1 mole of h3bo3...........6.023*10²³ each h and B and 0 so we will have 3hydrogen+ 1 B+3 oxygen = 7*6.023*10²³ atoms
1 mole .......7*6.023*10²³atoms
4 moles ........x atoms
x=4*7*6.023*10²³.
Answer: I'm sure it is c
Explanation: that's my answer
Atomic number
atomic mass
group|family
periods
and element symbols
Answer:
The mass of water to be added is 2 pounds
Explanation:
The given parameters are;
The mass of the given solution = 2 pounds
The concentration of the given solution = 30%
The desired concentration of the solution = 15%
The mass, m of the acetic acid in the given solution = 30% × 2 pounds
m = 30/100 × 2 pounds = 0.6 pounds
To make a 15% acetic acid solution of acetic acid, the mass X of the required volume, is given as follows;
15% of X = 0.6 pounds
15/100 × X = 3/20 × 0.6 pounds
∴ The mass of the solution required X = 0.6 × 20/3 = 4 pounds
The mass of the solution that will contain 0.6 pounds of acetic acid giving a 15% acetic acid solution is 4 pounds
Therefore, the mass of water to be added to the original solution to make the a 15% acetic acid solution is 2 pounds.
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.
