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3 years ago

A sample of gas occupies 7.80 liters at 425°C? What will be the volume of the gas at 35°C if the pressure does not change?

Chemistry

2 answers:

Lisa [10]3 years ago

6 0

The answer is 3.44 liters

Ksenya-84 [330]3 years ago

5 0

V₁/T₁=V₂/T₂

T₁=425+273.15=698.15 K
T₂=35+273.15=308.15 K

V₂=V₁T₂/T₁

V₂=7.80*308.15/698.15=3.44 L

3.44 liters

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How many moles of carbon are in the sample?

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3 years ago

The pressure of a gas contained in a cylinder with a movable

insens350 [35]

Answer:

B) 244.5

Explanation:

2.62 torr =349.3046 pascals.

349.3046*0.7 sq meter = 244.5 N

4 0

3 years ago

The pictures are the answers

Mice21 [21]

Answer:

Explanation:

If the students want to know at what percent of CO2 in the air the plant will grow at the fastest, then the percent of CO2 should be a different value for each plant in the table.

There are 2 tables that have different values for the CO2 - the tables in answer choices C and D.

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3 years ago

Automobile airbags contain solid sodium azide, NaN3, that reacts to produce nitrogen gas when heated, thus inflating the bag. 2N

Vitek1552 [10]

Answer : The value of work done for the system is 1144.69 J

Explanation :

First we have to calculate the moles of $NaN_3$

$\text{Moles of }NaN_3=\frac{\text{Mass of }NaN_3}{\text{Molar mass of }NaN_3}$

Molar mass of $NaN_3$ = 65.01 g/mole

$\text{Moles of }NaN_3=\frac{20.2g}{65.01g/mole}=0.311mole$

Now we have to calculate the moles of nitrogen gas.

The balanced chemical reaction is,

$2NaN_3(s)\rightarrow 2Na(s)+3N_2(g)$

From the balanced reaction we conclude that

As, 2 mole of $NaN_3$ react to give 3 mole of $N_2$

So, 0.311 moles of $NaN_3$ react to give $\frac{0.311}{2}\times 3=0.466$ moles of $N_2$

Now we have to calculate the volume of nitrogen gas.

Using ideal gas equation:

$PV=nRT$

where,

P = Pressure of $N_2$ gas = 1.00 atm

V = Volume of $N_2$ gas = ?

n = number of moles $N_2$ = 0.466 mole

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

T = Temperature of $N_2$ gas = $22^oC=273+22=295K$

Putting values in above equation, we get:

$1.00atm\times V=0.466mole\times (0.0821L.atm/mol.K)\times 295K$

$V=11.3L$

As initially no nitrogen was present. So,

Volume expanded = Volume of nitrogen evolved

Thus,

Expansion work = Pressure × Volume

Expansion work = 1.00 atm × 11.3 L

Expansion work = 11.3 L.atm

Conversion used : (1 L.atm = 101.3 J)

Expansion work = 11.3 × 101.3 = 1144.69 J

Therefore, the value of work done for the system is 1144.69 J

5 0

3 years ago

if i add water to 100 mL of a 0.75 M NaOH solution un til the final volume is 165mL what will the molarity of the diluted soluti

ehidna [41]

Answer is: <span>the molarity of the diluted solution 0,454 M.
</span>V₁(NaOH) = 100 mL ÷ 1000 mL/L = 0,1 L.
c₁(NaOH) = 0,75 M = 0,75 mol/L.
n₁(NaOH) = c₁(NaOH) · V₁(NaOH).
n₁(NaOH) = 0,75 mol/L · 0,1 L.
n₁(NaOH) = 0,075 mol
n₂(NaOH) = n₁(NaOH) = 0,075 mol.
V₂(NaOH) = 165 mL ÷ 1000 mL/L = 0,165 L.
c₂(NaOH) = n₂(NaOH) ÷ V₂(NaOH).
c₂(NaOH) = 0,075 mol ÷ 0,165 L.
c₂(NaOH) = 0,454 mol/L.

7 0

4 years ago