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anygoal [31]
3 years ago
9

Calculate the work (kJ) done during a reaction in which the internal volume contracts from 85 L to 12 L against an outside press

ure of 2.4 atm.
Chemistry
1 answer:
butalik [34]3 years ago
5 0

Answer:

The work done on the system is 17.75_KJ

Explanation:

To solve this question we need to know the required equations from the given variables, thus

Initial volume = 85L

Final volume = 12L

External pressure = 2.4 atm.

work done = - PΔV

2.4×(85-12) = 175.2L×atm

Converting from L•atm to KJ is given by

1 L•atm = 0.1013 kJ

(175.2 L•atm) * (0.1013 kJ / 1 L•atm)

= 17.75 kJ

The work done on the system is 17.75_KJ

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100.00 mL of 0.15 M nitrous acid (HNO2) are titrated with a 0.15 M NaOH solution. (a) Calculate the pH for the initial solution.
wolverine [178]

Answer:

a. pH = 2.04

b. pH = 3.85

c. pH = 8.06

d. pH = 11.56

Explanation:

The nitrous acid is a weak acid (Ka = 5.6x10⁻⁴) that reacts with NaOH as follows:

HNO₂ + NaOH → NaNO₂(aq) + H₂O(l)

a. At the beginning there is just a solution of 0.12M HNO₂. As Ka is:

Ka = [H⁺] [NO₂⁻] / [HNO₂]

Where [H⁺] and [NO₂⁻] ions comes from the same equilibrium ([H⁺] = [NO₂⁻] = X):

5.6x10⁻⁴ = X² / 0.15M

8.4x10⁻⁵ = X²

X = [H⁺] = 9.165x10⁻³M

As pH = -log [H⁺]

<h3>pH = 2.04</h3><h3 />

b. At this point we have HNO₂ and NaNO₂ (The weak acid and the conjugate base), a buffer. The pH of a buffer is obtained using H-H equation:

pH = pKa + log [NaNO₂] / [HNO₂]

<em>Where pH is the pH of the buffer,</em>

<em>pKa is -log Ka = 3.25</em>

<em>And [NaNO₂] [HNO₂] could be taken as the moles of each compound.</em>

<em />

The initial moles of HNO₂ are:

0.100L * (0.15mol / L) = 0.015moles

The moles of base added are:

0.0800L * (0.15mol / L) = 0.012moles

The moles of base added = Moles of NaNO₂ produced = 0.012moles.

And the moles of HNO₂ that remains are:

0.015moles - 0.012moles = 0.003moles

Replacing in H-H equation:

pH = 3.25 + log [0.012moles] / [0.003moles]

<h3>pH = 3.85</h3><h3 />

c. At equivalence point all HNO2 reacts producing NaNO₂. The volume added of NaOH must be 100mL. That means the concentration of the NaNO₂ is:

0.15M / 2 = 0.075M

The NaNO₂ is in equilibrium with water as follows:

NaNO₂(aq) + H₂O(l) ⇄ HNO₂(aq) + OH⁻(aq) + Na⁺

The equilibrium constant, kb, is:

Kb = Kw/Ka = 1x10⁻¹⁴ / 5.6x10⁻⁴ = 1.79x10⁻¹¹ = [OH⁻] [HNO₂] / [NaNO₂]

<em>Where [OH⁻] = [HNO₂] = x</em>

<em>[NaNO₂] = 0.075M</em>

<em />

1.79x10⁻¹¹ = [X] [X] / [0.075M]

1.34x10⁻¹² = X²

X = 1.16x10⁻⁶M = [OH⁻]

pOH = -log [OH-] = 5.94

pH = 14-pOH

<h3>pH = 8.06</h3><h3 />

d. At this point, 5mL of NaOH are added in excess, the moles are:

5mL = 5x10⁻³L * (0.15mol / L) =7.5x10⁻⁴moles NaOH

In 100mL + 105mL = 205mL = 0.205L. [NaOH] = 7.5x10⁻⁴moles NaOH / 0.205L =

3.66x10⁻³M = [OH⁻]

pOH = 2.44

pH = 14 - pOH

<h3>pH = 11.56</h3>
5 0
2 years ago
3. How many grams of potassium permanganate will react with 36 grams of iron(III)<br>hydroxide? ​
Mrac [35]

7 would be correct because I divide it

8 0
3 years ago
A city's water supply is contaminated with a toxin at a concentration of 0.63 mg/L. For the water to be safe for drinking, the c
Shalnov [3]

Answer:

Approximately 22.37 days, will it take for the water to be safe to drink.

Explanation:

Using integrated rate law for first order kinetics as:

[A_t]=[A_0]e^{-kt}

Where,  

[A_t] is the concentration at time t

[A_0] is the initial concentration

k is rate constant

Given that:- k = 0.27 (day)⁻¹

[A_0] = 0.63 mg/L

[A_t]=1.5\times 10^{-3} mg/L

Applying in the above equation as:-

1.5\times 10^{-3}=0.63e^{-0.27\times t}

63e^{-0.27t}=150\times \:10^{-3}

e^{-0.27t}=\frac{1}{420}

t=\frac{100\ln \left(420\right)}{27}=22.37

<u>Approximately 22.37 days, will it take for the water to be safe to drink.</u>

7 0
3 years ago
20 POINTS!!!!......
tester [92]
Density= mass/ volume

So density = 99/10= 9.9g/cm^3

Hope this helps!! xx
5 0
3 years ago
Read 2 more answers
What was the resulting molarity of your primary standard solution of potassium iodate kio3?
melamori03 [73]
Missing question: 0,535 gram of KIO₃ dissolved in 250 mL of de-ionized water to <span>make primary standard solution.
m(</span>KIO₃) = 0,535 g.
V(KIO₃) = 250 mL ÷ 1000 mL/L = 0,25 L.
n(KIO₃) = m(KIO₃) ÷ M(KIO₃).
n(KIO₃) = 0,535 g ÷ 214 g/mol.
n(KIO₃) = 0,0025 mol.
c(KIO₃) = n(KIO₃) ÷ V(KIO₃).
c(KIO₃) = 0,0025 mol ÷ 0,25 L.
c(KIO₃) = 0,01 mol/L = 0,01 M.
8 0
2 years ago
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