Ask question

Ask question

All categories

3 years ago

5

If it takes 3 hours to drive a distance of 192km on a freeway, what would be your average speed in km/h? You can use a calculato

r.

1 answer:

Bingel [31]3 years ago

4 0

Answer:

The average speed in km/h would be 64km per hour.

Speed = Distance /Time

192/3 = 64

Explanation:

I hope this helps

You might be interested in

How many molecules of O2 are in 153.9 g O2?

MrRissso [65]

Answer:

2.895*10^24

Explanation:

mass of Oxygen give = 153.9g

molar mass of O2 molecule = 16*2=32g/mol

n= mole

To find the mole

n= mass/ molar mass

n= 153.9/32

n=4.81mol.

To find the number of molecules of o

Nm= number of molecule

Nn = Number of mole

NA = number of Avogados

Nm= Nn * NA

Nm= 4.81 *6.02*10^23

Nm= 2.895*10^24

3 0

3 years ago

Is four grams of H2O more than, less than, or equal to one mole?<br>

geniusboy [140]

It is equal because they both weight the same

4 0

4 years ago

What is the gram of sulfur are in 8.5 moles of S?

love history [14]

Answer:

about 0.255

Explanation:

5 0

4 years ago

O2 + 2H2 --> 2H2O in this equation, the O2 is a

mart [117]

Answer:

this is too easy just do it yourself

7 0

3 years ago

Determine the mole fractions and partial pressures of CO2, CH4, and He in a sample of gas that contains 1.20 moles of CO2, 1.79

BARSIC [14]

Answer : The mole fraction and partial pressure of $CH_4,CO_2$ and $He$ gases are, 0.267, 0.179, 0.554 and 1.54, 1.03 and 3.20 atm respectively.

Explanation : Given,

Moles of $CH_4$ = 1.79 mole

Moles of $CO_2$ = 1.20 mole

Moles of $He$ = 3.71 mole

Now we have to calculate the mole fraction of $CH_4,CO_2$ and $He$ gases.

$\text{Mole fraction of }CH_4=\frac{\text{Moles of }CH_4}{\text{Moles of }CH_4+\text{Moles of }CO_2+\text{Moles of }He}$

$\text{Mole fraction of }CH_4=\frac{1.79}{1.79+1.20+3.71}=0.267$

and,

$\text{Mole fraction of }CO_2=\frac{\text{Moles of }CO_2}{\text{Moles of }CH_4+\text{Moles of }CO_2+\text{Moles of }He}$

$\text{Mole fraction of }CO_2=\frac{1.20}{1.79+1.20+3.71}=0.179$

and,

$\text{Mole fraction of }He=\frac{\text{Moles of }He}{\text{Moles of }CH_4+\text{Moles of }CO_2+\text{Moles of }He}$

$\text{Mole fraction of }He=\frac{3.71}{1.79+1.20+3.71}=0.554$

Thus, the mole fraction of $CH_4,CO_2$ and $He$ gases are, 0.267, 0.179 and 0.554 respectively.

Now we have to calculate the partial pressure of $CH_4,CO_2$ and $He$ gases.

According to the Raoult's law,

$p_i=X_i\times p_T$

where,

$p_i$ = partial pressure of gas

$p_T$ = total pressure of gas = 5.78 atm

$X_i$ = mole fraction of gas

$p_{CH_4}=X_{CH_4}\times p_T$

$p_{CH_4}=0.267\times 5.78atm=1.54atm$

and,

$p_{CO_2}=X_{CO_2}\times p_T$

$p_{CO_2}=0.179\times 5.78atm=1.03atm$

and,

$p_{He}=X_{He}\times p_T$

$p_{He}=0.554\times 5.78atm=3.20atm$

Thus, the partial pressure of $CH_4,CO_2$ and $He$ gases are, 1.54, 1.03 and 3.20 atm respectively.

4 0

3 years ago