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stiv31 [10]
3 years ago
12

Calculate how many grams of sodium chloride we would need to make a solution with the same molarity of seawater, which is about

0.598 M.
Chemistry
1 answer:
aliya0001 [1]3 years ago
7 0

Answer:

34.92 grams NaCl (in 1 L of solution)

Explanation:

The chemical formula of sodium chloride is NaCl. From the formula, we can calculate the molar mass of NaCl:

MM(NaCl)= MM(Na) + MM(Cl) = 23 g/mol + 35.4 g/mol = 58.4 g/mol

A solution of NaCl with a molarity of 0.598 M has 0.598 moles of NaCl per liter of solution. So, we multiply the moles by the molar mass of NaCl to calculate the mass we need:

mass of NaCl = 0.598 mol x 58.4 g/mol = 34.92 g NaCl

Therefore, we need 34.92 grams of NaCl to prepare 1 liter of a solution with a molarity of 0.598 M.

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The solubility of Z is 60 g/ 100 g water at 20 °C. How many grams of solution are produced when a saturated solution is prepared
V125BC [204]

Answer:

Saturated solution = 180 gram

Explanation:

Given:

Solubility of Z = 60 g / 100 g water

Given temperature =  20°C

Amount of water = 300 grams

Find:

Saturated solution

Computation:

Saturated solution = [Solubility of Z] × Amount of water

Saturated solution = [60 g / 100 g] × 300 grams

Saturated solution = [0.6] × 300 grams

Saturated solution = 180 gram

3 0
3 years ago
Please help me with these Radioactive Decay Problems ASAP! I need helppppp
Masteriza [31]

Answer:

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Explanation:

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5 0
3 years ago
What volume (in milliliters) of oxygen gas is required to react with 4.03 g of Mg at STP?
BigorU [14]
Mg reaction with O₂ gas will produce MgO so the equation will be
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The first step is to find the number of moles of Mg in 4.03g of Mg.  You can do this by dividing 4.03g Mg by its molar mass (which is 24.3g/mol) to get 0.1658mol Mg.  Then you have to find the number of moles of O₂ that will react with 0.1658mol Mg.  To do this you need to use the fact that 1mol O₂ will react with 2mol Mg (this reatio is from the chemical equation) so you have to multiply 0.1658mol Mg by (1mol O₂)/(2mol Mg) to get 0.0829mol O₂.  From here you would usually use PV=nRT and solve for V However, the question tells us that we are at STP, that means you can use the fact that 22.4L of gas is 1 mol of gas at STP.  Using that information we can find the volume of O₂ gas by mulitlying 0.0829mol O₂ by 22.4L/mol to get 1.857L which equals 1857mL.
therefore, 1857mL of O₂ gas will react with 4.03g of Mg.

I hope this helps. Let me know in the comments if anything is unclear.
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3 years ago
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