Answer:
a. 9.2
b. 4.4
c. 6.3
Explanation:
In order to calculate the pH of each solution, we will use the definition of pH.
pH = -log [H⁺]
(a) [H⁺] = 5.4 × 10⁻¹⁰ M
pH = -log [H⁺] = -log 5.4 × 10⁻¹⁰ = 9.2
Since pH > 7, the solution is basic.
(b) [H⁺] = 4.3 × 10⁻⁵ M
pH = -log [H⁺] = -log 4.3 × 10⁻⁵ = 4.4
Since pH < 7, the solution is acid.
(c) [H⁺] = 5.4 × 10⁻⁷ M
pH = -log [H⁺] = -log 5.4 × 10⁻⁷ = 6.3
Since pH < 7, the solution is acid.
Answer:
It depends on their melting and/or their boiling points, because the heat provides the particles with kinetic energy to break the electrosatic bonds in the substances, which can differ in strength
Explanation:
Answer:
Energy
Explanation:
A sugar group would be used for carbohydrates or nucleic acids. An adenosine group would be used in ATP formation if I recall correct. And disaccharides are just two monosaccharides linked together, so that would also be for carbohydrates. Therefore, energy is the answer.
Answer:
Coefficient of 
Coefficient of
=8
Explanation:
We are given that a reaction in which
reacts with 
We have to find the coefficient of each reactants in balanced reaction

Coefficient is defined the constant value multiplied with a reactant in a reaction.
Coefficient of
=3
Coefficient of 
Coefficient of 
Coefficient of 
Coefficient of 
Coefficient of KOH=2
Hence, Coefficient of
and coefficient of 
Molar mass of NH_3



We know.
No of moles=Given mass/Molar mass


Now
Lets write the balanced equation

- There is 2moles of Ammonia
- 3moles of H_2
- 1mole of N_2
Now

For Hydrogen



For Ammonia



For Nitrogen

