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choli [55]
3 years ago
12

When are conservation efforts most effective? a. Conservation efforts are most effective when multiple groups cooperate. b. Cons

ervation efforts are most effective when government is solely responsible for them. c. Conservation efforts are most effective with individual efforts. d. Conservation efforts are always effective.
Chemistry
2 answers:
alexandr402 [8]3 years ago
7 0
Conservation efforts most effective if <span>Conservation efforts are most effective when multiple groups cooperate. The answer is letter A. The rest of the choices do not answer the question above</span>
Natalka [10]3 years ago
3 0

Answer: Option (a) is the correct answer.

Explanation:

Conservation efforts means to carefully preserve or protect something. These efforts are fruitful when a majority of people or communities come together in order to preserve something good for the betterment of environment.

As when maximum number of people come lend their hands for help only then conservation efforts can be achieved.

For example, a city with maximum pollution can only be conserved when maximum groups of people decide to conserve plants or plant more number of trees.  

Thus, we can conclude that conservation efforts are most effective when conservation efforts are most effective when multiple groups cooperate.

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Answer the following questions as an example of physical or chemical property. (Ive answered the first 3 already)
natulia [17]
Physical properties is actually the measure of properties of a substance without producing any changes init. I.e colour, density melting and boling point. While chemical properties shows that either the specific substance will undergo the r/n or not.
So, gasoline is flammable is it’s chemical property.
5 0
3 years ago
Which letter represents the activated complex?<br> 1. A<br> 2. B<br> 3. F<br> 4. G
lana66690 [7]

Answer:

2. B.

Explanation:

  • The letter B represents the activated complex.
  • The activated complex is the intermediate that is formed between the states of reactants "F" and products "G".

  • Letter A represents the activation energy of the reaction, that is the difference in potential energy between the reactants "F" and activated complex "B".

  • Letter C represents the enthalpy change of the reaction, that is the difference in potential energy between products "G" and reactants "F".

  • Letter D represents the potential energy of products "G".

  • Letter E represents the potential energy of reactants "E".
4 0
3 years ago
1. The solubility of lead(II) chloride at some high temperature is 3.1 x 10-2 M. Find the Ksp of PbCl2 at this temperature.
solniwko [45]

Answer:

1) The solubility product of the lead(II) chloride is 1.2\times 10^{-4}.

2) The solubility of the aluminium hydroxide is 1.6\times 10^{-10} M.

3)The given statement is false.

Explanation:

1)

Solubility of lead chloride = S=3.1\times 10^-2M

PbCl_2(aq)\rightleftharpoons Pb^{2+}(aq)+2Cl^-(aq)

                            S     2S

The solubility product of the lead(II) chloride = K_{sp}

K_{sp}=[Pb^{2+}][Cl^-]^2

K_{sp}=S\times (2S)^2=4S^3=4\times (3.1\times 10^{-2})^3=1.2\times 10^{-4}

The solubility product of the lead(II) chloride is 1.2\times 10^{-4}.

2)

Concentration of aluminium nitrate = 0.000010 M

Concentration of aluminum ion =1\timed 0.000010 M=0.000010 M

Solubility of aluminium hydroxide in aluminum nitrate solution = S

Al(OH)_3(aq)\rightleftharpoons Al^{3+}(aq)+3OH^-(aq)

                            S     3S

The solubility product of the aluminium nitrate = K_{sp}=1.0\times 10^{-33}

K_{sp}=[Al^{3+}][OH^-]^3

1.0\times 10^{-33}=(0.000010+S)\times (3S)^3

S=1.6\times 10^{-10} M

The solubility of the aluminium hydroxide is 1.6\times 10^{-10} M.

3.

Molarity=\frac{Moles}{Volume (L)}

Mass of NaCl= 3.5 mg = 0.0035 g

1 mg = 0.001 g

Moles of NaCl = \frac{0.0035 g}{58.5 g/mol}=6.0\times 10^{-5} mol

Volume of the solution = 0.250 L

[NaCl]=\frac{6.0\times 10^{-5} mol}{0.250 L}=0.00024 M

1 mole of NaCl gives 1 mole of sodium ion and 1 mole of chloride ions.

[Cl^-]=[NaCl]=0.00024 M

Moles of lead (II) nitrate = n

Volume of the solution = 0.250 L

Molarity lead(II) nitrate = 0.12 M

n=0.12 M]\times 0.250 L=0.030 mol

1 mole of lead nitrate gives 1 mole of lead (II) ion and 2 moles of nitrate ions.

[Pb^{2+}]=[Pb(NO_2)_3]=0.030 M

PbCl_2(aq)\rightleftharpoons Pb^{2+}(aq)+2Cl^-(aq)

Solubility of lead(II) chloride = K_{sp}=1.2\times 10^{-4}

Ionic product of the lead chloride in solution :

Q_i=[Pb^{2+}][Cl^-]^2=0.030 M\times (0.00024 M)^2=1.7\times 10^{-9}

Q_i ( no precipitation)

The given statement is false.

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3 years ago
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