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gayaneshka [121]
3 years ago
5

Example of a neutral mutation?

Chemistry
2 answers:
Pavlova-9 [17]3 years ago
5 0

Answer:

D.a red flower with one part colored yellow

Explanation:

edg2020

Norma-Jean [14]3 years ago
3 0

The answer is D

Explanation: A RED FLOWER WITH ONE PART COLORED YELLOW

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I need help please!Perform the following conversion. Make sure your answer has the correct number of significant digits.
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3220000

Explanation:

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3.22×10^6

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Which statement is true of diffusion?
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Molecules move from areas of high concentration to areas of low concentration.

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If a solution containing 45.101 g of mercury(II) acetate is allowed to react completely with a solution containing 12.026 g of s
AnnyKZ [126]

Answer:

14.533 grams of solid precipitate of mercury(II) dichromate will form.

Explanation:

Hg(CH_3COO)_2(aq)+Na_2Cr_2O_7(aq)\rightarrow HgCr_2O_7(s)+2CH_3COONa(aq)

Moles of mercury(II) acetate = \frac{45.101 g}{318.70 g/mol}=0.14152 mol

Moles of sodium dichromate = \frac{12.026 g}{261.97 g/mol}=0.045906 mol

According to reaction , 1 mole of sodium dichromate reacts with 1 mole of mercury(II) acetate , then 0.045906 moles of sodium dichromate will recat with :

\frac{1}{1}\times 0.045906 mol=0.045906 mol of mercury(II) acetate

This means that mercury(II) acetate is present in an excess amount and sodium dichromate is present in limiting amount.So, amount of precipitate will depend upon moles of sodium dichromate.

According to reaction , 1 mole of sodium dichromate gives 1 mole of mercury(II) dichromate , then 0.045906 moles of sodium dichromate will give :

\frac{1}{1}\times 0.045906 mol=0.045906 mol of mercury(II) dichromate

Mass of 0.045906 moles of mercury(II) dichromate:

0.045906 mol × 316.59 g/mol = 14.533 g

14.533 grams of solid precipitate of mercury(II) dichromate will form.

3 0
3 years ago
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