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myrzilka [38]
3 years ago
15

Gas cylinders, each containing 30 kg of a mixture of 75 mol% acetylene (c2h2) and balance n2 are available. how many cylinders a

re needed to produce 1.5 m3 of the gas mixture at 200 atm (abs) and 150°c?
Chemistry
1 answer:
marta [7]3 years ago
3 0

Answer:- 8 cylinders are needed to produce the required amount of the gas.

Solution: First we calculate the moles of the gas for it's volume 1.5m^3 at 200 atm and 150 degree C using ideal gas law equation:

PV = nRT

n=\frac{PV}{RT}

V = 1.5m^3 = 1500 L

P = 200 atm

T = 150 + 273 = 423 K

R = 0.0821\frac{atm.L}{mol.K}

Let's plug in the values in the equation:

n=\frac{200*1500}{0.0821*423}

n = 8638.5 mol

Acetylene and nitrogen are 75 mol% and 25 mol% respectively in the cylinder.

So, moles of acetylene = \frac{8638.5*75}{100} = 6478.9 moles

moles of nitrogen = 8638.5 - 6478.9 = 2159.6 moles

mass of acetylene = 6478.9*26 = 168451.4 g

mass of nitrogen = 2159.6*28 = 60468.8 g

Total mass of mixture = 168451.4 g + 60468.8 g = 228920.2 g = 228.92 kg

number of cylinders = \frac{228.92}{30} = 7.63 = 8

So, 8 cylinders are needed.

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Answer:

4.17L

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The new volume of the gas is 4.17L

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2 years ago
How is the orbital configuration of neutral atoms related to the atoms chemical properties
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Answer:

Explanation:

Chemical properties of atoms relies solely on the number of electrons they contain, more particularly, the valence or outermost electrons in the orbit round the nucleus of an atom.

A neutral atom is one that has not gained or lost electron or even shared electrons with any other atom.

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7 0
3 years ago
Read 2 more answers
Calculate the ph of a 0.60 m h2so3, solution that has the stepwise dissociation constants ka1 = 1.5 × 10-2 and 1.82 1.06 1.02 2.
Vsevolod [243]
Missing in your question Ka2 =6.3x10^-8
From this reaction:
 H2SO3 + H2O ↔ H3O+  + HSO3-
by using the ICE table :
                H2SO3     ↔    H3O     +    HSO3- 
intial         0.6                     0                  0
change     -X                      +X                +X
Equ         (0.6-X)                  X                   X

when Ka1 = [H3O+][HSO3-]/[H2SO3]
So by substitution:
1.5X10^-2 = (X*X) / (0.6-X) by solving this equation for X
∴ X = 0.088
∴[H2SO3] = 0.6 - 0.088 = 0.512
[HSO3-] = [H3O+] = 0.088

by using the ICE table 2:
                 HSO3-     ↔   H3O     +     SO3-
initial        0.088              0.088              0
change    -X                      +X                   +X
Equ         (0.088-X)          (0.088+X)          X

Ka2= [H3O+] [SO3-] / [HSO3-]
we can assume [HSO3-] =  0.088 as the value of Ka2 is very small
 6.3x10^-8 = (0.088+X)*X / 0.088
X^2 +0.088 X - 5.5x10^-9= 0 by solving this equation for X
∴X= 6.3x10^-8
∴[H3O+] = 0.088 + 6.3x10^-8
               = 0.088 m ( because X is so small)
∴PH= -㏒[H3O+]
       = -㏒ 0.088 = 1.06 
5 0
3 years ago
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Answer: 770 g water are needed to dissolve 27.8 g of ammonium nitrate NH_4NO_3 in order to prepare a 0.452 m solution

Explanation:

Molality : It is defined as the number of moles of solute present per kg of solvent

Formula used :

Molality=\frac{n\times 1000}{W_s}

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n= moles of solute

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W_s = weight of the solvent in g = ?

0.452=\frac{0.348\times 1000}{W_s}

W_s=770g

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