From Boyle's law the volume of a fixed mass of a gas is inversely proportional to temperature at constant absolute temperature.
Thus, Vα1/P
= V = k/P where k is a constant
P1V1=P2V2
Therefore; V2 = P1V1/P2
= ( 6.0 ×10^-5 × 775) /622
= 7.476 × 10^-5 L
Hence, the new volume of the air mass is 7.476 × 10^-5 L
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.
Answer:
are present in solution.
Explanation:
Molarity of the solution = 0.210 M
Volume of the solution = 65.5 ml = 0.0655 L
Moles of aluminum iodide= n
n = 0.013755 moles of aluminum iodide
1 mole of aluminum iodide contains 3 moles of iodide ions:
Then 0.013755 moles of aluminum iodide will contain:
of iodide ions
Number of iodide ions in 0.041265 moles:
are present in solution.
Answer:- 1840 g.
Solution:- We have been given with 3.35 moles of 
and asked to calculate it's mass.
To convert the moles to grams we multiply the moles by the molar mass of the compound. Molar mass of the compound is the sum of atomic masses of all the atoms present in it.
molar mass of = atomic mass of Hg + 2(atomic mass of I) + 6(atomic mass of O)
= 200.59+2(126.90)+6(16.00)
= 200.59+253.80+96.00
= 550.39 gram per mol
Let's multiply the given moles by the molar mass:
= 1843.8 g
Since, there are three sig figs in the given moles of compound, we need to round the calculated my to three sig figs also. So, on rounding off to three sig figs the mass becomes 1840 g.
Answer:
[C] carbon solid
Explanation:
Pure solids and liquids are never included in the equilibrium constant expression because they do not affect the reactant amount at equilibrium in the reaction, thus since your equation has [C] as solid it will not be part of the equlibrium equation.
