A sample of 8.44 grams of NaOH is dissolved into 536 mL of aqueous 0.250 M NaOH (assume no volume change). This solution is then poured into 1.96 gallons of water. You may assume that the two volumes can be added.) What is the concentration of NaOH in the final solution?

Answer in units of M.

What volume of ethanol must be diluted with water to prepare 402 mL of 0.868 M solution? The density of ethanol is given as 0.789 g/mL
Answer in units of mL.

What weight of sodium nitrate (NaNO3) must be used to prepare 560mL of a 10.8Msolution of this salt in water?
Answer in units of g.

We dissolve 95 grams of sodium sulfate (Na2SO4) in 8 grams of water. What is the percent by mass of sodium sulfate in this solution?
Answer in units of %.

What volume of 0.558MK3PO4 is required to react with 47 mL of 0.343 M MgCl2 according to the equation 2K3PO4 + 3MgCl2 = Mg3(PO4)2 + 6KCl
Answer in units of mL.

We find that it takes 70.3 mL of NaOH to neutralize (react exactly with) 155 mL of a 0.300 molar HNO3 solution according to the reaction NaOH + HNO3 = NaNO3 + H2O.
What is the molarity of the NaOH solution?
Answer in units of M.

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To solve these questions, we'll need to use some basic concepts in chemistry, such as stoichiometry and concentration calculations. I'll break down the steps to find the answers to each question.

1. What is the concentration of NaOH in the final solution?

To find the concentration of NaOH in the final solution, we'll use the concept of dilution. First, we need to calculate the amount of NaOH added to the water:

mass of NaOH = mass of NaOH sample + mass of NaOH in the solution
mass of NaOH = 8.44 g + (536 mL * 0.250 M) * (40.00 g/mol)

Next, we calculate the final volume of the solution:

final volume = initial volume of water + volume of NaOH solution
final volume = 1.96 gallons + (536 mL/1000 mL per L) + (536 mL/1000 mL per L)

Finally, we can calculate the concentration of NaOH in the final solution:

concentration = mass of NaOH / final volume

2. What volume of ethanol must be diluted with water to prepare 402 mL of a 0.868 M solution?

To solve this problem, we'll use the concept of concentration and the equation: concentration = mass/volume.

First, we need to calculate the mass of ethanol needed:

mass of ethanol = concentration * volume
mass of ethanol = 0.868 M * 402 mL * 0.789 g/mL

Next, we can calculate the volume of ethanol needed:

volume of ethanol = mass of ethanol / density of ethanol

3. What weight of sodium nitrate (NaNO3) must be used to prepare a 10.8 M solution in 560 mL of water?

To solve this problem, we'll use the concept of concentration and the equation: concentration = mass/volume.

First, we need to calculate the mass of NaNO3 needed:

mass of NaNO3 = concentration * volume
mass of NaNO3 = 10.8 M * 560 mL * (62.00 g/mol + 14.01 g/mol + 3*16.00 g/mol)

4. What is the percent by mass of sodium sulfate in the solution?

To find the percent by mass of sodium sulfate in the solution, we'll use the equation: percent by mass = (mass of component / mass of solution) * 100.

First, we need to calculate the mass of sodium sulfate:

mass of Na2SO4 = 95 g

Next, we calculate the total mass of the solution:

mass of solution = mass of Na2SO4 + mass of water

Finally, we can calculate the percent by mass of sodium sulfate:

percent by mass = (mass of Na2SO4 / mass of solution) * 100

5. What volume of 0.558 M K3PO4 is required to react with 47 mL of 0.343 M MgCl2?

To solve this question, we'll use the stoichiometry of the balanced chemical equation:

2K3PO4 + 3MgCl2 → Mg3(PO4)2 + 6KCl

First, we need to determine the stoichiometric ratio between K3PO4 and MgCl2:

1 M K3PO4 reacts with 1.5 M MgCl2

Next, we can use the stoichiometry to calculate the volume of K3PO4 needed:

volume of K3PO4 = (47 mL * 0.343 M * 3 M K3PO4) / (0.558 M * 1.5 M MgCl2)

6. What is the molarity of the NaOH solution?

To find the molarity of the NaOH solution, we'll use the stoichiometry of the balanced chemical equation:

NaOH + HNO3 → NaNO3 + H2O

First, we need to determine the stoichiometric ratio between NaOH and HNO3:

1 M NaOH reacts with 1 M HNO3

Next, we can use the stoichiometry to calculate the molarity of the NaOH solution:

molarity of NaOH = (0.300 M HNO3 * 155 mL) / 70.3 mL