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amm1812
3 years ago
10

Acid-Base Reactions Custom Learning Goal: To learn to classify acids and bases and to predict the products of neutralization rea

ctions. Acids are substances that ionize to form free H ions in solution whereas bases are substances that combine with H ions. Strong acids and strong bases completely ionize but weak acid and weak bases only partially ionize. A salt is a term for an ionic compound such as NaCl or MgBrz When an acid and a base are mixed together, a neutralization reaction occurs. The products of an acid-base reaction do not have the chemical characteristics of either the acid or the base that originally reacted. Part A Classify each substance as a strong acid, strong base, weak acid, or weak base.
Chemistry
1 answer:
katovenus [111]3 years ago
5 0

Answer:

HCL, H2SO4 and HNO3 are strong acids.

NaOH, KOH and LiOH are strong bases.

H3PO4, CH3COOH and HF are weak acids.

NH4OH, N2H4 and Zn(OH)2 are weak bases.

Explanation:

HCL, H2SO4 and HNO3 are strong acids because they completely converted into ions and gives H ion in the solution while H3PO4, CH3COOH and HF are weak acids due to their half ionization in the solution. NaOH, KOH and LiOH are strong bases completely dissociate in the solution and give OH ion while NH4OH, N2H4 and Zn(OH)2 are weak bases which cannot dissociate completely and yield less amount of OH ions. when acid and base combine together, they exchange their partners produces salt and water.

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therefore there’s 3 electrons and 3 protons

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3 years ago
The photoelectric effect occurs when _____ are emitted from metal when the metal is struck by light of certain frequencies.
Ainat [17]

Answer:

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3 years ago
Question 26 Suppose a flask is filled with of and of . The following reaction becomes possible: The equilibrium constant for thi
Blizzard [7]

Answer:

0.36 M

Explanation:

There is some info missing. I think this is the complete question.

<em>Suppose a 250 mL flask is filled with 0.30 mol of N₂ and 0.70 mol of NO. The following reaction becomes possible: </em>

<em>N₂(g) +O₂(g) ⇄ 2 NO(g) </em>

<em>The equilibrium constant K for this reaction is 7.70 at the temperature of the flask.  Calculate the equilibrium molarity of O₂. Round your answer to two decimal places.</em>

<em />

Initially, there is no O₂, so the reaction can only proceed to the left to attain equilibrium. The initial concentrations of the other substances are:

[N₂] = 0.30 mol / 0.250 L = 1.2 M

[NO] = 0.70 mol / 0.250 L = 2.8 M

We can find the concentrations at equilibrium using an ICE Chart. We recognize 3 stages (Initial, Change, and Equilibrium) and complete each row with the concentration or change in the concentration.

    N₂(g) +O₂(g) ⇄ 2 NO(g)

I    1.2        0              2.8

C  +x         +x            -2x

E  1.2+x      x           2.8 - 2x

The equilibrium constant (K) is:

K=7.70=\frac{[NO]^{2}}{[N_{2}][O_{2}]} =\frac{(2.8-2x)^{2} }{(1.2+x).x}

Solving for x, the positive one is x = 0.3601 M

[O₂] = 0.3601 M ≈ 0.36 M

7 0
3 years ago
Guide Questions:
shtirl [24]

Answer:

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4 years ago
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