Answer:
2
Explanation:
First, find the hydronium ion concentration of the solution with a pH of 4.
[H₃O⁺] = 10^-pH
[H₃O⁺] = 10⁻⁴
[H₃O⁺] = 1 × 10⁻⁴
Next, multiple the hydronium ion concentration by 100 to find the hydronium ion concentration of the new solution.
[H₃O⁺] = 1.0 × 10⁻⁴ × 100 = 0.01
Lastly, find the pH.
pH = -log [H₃O⁺]
pH = -log (0.01)
pH = 2
The pH of a solution that has a hydronium ion concentration 100 times greater than a solution with a pH of 4 is 2.
Hope this helps.
The calculation is (Measurement in m³ x 1000³)
Therefore
Measurement in m³ x 1 000 000
Its not a fraction as there are 1000mm in a m, so to convert from m to mm you must multiply
Answer:
(A) is 0.0773 mol B2H6
(C) is 2.79 x 10^23 H atoms
Explanation:
Questions (A) and (B) are the same.
2.14 g B2H6 x (1 mol B2H6/27.668g B2H6) = 0.0773 mol B2H6 (A)
<u>27.668 is the molar mass of B2H6 calculated from the period table: </u>
(2 x 10.81) + (6 x 1.008) = 27.668
1.008 is the mass of H and 10.81 is the mass of B
(C)
0.0773 mol B2H6 x (6 mol H/ 1 mol B2H6) x (6.022 x 10^23 H atoms/1 mol H)
= 2.79 x 10^23 hydrogen atoms
Further Explanation:
- For every 1 mol of B2H6, there are 6 moles of H (indicated by the subscript)
- 6.022 x 10^23 is Avogrado's number and it equals to 1 mol of anything
- Avogrado's number can be in units of atoms, molecules, or particles
Answer:- 537 kJ of heat is released.
Solution:- For the given equation, 
is -657 kJ and the coefficient of 
in the balanced equation is 2. It means 657 kJ of heat is released when 2 moles of chlorine are used. We need to calculate the heat released when 116 g of are used.
Grams of chlorine are converted to moles and then multiplied by the value and divided by the coefficient of chlorine and the set could be shown using dimensional analysis as:
= 537.46 kJ
If we use the correct sig figs then it needs to be round off to three sig figs as the given grams of chlorine has only three sig figs. So, 537 kJ of heat is released.
I believe there would still be 3 protons in Lithium-8 because its usually the neutrons that are affected.
