A solution is prepared by dissolving sugar in water. The solution is 25.0%, by mass, sugar. How many grams of WATER are in 472 g
rams of this solution?
1 answer:
If the sugar makes up 25% of the solution, that means the water makes up 75%, or 3/4, of the solution.
Set up a proportion to solve:
![\frac{3}{4} = \frac{x}{472}](https://tex.z-dn.net/?f=%20%5Cfrac%7B3%7D%7B4%7D%20%3D%20%5Cfrac%7Bx%7D%7B472%7D)
Cross multiply:
![4x = 1416](https://tex.z-dn.net/?f=4x%20%3D%201416)
![x = 354](https://tex.z-dn.net/?f=x%20%3D%20354)
Therefore, there are 354 grams of water in the solution.
You might be interested in
The correct answer is heterogeneous.
The balanced chemical equation between HCl and
is:
![2HCl (aq) + Ba(OH)_{2}(aq) -->BaCl_{2}(aq) + 2 H_{2}O(l)](https://tex.z-dn.net/?f=%202HCl%20%28aq%29%20%2B%20Ba%28OH%29_%7B2%7D%28aq%29%20--%3EBaCl_%7B2%7D%28aq%29%20%2B%202%20H_%7B2%7DO%28l%29%20%20)
Moles of
= ![1.00 g Ba(OH)_{2} * \frac{1 mol Ba(OH)_{2}}{171.3 g Ba(OH)_{2}} = 0.00584 mol Ba(OH)_{2}](https://tex.z-dn.net/?f=%201.00%20g%20Ba%28OH%29_%7B2%7D%20%2A%20%5Cfrac%7B1%20mol%20Ba%28OH%29_%7B2%7D%7D%7B171.3%20g%20Ba%28OH%29_%7B2%7D%7D%20%3D%200.00584%20mol%20Ba%28OH%29_%7B2%7D%20%20)
Moles of HCl required to neutralize
:
![0.00584 mol Ba(OH)_{2} * \frac{ 2 mol HCl}{1 mol Ba(OH)_{2}} = 0.01168 mol HCl](https://tex.z-dn.net/?f=%200.00584%20mol%20Ba%28OH%29_%7B2%7D%20%2A%20%5Cfrac%7B%202%20mol%20HCl%7D%7B1%20mol%20Ba%28OH%29_%7B2%7D%7D%20%3D%20%20%200.01168%20mol%20HCl%20)
Calculating the volume of HCl from moles and molarity:
![0.01168 mol HCl * \frac{1 L}{0.100 mol} * \frac{1000 mL}{1 L} = 116.8 mL](https://tex.z-dn.net/?f=%200.01168%20mol%20HCl%20%2A%20%5Cfrac%7B1%20L%7D%7B0.100%20mol%7D%20%2A%20%5Cfrac%7B1000%20mL%7D%7B1%20L%7D%20%3D%20116.8%20mL%20)
An electron
proton=positive charge
electron=negative charge
Answer:
Letter C...........................
Explanation:
Letter C