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lisov135 [29]
2 years ago
15

How many moles of CO2 could fit in a 475 mL bag a -22° C and 855 mmHg?

Chemistry
1 answer:
Sergeeva-Olga [200]2 years ago
4 0

Answer:

The amount of moles of CO₂ is 0.026 moles

Explanation:

An ideal gas is a theoretical gas that is considered to be composed of point particles that move randomly and do not interact with each other. Gases in general are ideal when they are at high temperatures and low pressures.

The equation known as the ideal gas equation explains the relationship between the four variables P (Pressure), V (Volume), T (Temperature) and n (Amount of substance). The ideal gas law is expressed mathematically as:

P*V = n*R*T

where P represents the pressure of the gas, V its volume, n the number of moles of gas (which must remain constant), R the constant of the gases and T the temperature of the gas.

In this case:

  • P= 855 mmHg
  • V= 475 mL= 0.475 L
  • n= ?
  • R= 62.36367 \frac{mmHg*L}{mol*K}
  • T= -22°C= 251 °K

Replacing:

855 mmHg*0.475 L=n*62.36367 \frac{mmHg*L}{mol*K} *251 °K

Solving:

n=\frac{855 mmHg*0.475 L}{62.36367 \frac{mmHg*L}{mol*K} *251 K}

n= 0.026 moles

<u><em>The amount of moles of CO₂ is 0.026 moles</em></u>

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You wish to extract an organic compound from an aqueous phase into an organic layer (three to six extractions on a marco scale).
AVprozaik [17]

Answer:

Explanation:

It will be better to use solvents that are lighter than water, because their density has an influence on the miscibility . This will give you a better separation during extraction.

7 0
2 years ago
A particular first-order reaction has a rate constant of 1.35 × 102 s-1 at 25.0°c. what is the magnitude of k at 75.0°c if ea =
IgorC [24]
According to this formula:
K= A*(e^(-Ea/RT) when we have K =1.35X10^2 & T= 25+273= 298K &R=0.0821
Ea= 85.6 KJ/mol So by subsitution we can get A:
1.35x10^2 = A*(e^(-85.6/0.0821*298))
1.35x10^2 = A * 0.03
A= 4333
by substitution with the new value of T(75+273) = 348K & A to get the new K
∴K= 4333*(e^(-85.6/0.0821*348)
  = 2.16 x10^2
8 0
2 years ago
Exactly 1.0 mol N2O4 is placed in an empty 1.0-L container and allowed to reach equilibrium described by the equation N2O4(g) 2N
Amanda [17]

Answer : The correct option is, (a) 0.44

Explanation :

First we have to calculate the concentration of N_2O_4.

\text{Concentration of }N_2O_4=\frac{\text{Moles of }N_2O_4}{\text{Volume of solution}}

\text{Concentration of }N_2O_4=\frac{1.0moles}{1.0L}=1.0M

Now we have to calculate the dissociated concentration of N_2O_4.

The balanced equilibrium reaction is,

                             N_2O_4(g)\rightleftharpoons 2NO_2(aq)

Initial conc.           1.0 M          0

At eqm. conc.     (1.0-x) M    (2x) M

As we are given,

The percent of dissociation of N_2O_4 = \alpha = 28.0 %

So, the dissociate concentration of N_2O_4 = C\alpha=1.0M\times \frac{28.0}{100}=0.28M

The value of x = C\alpha = 0.28 M

Now we have to calculate the concentration of N_2O_4\text{ and }NO_2 at equilibrium.

Concentration of N_2O_4 = 1.0 - x  = 1.0 - 0.28 = 0.72 M

Concentration of NO_2 = 2x = 2 × 0.28 = 0.56 M

Now we have to calculate the equilibrium constant for the reaction.

The expression of equilibrium constant for the reaction will be:

K_c=\frac{[NO_2]^2}{[N_2O_4]}

Now put all the values in this expression, we get :

K_c=\frac{(0.56)^2}{0.72}=0.44

Therefore, the equilibrium constant K_c for the reaction is, 0.44

8 0
2 years ago
analysis of a compound indicates that it contains 1.04 grams K 0.70 g Cr and 0.86 g O. Find its empirical formula
MrMuchimi
1.04gK*1molK/39.01g K= 0.0267 mol K
0.70gCr*1mol/52.0g Cr = <span>0.0135 mol Cr   
0.86 gO* 1 mol/16.0 g O = 0.0538 mol O
</span>0.0267 mol K/0.0135 = 2 mol K
0.0135 mol Cr  /0.0135= 1 mol Cr
 0.0538 mol O/0.035= 4 mol Cr
K2CrO4
6 0
2 years ago
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White raven [17]
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20% m/v = 20 mass units / 100 volume units = 1 mass units / 5 volume units

Usually units of this are in g per L. So, it is equivalent to 1 g / 5 L
4 0
2 years ago
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