Which statement correctly describes the location and charge of the electrons in an atom?

For each pair of gases, select the one that most likely has the highest rate of effusion. Use the periodic table if necessary. O

madreJ [45]

According to Graham's law, the rate of effusion of a gas is inversely proportional to square root of its molecular weight.

This can be represented as follows.

Rate of effusion ∝ 1/√M

Therefore to find which gas has highest rate of effusion, we will find out the molar masses of the given compounds. The gas that is lighter in weight would have highest rate of effusion.

1) Molar mass of oxygen (O₂) is 32 g and that of H₂ is 2.01 g . Therefore H₂ would have highest rate of effusion

2) Molar mass of methane is 16.05 [12.01 + 4 (1.01)] g and that of CCl₄ [ 12 + 4(35.45) ] is 154 g. Therefore methane will have highest rate of effusion

3) Molar mass of N₂ is 28 g and molar mass of NH₃ is [ 14 + 3(3.01) ] = 17.03 g.

Therefore NH₃ will have highest rate of effusion.

4) Molar mass of F₂ is 38 g and that of Cl₂ is 71 g. Therefore F₂ will have highest rate of effusion

7 0

2 years ago

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Calculate the moles of camphor dissolved in 1.32 L of phenol. The molar mass of camphor is 152 g/mol and the molarity of phenol

LUCKY_DIMON [66]

Answer:

moles of camphor = 0.0522 moles

Explanation:

MW camphor= 152 g/mol

V solution = 1.32 L

M solution = 6.01 M

moles solute = ?

To calculate the moles of camphor, you must first know the grams of solute (camphor) that exist in the solution, this is calculated from the molarity equation:

$M=\frac{solute mass}{solution volume}$

From there the grams of the solute (camphor) are cleared:

$solutemass=M*solutionvolume=6.01M*1.32L=7.9332g$

Then by means of the molecular weight (MW) equation the moles can be obtained:

$MW=\frac{mass}{moles}$

$moles=\frac{mass}{MW} =\frac{7.9332g}{152\frac{g}{mol} } =0.0522 moles$

6 0

2 years ago

Rank the compounds below from slowest to Fastest rate of hydration. 1) formaldehyde 2) 3,3-dimethyIbutan-2-one 3) propanal 4)3-m

Makovka662 [10]

Answer:

5,3,4,1,2 this is your answer.

7 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

Which condition can cause excessive pressure on the high side of a self contained active recovery device

murzikaleks [220]

Answer:

What can cause excessive pressure on the high side of an active self-contained recovery device? A closed recovery tank inlet valve or excessive air or other non condensables in the recovery tank (either A or B) Portable refillable tanks or containers used to ship recovered refrigerants must meet what standard(s)?

Explanation:

please mark me as brainliest thank you

5 0

1 year ago