Watch the following videos on molecule mixing encapsulators:

How many moles of N are in 0.241 g of N2O?

gtnhenbr [62]

0.00547 moles of N.

6 0

3 years ago

Calculate the change in the entropy of the system and also the change in the entropy of the surroundings, and the resulting tota

Ghella [55]

Answer:

(a) Δ $S_{sys}$ = 2.881 J/K; Δ $S_{sur}$ = -2.881 J/K; total change in entropy = 0

(b)Δ $S_{sys}$ = 2.881 J/K; Δ $S_{sur}$ = 0 ; total change in entropy = 2.881 J/K

(c) Δ $S_{sys}$ = 0 ; Δ $S_{sur}$ = 0 ; total change in entropy = 0

Explanation:

In the given problem, we need to calculate the change in the entropy of the system and also the change in the entropy of the surroundings, and the resulting total change in entropy, when a sample of nitrogen gas of mass 14 g at 298 K and 1.00 bar doubles its volume. We have the following variable:

mass (m) = 14 g

Temperature = 298 K

Pressure = 1.00 bar

Initial volume = $V_{1}$

Final volume = $V_{2}$ = $2V_{1}$

(a) Change in entropy of the system Δ $S_{sys}$ = $nRIn\frac{V_{2} }{V_{1} }$

where R = 8.314 J/(mol*K)

n = number of moles = mass/molar mass = 14/ 28 = 0.5 moles

Δ $S_{sys}$ = 0.5*8.314*ln2 = 2.881 J/K

Change in entropy of the surrounding Δ $S_{sur}$ = -2.881 J/K

Therefore, for a reversible process, the total change in entropy = Δ $S_{sys}$ +Δ $S_{sur}$ = 2.881 - 2.881 = 0

(b) Because entropy is a state function, we use the same procedure as in part (a). Thus, Δ $S_{sys}$ = 2.881 J/K

Since surrounding does not change in this process Δ $S_{sur}$ = 0.

total change in entropy = Δ $S_{sys}$ +Δ $S_{sur}$ = 2.881 - 0 = 2.88 J/K

(c) For an adiabatic reversible expansion, q(rev) = 0, thus:

Δ $S_{sys}$ = 0

Since heat energy is not transferred from the system to the surrounding

Δ $S_{sur}$ = 0

total change in entropy = Δ $S_{sys}$ +Δ $S_{sur}$ = 0

6 0

3 years ago

A beaker with 155 mL of an acetic acid buffer with a pH of 5.000 is sitting on a benchtop. The total molarity of acid and conjug

OverLord2011 [107]

Answer:

ΔpH = 0.296

Explanation:

The equilibrium of acetic acid (CH₃COOH) in water is:

CH₃COOH ⇄ CH₃COO⁻ + H⁺

Henderson-Hasselbalch formula to find pH in a buffer is:

pH = pKa + log₁₀ [CH₃COO⁻] / [CH₃COOH]

Replacing with known values:

5.000 = 4.740 + log₁₀ [CH₃COO⁻] / [CH₃COOH]

0.260 = log₁₀ [CH₃COO⁻] / [CH₃COOH]

1.820 = [CH₃COO⁻] / [CH₃COOH] (1)

As total molarity of buffer is 0.100M:

[CH₃COO⁻] + [CH₃COOH] = 0.100M (2)

Replacing (2) in (1):

1.820 = 0.100M - [CH₃COOH] / [CH₃COOH]

1.820[CH₃COOH] = 0.100M - [CH₃COOH]

2.820[CH₃COOH] = 0.100M

[CH₃COOH] = 0.100M / 2.820

[CH₃COOH] = 0.035M

Thus: [CH₃COO⁻] = 0.100M - 0.035M = 0.065M

5.40 mL of a 0.490 M HCl are:

0.0054L × (0.490mol / L) = 2.646x10⁻³ moles HCl.

Moles of CH₃COO⁻ are: 0.155L × (0.065mol / L) = 0.0101 moles

HCl reacts with CH₃COO⁻ thus:

HCl + CH₃COO⁻ → CH₃COOH

After reaction, moles of CH₃COO⁻ are:

0.0101 moles - 2.646x10⁻³ moles = 7.429x10⁻³ moles of CH₃COO⁻

Moles of CH₃COOH before reaction are: 0.155L × (0.035mol / L) = 5.425x10⁻³ moles of CH₃COOH. As reaction produce 2.646x10⁻³ moles of CH₃COOH, final moles are:

5.425x10⁻³ moles + 2.646x10⁻³ moles = 8.071x10⁻³ moles of CH₃COOH. Replacing these values in Henderson-Hasselbalch formula:

pH = 4.740 + log₁₀ [7.429x10⁻³ moles] / [8.071x10⁻³ moles]

pH = 4.704

As initial pH was 5.000, change in pH is:

ΔpH = 5.000 - 4.740 = 0.296

4 0

3 years ago

HELPPPPPPP PLEASEEEE!!!!!!!

tester [92]

Answer:c

Explanation:

it increases by when it moves

4 0

3 years ago

Read 2 more answers

Using a value of Ksp = 1.8 x 10-2 for the reaction PbCl2 Pb+2(aq) + 2Cl -(aq). If the value of Ksp was determined to be only 1.2

ohaa [14]

Answer:

The correct answer is option D, that is, the ions are now combining to reduce their concentrations.

Explanation:

On the basis of the chemical equation:

PbCl₂ (s) ⇒ Pb²⁺ (aq) + 2Cl⁻ (aq)

It is mentioned that the value of Ksp is 1.8 × 10⁻²

As the solubility product is very less or negligible for this reaction that signifies that the ions produced are getting combined, resulting in the reduction of the concentration of the ions and enhancing the formation of the solid lead chloride. Therefore, they associate together to produce solid lead chloride.

4 0

3 years ago