How many grams of calcium nitrate, Ca(NO3)2 (molecular weight 164), contains 24 grams of oxygen atoms?

The energy is stored in the bonds between atoms. The bonds absorb energy as they are put under stress and release the energy as they relax (when the object returns to its original shape).

8 0

3 years ago

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In the preparation of a certain alkyl halide, 10 g of sodium bromide (NaBr), 10 mL distilled water (H20), and 9 mL 3-methyl-1-bu

Novosadov [1.4K]

Percentage yield shows the amount of reactants converted into products. The percentage yield of the reaction is 51.7%.

The equation of the reaction is sown in the image attached. The reaction is 1:1 as we can see.

Number of moles of NaBr = 10 g/103 g/mol = 0.097 moles

We can obtain the mass of 3-methyl-1-butanol from its density.

Mass = density × volume

Density of 3-methyl-1-butanol = 0.810 g/mL

Volume of 3-methyl-1-butanol = 9 mL

Mass of 3-methyl-1-butanol = 0.810 g/mL × 9 mL

Mass of 3-methyl-1-butanol = 7.29 g

Number of moles of 3-methyl-1-butanol = mass/molar mass = 7.29 g/88 g/mol = 0.083 moles

Since the reaction is 1:1 then the limiting reagent is 3-methyl-1-butanol

Mass of product 1-bromo-3-methylbutane = number of moles × molar mass

Molar mass of 1-bromo-3-methylbutane = 151 g/mol

Mass of product 1-bromo-3-methylbutane = 0.083 moles × 151 g/mol

= 12.53 g

Recall that % yield = actual yield/theoretical yield × 100

Actual yield of product = 6.48 g

Theoretical yield = 12.53 g

% yield = 6.48 g/12.53 g × 100

% yield = 51.7%

Learn more: brainly.com/question/5325004

7 0

2 years ago

Given that sodium chloride is 39.0% sodium by mass, how many grams of sodium chloride are needed to have 100.mg of Na present? G

seraphim [82]

<u>Answer:</u> The mass of sodium chloride solution present is 0.256 grams.

<u>Explanation:</u>

We are given:

39.0 % of sodium in sodium chloride solution

This means that 39.0 grams of sodium is present in 100 grams of sodium chloride solution

Mass of sodium given = 100 mg = 0.1 g (Conversion factor: 1 g = 1000 mg)

Applying unitary method:

If 39 grams of sodium metal is present in 100 grams of sodium chloride solution

So, if 0.1 grams of sodium metal will be present in = $\frac{100}{39}\times 0.1=0.256g$ of sodium chloride solution.

Hence, the mass of sodium chloride solution present is 0.256 grams.

8 0

3 years ago