To determine the volume of both concentration of vinegar, we need to set up two equations since we have two unknowns.
For the first equation, we do a mass balance:
mass of 100% vinegar + mass of 13% vinegar = mass of 42% vinegar
Assuming they have the same densities, then we can write this equation in terms of volume.
V(100%) + V(13%) = V(42%)
we let x = V(100%)
y = V(13%)
x + y = 150
For the second equation, we do a component balance:
1.00x + .13y = 150(.42)
x + .13y = 63
The two equations are
x + y = 150
x + .13y = 63
Solving for x and y,
x = 50
y = 100
Therefore, you need to mix 50 mL of the 100% vinegar and 100 mL of the 13% vinegar.
Explanation:
The given data is as follows.
Concentration = 0.1 
= 0.1 \frac{mol dm^{3}}{dm^{3}} \frac{10^{3}}{dm^{3}} \times \frac{6.022 \times 10^{23}}{1 mol} ions
= 
T = 
= (30 + 273) K = 303 K
Formula for electric double layer thickness (
) is as follows.
= 
where, 
= concentration =
Hence, putting the given values into the above equation as follows.
= 
= 
= 
m
or, = 
= 1 nm (approx)
Also, it is known that = 
Hence, we can conclude that addition of 0.1 of KCl in 0.1 of NaBr "" will decrease but not significantly.
D. amu
amu stands for atomic mass unit
Answer:
The answer to your question is
1.- 1.686 x 10²⁴ atoms
2.- 0.25 moles
Explanation:
1.-
1 mol ---------------- 6.023 x 10²³ atoms
2.8 moles ---------- x
x = (2.8 x 6.023 x 10²³) / 1
x = 1.686 x 10²⁴ atoms
2.- 1 mol ------------------ 6.023 x 10 ²³ molecules
x moles ------------- 1.50 x 10²³ molecules
x = (1.50 x 10²³ x 1) / 6.023 x 10²³
x = 0.25 moles
