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grin007 [14]
3 years ago
11

A sample of oxygen occupies 47.2 liters under a pressure of 1240 torr at 298K. What volume would it occupy at 303K if the pressu

re were decreased to 730 torr
Chemistry
1 answer:
kupik [55]3 years ago
3 0

Answer:

81.5 L

Explanation:

We can use the combined gas law equation that gives the relationship among pressure, temperature and volume of gases for a fixed amount of gas.

P1V1 / T1 = P2V2 / T2

where P1 - pressure, V1 - volume and T1 - temperature at the first instance

P2 - pressure, V2 - volume and T2 - temperature at the second instance

substituting the values in the equation

1240 Torr x 47.2 L / 298 K = 730 Torr x V2 / 303 K

V2 = 81.5 L

the new volume the gas would occupy when the conditions have changed is 81.5 L

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Helpful Knowledge: (We Know the amount in an object: 12g or C^12)

{A number of objects that are in a mole of objects?}

Well for the question it is pretty easy to answer because a number of objects in One mole would equal 6.02 × 10²³ 

Which 6.02 × 10²³ is an Avogadro's Number. 

So it depends on how many objects you have.

So for every object you have, One mole would equal 6.02 × 10²³. Or 62,000,000,000,000,0000,000,000. Big Number am I right. So that's why we just use 6.02 × 10²³.

Anywho, your answer would be 6.02 x 10²³ x n. 
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100 ml of a 0.300 m solution of agno3 reacts with 100 ml of a 0.300 m solution of hcl in a coffee-cup calorimeter and the temper
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Answer:

100 ml of a 0.300 m solution of agno3 reacts with 100 ml of a 0.300 m solution of hcl in a coffee-cup calorimeter and the temperature rises from 21.80 °c to 23.20 °c. Assuming the density and specific heat of the resulting solution is 1.00 g/ml and 4.18 j/g ∙ °c respectfully, what is the ΔH°rxn?

39.013 kJ/mol.

Explanation:

AgNO3(aq) + HCl(aq) --------------> AgCl(s) + HNO3(aq)

We can calculate the amount of heat (Q) released from the solution using the relation:

Q = m.c.ΔT,

Where, Q is the amount of heat released from the solution (Q = ??? J).

m is the mass of the solution (m of the solution = density of the solution x volume of the solution = (1.0 g/mL)(200 mL) = 200 g.

c is the specific heat capacity of the solution (c = 4.18 J/g∙°C).

ΔT is the difference in the T (ΔT = final temperature - initial temperature = 23.20 °C - 21.80 °C = 1.4 °C).

∴ Q = m.c.ΔT = (200 g)(4.18 J/g∙°C)(1.4 °C) = 1170.4 J.

∵ ΔH°rxn = Qrxn/(no. of moles of AgNO₃).

Molarity (M) is defined as the no. of moles of solute dissolved in a 1.0 L of the solution.

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               = (M)(Volume of the solution (L))

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           = Qrxn/(no. of moles of AgNO₃)

            = (1170.4 J)/(0.03 mol)

            = 39013.33 J/mol

           = 39.013 kJ/mol.

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