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Molodets [167]
3 years ago
14

Which of the following would not cause an increase in the pressure of a gaseous system in a container? A The container is made l

arger. Additional gas is added to the container. B A C The temperature in the container is increased. The volume of the container is decreased.​
Chemistry
2 answers:
ludmilkaskok [199]3 years ago
5 0

Answer:

The container is made larger

Explanation:

Delicious77 [7]3 years ago
5 0

Answer: The volume of the container is increased

Explanation:

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HELP ME PLS I NEED HELP
sesenic [268]
Lol your answer is c
3 0
3 years ago
If you are using 3.00% (mass/mass) hydrogen peroxide solution and you determine that the mass of solution required to reach the
Maurinko [17]

Answer:

0.004522 moles of hydrogen peroxide molecules are present.

Explanation:

Mass by mass percentage of hydrogen peroxide solution = w/w% = 3%

Mass of the solution , m= 5.125 g

Mass of the hydrogen peroxide = x

w/w\% = \frac{x}{m}\times 100

3\%=\frac{x}{5.125 g}\times 100

x=\frac{3\times 5.125 g}{100}=0.15375 g

Mass of hydregn pervade in the solution = 0.15375 g

Moles of hydregn pervade in the solution :

=\fraC{ 0.15375 g}{34 g/mol}=0.004522 mol

0.004522 moles of hydrogen peroxide molecules are present.

5 0
3 years ago
Determine the mass of 2.75 moles of CaSO4. Record your work and your answer.
maw [93]

Explanation:

RFM \:  = 160 \: g \\ 1 \: mole \: weighs \: 160 \: g \\ 2.75 \: moles \: weighs \:  \frac{2.75 \times 160}{1} g \\  = 440 \: g

6 0
3 years ago
How many grams of NH3 can be produced from 12.0g of H2?
RSB [31]

Answer:

Balanced reaction:

3 H2 (g)  + N2 (g)  → 2 NH3 (g)

Use stoichiometry to convert g of H2 to g of NH3.  The process would be:

g H2 → mol H2 → mol NH3 → g NH3

12.0 g H2 x (1 mol H2 / 2.02 g H2) x (2 mol NH3 / 3 mol H2) x (17.03 g NH3 / 1 mol NH3) = 67.4 g NH3

Explanation: See above

Hope this helps, friend.

8 0
2 years ago
Assuming that all the energy given off in the reaction goes to heating up only the air in the house, determine the mass of metha
nirvana33 [79]

Answer:

0.92 kg

Explanation:

The volume occupied by the air is:

35.0m\times 35.0m \times 3.2m \times \frac{10^{3}L }{1m^{3} } =3.9 \times 10^{6} L

The moles of air are:

3.9 \times 10^{6} L \times \frac{1.00mol}{22.4L} =1.7 \times 10^{5}mol

The heat required to heat the air by 10.0 °C (or 10.0 K) is:

1.7 \times 10^{5}mol \times \frac{30J}{K.mol} \times 10.0 K = 5.1 \times 10^{7}J

Methane's heat of combustion is 55.5 MJ/kg. The mass of methane required to heat the air is:

5.1 \times 10^{7}J \times \frac{1kgCH_{4}}{55.5 \times 10^{6} J } =0.92kgCH_{4}

3 0
3 years ago
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