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lora16 [44]
2 years ago
8

Try the same filtering method with the salt water. What happened? Taste the

Chemistry
2 answers:
enyata [817]2 years ago
3 0

Yeah, what? I don't think Brainly knows that. Maybe you should taste the filtered salt water and find out the answer.

Lerok [7]2 years ago
3 0

Answer:

It should, well, taste less salty or even like fresh water if it is filtered enough

Explanation:

Depending on how you filter it, the salt should separate out of the water

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Calculate the pH of each solution at 25∘C
JulijaS [17]

Answer: pH of HCl =5, HNO3 = 1,

NaOH = 9, KOH = 12

Explanation:

pH = -log [H+ ]

1. 1.0 x 10^-5 M HCl

pH = - log (1.0 x 10^-5)

= 5 - log 1 = 5

2. 0.1 M HNO3

pH = - log (1.0 x 10 ^ -1)

pH = 1 - log 1 = 1

3. 1.0 x 10^-5 NaOH

pOH = - log (1.0 x 10^-5)

pOH = 5 - log 1 = 5

pH + pOH = 14

Therefore , pH = 14 - 5 = 9

4. 0.01 M KOH

pOH = - log ( 1.0 x 10^ -2)

= 2 - log 1 = 2

pH + pOH = 14

Therefore, pH = 14 - 2 = 12

8 0
3 years ago
True or False When the attraction between the particles overcomes their motion the particles clump together to boil. ​
bixtya [17]

Answer:

False

Explanation:

It is wrong to claim that when the attraction between particles overcomes their motion, the particles will clump together to boil.

During boiling particles do no clump together, they tend to move apart more rapidly.

  • Boiling occurs when the vapor pressure overcomes the ambient atmospheric pressure.
  • The hotter part of the boiler close the heat source moves rapidly away because they have become less dense.
  • The colder and denser part sinks and this interaction sets up a convection cell.
6 0
3 years ago
A burner is placed below the right side of a beaker. Which image shows the motion of the liquid that will result?
matrenka [14]

Answer:

the result will be the whole surface of the liquid boilling

Explanation:

3 0
3 years ago
A titration of 25.0 mL of a solution of the weak base aniline, C6H5NH2, requires 25.67 mL of 0.175 M HCl to reach the equivalenc
Lorico [155]

Answer:

a. 0.180M of C₆H₅NH₂

b. 0.0887M C₆H₅NH₃⁺

c. pH = 2.83

Explanation:

a. Based in the chemical equation:

C₆H₅NH₂(aq) + HCl(aq) → C₆H₅NH₃⁺(aq) + Cl⁻(aq)

<em>1 mole of aniline reacts per mole of HCl</em>

Moles required to reach equivalence point are:

Moles HCl = 0.02567L ₓ (0.175mol / L) = 4.492x10⁻³ moles HCl = moles C₆H₅NH₂

As the original solution had a volume of 25.0mL = 0.0250L:

4.492x10⁻³ moles C₆H₅NH₂ / 0.0250L = 0.180M of C₆H₅NH₂

b. At equivalence point, moles of C₆H₅NH₃⁺ are equal to initial moles of C₆H₅NH₂, that is 4.492x10⁻³ moles

But now, volume is 25.0mL + 25.67mL = 50.67mL = 0.05067L. Thus, molar concentration of C₆H₅NH₃⁺ is:

[C₆H₅NH₃⁺] = 4.492x10⁻³ moles / 0.05067L = 0.0887M C₆H₅NH₃⁺

c. At equivalence point you have just 0.0887M C₆H₅NH₃⁺ in solution. C₆H₅NH₃⁺ has as equilibrium in water:

C₆H₅NH₃⁺(aq) + H₂O(l) → C₆H₅NH₂ + H₃O⁺

Where Ka = Kw / Kb = 1x10⁻¹⁴ / 4.0x10⁻¹⁰ =

<em>2.5x10⁻⁵ = [C₆H₅NH₂] [H₃O⁺] / [C₆H₅NH₃⁺]</em>

When the system reaches equilibrium, molar concentrations are:

[C₆H₅NH₃⁺] = 0.0887M - X

[C₆H₅NH₂] = X

[H₃O⁺] = X

Replacing in Ka formula:

2.5x10⁻⁵ = [X] [X] / [0.0887M - X]

2.2175x10⁻⁶ - 2.5x10⁻⁵X = X²

0 = X² + 2.5x10⁻⁵X - 2.2175x10⁻⁶

Solving for X:

X = -0.0015 → False solution. There is no negative concentrations.

X = 0.001477 → Right solution.

As [H₃O⁺] = X, [H₃O⁺] = 0.001477

Knowing pH = -log [H₃O⁺]

pH = -log 0.001477

<h3>pH = 2.83</h3>

7 0
3 years ago
What several factors help the veins push blood back to the heart
Tomtit [17]
A)Constant pumping from the heart which provides adequate pressure necessary to move the blood through the arteries and veins
B)Gravitational force
6 0
3 years ago
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