All categories

13 days ago

At 1.01 bar, how many moles of CO2 are released by raising the temperature of 1 litre of water from 20∘C to 25∘C

Chemistry

1 answer:

VMariaS [1K]13 days ago

5 0

Answer: 0.0007 moles of $CO_2$ are released when the temperature rises.

Explanation:

To determine the moles, we utilize the ideal gas law, as follows:

$PV=nRT$

where,

P = gas pressure = 1.01 bar

V = gas volume = 1L

R = gas constant = $0.08314\text{ L bar }mol^{-1}K^{-1}$

Calculated moles at T = 20° C

The gas temperature = 20° C = (273 + 20)K = 293K

Substituting values into the equation gives:

$1.01bar\times 1L=n_1\times 0.0814\text{ L bar }mol^{-1}K^{-1}\times 293K\\n_1=0.04146moles$

Calculated moles at T = 25° C

The gas temperature = 25° C = (273 + 25)K = 298K

Substituting values into the equation gives:

$1.01bar\times 1L=n_2\times 0.0814\text{ L bar }mol^{-1}K^{-1}\times 298K\\n_2=0.04076moles$

Released moles = $n_1-n_2=0.04146-0.04076=0.0007moles$

Therefore, 0.0007 moles of $CO_2$ are released when the temperature increases from 20° C to 25° C.

You might be interested in

Butane (c4h10) undergoes combustion in excess oxygen to generate gaseous carbon dioxide and water. given δh°f[c4h10(g)] = –124.7

KiRa [971]

The Δ H value for butane (g) is -124.7 kJ/mol.

The Δ H value for CO2 (g) is -393.5 kJ/mol.

The Δ H value for H2O (g) is -241.8 kJ/mol.

The mass of butane is 8.30 grams.

Butane has a molar mass of 58 g/mol.

Considering the reaction,

C₄H₁₀ + 6.5 O₂ = 4CO₂ + 5H₂O

To determine the Δ H° of the reaction:

ΔH°rxn = ∑nH° f (products) - ∑nH° f (reactants)

By substituting values, we find that

Δ H° rxn = 4 (-393.5) + 5 (-241.8) - (-124.7)

= -1574 -1209 + 124.7

= -2783 - 124.7

= -2658.3 kJ/mol

Now, we will calculate how many moles of butane are in 8.30 grams.

Number of moles = mass/molar mass

= 8.30 / 58

= 0.143 moles

Therefore, the total energy released during the reaction is given by,

Q = number of moles × ΔH° rxn

= 0.143 × (2658.3)

= 380.14 kJ

Thus, the total heat released in the reaction is 380.14 kJ.

6 0

6 days ago

The Michaelis‑Menten equation models the hyperbolic relationship between [S] and the initial reaction rate V 0 V0 for an enzyme‑

alisha [964]

Answer:

The equation formulated by Michaelis-Menten is expressed as

v₀ = Kcat × [E₀] × [S] / (Km + [S])

in which,

Kcat denotes the experimental reaction rate constant; [S] signifies the concentration of the substrate, and

Km represents the Michaelis-Menten constant.

Explanation:

Refer to the attached image for an in-depth clarification

3 0

9 days ago

Why the gross reading is needed when doing the titration?

Anarel [852]

Answer:

The response is provided below.

Explanation:

Numerous aspects can influence the actual results of titration. These factors vary from human error to misjudging measurements, a researcher's interpretation of color changes, and improper techniques during the experimental procedure.

Thus, to mitigate these errors, researchers must engage thoroughly throughout experimentation, and employing gross readings can assist in reducing mistakes when determining the final titre value.

7 0

7 days ago

3. For the reaction: 2X + 3Y 3Z, the combination of 2.00 moles of X with 2.00

lions [985]

Answer:

Explanation:

Considering the reaction: 2X + 3Y = 3Z, combining 2.00 moles of X with 2.00 moles of Y results in the production of 1.75 moles of Z.

2 mol 2 mol 1.75 mol

2X + 3Y = 3Z

2 mol is required with 3 mol to yield 3 mol.

3 mol Z / 3 mol Y = 1 to 1

should yield 2 mol Z

1.75 / 2 = 87.5 % production yield

3 0

1 day ago

Next we need to determine the mass of oleic acid in the monolayer. The concentration of the oleic acid/benzene solution is 0.02g

Alekssandra [968]

Response:

$m=1x10^{-6}g$

Clarification:

Hello,

In this scenario, since a single drop equates to 0.05 mL of the solution provided, with a concentration of 0.02 g/mL, the mass of oleic acid in one drop calculates to:

$m=0.02\frac{g}{L}*0.05mL*\frac{1L}{1000mL}\\ \\m=1x10^{-6}g$

Best wishes.

3 0

12 days ago