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Inga [223]
4 years ago
8

If the amount of dissolved solute in a solution at a given temperature is greater than the amount that can permanently remain in

solution at that temperature, the solution is said to be
Chemistry
1 answer:
Ksivusya [100]4 years ago
7 0
I think the answer is <span>supersaturated</span>
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1.The movement of particles from a less crowded area to a more crowded area requires
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3 years ago
Calculate the temperature of a 0.50 mol sample of a gas at 0.987 atm and a volume of 12 L.
Aleonysh [2.5K]

Answer: 15.5^0C

Explanation:

According to ideal gas equation:

PV=nRT

P = pressure of gas = 0.987 atm

V = Volume of gas = 12 L

n = number of moles = 0.50

R = gas constant =0.0821Latm/Kmol

T =temperature =  ?

T=\frac{PV}{nR}

T=\frac{0.987atm\times 12L}{0.0820 L atm/K mol\times 0.50mol}=288.5K=(288.5-273)^0C=15.5^0C

Thus the temperature of a 0.50 mol sample of a gas at 0.987 atm and a volume of 12 L is 15.5^0C

8 0
3 years ago
In the Haber reaction, patented by German chemist Fritz Haber in 1908, dinitrogen gas combines with dihydrogen gas to produce ga
Lady bird [3.3K]

Answer:

0.41kg/sec

Explanation:

PV= nRT

Given : V= 505 L

P=0.88 atm

R= 0.08206 Latm/K*mol

T= 172 .0C = 172+273 = 445 K

n = PV /RT = 0.88 * 505 / 0.08206 * 445 = 12.17 moles per sec of N2 are consumed

As per reaction : N2 + 3H2 ----> 2NH3

1 mole N2 is consumed to produce 2 moles NH3

moles of NH3 produced per sec :

(2 moles NH3/1mol N2) * 12.17 moles N2 = 24.34 moles NH3 per sec

grams of NH3 produced per sec =

24.34 moles NH3 per sec * molar mass NH3 = 24.34 moles NH3 per sec * 17.031 g/mol = 414.5 g NH3 per sec

rate in Kg/sec = 414.5 g NH3 per sec * (1kg /1000g) = 0.4145 Kg/sec

= 0.41kg/sec

5 0
3 years ago
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