Answer:
The products of self-ionization of water are OH⁻ and H⁺.
Explanation:
- The water is self ionized according to the equation:
<em>H₂O → OH⁻ + H⁺.
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The ionic product for water (Kw) = [OH⁻][H⁺] = 10⁻¹⁴.
Kw is also called "self-ionization constant" or "auto-ionization constant".
The balanced equation between NaOH and H₂SO₄ is as follows
2NaOH + H₂SO₄ ---> Na₂SO₄ + 2H₂O
stoichiometry of NaOH to H₂SO₄ is 2:1
number of moles of NaOH moles reacted = molarity of NaOH x volume
number of NaOH moles = 0.08964 mol/L x 27.86 x 10⁻³ L = 2.497 x 10⁻³ mol
according to molar ratio of 2:1
2 mol of NaOH reacts with 1 mol of H₂SO₄
therefore 2.497 x 10⁻³ mol of NaOH reacts with - 1/2 x 2.497 x 10⁻³ mol of H₂SO₄
number of moles of H₂SO₄ reacted - 1.249 x 10⁻³ mol
Number of H₂SO₄ moles in 34.53 mL - 1.249 x 10⁻³ mol
number of H₂SO₄ moles in 1000 mL - 1.249 x 10⁻³ mol / 34.53 x 10⁻³ L = 0.03617 mol
molarity of H₂SO₄ is 0.03617 M
<h3>
Answer:</h3>
1 x 10^13 stadiums
<h3>
Explanation:</h3>
From the question;
1 x 10^5 people can fill 1 stadium
We are given, 1 x 10^18 atoms of iron
We are required to determine the number of stadiums that 1 x 10^18 atoms of iron would occupy.
We are going to assume that a stadium would occupy a number of atoms equivalent to the number of people.
Therefore;
One stadium = 1 x 10^5 atoms
Then, to find the number of stadiums that will be occupied by 1 x 10^18 atoms;
No. of stadiums = Total number of atoms ÷ Atoms in a single stadium
= 1 x 10^18 atoms ÷ 1 x 10^5 atoms
= 1 x 10^13 stadiums
Therefore, 1 x 10^18 atoms of iron would occupy 1 x 10^13 stadiums
Create more of their species
(NH4)3PO4 :
N = 14 * 3 = 42
H = 1 * 12 = 12
P = 31 * 1 = 31
O = 16 * 4 = 64
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42+12+31+64 = 149 g / mol
Hope this helps!.
