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2 years ago

3.0 L of oxygen gas is heated from 100k to 250k. What is the new volume after the temperature has been increased???

Chemistry

1 answer:

alexandr402 [8]2 years ago

7 0

Answer:

7.5 L

Explanation:

Using Charles' law, which is V1/T1=V2/T2, we can plug in these numbers to find the answer. The law states that volume is directly proportional to temperature. 3.0L/100K = x L / 250 K. Solve for x to get 7.5 L. Hope this helps.

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Which of the following samples contains the greatest number of atoms?

mart [117]

Answer:

C. 11 moles of N2O

Explanation:

A. CO2 exists as a molecular compound. The number of atoms present = 1 atom of Carbon and 2 atoms oxygen = 3 atoms * 9 = 18 atoms

B. Xe exists as an atom. Number of atoms present = 10 * 1 atom = 10 atoms

C. N20 contains 3 atoms; 2 atoms of Nitrogen and 1 atom of oxygen.

Number of atoms present = 3 * 11 = 33 atoms

D. CO contains 2 atoms, 1 each of carbon and oxygen.

Number of atoms present = 12 * 2 = 24 atoms

7 0

2 years ago

Just wanted to make sure I have the right idea.

Rasek [7]

Explanation:

yes you got the right idea

4 0

2 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

2 years ago

Science has shown that air pollution can cause a variety of health issues in our society. Which of the following can be a health

ololo11 [35]

The correct one is Respiratory Distress !!

As rest are not concerned with air pollution !!

8 0

2 years ago

Read 2 more answers

How many grams of n2f4 can be produced from 225g f2?

3241004551 [841]

The answer is 615.91 grams of <span>n2f4

Solution:
225g F2 x [(1molF2)/(38gramsF2)] x [</span>(1molF2)/(1molN2F4)] x [(104.02 grams N2F4)/(1molN2F4)]
=615.91 grams

8 0

3 years ago