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Sati [7]
3 years ago
10

Circle the solution with the highest concentration

Chemistry
2 answers:
hammer [34]3 years ago
7 0
Explanation

5mill


Hope this helps
choli [55]3 years ago
5 0

Answer:

answer is option a..............

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Which of the following is a buffer system? Which of the following is a buffer system? H2CO3(aq) and KHCO3(aq) NaCl(aq) and NaOH(
Readme [11.4K]

Answer:

Explanation:

A buffer is defined as an aqueous mixture of a weak acid and its conjugate base or vice versa.

In the systems:

H₂CO₃(aq) and KHCO₃(aq): Carbonic acid, H₂CO₃, is a weak acid that, in solution with its conjugate pair, HCO₃⁻ make a <em>buffer system.</em>

NaCl(aq) and NaOH(aq): NaCl is a salt and NaOH is a strong base. Thus, this system <em>is not </em> a buffer system.

H₂O(l) and HCl(aq): Water is a solvent and HCl a strong acid. This <em>is not </em>a buffer system.

HCl(aq) and NaOH(aq): HCl is a strong acid and NaOH a strong base. This <em>is not </em>a buffer system.

NaCl(aq) and NaNO₃(aq): Both NaCl and NaNO₃ are salts and this system <em>is not </em>a buffer system.

3 0
3 years ago
Solve for b if R = x(A + B)
Tatiana [17]

This is what I got. Hope it helps :)

7 0
3 years ago
What is the term for a fine grained soil deposited by the wind
balandron [24]

I think it is called a loess. Please correct me if I am wrong.

4 0
3 years ago
Calculate the pka of hypochlorous acid. The ph of a 0.015 m solution of hypochlorous acid has a ph of 4.64.
o-na [289]

Answer:

  • pKa = 7.46

Explanation:

<u>1) Data:</u>

a) Hypochlorous acid = HClO

b) [HClO} = 0.015

c) pH = 4.64

d) pKa = ?

<u>2) Strategy:</u>

With the pH calculate [H₃O⁺], then use the equilibrium equation to calculate the equilibrium constant, Ka, and finally calculate pKa from the definition.

<u>3) Solution:</u>

a) pH

  • pH = - log [H₃O⁺]

  • 4.64 = - log [H₃O⁺]

  • [H_3O^+]= 10^{-4.64} = 2.29.10^{-5}

b) Equilibrium equation: HClO (aq) ⇄ ClO⁻ (aq) + H₃O⁺ (aq)

c) Equilibrium constant: Ka =  [ClO⁻] [H₃O⁺] / [HClO]

d) From the stoichiometry: [CLO⁻] = [H₃O⁺] = 2.29 × 10 ⁻⁵ M

e) By substitution: Ka = (2.29 × 10 ⁻⁵ M)² / 0.015M = 3.50 × 10⁻⁸ M

f) By definition: pKa = - log Ka = - log (3.50 × 10 ⁻⁸) = 7.46

5 0
3 years ago
How many moles of aluminum oxide would form if 12.5 moles of aluminum burned​
jarptica [38.1K]
<h3>Al + O2 -> Al2O3</h3>

Balance it:

<h3>2Al + 3O2 -> 2Al2O3</h3><h3 />

So you need 2 Al and 3 O2 to make 2 Al2O3 (aluminum oxide).

I'm going to assume you have all the O2 you need.

Since 2 mols of Al is needed to make 2 mols of the product, it's a 1:1 ratio. You get as much aluminum oxide for as much aluminum you burn.

So 12.5 mols if there is not a lack of the O2.

7 0
3 years ago
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