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Musya8 [376]
3 years ago
12

How do densities of the solids compare to those of the liquid?

Chemistry
1 answer:
Dmitriy789 [7]3 years ago
3 0

Generally, solids are denser than liquid, which is denser than gas of one particular substance. In liquid phase, molecules are freer and can move around, so it is less dense than solid, but denser than gas, in which the molecules move at high speed with large separation from one another.

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The wastewater solution from a factory containing high levels of salts needs to be diluted before it can be released into the en
Vesna [10]

Answer:

3.5 atm

Explanation:

As stated in the question pressure is required to counteract the natural tendency  for water to dilute the more concentrated solution. The difference in concentrations will give us the answer using the osmotic pressure equation.

π = ( n/v)  RT where n/v is the molarity (mol/L), R is the gas constant and T is the temperature.

The difference in osmotic pressure of the solutions is:

Δπ = Δ c RT where c is the difference in molar concentrations.

pressure required = Δπ = (0.190 - 0.048) M x 0.821 Latm/Kmol x 298 K

= 3.47 atm

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A chemist titrates 150.0 mL of a 0.2653 M carbonic acid (H2CO3) solution with 0.2196 M NaOH solution at 25 °C. Calculate the pH
xxTIMURxx [149]

Answer:

9.3

Explanation:

This is long and complicated so get ready

We are going to use the conjugate base of carbonic acid with water to make carbonic acid and OH- (Na is simply a spectator ion and is irrelavent here)

Let the conjugate base be A- and Carbonic acid be HA

A- + H20 ⇄ HA + OH-

To find the concentration of A- we must find the concentration of the reactants given. We know this will be equal because it is a strong base and all of it disassociates.

to get moles of acid we take the concentration and multiply by liters to cancel

.2653 x .150 = .039795 mol HA

Because it is at equivalence point we know the moles will be equal. We are given the concentration so we only have to solve for liters

We plug it into the equation and found: .181 L

Now use moles and combined volums to fins concentrarion which is .120 M

Now plug that use the Ka converted to Kb to find the cincentrations of HA and OH-

Ka is (10^-3.60) = 2.4E-4

Kb x Ka is 10^-14

Kb = 3.98E-11

Now we know Kb = [HA] [OH] / [A-]

Solve for this through algebra by using x for the values you dont know

youll find x^2 = 3.3E-10

X = 1.8 E -5

this is the OH- concentration

-log [oh] = pOH

pOH = 4.73

We know 14-pOH = ph so pH= 9.3

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3 years ago
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Answer:

something cyan colored?

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