Answer:
1230
Explanation:
1.20×1025=1230 is your answer
Answer: New pressure inside the flask would be 148.8 kPa.
Explanation: The combined gas law equation is given by:
As the flask is a closed flask, so the volume remains constant. Temperature is constant also.
So, the relation between pressure and number of moles becomes
- Final conditions: When additional 3 puffs of air is added
Putting the values, in above equation, we get
I think it would be Bromine and Mercury, hope that helps
Answer:
800.0 mL.
Explanation:
- To solve this problem; we must mention the rule states the no. of millimoles of a substance before and after dilution is the same.
<em>(MV)before dilution of HCl = (MV)after dilution of HCl</em>
M before dilution = 12.0 M, V before dilution = 100.0 mL.
M after dilution = 1.5 M, V after dilution = ??? mL.
∵ (MV)before dilution of HCl = (MV)after dilution of HCl
∴ (12.0 M)(100.0 mL) = (1.5 M)(V after dilution of HCl)
<em>∴ V after dilution of HCl = (12.0 M)(100.0 mL)/(1`.5 M) = 800.0 mL.</em>
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.