Answer:
K, the rate constant = 9.73 × 10^(-1)/s
Explanation:
r = K × [A]^x × [B]^y
r = Rate = 1.07 × 10^(-1)/s
K = Rate constant
A and B = Concentration in mol/dm^-3
A = 0.44M
B = 0.11M
x = Order of reaction with respect to A = 0
y = Order of reaction with respect to B = 1
Solving, we get
r/([A]^x × [B]^y) = K
K = 1.07 × 10^(-1)/s/(0.44^0 × 0.11^1)= 0.9727
K = 0.9727
We know that Weighted atomic mass of Element is Calculated based upon their existence of isotopes and the Relative abundance of these isotopes.
Given that the Element X is Composed of :
Isotope Relative abundance Atomic Mass
⁵⁵X 70% 55
⁵⁶X 20% 56
⁵⁷X 10% 57
Weighted Atomic Mass of Element X :
= (0.70)(55) + (0.20)(56) + (0.10)(57)
= 38.50 + 11.2 + 5.70
= 55.4
So, the Weighted Atomic Mass of Element X is 55.4
Answer:
2,2,3,3-tetrapropyloxirane
Explanation:
In this case, we have to know first the alkene that will react with the peroxyacid. So:
<u>What do we know about the unknown alkene? </u>
We know the product of the ozonolysis reaction (see figure 1). This reaction is an <u>oxidative rupture reaction</u>. Therefore, the double bond will be broken and we have to replace the carbons on each side of the double bond by oxygens. If 
is the only product we will have a symmetric molecule in this case 4,5-dipropyloct-4-ene.
<u>What is the product with the peroxyacid?</u>
This compound in the presence of alkenes will produce <u>peroxides.</u> Therefore we have to put a peroxide group in the carbons where the double bond was placed. So, we will have as product <u>2,2,3,3-tetrapropyloxirane.</u> (see figure 2)
False They can function as both. An example is Aluminium Oxide. These kind of substances are called "Amphoteric", they can behave as both acids and bases.
