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Blababa [14]
2 years ago
11

The percent yield of a chemical reaction is valuable to many industries. Low percent yields can result in the loss of large amou

nts of money. Which of the following would likely increase the percent yield of a reaction? Check all that apply. Using accurate measurements using pure chemicals performing the reaction under the most ideal conditions adding water to the experiment.
Chemistry
1 answer:
aev [14]2 years ago
7 0

Using accurate measurements, using pure chemicals and performing the reaction under the most ideal conditions is important to get a valuable percent yield.

<h3>How we calculate the percent yield?</h3>

Percent yield of any chemical reaction is define as the ratios of the actual yield to the theoretical yield of the product and multiply by the 100.

To get the high percent yield or actual yield of any reaction, we have to perform the reaction under ideal condition because if we not use the standard condition then we get the low rate of reaction. Reactants should be present in the pure form as impurity make unwanted products and reduce the productivity of main product and accurate amount of reactants also important for the spontaneous reaction.

Hence, options (a), (b) & (c) are correct.

To know more about percent yield, visit the below link:

brainly.com/question/8638404

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I am so confused that i don't even understand how to answer it
maksim [4K]

Answer:

What's the question so I can help?

I will just edit my answer when you have asked your question and post it here.

Explanation:

4 0
2 years ago
For the gas phase decomposition of 1-bromopropane, CH3CH2CH2BrCH3CH=CH2 + HBr the rate constant at 622 K is 6.43×10-4 /s and the
Radda [10]

Answer is: activation energy of this reaction is 212,01975 kJ/mol.<span>
Arrhenius equation: ln(k</span>₁/k₂) = Ea/R (1/T₂ - 1/T₁).<span>
k</span>₁ = 0,000643 1/s.<span>
k</span>₂ = 0,00828 1/s.

T₁ = 622 K.

T₂ = 666 K.

R = 8,3145 J/Kmol.

<span> 1/T</span>₁ = 1/622 K = 0,0016 1/K.<span>
1/T</span>₂ = 1/666 K = 0,0015 1/K.<span>
ln(0,000643/0,00828) = Ea/8,3145 J/Kmol · (-0,0001 1/K).
-2,55 = Ea/8,3145 J/Kmol </span>· (-0,0001 1/K).<span>
Ea = 212019,75 J/mol = 212,01975 kJ/mol.</span>

6 0
3 years ago
The measured voltage of an electrochemical cell consisting of pure nickel immersed in a solution of Ni2+ ions of unknown concent
Verizon [17]

Answer:

[Ni²⁺] = 1.33 M

Explanation:

To do this, we need to use the Nernst equation which (in standard conditions of temperature)

E = E° - RT/nF lnQ

However R and F are constant, and the reaction is taking place in 25 °C so we can assume the nernst equation like this:

E = E° - 0.05916/n logQ

As the nickel is in the cathode, this means that this element is being reducted while Cadmium is being oxidized, therefore the REDOX reaction would be:

Cd(s) + Ni²⁺(aq) --------> Cd²⁺(aq) + Ni(s)

With this, Q:

Q = [Cd²⁺] / [Ni²⁺]

Now, we need to know the value of the standard reduction potentials, which can be calculated with the semi equations of reduction and oxidation:

Cd(s) ------------> Cd²⁺ + 2e⁻       E°₁ = 0.40 V

Ni²⁺ + 2e⁻ -------------> Ni(s)        E°₂ = -0.25 V

E° = E°₁ + E°₂

E° = 0.40 - 0.25 = 0.15 V

Now that we have all the data, we can solve for the [Ni²⁺]:

0.133 = 0.15 - 0.05916/2 log(5/[Ni²⁺])

0.133 - 0.15 = -0.05916/2 log(5/[Ni²⁺])

-0.017 = -0.02958 log(5/[Ni²⁺])

-0.017/-0.02958 = log(5/[Ni²⁺])

0.5747 = log(5/[Ni²⁺])

10^(0.5747) = 5/[Ni²⁺]

[Ni²⁺] = 5/3.7558

<h2>[Ni²⁺] = 1.33 M</h2>
7 0
3 years ago
Why do thermistors increase in conductivity when heated? What happens in normal metals? Explain on the atomic level.
Marysya12 [62]

Metal conductivity generally goes down or resistivity goes up with temperature goes up.

7 0
2 years ago
How is calculating a weighted average
kirza4 [7]

Answer:

The Key difference between average vs weighted average is that simple average is nothing but simply adding up all the observation values and dividing the same by the total number of observations to calculate the average whereas weighted average is an average where each observation value will have a frequency assigned.

Explanation:

8 0
4 years ago
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