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2 years ago

If you have two pistons in a system, which exerts the larger force?

Chemistry

1 answer:

Alik [6]2 years ago

8 0

Answer:

The larger piston

Explanation:

(i just took the same test)

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Please help, I'm struggling with stoichiometry

rusak2 [61]

Answer:

a) HNO3

b) 26.8g (3 s.f.)

c) 1.29g (3 s.f.)

Please see the attached pictures for full solution.

To balance an equation, ensure that the number of atoms for each element is the same on both sides.

3 0

3 years ago

In the reaction n2 + 3h2 ---> 2nh3, how many grams of nh3 are produced if 25.0 g n2 reacts with excess h2? question 10 option

Lisa [10]

Molar mass of N2 = 28

Moles of N2 = 25 / 28 = 0.89

So, moles of NH3 produce = 2 x 0.89 = 1.78

Note: H2 is in excess. so no need to care about it.

8 0

3 years ago

The unbalanced equation below shows the combustion of methane.

Ratling [72]

B. 1, 2, 1, 2. You balance the equation by making sure there are equal elements on both sides.

3 0

2 years ago

Points)<br> Determine the molar mass of NaCl (the solute) in a 0.1M aqueous solution of NaCl

nalin [4]

Answer:

58.443 g/mol

Explanation:

The molar mass of NaCl is the sum of the molar masses of the individual atoms:

Na: 22.989770 g/mol

Cl: 35.453 g/mol

The total molar mass is ...

NaCl: 58.443 g/mol

The molar mass does not depend on whether the material is in solution or in any other form.

4 0

3 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