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2 years ago

How do this I’m in school really don’t know what to do

Chemistry

1 answer:

avanturin [10]2 years ago

4 0

Answer:

you gotta rotate it 90 degrees and get that screen fixed

Explanation:

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Consider the two gaseous equilibria: The values of the equilibrium constants K 1 and K 2 are related by

Sauron [17]

The question is missing parts. The complete question is as follows.

Consider the two gaseous equilibria involving SO2 and the corresponding equilibrium constants at 298K:

$SO_{2}_{(g)} + \frac{1}{2}O_{2}$ ⇔ $SO_{3}_{(g)}$ ; $K_{1}$

$2SO_{3}_{(g)}$ ⇔ $2SO_{2}_{(g)}+O_{2}_{(g)}; K_{2}$

The values of the equilibrium constants are related by:

a) $K_{1}$ = $K_{2}$

b) $K_{2} = K_{1}^{2}$

c) $K_{2} = \frac{1}{K_{1}^{2}}$

d) $K_{2}=\frac{1}{K_{1}}$

Answer: c) $K_{2} = \frac{1}{K_{1}^{2}}$

Explanation: <u>Equilibrium</u> <u>constant</u> is a value in which the rate of the reaction going towards the right is the same rate as the reaction going towards the left. It is represented by letter K and is calculated as:

$K=\frac{[products]^{n}}{[reagents]^{m}}$

The concentration of each product divided by the concentration of each reagent. The indices, m and n, represent the coefficient of each product and each reagent.

The equilibrium constants of each reaction are:

$SO_{2}_{(g)} + \frac{1}{2}O_{2}$ ⇔ $SO_{3}_{(g)}$

$K_{1}=\frac{[SO_{3}]}{[SO_{2}][O_{2}]^{1/2}}$

$2SO_{3}_{(g)}$ ⇔ $2SO_{2}_{(g)}+O_{2}_{(g)}$

$K_{2}=\frac{[SO_{2}]^{2}[O_{2}]}{[SO_{3}]^{2}}$

Now, analysing each constant, it is easy to see that $K_{1}$ is the inverse of $K_{2}$ .

If you doubled the first reaction, it will have the same coefficients of the second reaction. Since coefficients are "transformed" in power for the constant, the relationship is:

$K_{2}=\frac{1}{K_{1}^{2}}$

8 0

3 years ago

A bomb calorimeter has a heat capacity of 2.18 kJ/K. When a 0.176g sample of gas with a molar mass of 28.0 g/mol was burned in t

irina1246 [14]

Answer:

Heat of combustion =-7.35 *10² kJ

Explanation:

Step 1: Data given

Mass = 0.176 grams

Molar mass = 28.0 g/mol

Temperature increase 2.12 K

A bomb calorimeter has a heat capacity of 2.18 kJ/K

Step 2: Calculate heat of combustion

Energy released = mass *heat capacity* change in temperature

Q = c*∆T

Q = 2.18 kJ/K*2.12 K = 4.62 kJ released

Heat of combustion = (energy released)/(mole of substance)

Moles of gas = (0.176 g)/(28.0 g/mol) = 0.00629 moles

Heat of combustion = (energy released)/(mole of substance)

Heat of combustion = (4.62 kJ)/(0.00629 moles) = 7.35 *10² kJ

Heat of combustion =-7.35 *10² kJ (negative because heat is lost)

6 0

3 years ago

you are heating a solution and recording its temperature every 2 min during an experiment. which is your independent variable an

Zepler [3.9K]

An independent variable is the variable you are changing in order to measure the dependent variable, which is what you are measuring.

In this example, the

independent variable: chemicals in solution
dependent variable: temperature of solution

7 0

3 years ago

Please go to this link... I've been waiting for someone to answer for 4 hours...

Charra [1.4K]

On my way! to find the answer

6 0

4 years ago

How many rings does an alkane have if its formula is c10h16?

stepladder [879]

If it is saturated compound then we can calculate the double bond equivalent which will be equal to the number of rings in the compound

the double bond equivalent can be calculated using following formula

$DBE = (C + 1 -\frac{H}{2} - \frac{X}{2} +\frac{N}{2} )$

Where

H = number of Hydrogen atoms

C = number of carbon atoms

X= number of halogen atoms

N = number of nitrogen atoms

DBE = (10 + 1 - 16 / 2 ) = 3

Hence there are three rings in the compound

3 0

3 years ago