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3 years ago

Exactly one mole of an ideal gas is contained in a 2.00-liter container at 1,000 K. What is the pressure exerted by this gas?

Chemistry

2 answers:

levacccp [35]3 years ago

7 0

Use PV =nRT

so P = nRT/V

= 1 mole(0.08205 L atm/K mol)(1000K) / 2 L

= 41 atm

ziro4ka [17]3 years ago

6 0

Answer : The pressure of the gas is, 41.025 atm

Solution :

Using ideal gas equation :

$PV=nRT\\\\P=\frac{nRT}{V}$

where,

n = number of moles of gas = 1 mole

P = pressure of the gas = ?

T = temperature of the gas = 1000 K

R = gas constant = 0.08205 L.atm/mole.K

V = volume of gas = 2.00 L

Now put all the given values in the above equation, we get the pressure of the gas.

$P=\frac{nRT}{V}$

$P=\frac{1mole\times (0.08205L.atm/mole.K)\times 1000K}{2.00L}$

$P=41.025atm$

Therefore, the pressure of the gas is, 41.025 atm

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If the temperature of 15 grams of water changes from 21C to 24C, how many joules of heat were involved? Show work

goblinko [34]

Answer:

189 Joules

Explanation:

Applying,

Q = cm(t₂-t₁)............. equation 1

Where Q = Heat, c = specific heat capacity of water, m = mass of water, t₁ = Initial Temperature, t₂ = Final temperature.

From the question,

Given: m = 15 grams = 0.015 kg, t₁ = 21 °C, t₂ = 24 °C

Constant: c = 4200J/kg.°C

Substitute these values into equation 1

Q = 0.015×4200×(24-21)

Q = 0.015×4200×3

Q = 189 Joules

6 0

2 years ago

Yuri [45]

the answer is d hope this helps

8 0

3 years ago

Read 2 more answers

The metabolic oxidation of glucose, C6H12O6, in our bodies produces CO2, which is expelled from our lungs as a gas.

enot [183]

Answer:

$\large \boxed{\text{21.6 L}}$

Explanation:

We must do the conversions

mass of C₆H₁₂O₆ ⟶ moles of C₆H₁₂O₆ ⟶ moles of CO₂ ⟶ volume of CO₂

We will need a chemical equation with masses and molar masses, so, let's gather all the information in one place.

Mᵣ: 180.16

C₆H₁₂O₆ + 6O₂ ⟶ 6CO₂ + 6H₂O

m/g: 24.5

(a) Moles of C₆H₁₂O₆

$\text{Moles of C$_{6}$H$_{12}$O}_{6} = \text{24.5 g C$_{6}$H$_{12}$O}_{6}\times \dfrac{\text{1 mol C$_{6}$H$_{12}$O}_{6}}{\text{180.16 g C$_{6}$H$_{12}$O}_{6}}\\\\= \text{0.1360 mol C$_{6}$H$_{12}$O}_{6}$

(b) Moles of CO₂

$\text{Moles of CO}_{2} =\text{0.1360 mol C$_{6}$H$_{12}$O}_{6} \times \dfrac{\text{6 mol CO}_{2}}{\text{1 mol C$_{6}$H$_{12}$O}_{6}} = \text{0.8159 mol CO}_{2}$

(c) Volume of CO₂

We can use the Ideal Gas Law.

pV = nRT

Data:

p = 0.960 atm

n = 0.8159 mol

T = 37 °C

(i) Convert the temperature to kelvins

T = (37 + 273.15) K= 310.15 K

(ii) Calculate the volume

$\begin{array}{rcl}pV &=& nRT\\\text{0.960 atm} \times V & = & \text{0.8159 mol} \times \text{0.082 06 L}\cdot\text{atm}\cdot\text{K}^{-1}\text{mol}^{-1} \times \text{310.15 K}\\0.960V & = & \text{20.77 L}\\V & = & \textbf{21.6 L} \\\end{array}\\\text{The volume of carbon dioxide is $\large \boxed{\textbf{21.6 L}}$}$