All categories

3 years ago

A 2.5 L container holds a sample of hydrogen gas at 291 K and 180 kPa.

Chemistry

1 answer:

netineya [11]3 years ago

6 0

Answer:

The new temperature will be 565.83 K.

Explanation:

Gay Lussac's law establishes the relationship between the temperature and the pressure of a gas when the volume is constant. This law says that the pressure of the gas is directly proportional to its temperature. This means that if the temperature increases, the pressure will increase; or if the temperature decreases, the pressure will decrease.

In other words, Gay-Lussac's law states that when a gas undergoes a constant volume transformation, the ratio of the pressure exerted by the gas temperature remains constant:

$\frac{P}{T} =k$

When an ideal gas goes from a state 1 to a state 2, it is true:

$\frac{P1}{T1} =\frac{P2}{T2}$

In this case:

P1= 180 kPa
T1= 291 K
P2= 350 kPa
T2= ?

Replacing:

$\frac{180 kPa}{291 K} =\frac{350 kPa}{T2}$

Solving:

$T2=350 kPa*\frac{291 K}{180 kPa}$

T2= 565.83 K

The new temperature will be 565.83 K.

You might be interested in

Determine the limiting reactant (lr) and the mass (in g) of nitrogen that can be formed from 50.0 g n2o4 and 45.0 g n2h4. some p

Licemer1 [7]

N2O4(l) + 2 N2H4(l) → 3 N2(g) + 4 H2O(g)
1) to calculate the limiting reactant you need to pass grams to moles.
 moles is calculated by dividing mass by molar mass

mass of N2O4: 50.0 g
molar mass of N2O4 = 92.02 g/mol
molar mass of N2H4 = 32.05 g/mol.
mass of N2H4:45.0 g

moles N2O4=50.0/92.02 g/mol= 0,54 mol of N2O4
moles N2H4= 45/32.05 g/mol= 1,40 mol of N2H4

 2)
By looking at the balanced equation, you can see that 1 mol of N2O4 needs 2 moles of N2H4 to fully react . So to react 0,54 moles of N2O4, you need 2x0,54 moles of N2H4 moles
N2H4 needed = 1,08 moles.
You have more that 1,08 moles N2H4, so this means the limiting reagent is not N2H4, it's N2O4. The molecule that has molecules that are left is never the limiting reactant.

3) 1 mol of N2O4 reacting, will produce 3 mol of N2 (look at the equation)
There are 0,54 mol of N2O4 available to react, so how many moles will produce of N2?
1 mol N2O4------------3 mol of N2
0,54 mol N2O4--------x
x=1,62 mol of N2

4) the only thing left to do is convert the moles obtained, to grams.
We use the same formula as before, moles equal to mass divided by molar mass.
$moles= \frac{grams}{molar mass}$ (molar mass of N2= 28)
1,62 mol of N2= mass/ 28
mass of N2= 45,36 grams

4 0

3 years ago

What is condensation ?

fomenos

Answer:

gvapor or gas to liquid

Explanation:

water which collects as droplets on cold surface when humid air is in contact with it

3 0

3 years ago

Read 2 more answers

By pipet, 11.00 mL of a 0.823 MM stock solution of potassium permanganate (KMnO4) was transferred to a 50.00-mL volumetric flask

tangare [24]

Answer:

1) 0.18106 M is the molarity of the resulting solution.

2) 0.823 Molar is the molarity of the solution.

Explanation:

1) Volume of stock solution = $V_1=11.00 mL$

Concentration of stock solution = $M_1=0.823 M$

Volume of stock solution after dilution = $V_2=50.00 mL$

Concentration of stock solution after dilution = $M_2=?$

$M_1V_1=M_2V_2$ ( dilution )

$M_2=\frac{0.823 M\times 11.00 mL}{50 ,00 mL}=0.18106 M$

0.18106 M is the molarity of the resulting solution.

Molarity of the solution is the moles of compound in 1 Liter solutions.

$Molarity=\frac{\text{Mass of compound}}{\text{Molar mas of compound}\times Volume (L)}$

Mass of potassium permanganate = 13.0 g

Molar mass of potassium permangante = 158 g/mol

Volume of the solution = 100.00 mL = 0.100 L ( 1 mL=0.001 L)

$Molarity=\frac{13.0 g}{158 g/mol\times 0.100 L}=0.823 mol/L$

0.823 Molar is the molarity of the solution.

6 0

3 years ago

Which chemical class does phenobarbital belong to?

drek231 [11]

Phenobarbital is derivative of Barbituric Acid and Barbituric Acid is derivative of Urea. (structures shown in Fig below)

Urea has H₂N- group attached to Carbonyl Group (C=O), and such class of comounds conataining H₂N-C=O bond are called as Amides.

Result:
So, Phenobarbital belongs to Amides.

3 0

3 years ago

What volume of the stock solution (Part A) would contain the number of moles present in the diluted solution (Part B)?

vova2212 [387]

Idk what is is i am sorry

6 0

2 years ago