All categories

3 years ago

At the boiling point, the vapor pressure of a liquid is equal to the _______.

Chemistry

1 answer:

Marat540 [252]3 years ago

6 0

At the boiling point, the vapor pressure of a liquid is equal to the atmospheric pressure.

You might be interested in

If 10.0g of oxygen reacts with 10.0g of hydrogen, how many grams of water can be

rusak2 [61]

Answer:

the answer is 90g

Explanation:

2g of H2 produce 18g of H2O/10.0g of H2 to produce x the answer is 90g

8 0

3 years ago

What is the empirical formula of a compound that is 24.42 % calcium, 17.07 % nitrogen, and 58.5% oxygen?

REY [17]

Answer:

$CaN_{2} O_{6}$

Explanation:

When calculating an empirical formula from percentages, assume you have a 100g sample. This allows you to convert the percentages directly to grams, because X % of 100g is X grams.

So:

24.42 % = 24.42 g Ca, 17.07% = 17.07g N, 58.5% = 58.5g O

The next step is to divide each mass by their molar mass to convert your grams to moles.

24.42/40.08 = 0.6092 mol

17.07/14.01 = 1.218 mol

58.85/15.99 = 3.680 mol

Then you will divide all of your mol values by the SMALLEST number of moles. This gives you whole numbers that are the mole ratio (subcripts) of the empircal formula.

0.6092 mol/0.6092 mol = 1

1.218 mol/0.6092 mol = 2

3.680 mol/0.6092 mol = 6

So the empirical formula is $CaN_{2} O_{6}$

5 0

2 years ago

How many significant figures are in 136,000 ml?

WINSTONCH [101]

Answer:

There are 3 sig figs

Explanation:

Sig figs are the amount of figures counting from the left side of the number until you reach the first zero. Starting from the left, I counted one, three, and six. That's 3 numbers, and therefore, that's how we got our answer. The only exception is when there is a decimal. Then all the numbers before the decimal, despite whether or not they're zeros, are sig figs.

8 0

3 years ago

Read 2 more answers

Please help I’ll mark brainliest!

Anni [7]

Answer:

D because excited electrons fall back.....

6 0

3 years ago

Read 2 more answers

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