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Neko [114]
3 years ago
15

Peppered moths come in two colors, black and white. What did Kettlewell show, with regard to peppered moth populations and tree

coloring? (Recall that England at the time was heated mainly by coal in urban areas.) A) In smoggy areas, dark moths have a higher survival rate than white moths. B) White moths stand out in rural areas, and quickly disappear from those areas. C) The black moth and white moth forms were actually two different species of moths. D) Environmental conditions do not affect physical features of a species over a short time period.
Chemistry
2 answers:
Norma-Jean [14]3 years ago
5 0

Answer is: A) In smoggy areas, dark moths have a higher survival rate than white moths.

The light moths will be captured by predators more easily than the dark moths, and the population of dark moths will rise.

This is example of natural selection and adaptation.

Predators will easily capture white moths and their population will drope.

Moth populations can survive in a variety of different situations, because the environment is constantly changing and different alleles are favored.  

Vera_Pavlovna [14]3 years ago
4 0
A is the answer because the trees turned blackish from the coal so moths that were black could camouflage on those trees easier.
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2 years ago
Calculate the concentration expressed in percent, if 10 g of NaOH is diluted to 500 ml with water.
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If 10 g of NaOH is diluted to 500 ml with water then  the concentration expressed in percent is 0.5 mol/L .

Calculation ,

Given mass in gram = 10 g

Number of moles = given mass /molar mass = 10 g / 40 g/mol = 0.25mole

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Putting the value of mass and volume in equation i we get concentration expressed in percent .

C = number of moles ×100/ volume in liter = 0.25mole ×100/ 0.5 L

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1 year ago
When the following oxidation–reduction reaction in acidic solution is balanced, what is the lowest whole-number coefficient for
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Answer:

c. 8, product side

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Step 2: perform the mass balance adding H⁺ and H₂O where necessary

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Step 3: perform the electrical balance adding electrons where necessary.

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Step 4: multiply both half-reactions by numbers that secure that the number of electrons gained and lost are the same.

2 × (8 H⁺(aq) + MnO₄⁻(aq) + 5 e⁻ → Mn²⁺(aq) + 4 H₂O(l))

5 × (2 Br⁻(aq) → Br₂(l) + 2 e⁻)

Step 5: add both half-reactions side to side.

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a = Fnet / m

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