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4 years ago

Which layer of the epidermis is most actively involved in replacing exfoliated skin cells?

Chemistry

2 answers:

olya-2409 [2.1K]4 years ago

8 0

Answer: Basal layer

Explanation:

Epidermis of thick skin has five layers. Different layers of the skin serves different function. The outermost layer is basal which is generally removed by exfoliation.

There are several layers of keratinocytes at the surface of the epidermis. The topmost layer sheds usually because it is dead. It is replaced by stem cell that tends to divide in the basal layer.

iris [78.8K]4 years ago

5 0

The Basal layer is the answer

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Select the correct statement about isotopes. A. All the isotopes of an element are radioactive. B. Isotopes of the same element

frutty [35]

Answer: Option (B) is the correct answer.

Explanation:

Isotopes are the substances or species that contain same number of protons but different number of neutrons.

As it is known that atomic number means the total number of protons present in an atom. Hence, this also means that in isotopes atomic number of the species is same.

For example, isotopes of hydrogen are $^{1}_{1}H$ , $^{2}_{1}H$ , and $^{3}_{1}H$ .

Isotopes of only heavier elements are radioactive in nature as they have large difference in the number of protons and neutrons.

Thus, we can conclude that the statement isotopes of the same element have the same atomic number but differ in their atomic masses, is correct.

7 0

3 years ago

Type Calculations. Given the balanced equation: 2 Al + 3 H2SO4---> Al2(SO4)3 + 3 H2 Molar mass (g/mol): Al=26.98; H2SO4=98.08

Oxana [17]

Answer:

A. 1.88 mol H₂

B. 182 g Al₂(SO₄)₃

C. 54.8%

Explanation:

2 Al + 3 H₂SO₄ ⇒ Al₂(SO₄)₃ + 3 H₂

A. Convert grams of Al to moles. The molar mass is 26.98 g/mol.

(33.8 g)/(26.98 g/mol) = 1.253 mol Al

Use stoichiometry to convert moles of Al to moles of H₂. Looking at the equation, you can see that for every 2 mol of Al consumed, 3 moles of H₂ is produced. Use this relationship.

(1.253 mol Al) × (3 mol H₂)/(2 mol Al) = 1.879 mol H₂

You will produce 1.88 mol of H₂ gas.

B. Again, use stoichiometry. For every 3 moles of H₂SO₄ consumed, 1 mole of Al₂(SO₄)₃ is produced.

(1.60 mol H₂SO₄) × (1 mol Al₂(SO₄)₃/3 mol H₂SO₄) = 0.533 mol Al₂(SO₄)₃

Convert moles of Al₂(SO₄)₃ to grams. The molar mass is 342.15 g/mol.

(0.533 mol) × (342.15 g/mol) = 182.48 g Al₂(SO₄)₃

You will produce 182 g of Al₂(SO₄)₃.

C. Calculate percent yield by dividing the actual yield by the theoretical yield. Multiply by 100%.

(100.0/182.48) × 100% = 54.8%

The percent yield is 54.8%.

7 0

3 years ago

Calculate the pka of hypochlorous acid. The ph of a 0.015 m solution of hypochlorous acid has a ph of 4.64.

o-na [289]

Answer:

pKa = 7.46

Explanation:

1) Data:

a) Hypochlorous acid = HClO

b) [HClO} = 0.015

c) pH = 4.64

d) pKa = ?

2) Strategy:

With the pH calculate [H₃O⁺], then use the equilibrium equation to calculate the equilibrium constant, Ka, and finally calculate pKa from the definition.

3) Solution:

a) pH

pH = - log [H₃O⁺]

4.64 = - log [H₃O⁺]

$[H_3O^+]= 10^{-4.64} = 2.29.10^{-5}$

b) Equilibrium equation: HClO (aq) ⇄ ClO⁻ (aq) + H₃O⁺ (aq)

c) Equilibrium constant: Ka = [ClO⁻] [H₃O⁺] / [HClO]

d) From the stoichiometry: [CLO⁻] = [H₃O⁺] = 2.29 × 10 ⁻⁵ M

e) By substitution: Ka = (2.29 × 10 ⁻⁵ M)² / 0.015M = 3.50 × 10⁻⁸ M

f) By definition: pKa = - log Ka = - log (3.50 × 10 ⁻⁸) = 7.46

5 0

4 years ago

Colloids are true solutions. ____ True ____ False

Reptile [31]

This is false, they have intermediate properties between suspension and true solution

7 0

4 years ago

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Describe two observations you might make when a chemical change occurs

Sav [38]

The first one is color change
Second is the precipitate present

7 0

3 years ago