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3 years ago

How many grams of hydrogen are needed to react with 1.40 g of nitrogen to produce ammonia?

Chemistry

1 answer:

vazorg [7]3 years ago

8 0

Answer:

Write a balanced chemical reaction:

N2 + 3H2 ==> 2NH3

Looking at the mole ratios in this balanced equation you can see it takes 3 moles H2 to make 2 moles NH3. So, next calculate the moles of NH3 represented by 1.80 g and then convert to moles of H2 needed:

moles of NH3 = 1.80 g x 1 mole/17 g = 0.106 moles NH3

Moles H2 needed = 0.106 moles NH3 x 3 moles H2/2 moles NH3 = 0.159 moles H2 needed

Grams H2 needed = 0.159 moles x 2 g/mole = 0.318 grams H2 needed

Explanation:

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Buck's Turf Care Company mowed 233 lawns over a 11 week period. What is the average weekly rate of mowing lawns?

masya89 [10]

Answer: 21 lawns per week

Explanation:

The average weekly rate refers to how many lawns were mowed per week given that 233 were done in 11 weeks.

Rate will be given by;

= Lawns mowed / Weeks taken

= 233 / 11

= 21 lawns per week

6 0

3 years ago

How does an objects speed affect its kinetic energy?

rewona [7]

Because it’s kinetic energy INCREASES the speed

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2 years ago

Which physical property does the inventor need to consider when choosing a

adelina 88 [10]

Answer:

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Explanation:

Ductility

6 0

3 years ago

All of the following are forms of acid precipitation EXCEPT ________.

Aneli [31]

Dihydrogen oxide is the right answer. Dihydrogen oxide is just 2 hydrogen and 1 oxygen which is H2O or water.

6 0

3 years ago

Uranium-238 decays to lead-206 with a half-life of 4.5 x 109 yr. Determine how much uranium-238 decays in milligrams (to three s

Mamont248 [21]

Answer:

2.15 mg of uranium-238 decays

Explanation:

For decay of radioactive nuclide-

$N=N_{0}.(\frac{1}{2})^{\frac{t}{t_{\frac{1}{2}}}}$

where N is amount of radioactive nuclide after t time, $N_{0}$ is initial amount of radioactive nuclide and $t_{\frac{1}{2}}$ is half life of radioactive nuclide

Here $N_{0}=4.60 mg$ , $t=4.1\times 10^{9}yr$ and $t_{\frac{1}{2}}=4.5\times 10^{9}yr$

So, $N=(4.60mg)\times (\frac{1}{2})^{\frac{4.1\times 10^{9}}{4.5\times 10^{9}}}$

so, N = 2.446 mg

mass of uranium-238 decays = (4.60-2.446) mg = 2.15 mg

3 0

3 years ago