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melomori [17]
2 years ago
11

How does the presence of a nucleus provide a method of basic cell classification?

Chemistry
1 answer:
Zanzabum2 years ago
7 0

Answer:

Sample Respond: The nucleus maintains the integrity of genes and controls the activities of the cell by regulating gene expression—the nucleus is, therefore, the control center of the cell.

good luck

Explanation:

I got 100 mark as brainiest

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B, Frost Wedging. Have a good rest of your day!
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2 years ago
A sample of gas occupies a volume of 61.5 mL . As it expands, it does 130.1 J of work on its surroundings at a constant pressure
Lesechka [4]

Answer:

the final volume of the gas is V_2 = 1311.5 mL

Explanation:

Given that:

a sample gas has an initial volume of 61.5 mL

The workdone = 130.1 J

Pressure = 783 torr

The objective is to determine the final volume of the gas.

Since the process does 130.1 J of work on its surroundings at a constant pressure of 783 Torr. Then, the pressure is external.

Converting the external pressure to atm ; we have

External Pressure P_{ext}:

P_{ext} = 783 \ torr \times \dfrac{1 \ atm}{760 \ torr}

P_{ext} = 1.03 \ atm

The workdone W = P_{ext}V

The change in volume ΔV= \dfrac{W}{P_{ext}}

ΔV = \dfrac{130.1 \ J  \times \dfrac{1 \ L  \ atm}{ 101.325 \ J}  }{1.03 \ atm }

ΔV = \dfrac{1.28398717 }{1.03  }

ΔV = 1.25 L

ΔV = 1250 mL

Recall that the initial  volume = 61.5 mL

The change in volume V is \Delta V = V_2 -V_1

-  V_2= -  \Delta V  -V_1

multiply through by (-), we have:

V_2=   \Delta V+V_1

V_2 =  1250 mL + 61.5 mL

V_2 = 1311.5 mL

∴ the final volume of the gas is V_2 = 1311.5 mL

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3 years ago
Methyl violet is an indicator that changes color over a range from pH=0.0 to pH=1.6. What is Ka of methyl violet?
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If 500.0 mL of 0.450 M sodium phosphate is reacted with an excess of iron (II) nitrate solution, how many grams of iron (II) pho
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Answer:

If 500.0 mL of 0.450 M sodium phosphate is reacted with an excess of iron (II) nitrate solution, how many grams of iron (II) phosphate are produced?

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Explanation:

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A 367.8 g sample of potassium chlorate was decomposed according to the following equation:
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The answer is A i did this before hope this helps
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