Can u help me pls.

A flask contains a gas mixture of hydrogen, nitrogen and methane. The total pressure of the mix is 3.0 atm. The partial pressure

ollegr [7]

Answer:

$P_{N_2}=1.0atm$

Explanation:

Hello,

Considering the Dalton's law which states that the total pressure of a gaseous system is defined by the summation of the the partial pressures of the present gases:

$P_T=\Sigma P_i$

For the given system:

$P_T=P_{H_2}+P_{N_2}+P_{CH_4}$

Solving for the partial pressure of nitrogen we obtain:

$P_{N_2}=3.0atm-0.5atm-1.5atm=1.0atm$

Best regards.

3 0

3 years ago

Good morning to all Indians

zimovet [89]

<h3>Good Morning have a nice day </h3>

<h2>#itz mishika here#</h2>

3 0

2 years ago

Read 2 more answers

How many atoms of hydrogen are in 100 g of hydrogen peroxide (h2o2)?

Maurinko [17]

Hello friends..

find the MW of HP
calculate the # of mols per 100g of HP
take the # of mols times 6.023 x 10^23 times 2 (2 H per molecule)

Hope it helps you..

7 0

3 years ago

Select the correct answer. An adult human lung has a volume of 2. 9 liters. If its volume decreases to 1. 2 liters and it contai

maria [59]

Moles are the ratio of the mass and the molar mass of the substance given and is estimated in mol. The moles that the lungs contain when it is full is 0.12 moles.

<h3>What is the relation of the volume and moles?</h3>

Moles of the substances are directly proportional to the volume of the substance and can be shown as,

$\rm \dfrac{V_{1}}{V_{2}}= \dfrac{n_{1}}{n_{2}}$

Here,

Initial volume $\rm (V_{1})$ = 2.9 L

Final volume $\rm (V_{2})$ = 1.2 L

Initial moles = $\rm n_{1}$

Final moles $\rm (n_{2})$ = 0.049 mol

Substituting values in the equation:

$\begin{aligned} \dfrac{2.9}{1.2} &=\rm \dfrac{n_{1}}{0.049}\\\\&= \dfrac{2.9\times 0.049}{1.2}\\\\&= 0.12\;\rm mol\end{aligned}$

Therefore, the number of moles in the lungs are 0.12 moles.

Learn more about moles and volumes here:

brainly.com/question/13909347

6 0

2 years ago

15.5 g of an unknown metal at 165.0°C is dropped into 150.0mL of H2O at 23.0°C in a coffee cup calorimeter. The metal and H2O re

tino4ka555 [31]

Answer:

Specific heat capacity of metal is 2.09 j/g.°C.

Explanation:

Specific heat capacity:

It is the amount of heat required to raise the temperature of one gram of substance by one degree.

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

Given data:

Mass of metal = 15.5 g

Initial temperature = 165.0°C

Initial temperature of water = 23.0°C

Final temperature = 30.0°C

Specific heat capacity of metal = ?

Specific heat capacity of water = 4.184 J/g°C

Volume of water = 150.0 mL or 150.0 g

Solution:

Formula:

- Qm = +Qw

Now we will put the values in formula.

-15.5 g × c × [ 30.0°C - 165.0°C] = 150 g × 4.184 J/g°C × [ 30.0°C - 23.0°C]

15.5 g × c × 135°C = 4393.2 j

2092.5 g.°C × c = 4393.2 j

c = 4393.2 j/2092.5 g.°C

c = 2.09 j/g.°C

4 0

3 years ago

Can u help me pls. ​

Can u help me pls.