Answer:
Given info : 500 cc of 2N Na2CO3 are mixed with 400 cc of 3N H2SO4 and volume was diluted to one litre. To find : will the resulting solution is acidic , basic or neutral ? Calculate the molarity of the dilute solution. solution : no of moles of Na2CO3 = normality/n %3D - factor x volume 2/2 x 500/1000 = 0.5 mol %D no of moles of H2SO4 = 3/2 x 400/100O = 0.6 mol %3D We see, Na2CO3 + H2S04 => Na2S04 + CO2 + H2O Here one mol of Na2C03 reacts with one mole of H2SO4. So, 0.5 mol of Na2CO3 reacts with 0.5 mol of H2SO4. so, remaining 0.1 mol of H2SO4 makes solution acidic. Now molarity of solution = remaining no of moles of H2SO4/volume of solution= 0.1/1 = %3D 0.1M
<h3>
Answer:</h3>
266.325 g
<h3>
Explanation:</h3>
We are given the balanced equation;
2NaOH + H₂SO₄ → H₂O + Na₂SO₄
We are required to determine the mass of Na₂SO₄ that will be formed.
<h3>Step 1: Determine the number of moles of NaOH</h3>
Moles = Mass ÷ molar mass
Molar mass of NaOH is 40.0 g/mol
Therefore;
Moles of NaOH = 150 g ÷ 40 g/mol
= 3.75 moles
<h3>Step 2: Determine the number of moles of sodium sulfate formed</h3>
- From the equation 2 moles of NaOH reacts with sulfuric acid to form 1 mole of sodium sulfate.
- Therefore; mole ratio of NaOH : Na₂SO₄ is 2 : 1
Thus, moles of Na₂SO₄ = Moles of NaOH ÷ 2
= 3.75 moles ÷ 2
= 1.875 moles
<h3>Step 3: Determine the mass of Na₂SO₄ produced.</h3>
we know that;
Mass = Moles × Molar mass
Molar mass of Na₂SO₄ is 142.04 g/mol
Therefore;
Mass of Na₂SO₄ = 1.875 moles × 142.04 g/mol
= 266.325 g
Thus, the mass of sodium sulfate formed 266.325 g
Answer:
5Atm
Explanation:
I just guess and it’s right
Answer: B. a chemical change
Explanation:
B ia the correct one .
Hydrogen only uses the first energy shell, which holds 2 electrons, not 8.