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vladimir2022 [97]
3 years ago
10

Consider the following reaction: CO(g)+2H2(g) <--> CH3OH(g).

Chemistry
1 answer:
Andrew [12]3 years ago
4 0

Answer:

123 is the equilibrium constant for the reaction.

Explanation:

CO(g)+2H_2(g)\rightleftharpoons CH_3OH(g)

Equilibrium concentration of reactants :

[CO]=0.115 M,[H_2]=0.116 M

Equilibrium concentration of products:

[CH_3OH]=0.190 M

The expression of an equilibrium constant is given by :

K_c=\frac{[CH_3OH]}{[CO][H_2]^2}

K_c=\frac{0.190 M}{0.115 M\times (0.116 M)^2}

K_c=122.78 \approx 123

123 is the equilibrium constant for the reaction.

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In an experiment, students were given an unknown mineral. The unknown mineral was placed in 150 ml of water. Once in the water,
Dahasolnce [82]

Answer:

<h2>15 g/mL</h2>

Explanation:

The density of a substance can be found by using the formula

density =  \frac{mass}{volume}  \\

But from the question

volume = final volume of water - initial volume of water

volume = 165 - 150 = 15 mL

We have

density =  \frac{225}{15}  = 15 \\

We have the final answer as

<h3>15 g/mL</h3>

Hope this helps you

5 0
2 years ago
The density of water at 40°c is 0.992 g/ml. what is the volume of 3.45 g of water at this temperature?
wariber [46]
3.45÷0.992=3.48ml
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8 0
3 years ago
If an ice cube weighing 25.0 g with an initial
riadik2000 [5.3K]

Answer:

11

∘

C

Explanation:

As far as solving this problem goes, it is very important that you do not forget to account for the phase change underwent by the solid water at

0

∘

C

to liquid at

0

∘

C

.

The heat needed to melt the solid at its melting point will come from the warmer water sample. This means that you have

q

1

+

q

2

=

−

q

3

(

1

)

, where

q

1

- the heat absorbed by the solid at

0

∘

C

q

2

- the heat absorbed by the liquid at

0

∘

C

q

3

- the heat lost by the warmer water sample

The two equations that you will use are

q

=

m

⋅

c

⋅

Δ

T

, where

q

- heat absorbed/lost

m

- the mass of the sample

c

- the specific heat of water, equal to

4.18

J

g

∘

C

Δ

T

- the change in temperature, defined as final temperature minus initial temperature

and

q

=

n

⋅

Δ

H

fus

, where

q

- heat absorbed

n

- the number of moles of water

Δ

H

fus

- the molar heat of fusion of water, equal to

6.01 kJ/mol

Use water's molar mass to find how many moles of water you have in the

100.0-g

sample

100.0

g

⋅

1 mole H

2

O

18.015

g

=

5.551 moles H

2

O

So, how much heat is needed to allow the sample to go from solid at

0

∘

C

to liquid at

0

∘

C

?

q

1

=

5.551

moles

⋅

6.01

kJ

mole

=

33.36 kJ

This means that equation

(

1

)

becomes

33.36 kJ

+

q

2

=

−

q

3

The minus sign for

q

3

is used because heat lost carries a negative sign.

So, if

T

f

is the final temperature of the water, you can say that

33.36 kJ

+

m

sample

⋅

c

⋅

Δ

T

sample

=

−

m

water

⋅

c

⋅

Δ

T

water

More specifically, you have

33.36 kJ

+

100.0

g

⋅

4.18

J

g

∘

C

⋅

(

T

f

−

0

)

∘

C

=

−

650

g

⋅

4.18

J

g

∘

C

⋅

(

T

f

−

25

)

∘

C

33.36 kJ

+

418 J

⋅

(

T

f

−

0

)

=

−

2717 J

⋅

(

T

f

−

25

)

Convert the joules to kilojoules to get

33.36

kJ

+

0.418

kJ

⋅

T

f

=

−

2.717

kJ

⋅

(

T

f

−

25

)

This is equivalent to

0.418

⋅

T

f

+

2.717

⋅

T

f

=

67.925

−

33.36

T

f

=

34.565

0.418

+

2.717

=

11.026

∘

C

Rounded to two sig figs, the number of sig figs you have for the mass of warmer water, the answer will be

T

f

=

11

∘

C

Explanation:

3 0
2 years ago
How many moles of CO2 must dissolve in excess water to produce 12 moles of<br><br> H2CO3?
vodka [1.7K]

Answer:

12 moles of CO₂.

Explanation:

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From the balanced equation above,

1 mole of CO₂ dissolves in water to produce 1 mole of H₂CO₃.

Finally, we shall determine the number of moles of CO₂ that will dissolve in water to produce 12 moles of H₂CO₃. This can be obtained as follow:

From the balanced equation above,

1 mole of CO₂ dissolves in water to produce 1 mole of H₂CO₃.

Therefore, 12 moles of CO₂ will also dissolve in water to produce 12 moles of H₂CO₃.

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