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3 years ago

Which of the following is the best definition of ethics?

Chemistry

1 answer:

Andrej [43]3 years ago

4 0

Answer:

Explanation:

I believe the answer is C.

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lactic acid, which consists of C, H, and O, has long been thought to be responsible for muscle soreness following strenuous exer

Minchanka [31]

Answer:

C3H6O3

Explanation:

I got you babe

3 0

3 years ago

MULTIPLE CHOICE

maxonik [38]

It would be True and why becuase I have found this app called quizlet hope this helps sorry if I’m wrong.

3 0

3 years ago

Read 2 more answers

What substance is acting as the Brønsted-Lowry base in the forward reaction below?

asambeis [7]

The question is asking to state the substance that is acting as the Bronsted Lowry base in the forward reaction, base on my research, I would say that the substance acting to said formulas is the NH3. I hope you are satisfied with my answer and feel free to ask for more if you have question and further clarification

3 0

3 years ago

What is the main source of energy for the water cycle.

viktelen [127]

Answer:

The main source of energy is the sun. The sun has a big part in the water cycle because it turns the water into water vapor. This cause the water cycle to start over again once it rains. Therefore, making the sun ne of the biggest parts in the water cycle.

7 0

2 years ago

Determine the theoretical maximum moles of hydroquinone, , that could be produced in this experiment. The reactant, quinone, is

dedylja [7]

Answer:

Theoretical maximum moles of hydroquinone: 0.2167 mol.

Explanation:

Hello,

In this case, the undergoing chemical reaction is like:

$Quinone\rightarrow Hydroquinone$

In such a way, since the molar mass of quinone is 108.1 g/mol and it is in a 1:1 molar ratio with hydroquinone, we can easily compute the theoretical maximum moles of hydroquinone by stoichiometry:

$n_{hydroquinone}=23.4g\ quinone*\frac{1mol\ quinone}{108.1 g\ quinone}*\frac{1mol\ hydroquinone}{1mol\ quinone} \\\\n_{hydroquinone}=0.2167mol$

Clearly, this is the theoretical yield which in grams is:

$m_{hydroquinone}=0.2167mol*\frac{110.11g}{1mol} =23.84g$

Which allows us to compute the percent yield as well since the obtained mass of the product is 13.0 g:

$Y=\frac{13.0g}{23.84g}*100\%=54.5\%$

Best regards.

8 0

3 years ago