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Vladimir [108]
3 years ago
12

The layer of skin that we see when we

Chemistry
1 answer:
belka [17]3 years ago
7 0

Answer:stratum corneum

Explanation:

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The products
alexandr1967 [171]

Answer:

The products  of self-ionization of water are OH⁻ and H⁺.

Explanation:

  • The water is self ionized according to the equation:

<em>H₂O → OH⁻ + H⁺. </em>

<em></em>

The ionic product for water (Kw) = [OH⁻][H⁺] = 10⁻¹⁴.

Kw is also called "self-ionization constant" or "auto-ionization constant".

6 0
3 years ago
Can animals detect radio waves
Svetach [21]

Answer:

No.

Explanation:

No organism can detect X-rays or radio waves

4 0
2 years ago
Read 2 more answers
Who are the scientists that contributed in arranging of the periodic table?
Irina-Kira [14]
In 1869 Russian chemist DIMITRI MENDELEEV started the development of the periodic table,arranging chemical elements by atomic mass. He predicted the discovery of other elements and left spaces open in his periodic table for them. HOPE THIS HELPSS HAVE A GREAT DAY <333
7 0
3 years ago
A sample of 3.62 moles of diphosphorous trioxide is
Sindrei [870]

H_3PO_32H_3PO_32H_3PO_3P_2O_3Answer:

B

Explanation:

This question is about stoichiometry. From the balanced equation P_2O_3 + 3H_2O⇒2H_3PO_3, we see that 3 moles of water is needed to react with 1 mole of P_2O_3.

This means that, to fully react 3.62 moles of P_2O_3, we would need 3*3.62 or  10.86 moles of water. However, we only have 6.31 moles, so water is the limiting reactant.

Since 3 moles of water react with 1 mole of P_2O_3, 6.31 moles of water can fully react with 6.31÷3 or 2.1033 moles of P_2O_3.

From the balanced equation, we see that every mole of P_2O_3 reacted gets you 2 moles of 2H_3PO_3. Therefore, 2.1033 moles of P_2O_3 would give you approximately 4.21 moles of H_3PO_3.

5 0
2 years ago
Consider an amphoteric hydroxide, m(oh)2(s), where m is a generic metal. estimate the solubility of m(oh)2 in a solution buffere
Colt1911 [192]
Missing data in your question: (please check the attached photo)
from this balanced equation:
M(OH)2(s) ↔ M2+(aq) + 2OH-(aq) and when we have Ksp = 2x10^-16
∴Ksp = [M2+][OH]^2
2x10^-16 = [M2+][OH]^2
a) SO at PH = 7 
∴POH = 14-PH = 14- 7 = 7
when POH = -㏒[OH]
7= -㏒[OH]
∴[OH] = 1x10^-7 m by substitution with this value in the Ksp formula,
∴[M2+] =Ksp /[OH]^2
            = (2x10^-16)/(1x10^-7)^2
             = 0.02 M
b) at PH =10
when POH = 14- PH = 14-10 = 4 
when POH = -㏒[OH-]
            4  = -㏒[OH-]
∴[OH] = 1x10^-4 ,by substitution with this value in the Ksp formula
[M2+] = Ksp/ [OH]^2
          = 2x10^-16 / (1x10^-4)^2
          = 2x10^-8 M
c) at PH= 14 
when POH = 14-PH
                   = 14 - 14 
                   = 0
when POH = -㏒[OH]
              0 = - ㏒[OH]
∴[OH] = 1 m 
by substitution with this value in Ksp formula :
[M2+] = Ksp / [OH]^2
          = (2x10^-16) / 1^2
          = 2x10^-16 M


8 0
3 years ago
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