How much light would be absorbed by the different drinks?

The temperature is -14°C the air pressure in an automobile tire is 149K PA if the volume does not change what is the pressure af

Jet001 [13]

Answer: The pressure after the tire is heated to 17.3°C is 167 kPa

Explanation:

To calculate the final temperature of the system, we use the equation given by Gay-Lussac Law. This law states that pressure of the gas is directly proportional to the temperature of the gas at constant pressure.

Mathematically,

$\frac{P_1}{T_1}=\frac{P_2}{T_2}$

where,

$P_1\text{ and }T_1$ are the initial pressure and temperature of the gas.

$P_2\text{ and }T_2$ are the final pressure and temperature of the gas.

We are given:

$P_1=149kPa\\T_1=-14^0C=(273-14)=259K\\P_2=?=27.5psi\\T_2=17.3^0C=(273+17.3)=290.3K$

Putting values in above equation, we get:

$\frac{149}{259}=\frac{P_2}{290.3}\\\\P_2=167kPa$

Hence, the pressure after the tire is heated to 17.3°C is 167 kPa

4 0

4 years ago

CH4 with pressure 1 atm and volume 10 liter at 27°C is passed into a reactor with 20% excess oxygen, how many moles of oxygen is

BaLLatris [955]

Answer : The moles of $O_2$ left in the products are 0.16 moles.

Explanation :

First we have to calculate the moles of $CH_4$ .

Using ideal gas equation:

$PV=nRT$

where,

P = pressure of gas = 1 atm

V = volume of gas = 10 L

T = temperature of gas = $27^oC=273+27=300K$

n = number of moles of gas = ?

R = gas constant = 0.0821 L.atm/mol.K

Now put all the given values in the ideal gas equation, we get:

$(1atm)\times (10L)=n\times (0.0821L.atm/mol.K)\times (300K)$

$n=0.406mole$

Now we have to calculate the moles of $O_2$ .

The balanced chemical reaction will be:

$CH_4+2O_2\rightarrow CO_2+2H_2O$

From the balanced reaction we conclude that,

As, 1 mole of $CH_4$ react with 2 moles of $O_2$

So, 0.406 mole of $CH_4$ react with $2\times 0.406=0.812$ moles of $O_2$

Now we have to calculate the excess moles of $O_2$ .

$O_2$ is 20 % excess. That means,

Excess moles of $O_2$ = $\frac{(100 + 20)}{100}$ × Required moles of $O_2$

Excess moles of $O_2$ = 1.2 × Required moles of $O_2$

Excess moles of $O_2$ = 1.2 × 0.812 = 0.97 mole

Now we have to calculate the moles of $O_2$ left in the products.

Moles of $O_2$ left in the products = Excess moles of $O_2$ - Required moles of $O_2$

Moles of $O_2$ left in the products = 0.97 - 0.812 = 0.16 mole

Therefore, the moles of $O_2$ left in the products are 0.16 moles.

7 0

3 years ago

1. Mg + N2 —> MgN2

quester [9]

(1) IS THE BALANCED EQUATION AND REACTANT-> MG AND N2 AND PRODUCT ->MGN2.

(2) IS NOT BALANCED AND REACTANT->CF4 AND BR AND PRODUCT IS CBR4 AND F2

8 0

3 years ago

You probably already know the names of some molecules. Identify which compound (water, salt, carbon dioxide, or methane) corresp

vladimir1956 [14]

CO2 is carbon

H2O is water

CH4 is methane

NaCI is salt

7 0

3 years ago

Read 2 more answers

Which of these reactions is an addition reaction?

Akimi4 [234]

Answer:

https://smartlyhelper.com/chemistry/question14062929

I found the answer in this link

6 0

3 years ago