The normality of the H₂SO₄ that reacted with 25cc of 5 % NaOH solution is 1.1 N.
<h3>What is the molarity of 5% NaOH?</h3>
The molarity of 5% NaOH is 1.32 M
25 cc of NaOH neutralized 30cc of H₂SO₄ solution.
Equation of reaction is given below:
- 2 NaOH + H₂SO₄ ---> Na₂SO₄ + 2 H₂O
Molarity of H₂SO₄ = 1.32 x 1 x 25/(30 x 2) = 0.55 M
- Normality = Molarity × moles of H⁺ ions per mole of acid
moles of H⁺ ions per mole of H₂SO₄ = 2
Normality of H₂SO₄ = 0.55 x 2 = 1.1 N
In conclusion, the normality of an acid is determined from the molarity and the moles of H⁺ ions per mole of acid.
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Answer:
2NaOH (aq) + CaCl2 (aq) -> 2NaCl(aq) + Ca(OH)2(s)
Formula of precipitate: Ca(OH)2 <em>(s)</em>
Explanation:
First, we do the double replacement reaction to determine our chemical equation between the reactants and products. Once we have our products, with a solubility chart (I added one below) we can determine which of the products is soluble or insoluble.
In this case NaCl is soluble or aqueous (meaning it can dissolve in water) and Ca(OH)2 is insoluble (meaning that when the reactions takes place, these two will form a solid/precipitate)
Answer:
i think the answer is yess
The value of equilibrium constant is equal to the quotient of the products raised to its stoichiometric coefficient over the reaction's reactants raised to its respective stoichiometric coeff. The equation is Kc=[SO2][Cl2]/[SO2Cl2]= [1.3*10^-2][1.3*10^-2]/[2.2*10^-2-<span>1.3*10^-2]=0.0188. The final answer is Kc=0.0188.</span>
Answer:
ways it rust are
1.you put water on it
2.let it sit out side for about a month or 2
or
you put the iron in water still wait a month or 2 then
you got the full or sum rusty parts
Explanation:
ether way you need water, air ,and the main thing... IRON
