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3 years ago

A gas at STP has a volume of 1.5 L. What will the pressure be if it is at 26° C occupying .75 L

Chemistry

1 answer:

andrey2020 [161]3 years ago

6 0

Answer:

P₂ = 2.19 atm

Explanation:

Given data:

Initial volume = 1.5 L

Initial pressure =1 atm

Initial temperature = 273K

Final temperature = 26°C (26+273 = 299 K)

Final volume = 0.75 L

Final pressure = ?

Formula:

P₁V₁/T₁ = P₂V₂/T₂

P₁ = Initial pressure

V₁ = Initial volume

T₁ = Initial temperature

P₂ = Final pressure

V₂ = Final volume

T₂ = Final temperature

Solution:

P₂ = P₁V₁ T₂/ T₁ V₂

P₂ = 1 atm × 1.5 L × 299 K / 273 K × 0.75 L

P₂ = 448.5 atm .L. K / 204.75 K.L

P₂ = 2.19 atm

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sesenic [268]

The balanced equation for the above reaction is as follows;
2C₈H₁₈ + 25O₂ ---> 16CO₂ + 18H₂O
stoichiometry of octane to CO₂ is 2:16
number of C₈H₁₈ moles reacted - 191.6 g / 114 g/mol = 1.68 mol
when 2 mol of octane reacts it forms 16 mol of CO₂
therefore when 1.68 mol of octane reacts - it forms 16/2 x 1.68 = 13.45 mol of CO₂
number of CO₂ moles formed - 13.45 mol
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mass of CO₂ formed is 591.8 g

3 0

3 years ago

If a volume of gas 177.6mL was collected at a temperature of 25.8C and the pressure is 799.7 torr, what is the original concentr

Step2247 [10]

Answer:

The original concentration of the acid was 0.605 M

Explanation:

Step 1: Data given

Volume of gas = 177.6 mL = 0.1176 L

Temperature = 25.8 °C = 298.95 K

Pressure = 799.7 torr = 799.7/ 760 = 1.0522368 atm

Volume of acid needed to react = 12.6 mL = 0.0126 L

Step 2: Calculate moles

p*V = n*R*T

n = (p*V)/(R*T)

⇒with n = the number of moles = TO BE DETERMINED

⇒with p = the pressure of the gas = 799.7 torr = 1.0522368 atm

⇒with V = the volume of the gas = 177.6 mL = 0.1776 L

⇒with R = the gas constant = 0.08206 L*atm/mol*K

⇒with T = the temperature = 25.8 °C = 298.95 K

n = 0.007618 moles

Step 3: Calculate original concentration

We need 0.007618 moles of acid to react with the same amount of moles gas

Concentration acid = moles / volume

Concentration acid = 0.007618 moles / 0.0126 L

Concentration acid = 0.605 M

The original concentration of the acid was 0.605 M

5 0

3 years ago

Which describes an impermeable rock where oil and natural gas deposits are found?

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Answer:

Explanation:

Groundwater is stored in the open spaces within rocks and within unconsolidated sediments. Rocks and sediments near the surface are under less pressure than those at significant depth and therefore tend to have more open space. For this reason, and because it’s expensive to drill deep wells, most of the groundwater that is accessed by individual users is within the first 100 m of the surface. Some municipal, agricultural, and industrial groundwater users get their water from greater depth, but deeper groundwater tends to be of lower quality than shallow groundwater, so there is a limit as to how deep we can go.

5 0

3 years ago

Calculate the freezing point of a solution containing 5.0 grams of KCl and 550.0 grams of water. The molal-freezing-point-depres

yulyashka [42]

<u>Answer:</u> The freezing point of solution is -0.454°C

<u>Explanation:</u>

Depression in freezing point is defined as the difference in the freezing point of pure solution and freezing point of solution.

The equation used to calculate depression in freezing point follows:

$\Delta T_f=\text{Freezing point of pure solution}-\text{Freezing point of solution}$

To calculate the depression in freezing point, we use the equation:

$\Delta T_f=iK_fm$

Or,

$\text{Freezing point of pure solution}-\text{Freezing point of solution}=i\times K_f\times \frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams)}}$

where,

Freezing point of pure solution = 0°C

i = Vant hoff factor = 2

$K_f$ = molal freezing point elevation constant = 1.86°C/m

$m_{solute}$ = Given mass of solute (KCl) = 5.0 g

$M_{solute}$ = Molar mass of solute (KCl) = 74.55 g/mol

$W_{solvent}$ = Mass of solvent (water) = 550.0 g

Putting values in above equation, we get:

$0-\text{Freezing point of solution}=2\times 1.86^oC/m\times \frac{5\times 1000}{74.55g/mol\times 550}\\\\\text{Freezing point of solution}=-0.454^oC$

Hence, the freezing point of solution is -0.454°C

3 0

3 years ago

Need help read the comment below.

BartSMP [9]

Answer:

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Explanation:

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3 years ago

A gas at STP has a volume of 1.5 L. What will the pressure be if it is at 26° C occupying .75 L​

A gas at STP has a volume of 1.5 L. What will the pressure be if it is at 26° C occupying .75 L