All categories

3 years ago

Among the gases listed, which would have the fastest rate of effusion?

Chemistry

2 answers:

Minchanka [31]3 years ago

7 0

CH4
I got this question right on the test

Delvig [45]3 years ago

3 0

"CH4" is the one gas among the choices given in the question that would have the fastest rate of effusion. The correct option among all the options that are given in the question is the second option or option "B". I hope that this is the answer that has actually come to your great help.

You might be interested in

Convert 5.00 moles of carbon to grams. Be sure to report your answer with the correct significant figures.

sergij07 [2.7K]

1 moles Carbon to grams = 12.0107 grams

2 moles Carbon to grams = 24.0214 grams

3 moles Carbon to grams = 36.0321 grams

4 moles Carbon to grams = 48.0428 grams

5 moles Carbon to grams = 60.0535 grams

6 moles Carbon to grams = 72.0642 grams

7 moles Carbon to grams = 84.0749 grams

8 moles Carbon to grams = 96.0856 grams

9 moles Carbon to grams = 108.0963 grams

10 moles Carbon to grams = 120.107 grams

8 0

2 years ago

The vapor pressure of water at 65oC is 187.54 mmHg. What is the vapor pressure of a ethylene glycol (CH2(OH)CH2(OH)) solution ma

Pavlova-9 [17]

Answer:

173.83 mmHg is the vapor pressure of a ethylene glycol solution.

Explanation:

Vapor pressure of water at 65 °C= $p_o= 187.54 mmHg$

Vapor pressure of the solution at 65 °C= $p_s$

The relative lowering of vapor pressure of solution in which non volatile solute is dissolved is equal to mole fraction of solute in the solution.

Mass of ethylene glycol = 22.37 g

Mass of water in a solution = 82.21 g

Moles of water= $n_1=\frac{82.21 g}{18 g/mol}=4.5672 mol$

Moles of ethylene glycol= $n_2=\frac{22.37 g}{62.07 g/mol}=0.3603 mol$

$\frac{p_o-p_s}{p_o}=\frac{n_2}{n_1+n_2}$

$\frac{187.54 mmHg-p_s}{187.54 mmHg}=\frac{0.3603 mol}{0.3603 mol+4.5672 mol}$

$p_s=173.83 mmHg$

173.83 mmHg is the vapor pressure of a ethylene glycol solution.

6 0

3 years ago

Determine the theoretical yield of HCl if 60.0 g of BC13 and 37.5 g of H20 are reacted according to the following balanced react

Pie

Answer : The theoretical yield of HCl is, 56.1735 grams

Explanation : Given,

Mass of $BCl_3$ = 60 g

Mass of $H_2O$ = 37.5 g

Molar mass of $BCl_3$ = 117 g/mole

Molar mass of $H_2O$ = 18 g/mole

Molar mass of $HCl$ = 36.5 g/mole

First we have to calculate the moles of $BCl_3$ and $H_2O$ .

$\text{Moles of }BCl_3=\frac{\text{Mass of }BCl_3}{\text{Molar mass of }BCl_3}=\frac{60g}{117g/mole}=0.513moles$

$\text{Moles of }H_2O=\frac{\text{Mass of }H_2O}{\text{Molar mass of }H_2O}=\frac{37.5g}{18g/mole}=2.083moles$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$BCl_3(g)+3H_2O(l)\rightarrow H_3BO_3(s)+3HCl(g)$

From the balanced reaction we conclude that

As, 1 mole of $BCl_3$ react with 3 mole of $H_2O$

So, 0.513 moles of $BCl_3$ react with $3\times 0.513=1.539$ moles of $H_2O$

From this we conclude that, $H_2O$ is an excess reagent because the given moles are greater than the required moles and $BCl_3$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $HCl$ .

As, 1 mole of $BCl_3$ react to give 3 moles of $HCl$

So, 0.513 moles of $BCl_3$ react to give $3\times 0.513=1.539$ moles of $HCl$

Now we have to calculate the mass of $HCl$ .

$\text{Mass of }HCl=\text{Moles of }HCl\times \text{Molar mass of }HCl$

$\text{Mass of }HCl=(1.539mole)\times (36.5g/mole)=56.1735g$

Therefore, the theoretical yield of HCl is, 56.1735 grams

7 0

3 years ago

What would happen if an alkali metal was combined with a halogen?explain

RideAnS [48]

All alkail metals react with halogens vigorously to produce salt. The alkali metal would loose an electron to form and ion with the halogen

7 0

3 years ago

The isomerization of cyclopropane to propylene is a first-order process with a half-life of 19 min at 500°c. the time it takes f

amid [387]

As mentioned, the isomerization of cyclopropane to propylene is a first-order process with a half-life of 19 min at 500°c. A first-order reaction kinetic rates means that the rate is constant throughout the reaction.

Thus, the time it takes for the partial pressure of cyclopropane to decrease from 1 atm to 0.125 atm at 500°c is 57 minutes.

3 0

3 years ago