All categories

2 years ago

How many atoms are in 15.1g of potassium?

Chemistry

1 answer:

Natalka [10]2 years ago

5 0

Answer:

2.32 x10^23 atoms

Explanation:

15.1/39.10 x (6.02 x10^23)

= 2.32 x10^23 atoms

First, you divide 15.1 g of potassium by its molar mass. (Molar mass of potassium is 39.10 g). Then you multiply using Avogadro's number (6.02 x10^23) to get your answer.

You might be interested in

An object has a density of 0.73 g/cm3. Will the object sink or float in water?

Galina-37 [17]

Answer:

<h2>It'll float</h2><h2>Less than 1g/cm3 it'll float in water </h2>

4 0

3 years ago

What could have caused the mass to be higher than the initial mass

ivann1987 [24]

Answer:

Possibly the addition of oxygen from the atmosphere

Explanation:

6 0

3 years ago

Read 2 more answers

I need help on my chemistry quiz

slamgirl [31]

Answer:

it could be 25degrees c... not sure

4 0

3 years ago

2.61 kilograms of water in a container have a pressure of 200 kPa and temperature of 200°C . What is the volume of this containe

Arisa [49]

<u>Answer:</u> The volume of the container is $2.8497m^3$

<u>Explanation:</u>

To calculate the volume of water, we use the equation given by ideal gas, which is:

$PV=nRT$

or,

$PV=\frac{m}{M}RT$

where,

P = pressure of container = 200 kPa

V = volume of container = ? L

m = Given mass of water = 2.61 kg = 2610 g (Conversion factor: 1kg = 1000 g)

M = Molar mass of water = 18 g/mol

R = Gas constant = $8.31\text{L kPa }mol^{-1}K^{-1}$

T = temperature of container = $200^oC=[200+273]K=473K$

Putting values in above equation, we get:

$200kPa\times V=\frac{2610g}{18g/mol}\times 8.31\text{L kPa }\times 473K\\\\V=2849.7L$

Converting this into cubic meter, we use the conversion factor:

$1m^3=1000L$

So, $\Rightarrow \frac{1m^3}{1000L}\times 2849.7L$

$\Rightarrow 2.8497m^3$

Hence, the volume of the container is $2.8497m^3$

6 0

3 years ago

A 1.25g sample of dry ice is added to a 765mL flask containing nitrogen gas at a temperature of 25.0°C and a pressure of 725 mmH

erica [24]

Answer : The total pressure in the flask is 1.86 atm.

Explanation :

First we have to calculate the pressure of $CO_2$ gas.

Using ideal gas equation :

$PV=nRT\\\\P_{CO_2}=\frac{w}{M}\frac{RT}{V}$

where,

P = Pressure of $CO_2$ gas = ?

V = Volume of $CO_2$ gas = 765 mL = 0.765 L (1 L = 1000 mL)

n = number of moles

w = mass of $CO_2$ gas = 1.25 g

M = molar mass of $CO_2$ gas = 44 g/mol

R = Gas constant = $0.0821L.atm/mol.K$

T = Temperature of $CO_2$ gas = $25.0^oC=273+25.0=298K$

Putting values in above equation, we get:

$P_{CO_2}=\frac{w}{M}\frac{RT}{V}$

$P_{CO_2}=\frac{1.25g}{44g/mol}\frac{(0.0821L.atm/mol.K)\times 298K}{0.765L}=0.909atm$

Now we have to calculate the total pressure in the flask.

$P_T=P_{N_2}+P_{CO_2}$

Given :

$P_{CO_2}=0.909atm$

$P_{N_2}=725mmHg=\frac{725}{760}=0.954atm$

conversion used : (1 atm = 760 mmHg)

Now put all the given values in the above expression, we get:

$P_T=0.954atm+0.909atm=1.86atm$

Therefore, the total pressure in the flask is 1.86 atm.

5 0

3 years ago