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Zinaida [17]
3 years ago
5

How are mutations different? Provide examples.

Chemistry
2 answers:
Murljashka [212]3 years ago
4 0

Answer:

sdoawjdiowadawoi siokdwajsiokwjDIWIAdawidjaskmdnkawdjad kadakwdkawdawhdaw

Explanation: im smart

mr_godi [17]3 years ago
4 0
Mutation is a permanent, heritable change in the nucleotide sequence or the process by which such a change occurs in a gene or in a chromosome.
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Acid rain is a term that refers to a mixture of wet and dry deposited material from the atmosphere containing higher than normal
sattari [20]

Answer: Volcanic Eruptions

Explanation:

3 0
3 years ago
Atomic size of inert gases do not affect you inertness why​
Tresset [83]

Answer:

Because your body has built-in resistance to certain gases, no matter the size of the gas cloud.

That is why we are able to stay non-inert to these types of gases, like Carbon dioxide.

6 0
2 years ago
How much time would it take for 336 mg of copper to be plated at a current of 5.6 A ? Express your answer using two significant
schepotkina [342]

Answer:

1.8 × 10² s

Explanation:

Let's consider the reduction that occurs upon the electroplating of copper.

Cu²⁺(aq) + 2 e⁻ ⇒ Cu(s)

We will establish the following relationships:

  • 1 g = 1,000 mg
  • The molar mass of Cu is 63.55 g/mol
  • When 1 mole of Cu is deposited, 2 moles of electrons circulate.
  • The charge of 1 mole of electrons is 96,486 C (Faraday's constant).
  • 1 A = 1 C/s

The time  that it would take for 336 mg of copper to be plated at a current of 5.6 A is:

336mgCu \times \frac{1gCu}{1,000mgCu} \times \frac{1molCu}{63.55gCu} \times \frac{2mole^{-} }{1molCu} \times \frac{94,486C}{1mole^{-}} \times \frac{1s}{5.6C} = 1.8 \times 10^{2} s

3 0
3 years ago
What type of solid is candle wax
son4ous [18]

Hi Sara


I would say it is an amorphous solid.


Because it's gradually loses shape.


I hope that's help:)

8 0
3 years ago
Read 2 more answers
A gas has a volume of 30 L at 105 kPa. What pressure is required to compress the gas to 21 L?
Aloiza [94]

Answer: 150 kPa

Explanation:

Given that,

Original volume of gas V1 = 30L

Original pressure of gas P1 = 105 kPa

New pressure of gas P2 = ?

New volume of gas V2 = 21L

Since pressure and volume are given while temperature is constant, apply the formula for Boyle's law

P1V1 = P2V2

105 kPa x 30L = P2 x 21L

3150 kPa L = P2 x 21L

P2 = 3150 kPa L / 21 L

P2 = 150 kPa

Thus, 150 kPa of pressure is required to compress the gas

8 0
3 years ago
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