Answer:
melting point,boiling point ,density,refractive index.
Explanation:
Answer:
Explanation:
Hello!
In this case, according to the Charles' law as a directly proportional relationship between volume and temperature:
Thus, in order to compute the final volume, V2, we obtain the following expression:
Best regards!
Answer:
Explanation:
The ΔH for 6 moles is +84 J. So you just divide +84 J by 6 moles to get +14 J/mol (ΔΗreaction).
For this problem, let's use the approach of dimensional analysis. This technique is done by cancelling out like units that appear both on the numerator and the denominator side. As a result, this technique will let you know that your final answer conforms to what parameter is asked. In this case, the final answer should be in mL. The solution is as follows:
14.5 mg * 1 mL/25 mg = 0.58 mL
Answer:
16.95 g of LiCl and 28.4 g of Na₂SO₄.
Explanation:
- The reaction between NaCl and Li₂SO₄ is represented as:
<em>2NaCl + Li₂SO₄ → 2LiCl + Na₂SO₄.</em>
- it is clear that 2.0 moles of NaCl react with 1.0 mole of Li₂SO₄ to produce 2.0 mole of LiCl and 1.0 mole of Na₂SO₄.
- We need to calculate the no. of moles of NaCl (23.45 g) and Li₂SO₄ (56.23 g) using the relation:
<em>n = mass/molar mass,</em>
n of NaCl = mass/molar mass = (23.45 g)/(58.44 g/mol) = 0.40 mol.
n of Li₂SO₄ = mass/molar mass = (56.23 g)/(109.94 g/mol) = 0.51 mol.
- <em>Since, every 1.0 mole of Li₂SO₄ reacts with 2.0 moles of NaCl, so 0.2 mole of Li₂SO₄ will react with 0.4 mol of NaCl.</em>
<em>∴ NaCl is the limiting reactant and Li₂SO₄ is in excess.</em>
- So, the products will be LiCl (0.4 mol) and Na₂SO₄ (0.2 mol).
- Now, we can get the masses of each product using the relation:
<em>mass = n x molar mass.</em>
∴ mass of LiCl = n x molar mass = (0.4 mol)(42.394 g/mol) = 16.95 g.
∴ mass of Na₂SO₄ = n x molar mass = (0.2 mol)(142.04 g/mol) = 28.4 g.