Sodium carbonate is used to neutralised sulfuric acid, H₂SO₄. Sodium carbonate is the salt of stron base (NaOH) and weak acid (H₂CO₃). The balanced chemical reaction for neutralization is as follows:
Na₂CO₃ + H₂SO₄→ Na₂SO₄ + H₂CO₃
From balanced chemical equation, it is clear that one mole of Na₂CO₃ is required to neutralize one mole of H₂SO₄. Molar mass of Na₂CO₃= 106 g/mol=0.106 kg/mol and molar mass of H₂SO₄= 98 g/mol=0.098 kg/mol.
To neutralize 0.098 kg of H₂SO₄ amount of Na₂CO₃ required is 0.106 kg, so, to neutralize 5.04×10³ kg of H₂SO₄, Na₂CO₃ required is=
kg= 5.451 X 10³ kg.
Answer: 0.86g/mL
Explanation:
Mass of empty cylinder = 23.731g
Mass of cylinder + liquid = 26.414g
Mass of the liquid = 26.414 — 23.731
= 2.683g
Volume of the liquid = 3.12mL
Density = Mass / volume
Density = 2.683g / 3.12mL
Density = 0.86g/mL
Triprotic acid is a class of Arrhenius acids that are capable of donating three protons per molecule when dissociating in aqueous solutions. So the chemical reaction as described in the question, at the third equivalence point, can be show as: H3R + 3NaOH ⇒ Na3R + 3H2O, where R is the counter ion of the triprotic acid. Therefore, the ratio between the reacted acid and base at the third equivalence point is 1:3.
The moles of NaOH is 0.106M*0.0352L = 0.003731 mole. So the moles of H3R is 0.003731mole/3=0.001244mole.
The molar mass of the acid can be calculated: 0.307g/0.001244mole=247 g/mol.
V
1
/T
1
=V
2
/T
2
(900.0 mL) / (300.0 K) = (x) / (405.0 K); x = 1215 mL.
Change the 900 to 800, and the 300 to 27, then change the 405 to 132. And solve
Answer:
1) 1,... 2
2) 18
3) n= 3 and I=1
Explanation:
1) when l= 0, its an s-sub-level, and only 1 orbital is possible which can carry only 2-electrons
2) the maximum number of electron is given by 2n^2= 2×3^2= 18
3) in 3p, the coefficient of p is the value of n= 3 and l-value of P is 1
