<h3>
Answer:</h3>
0.10 L
<h3>
Explanation:</h3>
The concentration of glucose is given as 180 g/L
The mass of glucose is 18 g
- Concentration in g/L is calculated by dividing mass of the solute by the volume of the solution.
- When calculating molarity on the other hand, we divide number of moles of the solute by the volume of the solution.
- Concentration in g/L = Mass of solute ÷ Volume
Rearranging the formula,
Volume = Mass of the solute ÷ concentration
= 18 g ÷ 180 g/L
= 0.10 L
Therefore, volume of water is 0.10 L
Answer : The vapor pressure of the dry oxygen gas is, 736.2 torr
Explanation : Given,
Volume of sample of oxygen = 500 mL
Vapor pressure of oxygen + water = 760 torr
Vapor pressure of water = 23.8 torr
Now we have to determine the vapor pressure of the dry oxygen gas.
Vapor pressure of the dry oxygen gas = Vapor pressure of (oxygen + water) - Vapor pressure of water
Vapor pressure of the dry oxygen gas = 760 torr - 23.8 torr
Vapor pressure of the dry oxygen gas = 736.2 torr
Thus, the vapor pressure of the dry oxygen gas is, 736.2 torr
Cave rock neutralizes pollutants in groundwater.
<h3>
Answer:</h3>
0.5 moles KF
<h3>
Explanation:</h3>
Mass of the compound = 29 g KF(Potassium chloride)
We need to get the moles of KF
Moles are given by dividing the mass of a compound by its molar mass.
In this case,
Molar mass of potassium chloride = 58.0967 g/mol
Therefore;
Moles = 29 g ÷ 58.0967 g/mol
= 0.5 moles
Therefore, 29 g of KF contains 0.5 moles
Hydrogen and carbon because they are the most common element of canyon