The naturally occurring isotopes of Li are Li-6 of mass 6.015121 amu and Li-7 of mass 7.016003 amu. The atomic mass of Li is 6.9409 amu, the percent abundance can be calculated using the following relation.
Atomic mass=m(Li-6 )×%(Li-6 )+m(Li-7 )×%(Li-7 )
Let the percent abundance of Li-6 be X thus, that of Li-7 will be 1-X, putting the values,
Or,
Or,
X=0.075
Thus, 
Thus, percent abundance of Li-6 is 0.075 or 7.5 % and that of Li-7 is 0.925 or 92.5%.
Answer:
0.7μM = 0.6 μM = 0.5 μM > 0.4 μM > 0.3 μM > 0.2 μM
Explanation:
An enzyme solution is saturated when all the active sites of the enzyme molecule are full. When an enzyme solution is saturated, the reaction is occurring at the maximum rate.
From the given information, an enzyme concentration of 1.0 μM Y can convert a maximum of 0.5 μM AB to the products A and B per second means that a 1.0 M Y solution is saturated when an AB concentration of 0.5 M or greater is present.
The addition of more substrate to a solution that contains the enzyme required for its catalysis will generally increase the rate of the reaction. However, if the enzyme is saturated with substrate, the addition of more substrate will have no effect on the rate of reaction.
<em>Therefore the reaction rates at substrate concentrations of 0.7μM, 0.6 μM, and 0.5 μM are equal. But the reaction rate at substrate concentrations of 0.2 μM is lower than at 0.3 μM, 0.3 μM is lower than 0.4 μM and 0.4 μM is lower than 0.5 μM, 0.6 μM and 0.7 μM.</em>
Answer:
Double=4 and triple=6
Explanation:
This is because double bonds are two pairs of electrons are shared between atoms and triple bonds are three pairs, and one pair of electrons is 2, so 2 x 2=4 and 2 x 3=6.
Answer: Sulphur and Oxygen
Explanation: Sulphate ion is 
. It is polyatomic ion which is formed by the combination of sulphur and oxygen elements in the ratio of 1:4.
It is negatively charged species and thus is named by the name _ate at the end.
It usually exists with positively charged species to get stable. eg: ferrous sulphate (
)
