Answer: 0.20 M
Explanation:
According to the dilution law,

where,
= molarity of stock solution = 1.40 M
= volume of stock solution = 72.0 ml
= molarity of diluted solution = m
= volume of diluted solution = 248 ml


Now 124 mL portion of this prepared solution is diluted by adding 133 mL of water.
According to the dilution law,

where,
= molarity of stock solution = 0.41 M
= volume of stock solution = 124 ml
= molarity of diluted solution = m
= volume of diluted solution = (124 +133) ml = 257 ml


Thus the final concentration of the solution is 0.20 M.
Answer:
the answer is c I took the test
As you increase in elevation, there is less air above you thus the pressure decreases. As the pressure decreases, air molecules spread out further (i.e. air expands) and the temperature decreases. If the humidity is at 100 percent (because it's snowing), the temperature decreases more slowly with height.
Students performed a procedure similar to Part II of this
experiment (Analyzing Juices for Vitamin C Content) as described in the
procedure section. Given that molarity of DCP is 9.98x10-4 M, it took 16.34 ml
of DCP to titrate 10 mL of sample.
Amount of ascorbic acid = 0.050 L sample (0.01634 L DCP/0.01
L sample)( 9.98x10-4 mol DCP/L DCP)(1 mol Ascorbic acid/ 1mol DCP)(176.124
g/mol)(1000mg/1g)= 14.36 mg ascorbic acid
Answer:
Option C = object B by 1 gram per cubic cm.
Explanation:
Given data:
Mass of object A = 12 g
Volume of object A = 8 cm³
Mass of object B = 20 g
Volume of object B = 8 cm³
Densities = ?
Solution:
Density:
Density is equal to the mass of substance divided by its volume.
Units:
SI unit of density is Kg/m3.
Other units are given below,
g/cm3, g/mL , kg/L
Formula:
D=m/v
D= density
m=mass
V=volume
Symbol:
The symbol used for density is called rho. It is represented by ρ. However letter D can also be used to represent the density.
Density of object A:
d = m/v
d = 12 g/ 8 cm³
d = 1.5 g/cm³
Density of object B:
d = m/v
d = 20 g/ 8 cm³
d = 2.5 g/cm³
object b has high density.