Answer:
We have to take 37.5 mL of a 0.400 M solution
Explanation:
Step 1: Data given
Stock volume = 100 mL = 0.100L
Stock concentration 0.400 M
Volume of solution he wants to make = 100 mL = 0.100L
Concentration of solution he wants to make = 0.150 M
Step 2: Calculate the volume of 0.400 M CuSO4 needed
C1*V1 = C2*V2
⇒with C1 = the stock concentration = 0.400M
⇒with V1 = the volume of the stock = TO BE DETERMINED
⇒with C2 = the concentration of the solution he wants to make = 0.150 M
⇒with V2 = the volume of the solution made = 0.100 L
0.400 M * V1 = 0.150M * 0.100L
V1 = (0.150M*0.100L) / 0.400 M
V1 = 0.0375 L = 37.5 mL
We have to take 37.5 mL of a 0.400 M solution
The density of human fat is
. The mass of the pure fat is 11.1 lbs.
First convert mass from lb to g as follows:
1 lb=453.592 g
Thus,

Density of a substance is defined as mass per unit volume, thus volume can be calculated as:

Putting the values,

Therefore, volume gained by person will be
.
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Initially when we dissolve a solute , rate of dissolution is far exceeding the rate of deposition. But as the concentration of solution increases , the rate of deposition also increases and a situation comes when rate of dissolution becomes equal to rate of deposition that situation is called dynamic equilibrium.
Saturated solution:
It is the solution in which dissolved solute is in dynamic equilibrium with undissolved solute, if we dissolve more solute into it it will not dissolve.
Unsaturated solution:
This solution contains less amount of solute than the equilibrium amount of it. If we dissolve more solute into it , it will dissolve.
Supersaturated solution:
This solution contains more amount of solute than its equilibrium concentration. These solution are unstable.
Answer:
70.0°C
Explanation:
We are given;
- Amount of heat generated by propane as 104.6 kJ or 104600 Joules
- Mass of water is 500 g
- Initial temperature as 20.0 ° C
We are required to determine the final temperature of water;
Taking the initial temperature is x°C
We know that the specific heat of water is 4.18 J/g°C
Quantity of heat = Mass × specific heat × change in temperature
In this case;
Change in temp =(x-20)° C
Therefore;
104600 J = 500 g × 4.18 J/g°C × (x-20)
104600 J = 2090x -41800
146400 = 2090 x
x = 70.0479
=70.0 °C
Thus, the final temperature of water is 70.0°C