The chemicals are reactive with one another
5.451 X 10³ kg of sodium carbonate must be added to neutralize 5.04×103 kg of sulfuric acid solution.
<u>Explanation</u>:
- Sodium carbonate is used to neutralized sulfuric acid, H₂SO₄. Sodium carbonate is the salt of a strong base (NaOH) and weak acid (H₂CO₃). The balanced chemical reaction for neutralization is as follows:
Na₂CO₃ + H₂SO₄ ----> Na₂SO₄ + H₂CO₃
- From a 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 = 5.451 X 10³ kg.
For example we are going to use this unbalanced chemical reaction:
H₂ + O₂ → H₂O.
First, calculate number of atoms (hydrogens and oxygens) on left and right. There is two oxygen and two hydrogen on left and two hydrogen and one oxygen on right.
You can not change molecular formula of compound, only you can put coefficient in fron of compound to balance reaction.
Put 2 in front water to balance oxygen (now you have two oxygens on left and right). But now you have four hydrogens on right, so you must put 2 in fron hydrogen on the left.
2H₂ + O₂ → 2H₂O.
Answer:
18.3 kilopascals
Explanation:
We are given that the volume of this container is 0.0372 meters^3, that the mass of water is 4.65 grams, and that the temperature of this water vapor ( over time ) is 368 degrees Kelvins. This is a problem where the ideal gas law is an " ideal " application.
_______________________________________________________
First calculate the number of moles present in the water ( H2O ). Water has a mass of 18, so it should be that n, in the ideal gas law - PV = nRT, is equal to 4 / 18. It is the amount of the substance.
We now have enough information to solve for P in PV = nRT,
P( 0.0372 ) = 4 / 18( 8.314 )( 368 ),
P ≈ 18,276.9
Pressure ≈ 18.3 kilopascals
<u><em>Hope that helps!</em></u>