Answer:
The minimum molarity of acetic acid in vinegar according to given standards is 0.6660 mol/L.
Explanation:
4% acetic acid by mass means that 4 gram of acetic acid in 100 g solution.
Given that density of the vinegar is same is that of water = 1 g/mL
Mass of the vinegar solution = 100 g
Volume of the vinegar solution = V


V = 100 mL = 0.1 L
Moles of acetic acid =


The minimum molarity of acetic acid in vinegar according to given standards is 0.6660 mol/L.
Answer : The correct option is, (3) change states of matter.
Explanation :
Latent heat : It is defined as the heat required to convert the solid into liquid or vapor and a liquid into a vapor without changing the temperature.
There are two types of latent heat.
(1) Latent heat of fusion
(2) Latent heat of vaporization
Latent heat of fusion : It is defined as the amount of heat energy released or absorbed when the solid converted to liquid at atmospheric pressure at its melting point.
Latent heat of vaporization : It is defined as the amount of heat energy released or absorbed when the liquid converted to vapor at atmospheric pressure at its boiling point.
Hence, latent heat is used to change states of matter.
Answer:
1) 0,081 ft/s
2) 0,746 lb/s
Explanation:
The relation between flow and velocity of a fluid is given by:
Q=Av
where:
- Q, flow [ft3/s]
- A, cross section of the pipe [ft2]
- v, velocity of the fluid [ft/s]
1)
To convert our data to appropiate units, we use the following convertion factors:
1 ft=12 inches
1 ft3=7,48 gallons
1 minute=60 seconds
So,

As the pipe has a circular section, we use A=πd^2/4:

Finally:
Q=vA......................v=Q/A

2)
The following formula is used to calculate the specific gravity of a material:
SG = ρ / ρW
where:
- ρ = density of the material [lb/ft3]
- ρW = density of water [lb/ft3] = 62.4 lbs/ft3
then:
ρ = SG*ρW = 1,49* 62,4 lb/ft3 = 93 lb/ft3
To calculate the mass flow, we just use the density of the chloroform in lb/ft3 to relate mass and volume:

The specific gravity of a sample is the ratio of the density of the sample with respect to one standard sample. The standard sample used in specific gravity calculation is water whose density is 1 g/mL. The solution having specific gravity 1.30 is the density of the sample that is 1.30 g/mL. Thus the weight of the 30 mL sample is (30×1.30) = 39 g.
Now the mass of the 10 mL of water is 10 g as density of water is 10 g/mL. Thus after addition the total mass of the solution is (39 + 10) = 49g and the volume is (30 + 10) = 40 mL. Thus the density of the mixture will be
g/mL. Thus the specific gravity of the mixed sample will be 1.225 g/mL.