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maw [93]
1 year ago
10

If you run this reaction with 15 g of CuCl2 and 20 g NaNO3, what is the theoretical yield of NaCl?

Chemistry
1 answer:
Margaret [11]1 year ago
7 0

The theoretical yield of a substance is the amount of the mass that is calculated from stoichiometry. The mass of sodium chloride produced is 13.03 gms.

<h3>What is mass?</h3>

Mass of the substance is a multiplication of the molar mass and moles.

The balanced reaction is given as:

CuCl₂ + 2NaNO₃ ⇒ 2NaCl + Cu(NO₃)₂

Moles of copper chloride are calculated as: 15 ÷ 134.45 = 0.115 moles

Moles of sodium nitrate are calculated as: 20 ÷ 84.99 = 0.2353 moles

From the reaction, 1 mole of copper chloride gives 2 moles of sodium chloride so, 0.115 moles of copper chloride will produce 0.223 moles.

Similarly, 2 moles of sodium nitrate produce 2 moles of sodium chloride so, 0.2353 moles produce 0.4706 moles.

From this, it can be concluded that copper chloride is a limiting reagent and will be used to calculate the mass of NaCl.

Mass of NaCl is calculated as: Mass = 0.223 moles × 58.44 = 13.03 gms.

Therefore, 13.03 gms is the theoretical yield of sodium chloride.

Learn more about theoretical yield here:

brainly.com/question/1419024

#SPJ1

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) Do you think the pH of 1,0 M tri-methyl ammonium (CH3)3NH+, pKa = 9.80, will be higher or lower than that of 1.0 M phenol, C6H
Elanso [62]

Answer:

1. The pH of 1.0 M trimethyl ammonium (pH = 1.01) is lower than the pH of 0.1 M phenol (5.00).

2. The difference in pH values is 4.95.

Explanation:

1. The pH of a compound can be found using the following equation:

pH = -log([H_{3}O^{+}])

First, we need to find [H₃O⁺] for trimethyl ammonium and for phenol.

<u>Trimethyl ammonium</u>:

We can calculate [H₃O⁺] using the Ka as follows:

(CH₃)₃NH⁺ + H₂O  →  (CH₃)₃N + H₃O⁺    

1.0 - x                               x           x  

Ka = \frac{[(CH_{3})_{3}N][H_{3}O^{+}]}{[(CH_{3})_{3}NH^{+}]}

10^{-pKa} = \frac{x*x}{1.0 - x}

10^{-9.80}(1.0 - x) - x^{2} = 0    

By solving the above equation for x we have:  

x = 0.097 = [H₃O⁺]

pH = -log([H_{3}O^{+}]) = -log(0.097) = 1.01                                      

<u>Phenol</u>:

C₆H₅OH + H₂O → C₆H₅O⁻ + H₃O⁺

1.0 - x                        x             x

Ka = \frac{[C_{6}H_{5}O^{-}][H_{3}O^{+}]}{[C_{6}H_{5}OH]}

10^{-10} = \frac{x^{2}}{1.0 - x}

1.0 \cdot 10^{-10}(1.0 - x) - x^{2} = 0

Solving the above equation for x we have:

x = 9.96x10⁻⁶ = [H₃O⁺]

pH = -log([H_{3}O^{+}]) = -log(9.99 \cdot 10^{-6}) = 5.00

Hence, the pH of 1.0 M trimethyl ammonium is lower than the pH of 0.1 M phenol.

2. The difference in pH values for the two acids is:

\Delta pH = pH_{C_{6}H_{5}OH} - pH_{(CH_{3})_{3}NH^{+}} = 5.00 - 1.01 = 4.95

Therefore, the difference in pH values is 4.95.

I hope it helps you!

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WILL GIVE BRAINLIEST
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Answer:

In order to be able to solve this problem, you will need to know the value of water's specific heat, which is listed as

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In water's case, you need to provide 4.18 J of heat per gram of water to increase its temperature by 1∘C.

What if you wanted to increase the temperature of 1 g of water by 2∘C ?

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And there you have it. The equation that describes all this will thus be

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In your case, you will have

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