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How many meters are their in one kilometer

tatuchka [14]

There are 1000 meters in 1 Kilometer.

Hope this helps you. :)

5 0

4 years ago

Read 2 more answers

A mixture 21.7 g NaCl 3.74 g kcl and 9.76 g licl how many moles of nacl are in this mixture

slava [35]

Moles (mol) = mass (g) / molar mass (g/mol)

Mass of NaCl = 21.7 g
Molar mass of NaCl = <span>58.4 g/mol
Hence, moles of NaCl = </span>21.7 g / 58.4 g/mol = 0.372 mol

Hence moles of NaCl in the mixture is 0.372 mol.

Let's assume that mixture has only given compounds and free of impurities.
Then, we can present this as a mole percentage.

mole % = (moles of desired substance / Total moles of the mixture) x 100%

Hence,
mole % of NaCl = (moles of NaCl / Total moles of the mixture) x 100%

Total moles of mixture = moles of NaCl + KCl + LiCl

Mass of KCl = 3.74 g
Molar mass of NaCl = 74.6 g/mol
Hence, moles of NaCl = 3.74 g / 74.6 g/mol = 0.050 mol

Mass of NaCl = <span>9.76 g
</span>Molar mass of NaCl = 42.4 g/mol
Hence, moles of NaCl = 9.76 g / 42.4 g/mol = 0.230 mol

Total moles = 0.372 mol + 0.050 mol + 0.230 mol = 0.652 mol

mole % of NaCl = (moles of NaCl / Total moles of the mixture) x 100%
= (0.372 mol / 0.652 mol) x 100%
= 57.06%

Hence, mixture has 57.06% of NaCl as the mole percentage.

8 0

3 years ago

If 3.0 moles of hydrogen react, how many grams of nitrogen will be used? Round your answer to the nearest whole number. N2 + 3H2

Iteru [2.4K]

Answer:

28g

Explanation:

Given parameters:

Number of moles of hydrogen = 3moles

Unknown:

Mass of nitrogen used = ?

Solution:

To solve this problem, let us establish the balanced reaction equation first;

N₂ + 3H₂ → 2NH₃

From the balanced reaction equation;

3 mole of H₂ will combine with 1 mole of N₂

Now,

Mass of Nitrogen = number of moles x molar mass

Molar mass of N₂ = 2(14) = 28g/mol

Mass of Nitrogen = 1 mole x 28g/mole = 28g

8 0

3 years ago

How does a chemical equation model a chemical reaction

ale4655 [162]

Answer:

reactant + reactant = product + product

Explanation:

(pretend = is the arrow)

the reaction is shown in the equation by modelling how the reactants react with eachother to form the products on the other side of the arrow

4 0

3 years ago

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

3 years ago