The mass percent of sulfurous acid in the new solution : 38.9%
<h3>Further explanation</h3>
<em>In a container you have 800 g of a 35% by mass solution of sulfurous acid, from which 80 ml of water evaporates. What is the mass percent of sulfurous acid in the new solution? data: density of water is 1g / ml.</em>
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solution 1
composition :


solution 2(new solution)
composition :

- Total mass of new solution after water evaporated

- %mass of acid in a new solution

Answer:
V = 34430 mL
Explanation:
Given data:
Volume in mL = ?
Number of moles of gas = 2.00 mol
Temperature = 36°C (36+273= 309K)
Pressure of gas = 1120 torr
Solution:
Formula:
PV = nRT
V = nRT/P
V = 2.00 mol ×62.4 torr • L/mol · K × 309K / 1120 torr
V = 38563.2 torr • L / 1120 torr
V = 34.43 L
L to mL
34.43 L ×1000 mL / 1 L
34430 mL
Answer:
Explanation:
2 Na+ 2 H2O → 2 NaOH + 1 H2
Arranged in groups and periods.
Group 1 in the same place.