Molar mass of  Mg(OH)2 = 58.32 g mol−1) Flag question: Question 3Question 3Tips1 ptsIf 0.86 mol of a substance has a mass of 52.10 g, what is its molar mass?Give your answer in g mol-1 and to two decimal places. Flag question: Question 4Question 4Tips1 ptsWhat is the percentage yield (no decimals in answer) where you obtained 16.6 grams from a reaction in which the theoretical yield was 25 g? Flag question: Question 5Question 5Tips2 ptsYou dissolve 8.18 g of sodium chloride in 500 mL of water, what is the molarity of sodium cations in the solution as moles per litre? Give your answer to two decimal places. Flag question: Question 6Question 6Tips1 ptsIf you obtained a percentage yield of 12.7% from a reaction where the theoretical yield was 84 g, what was the actual amount of material you obtained in grams?  Flag question: Question 7Question 72 ptsAs you move down a group (e.g. moving from Al to Ga), the number of electrons/filled orbitals means that the atomic radius of the atom  Flag question: Question 8Question 81 ptsFor the following reaction, which would be the limiting reagent?  2H2 +O2 --> 2H2O You add 0.225 mol of hydrogen gas (H2) and 0.125 mol of oxygen gas (O2)Group of answer choicesWater (H2O)Hydrogen Peroxide (H2O2)Hydrogen Gas (H2)Oxygen Gas (O2)

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Question 2: First, we need to convert the mass of Mg(OH)2 to moles using its molar mass.

Moles = mass / molar mass = 2.286 g / 58.32 g/mol = 0.039 mol

Since there are 2 moles of OH- for every mole of Mg(OH)2, we multiply the moles of Mg(OH)2 by 2 to get the moles of OH-.

Moles of OH- = 0.039 mol * 2 = 0.078 mol

So, there are 0.078 moles of OH- in 2.286 g of Mg(OH)2.

Question 3: Molar mass is calculated by dividing the mass of the substance by the number of moles.

Molar mass = mass / moles = 52.10 g / 0.86 mol = 60.58 g/mol

So, the molar mass of the substance is 60.58 g/mol.

Question 4: Percentage yield is calculated by dividing the actual yield by the theoretical yield and multiplying by 100.

Percentage yield = (actual yield / theoretical yield) * 100 = (16.6 g / 25 g) * 100 = 66%

So, the percentage yield is 66%.

Question 5: First, we need to convert the mass of NaCl to moles using its molar mass (58.44 g/mol).

Moles = mass / molar mass = 8.18 g / 58.44 g/mol = 0.14 mol

Since the volume of the solution is 500 mL, we convert this to liters (0.5 L) and then calculate the molarity.

Molarity = moles / volume = 0.14 mol / 0.5 L = 0.28 M

So, the molarity of sodium cations in the solution is 0.28 M.

Question 6: The actual yield is calculated by multiplying the theoretical yield by the percentage yield.

Actual yield = theoretical yield * (percentage yield / 100) = 84 g * (12.7 / 100) = 10.67 g

So, the actual amount of material obtained was 10.67 g.

Question 7: As you move down a group, the number of electrons/filled orbitals increases, which means that the atomic radius of the atom increases.

Question 8: The limiting reagent is the one that is completely consumed first in a chemical reaction. In this case, the reaction requires 2 moles of H2 for every mole of O2. Since we have more than enough H2 (0.225 mol) for the amount of O2 (0.125 mol), the limiting reagent is O2.