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stich3 [128]
2 years ago
9

Diagram of Krebs cycle. ​

Chemistry
1 answer:
Dmitry [639]2 years ago
6 0

Answer:

The drawing is this . Cannot you draw yourself

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En que flota el osmio
Valentin [98]
Translate to english
6 0
3 years ago
A chemist prepares a solution of mercury(I) chloride Hg2Cl2 by measuring out
Genrish500 [490]

Answer:

1.26 × 10^-8 M

Explanation:

We are given;

Number of moles of mercury (i) chloride as 0.000126 μmol

Volume is 100 mL

We are required to calculate the concentration of the solution.

We need to know that;

Concentration is also known as molarity is given by;

Molarity = Number of moles ÷ Volume

Number of moles = 1.26 × 10^-10 Moles

Volume = 0.01 L

Therefore;

Concentration = 1.26 × 10^-10 Moles ÷ 0.01 L

                       = 1.26 × 10^-8 M

Thus, the molarity of the solution is 1.26 × 10^-8 M

6 0
3 years ago
An average human being has about 5.0 L of blood in his or her body. If an average person were to eat 37.7 g of sugar (sucrose, ,
Viktor [21]

Answer: 0.0220275 M

Explanation:

So, we are given the following data or parameters which are going to help in solving this particular Question/problem.

=> Averagely, we have the volume = 5.0 L of blood in human body .

=> Mass of sugar eaten = 37.7 g of sugar (sucrose, 342.30 g/mol).

Therefore, the molarity of the blood sugar change can be calculated as below:

The molarity of the blood sugar change = (1/ volume) × mass/molar mass.

Thus, the molarity of the blood sugar change = (1/5) × 37.7/342.30 = 0.0220275 M.

6 0
2 years ago
A buffer solution contains 0.345 M acetic acid and 0.377 M sodium acetate . If 0.0613 moles of potassium hydroxide are added to
melamori03 [73]

Answer:

pH = 5.54

Explanation:

The pH of a buffer solution is given by the <em>Henderson-Hasselbach (H-H) equation</em>:

  • pH = pKa + log\frac{[CH_3COO^-]}{[CH_3COOH]}

For acetic acid, pKa = 4.75.

We <u>calculate the original number of moles for acetic acid and acetate</u>, using the <em>given concentrations and volume</em>:

  • CH₃COO⁻ ⇒ 0.377 M * 0.250 L = 0.0942 mol CH₃COO⁻
  • CH₃COOH ⇒ 0.345 M * 0.250 L = 0.0862 mol CH₃COOH

The number of CH₃COO⁻ moles will increase with the added moles of KOH while the number of CH₃COOH moles will decrease by the same amount.

Now we use the H-H equation to <u>calculate the new pH</u>, by using the <em>new concentrations</em>:

  • pH = 4.75 + log\frac{(0.0942+0.0613)mol/0.250L}{(0.0862-0.0613)mol/0.250L} = 5.54
6 0
3 years ago
Sorry if it is blurry
Zina [86]
I really cant read it sorry i tried
7 0
3 years ago
Read 2 more answers
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