Answer:
D. 40 % increase
Step-by-step explanation:
r = k[A]²/[B]
The rate is inversely proportional to [B]. If [B] is doubled, the rate is halved.
We must double this rate to get back to the original.
The rate is directly proportional to [A]².
2 = [A]₂/[A]₁² Take the square root of each side
√2 = [A]₂/[A]₁ Multiply each side by [A]₁
[A]₂ = √2[A]₁
[A]₂ = 1.41[A]₁
We must increase [A] by 41 %.
Answer:
The water lost is 36% of the total mass of the hydrate
Explanation:
<u>Step 1:</u> Data given
Molar mass of CuSO4*5H2O = 250 g/mol
Molar mass of CuSO4 = 160 g/mol
<u>Step 2:</u> Calculate mass of water lost
Mass of water lost = 250 - 160 = 90 grams
<u>Step 3:</u> Calculate % water
% water = (mass water / total mass of hydrate)*100 %
% water = (90 grams / 250 grams )*100% = 36 %
We can control this by the following equation
The hydrate has 5 moles of H2O
5*18. = 90 grams
(90/250)*100% = 36%
(160/250)*100% = 64 %
The water lost is 36% of the total mass of the hydrate
Answer:
Group 7A
Explanation:
The group 7A elements consists of the most reactive non-metals on the periodic table.
This group is known as the group of halogens. They consist of element fluorine, chlorine, bromine, iodine and astatine.
- The elements in this group have the highest electronegativity values.
- They have 7 valence electrons and requires just one electron to complete their octets.
- This way, they are highly reactive in their search for that single electron.