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4 years ago

If a compound contains an anion which is the conjugate base of a weak acid, the addition of hydronium ion solubility does what ?

Chemistry

1 answer:

wel4 years ago

3 0

Answer:

The addition of hydronium ion increases the solubility of the salt

Explanation:

When a weak acid, MA dissociates in an aqueous solution, the following products in equilibrium are obtained;

MA ----> M+ + A-

The anion which is a conjugate base will be removed from the solution with the addition of a hydronium ion from an acid: 2H3O+ + 2A- ----> 2HA + 2H20

This distorts the reaction equilibrium. According to Le Chatelier’s principle, more MA will dissolve until in order to restore the previous equilibrium of the reaction.

Therefore, an acidic pH increases the solubility of almost all sparingly soluble salts whose anion is the conjugate base of a weak acid.

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17. D i don’t know if i’m writing it wrong or don’t know how to balance but ye

leva [86]

Answer:

2 KL + Pb(NO3)2 = 2 KNO3 + PbL2

Explanation:

KL + Pb(NO3)2 = KNO3 + PbL2 is unballanced equation, just balance from there :)

8 0

3 years ago

Complete this nuclear reaction by selecting which particle would go in the

kompoz [17]

Answer:

Missing particles: $^{0}_{1}e^{+}$ (a positron) and an electron neutrino $\nu_{\rm e}$ .

The nuclear equation would be:

${\rm ^{19}_{10} Ne} \to ^{0}_{1}e^{+} + {\rm ^{19}_{\phantom{1}9}F }+ \nu_{e}$ .

Explanation:

The mass number of a particle is the number on the top-right corner of its symbol.

The atomic number of a particle is the number on the lower-right corner of its symbol.

The nuclear reaction here resembles a beta-plus decay. The mass numbers of the two nuclei are equal. However, the atomic number of the product nucleus is lower than that of the reactant nucleus by $1$ .

A beta decay may either be a beta-plus decay or a beta-minus decay. In a beta-plus decay, a positively-charged positron $^{0}_{1}e^{+}$ and an electron neutrino $\nu_e$ would be released. On the other hand, in a beta-minus decay, a negatively-charged electron $\rm ^{0}_{1}e^{-}$ and an electron antineutrino $\overline{\nu}_e$ would be released.

Electric charge needs to be conserved in nuclear reactions, including this one.

The atomic number of the $\rm Ne$ nucleus on the left-hand side is $10$ , meaning that the nucleus has a charge of $+10$ . On the other hand, the atomic number of the $\rm F$ nucleus on the right-hand side shows that this nucleus carries a charge of only $+9$ .

By the conservation of electric charge, the particles on the right-hand side must carry a positive charge of $+1$ . That rules out the possibility of the combination of one negatively-charged electron $\rm ^{0}_{1}e^{-}$ (with a charge of $-1$ ) and an electron antineutrino $\overline{\nu}_e$ (with no electric charge at all.)

Hence, the only possibility is that the missing particle is a positron (and an electron neutrino $\nu_e$ , which carries no electric charge.)

4 0

3 years ago

What is acid? 2.define an atom?

Wittaler [7]

Acid is an substance that ionized in water to produce hydrogen ion
Atom is the smallest neutral particle in the element

6 0

4 years ago

Determine the oxidation state of the metal ion in [cr(h2o)6]3+.

Hunter-Best [27]

Co2 is correct buddy

7 0

4 years ago

Consider the elements: Na, Mg, Al, Si, P.

Olegator [25]

Answer:

The elements mentioned in the series are Na, Mg, Al, Si, and P. It can be seen that all these elements are located in the same period. The atomic number of the mentioned elements are Na-11, Mg-12, Al-13, Si-14, and P-15.

a) There will be an increase in the ionization energy with the increase in the element's atomic number across a period. More energy is needed to withdraw an electron from a completely occupied shell in comparison to an incompletely occupied shell.

The atomic number of Na is 11. When one electron is withdrawn from Na, it gets converted into the inert gas configuration of Ne. Thus, it will require more energy to withdraw the second electron from Na. Hence, Na exhibits the lowest second ionization energy.

b) Across the period, with an increase in the element's atomic number, the atomic radii reduces from left to right. Thus, P exhibits the smallest atomic radius.

c) The metallic nature of the elements reduces from left to right across a period in the periodic table. Thus, P is the least metallic element.

d) Diamagnetic signifies towards the element that exhibits pair electrons in its sub-shells. The electronic configuration of Mg is,

1s2 2s2 2p6 3s2

In Mg, no unpaired electrons are present, while all the remaining elements mentioned exhibit unpaired electrons in their valence shell. Thus, Mg is the diamagnetic element.

8 0

3 years ago