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Setler79 [48]
2 years ago
8

No links plz!!CaOCl2 parent acid and base???​

Chemistry
1 answer:
Vesnalui [34]2 years ago
4 0

Answer:

this is not and reaction of acid and base.(CaOCL2)

Explanation:

Ca2, Cl, are base and O is a neutral.

so how can that be a acid and base reaction

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How much heat is gained when a 50.32g piece of aluminum is heated from 9.0°c to 16°c
Rashid [163]

Answer: 317 joules

Explanation:

The quantity of heat energy (Q) gained by aluminium depends on its Mass (M), specific heat capacity (C) and change in temperature (Φ)

Thus, Q = MCΦ

In this case,

Q = ?

Mass of aluminium = 50.32g

C = 0.90J/g°C

Φ = (Final temperature - Initial temperature)

= 16°C - 9°C = 7°C

Then, Q = MCΦ

Q = 50.32g x 0.90J/g°C x 7°C

Q = 317 joules

Thus, 317 joules of heat is gained.

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3 years ago
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olga_2 [115]

Answer:

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Explanation:

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3 years ago
The equilibrium constant for the reaction
Hitman42 [59]

The question is incomplete, here is the complete question:

The equilibrium constant for the reaction

N₂O₄(g)⇌2NO₂ at 2°C is Kc = 2.0

If each yellow sphere represents 1 mol of N₂O₄(g) and each gray sphere 1 mol of NO₂ which of the following 1.0 L containers represents the equilibrium mixture at 2°C?

The image is attached below.

<u>Answer:</u> The system which represents the equilibrium having value of K_c=2.0 is system (b)

<u>Explanation:</u>

Equilibrium constant in terms of concentration is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric ratios. It is expressed as K_c

For a general chemical reaction:

aA+bB\rightarrow cC+dD

The expression for K_{c} is written as:

K_{c}=\frac{[C]^c[D]^d}{[A]^a[B]^b}

For the given chemical equation:

N_2O_4(g)\rightleftharpoons 2NO_2

The expression of K_c for above equation follows:

K_c=\frac{[NO_2]^2}{[N_2O_4]}      .......(1)

We are given:

Volume of the container = 1.0 L

Value of K_c = 2.0

Molarity of the substance is calculated by using the equation:

\text{Molarity}=\frac{\text{Number of moles}}{\text{Volume}}

For the given images:

  • <u>For a:</u>

Number of Gray spheres = 8 moles

Number of yellow spheres = 4 moles

Putting values in expression 1, we get:

K_c=\frac{(8/1)^2}{(4/1)}\\\\K_c=16

  • <u>For b:</u>

Number of Gray spheres = 4 moles

Number of yellow spheres = 8 moles

Putting values in expression 1, we get:

K_c=\frac{(4/1)^2}{(8/1)}\\\\K_c=2

  • <u>For c:</u>

Number of Gray spheres = 6 moles

Number of yellow spheres = 6 moles

Putting values in expression 1, we get:

K_c=\frac{(6/1)^2}{(6/1)}\\\\K_c=6

  • <u>For d:</u>

Number of Gray spheres = 2 moles

Number of yellow spheres = 8 moles

Putting values in expression 1, we get:

K_c=\frac{(2/1)^2}{(8/1)}\\\\K_c=\frac{1}{2}

Hence, the system which represents the equilibrium having value of K_c=2.0 is system (b)

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How is hydrogen in heavy water different from hydrogen in normal water?
tatyana61 [14]

Answer:

I'd say the correct answer is A

Explanation:

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