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3 years ago

How many moles are in 12 liters of Cl2?

Chemistry

1 answer:

RoseWind [281]3 years ago

8 0

Answer:

$\boxed {\boxed {\sf 0.54 \ mol \ Cl_2}}$

Explanation:

A mole is any quantity of a substance that contains 6.02 × 10²³ particles. At standard temperature and pressure, or STP, 1 mole of as is equal to 22.4 liters. This is true for any gas, regardless of the specific kind.

Although it is not specified, we can assume this gas is at STP. Let's set up a ratio using this information: 22.4 L/mol

$\frac {22.4 \ L \ Cl_2}{1 \ mol \ Cl_2}$

Multiply by the given number of liters: 12

$12 \ L \ Cl_2 *\frac {22.4 \ L \ Cl_2}{1 \ mol \ Cl_2}$

Flip the ratio so the liters of chlorine cancel.

$12 \ L \ Cl_2 * \frac {1 \ mol \ Cl_2}{22.4 \ L \ Cl_2}$

$12 * \frac {1 \ mol \ Cl_2}{22.4 }$

$\frac {12}{22.4 } \ mol \ Cl_2$

$0.53571428571 \ mol \ Cl_2$

The original measurement of liters has 2 significant figures, so our answer must have the same.

For the number we found, that is the hundredth place.

0.53571428571

The 5 in the thousandth place tells us to round the 3 up to a 4.

$0.54 \ mol \ Cl_2$

12 liters of chlorine gas at STP is approximately 0.54 moles of chlorine gas.

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A mixture of gases with a pressure of 800.0 mm hg contains 60% nitrogen and 40% oxygen by volume. what is the partial pressure o

klasskru [66]

Hello!

We have the following statement data:

Data:
$P_{Total} = 800 mmHg$
$P\% N_{2} = 60\%$
$P\% O_{2} = 40\%$
$P_{partial} = ? (mmHg)$

As the percentage is the mole fraction multiplied by 100:

 $P = X_{ O_{2} }*100$

The mole fraction will be the percentage divided by 100, thus:
What is the partial pressure of oxygen in this mixture?

$X_{ O_{2} } = \frac{P}{100}$
$X_{ O_{2}} = \frac{40}{100}$
$\boxed{X_{ O_{2}} = 0.4}$

To calculate the partial pressure of the oxygen gas, it is enough to use the formula that involves the pressures (total and partial) and the fraction in quantity of matter:

In relation to $O_{2}$ :

$\frac{P O_{2} }{P_{total}} = X_O_{2}$
$\frac{P O_{2} }{800} = 0.4$
$P_O_{2} = 0.4*800$
$\boxed{\boxed{P_O_{2} = 320\:mmHg}}\end{array}}\qquad\quad\checkmark$

Answer:
b. 320.0 mm hg

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3 years ago

How would you describe the motion of plates in a divergent boundary?

Nutka1998 [239]

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14. An enzyme-catalyzed reaction was carried out with the substrate concentration initially a thousand times greater than the Km

emmainna [20.7K]

Answer:

27 min

Explanation:

The kinetics of an enzyme-catalyzed reaction can be determined by the equation of Michaelis-Menten:

$v = \frac{vmax[S]}{Km + [S]}$

Where v is the velocity in the equilibrium, vmax is the maximum velocity of the reaction (which is directed proportionally of the amount of the enzyme), Km is the equilibrium constant and [S] is the concentration of the substrate.

So, initially, the velocity of the formation of the substrate is 12μmol/9min = 1.33 μmol/min

If Km is a thousand times smaller then [S], then

v = vmax[S]/[S]

v = vmax

vmax = 1.33 μmol/min

For the new experiment, with one-third of the enzyme, the maximum velocity must be one third too, so:

vmax = 1.33/3 = 0.443 μmol/min

Km will still be much smaller then [S], so

v = vmax

v = 0.443 μmol/min

For 12 μmol formed:

0.443 = 12/t

t = 12/0.443

t = 27 min

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