Answer:
a) 2 (H+) ions
b) 1 (SO3²-) ions
c) 1.36 × 10^-22 grams.
Explanation:
According to this question, sulfurous acid has a chemical formula; H2SO3. It is made up of hydrogen and sulfite ion. Hydrogen ion (H+) is the cation while sulfite ion (SO32-) is the anion.
Based on the chemical formula, there are 2 moles of hydrogen ions that reacts with 1 mole of sulfite ion as follows:
2H+ + SO3²- → H2SO3
Hence;
- there are 2 hydrogen ions (2H+) present in H2SO3.
- there is 1 sulfite ion (SO3²-) present in H2SO3.
c) The mass of one formula unit of H2SO3 is calculated thus:
= 1.008 (2) + 32.065 + 15.999(3)
= 2.016 + 32.065 + 47.997
= 82.08 a.m.u
Since, 1 gram is = 6.02 x 10^23 a.m.u
82.08 a.m.u = 82.08/6.02 × 10^-23
= 13.6 × 10^-23
= 1.36 × 10^-22 grams.
Answer:
7.32g of HNO3 are required.
Explanation:
1st) From the balanced reaction we know that 2 moles of HNO3 react with 1 mole of Ca(OH)2 to produce 2 moles of H2O and 1 mole of Ca(NO3)2.
From this, we find that the relation between HNO3 and Ca(OH)2 is that 2 moles of HNO3 react with 1 mole of Ca(OH)2.
2nd) This is the order of the relations that we have to use in the equation to calculate the grams of nitric acid:
• starting with the 4.30 grams of Ca(OH)2.
,
• using the molar mass of Ca(OH)2 (74g/mol).
,
• relation of the 2 moles of HNO3 that react with 1 mole of Ca(OH)2 .
,
• using the molar mass of HNO3 (63.02g/mol).
So, 7.32g of HNO3 are required.
Original molarity was 1.7 moles of NaCl
Final molarity was 0.36 moles of NaCl
Given Information:
Original (concentrated) solution: 25 g NaCl in a 250 mL solution, solve for molarity
Final (diluted) solution: More water is added to make the new total volume 1.2 liters, solve for the new molarity
1. Solve for the molarity of the original (concentrated) solution.
Molarity (M) = moles of solute (mol) / liters of solution (L)
Convert the given information to the appropriate units before plugging in and solving for molarity.
Molarity (M) = 0.43 mol NaCl solute / 0.250 L solution = 1.7 M NaCl (original solution)
2. Solve for the molarity of the final (diluted) solution.
Remember that the amount of solute remains constant in a dilution problem; it is just the total volume of the solution that changes due to the addition of solvent.
Molarity (M) = 0.43 mol NaCl solute / 1.2 L solution
Molarity (M) of the final solution = 0.36 M NaCl
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Answer:
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