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lorasvet [3.4K]
4 years ago
11

approximately how many grams of sodium chloride (NaCl) are required to prepare 500. mL of 3.00 M solution

Chemistry
1 answer:
WITCHER [35]4 years ago
4 0
3.00 M  means 3.00 mol/L. So if you want 500 mL=0.5 L, it needs 3.00 mol/L * 0.5 L = 1.50 mol. For NaCl, the molar mass is 23+35.5=58.5 g/mol. So the amount of NaCl needed is 1.50*58.5=87.75 g. So the answer is 87.75 grams.
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The prefix 'kilo-' means a thousand. So a kilo-gram is a thousand grams. 1 kilogram=1,000 grams How many kilograms is 1,008 gram
lesantik [10]

<u>Answer:</u>  1.0 kilograms.

<u>Explanation:</u>

One kilogram is equal to a thousand grams.

Supposing x to be the number of kilograms equal to one thousand and eight grams, we can write it as:

1 kg = 1000 grams

x kg = 1008 grams

To solve for x, we can simply divide 1008 grams by 1000 thousand grams to get the answer.

x = 1008 / 1000

x = 1.008

Rounding this value to the nearest tenth, it will become 1.0 kilograms.

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4 years ago
Butane, C4H10 burns in oxygen. How many grams of water vapor, H2O, are produced by the combustion of 580 grams of butane at stan
zepelin [54]

Answer:

The answer to your question is 900 g of water vapor

Explanation:

Data

mass of H₂O = ?

mass of butane = 580 g

Balanced chemical reaction

             2C₄H₁₀ + 13O₂  ⇒  8CO₂  +  10H₂O

Process

1.- Calculate the molar weight of butane and water

Butane (C₄H₁₀) = 2[(12 x 4) + (1 x 10)]

                         = 2[48 + 10]

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Water (H₂O) = 10[(1 x 2) + (1 x 16)]

                    = 10[2 + 16]

                    = 10[18]

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2.- Use proportions and cross multiplication to find the mass of water vapor

             116 g of butane ------------- 180 g of water

             580 g of butane  ----------  x

                x = (580 x 180) / 116

                x = 900 g of water vapor

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Answer:

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