Answer:
2.90 x 10⁻¹¹moldm⁻³
Explanation:
Given parameters:
[H⁺] = 3.5 x 10⁻⁴mol/dm³
Unknown
[OH⁻] = ?
Solution;
The ionic product of water can be used to solve this problem. It has been experimentally determined to be 1 x 10⁻¹⁴mol² dm⁻⁶
[H⁺] [OH⁻] = 1 x 10⁻¹⁴
Therefore;
[OH⁻] = = = 0.29 x 10⁻¹⁰moldm⁻³
= 2.90 x 10⁻¹¹moldm⁻³
Answer:
The correct option is
C) Trial 1 will have the same calculated empirical formula as trial 2.
Explanation:
The empirical formula is the formula of a chemical compound that states the simplest whole number ratio of each of the atoms included in the compound. It is obtained by dividing the mass of an element present in the compound by the element's molar mass to find the mole ratio of the elements. The obtained mole value for each element is then divided by the smallest number of moles obtained in the division.
By definition the composition and ratio of elements combined in a chemical compound is fixed, therefore trial 1 will have the same calculated empirical formula as trial 2.
Answer:
There is nothing here for us to help you with.
Explanation:
Cofactors:
A. Coenzyme A (CoA-SH)
B. NAD+
C. Thiamine pyrophosphate (TPP)
D. FAD
E. Lipoic acid in oxidized form
Roles:
E... Attacks and attaches to the central carbon in pyruvate.
A...Oxidizes FADH2.
C...Accepts the acetyl group from reduced lipoic acid.
D... Oxidizes the reduced form of lipoic acid.
B... Initial electron acceptor in oxidation of pyruvate.
Answer: option 2
Explanation: each nitrogen is reduced from zero to -3 and Each H is Oxidised from 0 to +1
Since oxidation no of Nitrogen in molecular N2 is always zero and for H2 also zero.
In NH3, x+3(+1)=0
X+3=0
X=-3
And and hydrogen shows +1 Oxidation no when it is combined with non metal(like N,O,cl) and with -2 with metal like Na,Ca etc.
And zero in molecular form H2.