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3 years ago

2 PUIS

Chemistry

2 answers:

Vsevolod [243]3 years ago

6 0

The answer is C the study of living things

avanturin [10]3 years ago

5 0

Answer: C the study of living things

Explanation:

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What does a very small equilibrium constant (K≪ 1) tell you about the reaction? View Available Hint(s) What does a very small eq

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Answer: The reaction mixture contains large amounts of reactants

Explanation:

The equilibrium constant (K) tells the degree to which a chemical reaction will progress.

So, a very small equilibrium constant (K≪ 1) tells of a reaction that has the concentration of the products as small, while the concentration of the reactants is large.

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3 years ago

Consider the substitution reaction that rakes place when (R)-3-bromo 3-methylhexane is treated with methanol. Which of the follo

Otrada [13]

Answer:

The reaction would take place with racemization.

Explanation:

The compound (R)-3-bromo 3-methylhexane contains has the leaving group attached to a tertiary carbon atom. When the leaving group is attached to a tertiary carbon atom, the compound undergoes nucleophilic substitution via SN1 mechanism. This mechanism involves the formation of a planar carbocation intermediate. The nucleophile may attack this intermediate from either faces, leading to racemization of the product.

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ketch the following intermolecular interactions: a.A hydrogen-bonding interaction between thymol and the surface of a silica par

san4es73 [151]

Answer:

answer is attached

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3 years ago

The ksp of manganese(ii) carbonate, mnco3, is 2.42 × 10-11. calculate the solubility of this compound in g/l.

Vikki [24]

Answer : The solubility of this compound in g/L is $565.414\times 10^{-6}g/L$ .

Solution : Given,

$K_{sp}=2.42\times 10^{-11}$

Molar mass of $MnCO_3$ = 114.945g/mole

The balanced equilibrium reaction is,

$MnCO_3\rightleftharpoons Mn^{2+}+CO^{2-}_3$

At equilibrium s s

The expression for solubility constant is,

$K_{sp}=[Mn^{2+}][CO^{2-}_3]$

Now put the given values in this expression, we get

$2.42\times 10^{-11}=(s)(s)\\2.42\times 10^{-11}=s^2\\s=0.4919\times 10^{-5}=4.919\times 10^{-6}moles/L$

The value of 's' is the molar concentration of manganese ion and carbonate ion.

Now we have to calculate the solubility in terms of g/L multiplying by the Molar mass of the given compound.

$s=4.919\times 10^{-6}moles/L\times 114.945g/mole=565.414\times 10^{-6}g/L$

Therefore, the solubility of this compound in g/L is $565.414\times 10^{-6}g/L$ .

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3 years ago

Consider the following reactions. (Note: (s) = solid, (l) = liquid, and (g) = gas.) ½H2(g) + ½I2(g) → HI(g), ΔH = +6.2 kcal/mole

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½H2(g) + ½I2(g) → HI(g) ΔH = +6.2 kcal/mol
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½H2(g) + ½I2(g) + 6,2kcal/mole → HI(g)
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21.0 kcal/mole + C(s) + 2S(s) → CS2(l)
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C(s) + 2S(s) → CS2(l) ΔH = +2,1 kcal/mole
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ΔH > 0 ----------->>> ENDOTHERMIC REACTIONS

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3 years ago

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