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3 years ago

What is the molarity of a Sodium Chloride solution if 86.00g of NaCl was dissolved in enough water to form 1000.mL of solution?

Chemistry

1 answer:

Vladimir79 [104]3 years ago

4 0

Answer:

1.47 M

Explanation:

First we <u>convert 86.00 g of NaCl into moles of NaCl</u>, using the <em>molar mass of NaCl</em>:

86.00 g ÷ 58.44 g/mol = 1.472 mol NaCl

Then we <u>convert 1000 mL into liters:</u>

1000 mL / 1000 = 1.0 L

Finally we can <u>calculate the molarity of the solution</u>:

Molarity = moles / liters

1.472 mol / 1.0 L = 1.472 M

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The molar mass of h2o is 18.01 g/mom . The molar mass of o2 is 32.00 g/moo .what mass of h2o on grams must react to produce 50.0

Julli [10]

Oxygen can be obtained from water using electrolysis process as follows:
2 H2O .............> 2H2 + O2

It is given that:
molar mass of water = 10.01 grams and molar mass of O2 = 32 grams

From the balanced chemical equation, we can conclude that:
2 moles of water produce 1 mole of oxygen
2 x 18.01 = 36.02 grams of water produce 32 grams of oxygen.

To calculate how many grams of water must react to produce 50 grams of oxygen, we can use cross multiplication as follows:
mass of required water = 56.28125 grams

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3 years ago

(4 points) The following lead compound for a pharmaceutical drug contains a rotatable bond. Using the principles of rigidificati

masya89 [10]

Answer:

Explanation:

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3 0

2 years ago

When nahco3 completely decomposes, it can follow this balanced chemical equation: 2nahco3 → na2co3 h2co3 determine the theoretic

BigorU [14]

Theoretical yield = 2.397

The product could be sodium carbonate

percent yield = 98.456%

When nahco3 completely decomposes, it can follow this balanced chemical equation:

2nahco3 → na2co3 h2co3

If the mass of the NaHCO3 sample is 3.80 g, we must use stoichiometry to calculate the theoretical yields of each of the products.

mass of NaHCO₃ = 3.80 g

molar mass of NaHCO₃ = 84 g/mol

so the no of moles of NaHCO₃ = 3.80/84 = 0.0452 mol

You see, one mole of sodium carbonate and one mole of hydrogen carbonate are produced from two moles of sodium bicarbonate.

so, the no of moles of sodium carbonate = 0.0452/2 = 0.0226 mol

∴ mass of sodium carbonate ( Na₂CO₃) = no of moles of Na₂CO₃ × molar mass of Na₂CO₃

= 0.0226 × 106 ≈ 2.397 g

no of moles of hydrogen carbonate = 0.0452/2 = 0.0226 mol

mass of the hydrogen carbonate ( H₂CO₃) = no of moles of H₂CO₃ × molar mass of H₂CO₃

= 0.0226 × 62 g = 1.401 g

mass of one of the products was measured to be 2.36 g , from above data, we can say it must be sodium carbonate because value is the nearest of 2.397 g.

percentage yield = experimental yield/theoretical yield × 100

here experimental yield of Na₂CO₃ = 2.36 g

and theoretical yield of Na₂CO₃ = 2.397 g

∴ % yield = 2.36/2.397 × 100 ≈ 98.456%

Therefore the percentage yield of the product is 98.456%

To learn more about percentage yield visit:

brainly.com/question/22257659

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6 0

1 year ago

Determine how many ml of water you need to remove, by evaporation, if you have a 500 ml of 10.20 M HNO3 dilute solution and you

Ludmilka [50]

The total volume of water that would be removed will be 75 mL

<h3>Dilution equation</h3>

Using the dilution equation:

M1V1 = M2V2

In this case, M1 = 500 mL, V1 = 10.20 M, M2 = 12 M

Substitute:

V2 = 500 x 10.20/12

= 425 mL

The final volume in order to arrive at 12 M HNO3 would be 425 mL from the initial 500 mL. Thus, the total amount of water that will be removed by evaporation can be calculated as:

500 - 425 = 75 mL

More on dilution can be found here: brainly.com/question/7208939

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2 years ago

H4X - pKa1 =1.5

Nataly [62]

Othu['oi;eejlktmkolhkjdfm;hojnor;'mor;'mor;'mor;'mor;'mor;'mhim

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2 years ago