The illustration would be that of a double replacement reaction.
<h3>What are double replacement reactions?</h3>
They are reactions in which 2 ionic compounds exchange ions to form two new products.
Thus, in the reaction: ab + cd ----------> ad + cb
ab and cd are two ionic compounds. The b in ab is replaced by the d in cd while the d in cd itself is replaced by the b in ab. Hence, new products, ad and cd, are formed.
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Answer:
Explanation:
70% (vol/vol) means
cotnaimns 70 %(vol/vol) 70 ml of isoprapnol is there in 100 ml of Rubbing sold alcohol.
if it is 200 ml then obvouly it has the 70*2 =140 ml of isoproanol required.
Answer:
Approximately .
Explanation:
Make use of the molar mass data () to calculate the number of moles of molecules in that of :
.
Make sure that the equation for this reaction is balanced.
Coefficient of in this equation: .
Coefficient of in this equation: .
In other words, for every two moles of that this reaction consumes, two moles of would be produced.
Equivalently, for every mole of that this reaction consumes, one mole of would be produced.
Hence the ratio: .
Apply this ratio to find the number of moles of that this reaction would have produced:
.
Answer: -
D. Network
Explanation: -
Diamond is an allotrope of carbon. In diamond each carbon atom makes four bonds to other carbon atoms.
They exist in tetrahedral shape.
Diamond has strong covalent bonds. They extend in all the three dimensions
Such covalent bonds are called network covalent bonds. They require significant amounts of energy to break.
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 =10when 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 Mc) 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