9. A Toyota Prius hybrid gets 21 kilometers per liter in highway driving. What is

A sample of pure NO2 is heated to 337?C at which temperature it partially dissociates according to the equation2NO2(g)?2NO(g)+O2

tatyana61 [14]

Answer: Kc =5.915. $10^{-3}$

Explanation: Kc is an equilibrium constant of a chemical reaction and is dependent upon the reagents and products concentration. For this reaction, Kc is: Kc = $\frac{[NO]^{2} .[O2]}{[NO2]^{2} }$

To determine each concentration, we have to find each relation:

From the reaction, we know that [NO] = 2[O2] (1)

From the Ideal Gas Law,

P·V=ntotal·R·T

$\frac{P}{RT}=\frac{ntotal}{V}$

$\frac{P}{RT} = \frac{n(NO2)}{V} + \frac{n(NO)}{V} + \frac{n(O2)}{V}$

[NO2]+{NO}+[O2] = $\frac{0.745}{0.082.610}$ = 0.015 mol/L

As [NO] = 2[O2]

[NO2]+2[O2]+[O2]=0.015

[NO2] = 0.015 - 3[O2] (2)

To resolve this equation, we will turn towards density of the mixture:

ρ = $\frac{m(NO2)+m(NO)+m(O2)}{V}$

Substituting mass for molar mass and number of molar,

ρ = M(NO2)· $\frac{n(NO2)}{V}$ +M(NO)· $\frac{n(NO)}{V}$ +M(O2)· $\frac{n(O2)}{V}$

Knowing the molar mass of each molecule:

ρ = 46·[NO2]+30·[NO]+32·[O2] = 0.525 g/L (3)

Substituting (1), (2) and (3) we have:

46(0.015 - 3[O2])+30(2[O2])+32.[O2]=0.525

32[O2]+60[O2]-138[O2]=0.0525-0.69

-48[O2]= -0.6375

[O2]=13.3. $10^{-3}$ M

Calculating for [NO]

[NO]=2[O2]

[NO]=2.13.3. $10^{-3}$

[NO]=26.6. $10^{-3}$ M

And finding [NO2],

[NO2]=0.015 - 3[O2]

[NO2]=0.015 - 3.13.3. $10^{-3}$

[NO2]=39.885. $10^{-3}$ M

So, to calculate Kc:

Kc = (26.6. $10^{-3}$ )²· 13.3. $10^{-3}$ / (39.885. $10^{-3}$ )²

Kc= 5.915. $10^{-3}$

The Kc is 5.915. $10^{-3}$ .

6 0

2 years ago

How does a conductivity apparatus test whether a solution has ionic or covalent substances in it? It fails to light up except in

larisa [96]

Covalent compounds are those which are made up by sharing of electrons between them and the electronegativity of the elements which form covalent compounds are nearly same that is they are generally non-polar in nature whereas ionic compounds are formed by complete transfer of electrons from one element to the other and thus resulting in formation of ions of opposite charges that is cation (with positive charge) and anion (with negative charge). There is a large difference in electronegativity of the element involved in the formation of ionic compounds. They are held together by electrostatic force of attraction between them.

The conductivity of a substance is determined by the flow of electric charge. When the charged particles move that are present in the ionic compounds only in the conductivity apparatus they complete the circuit by the flow of electrons.

Hence, a conductivity apparatus test whether a solution has ionic or covalent substances in it as ions complete the flow of electrons to form a circuit.

7 0

2 years ago

Read 2 more answers

What happens during a mosquito 6 chromosomes to form eggs.

lbvjy [14]

Since the diploid number of mosquitoes is six, that means that there are 6 chromosomes in every somatic (non-reproductive) cell,including the cells that make up the stomach.The six chromosomes represent two three-chromosome sets the mosquito received one set of three chromosomes from the egg cell of the mother and a similar set of three chromosomes from the sperm cell of the father.

7 0

2 years ago

Calculate the number of kilojoules of energy required to convert 50.0 grams of solid DMSO initially at a temperature of 19.0°C t

GuDViN [60]

Answer:

20.79 kilojoules

Explanation:

Using Q = m×c×∆T

Where;

Q = Quantity of heat (J)

c = specific heat capacity of solid DMSO (1.80 J/g°C)

m = mass of DMSO

∆T = change in temperature

According to the provided information, m= 50g, initial temperature = 19.0°C, final temperature= 250.0°C

Q = m×c×∆T

Q = 50 × 1.80 × (250°C - 19°C)

Q = 90 × 231

Q = 20790 Joules

To convert Joules to kilojoules, we divide by 1000 i.e.

20790/1000

= 20.79 kilojoules

Hence, 20.79 kilojoules of energy is required to convert 50.0 grams of solid DMSO to gas.

4 0

2 years ago

Stu Dent has finished his titration, and he comes to you for help with the calculations. He tells you that 20.00 mL of unknown c

Alexus [3.1K]

Answer:

0.3229 M HBr(aq)

0.08436M H₂SO₄(aq)

Explanation:

Stu Dent has finished his titration, and he comes to you for help with the calculations. He tells you that 20.00 mL of unknown concentration HBr(aq) required 18.45 mL of 0.3500 M NaOH(aq) to neutralize it, to the point where thymol blue indicator changed from pale yellow to very pale blue. Calculate the concentration (molarity) of Stu's HBr(aq) sample.

Let's consider the balanced equation for the reaction between HBr(aq) and NaOH(aq).

NaOH(aq) + HBr(aq) ⇄ NaBr(aq) + H₂O(l)

When the neutralization is complete, all the HBr present reacts with NaOH in a 1:1 molar ratio.

$18.45 \times 10^{-3} L NaOH.\frac{0.3500molNaOH}{1LNaOH} .\frac{1molHBr}{1molNaOH} .\frac{1}{20.00 \times 10^{-3} LHBr} =\frac{0.3229molHBr}{1LHBr} =0.3229M$

Kemmi Major also does a titration. She measures 25.00 mL of unknown concentration H₂SO₄(aq) and titrates it with 0.1000 M NaOH(aq). When she has added 42.18 mL of the base, her phenolphthalein indicator turns light pink. What is the concentration (molarity) of Kemmi's H₂SO₄(aq) sample?

Let's consider the balanced equation for the reaction between H₂SO₄(aq) and NaOH(aq).

2 NaOH(aq) + H₂SO₄(aq) ⇄ Na₂SO₄(aq) + 2 H₂O(l)

When the neutralization is complete, all the H₂SO₄ present reacts with NaOH in a 1:2 molar ratio.

$42.18 \times 10^{-3} LNaOH.\frac{0.1000molNaOH}{1LNaOH} .\frac{1molH_{2}SO_{4}}{2molNaOH} .\frac{1}{25.00\times 10^{-3}LH_{2}SO_{4}} =\frac{0.08436molH_{2}SO_{4}}{1LH_{2}SO_{4}} =0.08436M$

6 0

2 years ago