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
PH + pOH = 14
11.8 + pOH = 14
pOH = 14 - 11.8
pOH = 2.2
[OH-] = 10 ^- pOH
[OH-] = 10 ^- 2.2
[OH-] = <span>6.33 x 10^-3 M
</span>
Answer B
hope this helps!
Answer:
0.55 atm
Explanation:
First of all, we need to calculate the number of moles corresponding to 1.00 g of carbon dioxide. This is given by
where
m = 1.00 g is the mass of the gas
Mm = 44.0 g/mol is the molar mass of the gas
Substituting,
Now we can find the pressure of the gas by using the ideal gas law:
where
p is the gas pressure
V = 1.00 L is the volume
n = 0.0227 mol is the number of moles
R = 0.082 L/(atm K mol) is the gas constant
T = 25.0 C + 273 = 298 K is the temperature of the gas
Solving the formula for p, we find
220 grams of sugar would be in 2 liters of orange juice
This might be right. Im not quite sure. This is what my 5th grade science teacher told me. 'Look at the 2 LR's and add them together. Then look at the total amount which is 32. When you add the 2 LR's you get 17. So subtract 17 from 32 and you get 15. So: C:15 is your answer." (LR's stands for liquid reactants)
10+17=17
Total amount is 32
32-17=15
15 is you mass
Hope this Helps
