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vladimir2022 [97]
3 years ago
5

How many formula units are in 2.58 moles of Caci.?

Chemistry
1 answer:
mafiozo [28]3 years ago
3 0

Answer:

1.55 x 10²⁴ formula units

Explanation:

Given parameters:

Number of moles of CaCl₂   = 2.58moles

Unknown:

Formula units in the compound  = ?

Solution:

To solve this problem, we must understand that;

    1 mole of a substance contains 6.02 x 10²³ formula units

   2.58 moles will contain 2.58 x  6.02 x 10²³ formula units  

                                            = 1.55 x 10²⁴ formula units

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If you have 85.0 grams of CO₂, calculate how many moles of CO₂ would you have?
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Indicate whether each of the following statements is correct or incorrect.1)Every Bronsted-Lowry acid is also a Lewis acid2)Ever
Hunter-Best [27]

Answer:

1) correct

2) incorrect

3) correct

4)incorrect

Explanation:

1) A Lewis acid is a substance that accepts a nonbonding pair of electrons.

A Bronsted-Lowry acid is a substance that donates a proton H⁺

Since the donation of a proton involves the acceptance of a pair of electrons, every Bronsted-Lowry acid is also a Lewis acid.

2)A Lewis acid not necessarily needs to have a proton to be donated.

3) Conjugated acids of weak bases are strong acids and conjugated acids of strong bases are weak acids.

4)K⁺ comes from a strong base, therefore is does not have an acidic behaviour.

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The radioactive isotope I-131 is used in
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At 2000°C, the equilibrium constant for the reaction below is Kc = 4.10 ´ 10–4 . If 0.600 moles of NO is placed in a 1.0-L react
erastova [34]

Answer:

At equilibrium, the concentration of N_{2 (g)} is going to be 0.30M

Explanation:

We first need the reaction.

With the information given we can assume that is:

N_{2 (g)} + O_{2 (g)} ⇄ 2NO_{(g)}

If there is placed 0.600 moles of NO in a 1.0-L vessel, we have a initial concentration of 0.60 M NO; and no N_{2 (g)} nor  O_{2 (g)} present. Immediately, N_{2 (g)} andO_{2 (g)} are going to be produced until equilibrium is reached.

By the ICE (initial, change, equilibrium) analysis:

I: [N_{2 (g)}]=0   ;     [O_{2 (g)} ]= 0    ; [NO_{(g)}]=0.60M

C: [N_{2 (g)}]=+x   ;     [O_{2 (g)} ]= +x    ; [NO_{(g)}]=-2x

E: [N_{2 (g)}]=0+x   ;     [O_{2 (g)} ]= 0+x   ; [NO_{(g)}]=0.60-2x

Now we can use the constant information:

K_{c}=\frac{[products]^{stoichiometric coefficient} }{[reactants]^{stoichiometric coefficient} }

4.10* 10^{-4} =\frac{(0.60-2x)^{2}}{(x)*(x)}

4.10* 10^{-4}= \frac{(0.60-2x)^{2}}{x^{2} }

4.10* 10^{-4} * x^{2}= (0.60-2x)^{2}}

\sqrt{4.10* 10^{-4} * x^{2}}= \sqrt{(0.60-2x)^{2}}}

0.0202 x =0.60 - 2x

2x+0.0202x=0.60

x=\frac{0.60}{2.0202}= 0.30

At equilibrium, the concentration of N_{2 (g)} is going to be 0.30M

3 0
3 years ago
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