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.
A polyatomic ion is a group of elements that has a charge that is not 0. for example the charge of P04 is -3
Answer:
MgCl2-> Mg+Cl
mass 5.00g Mass=1.263g
RFM Mg=24+35.5 ×2= 95 RFM= 24
moles= 5.00÷ 95= 0.0526 Moles=0.0526
Explanation:
mas of Mg= 1.263 grams
Answer:
It will feel hot.
Explanation:
The part of the metal that has been sitting in the sun will heat up, and because metal is a good conductor,<em> the part sitting in the sun will heat up the part sitting in the dark</em>.
We say metal is a good conductor because it has<em> high thermal conductivity</em>, which mean heat will flow from the hot section of the bar to the cold section of the bat.
Explanation:
An ionic equation will be the one in which all the participating species will be present as ions.
The given reaction will be as follows.

Balancing this equation by multiplying
by 2 and
by 3 on reactant side. Whereas multiply KBr by 6 on product side.

Hence, the net ionic equation will be as follows.

As both
and
are spectator ions. Hence, the net ionic equation will be as follows.