a) Group 2 elements have 2 electrons on their outer shell, so they form a 2+ charge.
b) they lose 2 electrons as they are transferred to the non metal.
c)They obtain this charge as when they are made into an ionic compound the 2 electrons on the outer shell are transferred to the non metal, meaning there are 2 more protons that electrons, giving it a positive charge.
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Change in temperature affects the rate of reaction since it causes a change in the number of collisions per unit time. These collisions cause the breaking of bonds and formation of new ones giving out new products. An increase in temperature increases the rate of collisions hence increasing the rate of reaction while a decrease in temperature leads to a decrease in the rate of reaction due to the decreased number of collisions per unit time. thus the correct choice for blank A is: B. the number of collisions between molecules and for blank B: decrease.
Answer:
a) Kb = 10^-9
b) pH = 3.02
Explanation:
a) pH 5.0 titration with a 100 mL sample containing 500 mL of 0.10 M HCl, or 0.05 moles of HCl. Therefore we have the following:
[NaA] and [A-] = 0.05/0.6 = 0.083 M
Kb = Kw/Ka = 10^-14/[H+] = 10^-14/10^-5 = 10^-9
b) For the stoichiometric point in the titration, 0.100 moles of NaA have to be found in a 1.1L solution, and this is equal to:
[A-] = [H+] = (0.1 L)*(1 M)/1.1 L = 0.091 M
pKb = 10^-9
Ka = 10^-5
HA = H+ + A-
Ka = 10^-5 = ([H+]*[A-])/[HA] = [H+]^2/(0.091 - [H+])
[H+]^2 + 10^5 * [H+] - 10^-5 * 0.091 = 0
Clearing [H+]:
[H+] = 0.00095 M
pH = -log([H+]) = -log(0.00095) = 3.02
Answer:
0.9 mole of Fe(OH)3.
Explanation:
We'll begin by writing the balanced equation for the reaction. This is given below:
Fe(NO3)3 + 3NaOH —> Fe(OH)3 + 3NaNO3
Now, we can determine the moles of iron (III) hydroxide formed from the reaction as follow:
From the balanced equation above,
3 moles of NaOH reacted to produce 1 mole of Fe(OH)3.
Therefore, 2.7 moles of NaOH will react to produce = 2.7/3 = 0.9 mole of Fe(OH)3.
Therefore, 0.9 mole of Fe(OH)3 is produced from the reaction.
First, we write the reaction for CH3OH combustion
CH3OH+3/2O2--->CO2+2H2O
for 1 mole of methanol, we get 1 mole of CO2, therefore for 5,25 moles of methanol we will get 5,25 moles of CO2
