Need help on all the questions and answers, thanks! QUESTION 1 Consider a manome

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Need help on all the questions and answers, thanks!

QUESTION 1

Consider a manometer (a barometer-like device used for measuring pressure) constructed using ethyl alcohol (? =0.789 g/mL). What would be the column height if the pressure is 23.8 mm Hg? The density of Hg is 13.56 g/mL)

h = ____ cm

QUESTION 2

Consider a sample of gas that occupies 23.8 L at 1.87 atm. What volume will the same sample occupy if the pressure is increased to 3.23 atm while the temperature is held constant?

V = ___ L

QUESTION 3

A fixed sample of gas occupies 25.5 mL at 210.7 K. What volume will the gas occupy at 294.2 K?

V = ___ mL

QUESTION 4

A fixed sample of gas exerts a pressure of 624 Torr at 354 K. What will the pressure be (in atm) if the temperature is reduced to 191 K (assuming the volume is held constant)?

p = ___ atm

QUESTION 5

A fixed sample of gas occupies 12.2 L at 50.5 oC and 0.741 atm. If the gas expands to 39.4 L and the temperature drops to 18.3 oC, what will the new pressure be?

p = ____ atm

QUESTION 6

How many moles of gas are contained in a sample of gas that occupies 24.4 L at 1.09 atm and 335.2 K?

QUESTION 7

H2S (g) can be generated by the following reaction:

2 HCl(aq) + FeS(s) ? FeCl2(aq) + H2S(g)

What volume (at STP) of H2S gas can be generated from 27.6 mL of 0.291 M HCl and an excess of FeS?

V = ___ mL

Explanation / Answer

1)

Pressure = height x density

height = Pressure (N/m^2)/density(kg/m^3)

Given,

Pressure = 23.8 mmHg x 133.32 = 3.2 x 10^3 N/m^2

density of ethyl alcohol = 0.786 g/ml = 0.786 x 10^3 kg/m^3

So,

height of column with ethyl alcohol = 3.2 x 10^3/0.786 x 10^3 = 4.1 m = 407.12 cm

---

2) with,

P1 = 1.87 atm

V1 = 23.8 L

P2 = 3.23 atm

V2 = ?

Using,

P1V1 = P2V2

or,

Volume occupied by gas at 3.23 atm (V2) = 1.87 x 23.8/3.23 = 13.8 L

---

3) Given,

V1 = 25.5 ml

T1 = 210.7 K

V2 = ?

T2 = 294.7 K

Using,

V1/T1 = V2/T2

So,

volume occupied by gas at 294.7 K (V2) = 25.5 x 294.7/210.7 = 35.7 ml

---

4) Given,

P1 = 624 torr = 0.821 atm

T1 = 354 K

P2 = ?

T2 = 191 K

Using,

P1/T1 = P2/T2

So,

Pressure occupied by gas at 191 K (P2) = 0.821 x 191/354 = 0.443 atm

---

5) moles of gas (n) = PV/RT

with,

P = 0.741 L

V = 12.2 L

R = gas constant

T = 50.5 + 273 = 323.5 K

So,

moles of gas (n) = 0.741 x 12.2/0.08205 x 323.5 = 0.3406 moles

When,

V = 39.4 L

T = 18.3 + 273 = 291.3 K

new pressure (P) = nRT/V

                            = 0.3406 x 0.08205 x 291.3/39.4 = 0.207 atm

---

6) moles of gas (n) = PV/RT

with given values,

moles of gas (n) = 1.09 x 24.4/0.08205 x 335.2 = 0.97 moles

---

7) moles of HCl used = 0.291 M x 0.0276 L = 8.032 x 10^-3 moles

moles of H2S formed = 8.032 x 10^-3/2 = 4.016 x 10^-3 moles

volume of H2S (at STP) generated = 4.016 x 10^-3 moles x 22.4 L = 0.09 L

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