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s2008m [1.1K]
3 years ago
14

Relate the strength of a weak acid to the strength of a conjugate base.

Chemistry
1 answer:
Elenna [48]3 years ago
4 0

Answer:

There is a relationship between the strength of an acid (or base) and the strength of its conjugate base (or conjugate acid): The stronger the acid, the weaker its conjugate base.  The weaker the acid, the stronger its conjugate base.  The stronger the base, the weaker its conjugate acid.

explanation

The strength of an acid and a base is determined by how completely they dissociate in water. Strong acids (like stomach acid) break down or dissociate in water. Weak acids maintains their protons in water.

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saveliy_v [14]
<span>M(NO3)2 ==> [M2+] + 2 [NO3-] 0.202 M ==> 0.202 M M(OH)2 ==> [M2+] + 2[OH-] 5.05*10^-18 ===> s + [2s]^2 5.05*10^-18 ===> 0.202 + [2s]^2 5.05*10^-18 = 0.202 * 4s^2 4s^2 = 25*10^-18 s^2 = 6.25*10^-18 s = 2.5*10^-9 So, the solubility is 2.5*10^-9</span>
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3 years ago
Which of the following is NOT a property of an acid?
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7 0
3 years ago
60 kg of fuel was completely burnt for an experiment. The amount of heat energy was found to be 180000KJ. Calculate calorific va
FrozenT [24]

Answer:

3000 kJ/kg

Explanation:

The calorific value of a substance is the amount of heat produced per unit mass by the combustion of the substance.

It is given by:

C=\frac{Q}{m}

where

Q is the amount of heat released

m is the mass of the fuel

In this problem, we have:

m = 60 kg is the mass of fuel

Q=180,000 kJ is the amount of heat released

Therefore, the calorific value of the fuel is:

C=\frac{180,000}{60}=3000 kJ/kg

6 0
3 years ago
Ionization of sodium dichromate​
inessss [21]

Explanation:

this is the answer of your question .

hope it helped you

5 0
2 years ago
Consider the reaction at 500 ° C 500°C . N 2 ( g ) + 3 H 2 ( g ) − ⇀ ↽ − 2 NH 3 ( g ) K c = 0.061 N2(g)+3H2(g)↽−−⇀2NH3(g)Kc=0.06
Sav [38]

Answer:

Q = 0.061 = Kc

Explanation:

Step 1: Data given

Temperature = 500 °C

Kc=0.061

1.14 mol/L  N2

5.52 mol/L H2

3.42 mol/L NH3

Step 2: Calculate Q

Q=[products]/[reactants]=[NH3]²/ [N2][H2]³

If Qc=Kc then the reaction is at equilibrium.  

If Qc<Kc then the reaction will shift right to reach equilibrium.

If Qc>Kc then the reaction will shift left to reach equilibrium.  

Q = (3.42)² / (1.14 * 5.52³)

Q = 11.6964/191.744

Q = 0.061

Q = Kc the reaction is at equilibrium.  

4 0
3 years ago
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