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4 years ago

How many moles of coF3 are produced when 1.25 moles of 3co reacts

Chemistry

2 answers:

oee [108]4 years ago

7 0

Answer:

1.25 moles

Explanation:

First, we need to balance the equation. Essentially, this means making sure we have the same number of each atom on each side.

On the left side, we currently have:

- 1 Co atom

- 2 F aomts

On the right side, we have:

- 1 Co atom

- 3 F atoms

To balance it, add a 2 to Co on the left, 3 to F2 on the left, and 2 to CoF3 on the right:

$2Co+3F_2$ → $2CoF_3$

Now, we have 1.25 moles of Co, and since the ratio between Co and CoF3 is 1:1, we also have 1.25 moles of CoF3.

Thus, the answer is 1.25 moles.

Sloan [31]4 years ago

3 0

Answer:

1.25 moles

Explanation:

Balancing the equation:

2Co + 3F2 --> 2CoF3

Co : CoF3

2 : 2

1.25 : X

X/1.25 = 2/2

X/1.25 = 1

X = 1.25 moles

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Write the balanced equation for the equilibrium reaction for the dissociation ofsilver chloride in water, and write the K expres

Marizza181 [45]

Answer:

See explanation

Explanation:

Hello there!

In this case, since the the concentrations are not given, and not even the Ksp, we can solve this problem by setting up the chemical equation, the equilibrium constant expression and the ICE table only:

$AgCl(s)\rightleftharpoons Ag^+(aq)+Cl^-(aq)$

Next, the equilibrium expression according to the produced aqueous species as the solid silver chloride is not involved in there:

$Ksp=[Ag^+][Cl^-]$

And therefore, the ICE table, in which x stands for the molar solubility of the silver chloride:

$\ \ \ \ \ \ \ \ \ \ \ \ \ \ AgCl(s)\rightleftharpoons Ag^+(aq)+Cl^-(aq)$

I - 0 0

C - +x +x

E - x x

Which leads to the following modified equilibrium expression:

$Ksp=x^2$

Unfortunately, values were not given, and they cannot be arbitrarily assigned or assumed.

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3 years ago

8. The standard enthalpy of formation of RbF(s) is –557.7kJ/mol and the standard enthalpy of formation of RbF(aq, 1 m) is –583.8

garri49 [273]

Explanation:

Given

The enthalpy of formation of RbF (s) is –557.7kJ/mol

The standard enthalpy of formation of RbF (aq, 1 m) is –583.8 kJ/mol

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= -583.8 - (-557.7) kJ/mol

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The enthalpy is negative which means that the temperature will rise when RbF is dissolved.

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If you need to translate, the answer is
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Answer:

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Explanation: hope this helped

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