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dsp73
3 years ago
10

A gaseous compound has a mass of 8.00 g at 2.81 atm and 300. K in a 1.00 L container.

Chemistry
2 answers:
Nimfa-mama [501]3 years ago
6 0

Answer:

The molar mass is the mass of a given chemical element or chemical compound (g) divided by the amount of substance (mol). The molar mass of a compound can be calculated by adding the standard atomic masses (in g/mol) of the constituent atoms.

Explanation:

To find density, we have to solve the equation for volume, or V. V = nRT / P. To incorporate mass, we can use the number of moles, or n. The number of moles equals the mass of the gas10.19 The typical atmospheric pressure on top of Mt. Everest (29028 ft) is about 265

torr. Convert this pressure to

(a) atm 265 torr ( 1 atm / 760 torr) = 0.349 atm

(b) mm Hg 265 torr ( 760 mm Hg / 760 torr) = 265 mm Hg

10.39 A scuba divers tank contains 0.29 kg of O2 compressed into a volume of 2.3 L.

(a) Calculate the gas pressure inside the tank at 9 °C.

PV = nRT so P = nRT / V

convert 0.29 kg to 290 g O2

convert 9 °C to K [9 °C + 273.15 = 282.15 K ] (do not pre-round)

290 g O2 ( 1 mol O2 / 32.00 g ) (0.0821 L atm / mol K) (282.15 K)

P = ------------------------------------------------------------------------------------- = 91 atm

2.3 L ( 2 SF)

10.41 Chlorine is widely used to purify municipal water supplies and to treat swimming

pool waters. Suppose that the volume of particular sample of Cl2 gas is 8.70 L at 895 torr divided by the molecular mass.

Maurinko [17]3 years ago
4 0

Answer:

The question asks for moles, which can be obtained from P-V-T data using the ideal gas

equation: n =

RT

PV .

Now use the rearranged gas law to determine the number of moles in the sample:

!

n = PV

RT = (6.47 x 105 Pa)(5.65 x 10-4m3

)

(8.314 J

mol K )(21.7 + 273.15 K) = 0.149 mol.

All conditions except the pressure and volume are fixed, so P1V1 = P2V2 can be used: 3.62 L

101kPa

(647 kPa)(0.565 L)

2

1 1

2 = =

P

PV V

Explanation:

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Explanation:

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It involve the elements of group 5th A.

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2 years ago
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2 years ago
Calculate the ph of a solution formed by mixing 200.0 ml of 0.30 m hclo with 300.0 ml of 0.20 m kclo. the ka for hclo is 2.9 × 1
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C(HClO) = 0,3 M.
<span>V(HClO) = 200 mL = 0,2 L.
n(HClO) = </span>c(HClO) · V(HClO).
n(HClO) = 0,06 mol.<span>
c(KClO</span>) = 0,2 M.
<span>V(KClO) = 0,3 L.
n(KClO) = 0,06 mol.
V(buffer solution) = 0,2 L + 0,3 L = 0,5 L.
ck</span>(HClO) = 0,06 mol ÷ 0,5 L = 0,12 M.
cs(KClO) = 0,06 mol ÷ 0,5 L = 0,12 M.<span>
Ka(HClO</span>) = 2,9·10⁻⁸.<span>
This is buffer solution, so use Henderson–Hasselbalch equation: 
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