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Nutka1998 [239]
3 years ago
7

Consider the following thermochemical equations.

Chemistry
1 answer:
Yakvenalex [24]3 years ago
5 0

<u>Answer:</u> The \Delta H^o_{rxn} for the reaction is -749.8 kJ.

<u>Explanation:</u>

Hess’s law of constant heat summation states that the amount of heat absorbed or evolved in a given chemical equation remains the same whether the process occurs in one step or several steps.

According to this law, the chemical equation is treated as ordinary algebraic expressions and can be added or subtracted to yield the required equation. This means that the enthalpy change of the overall reaction is equal to the sum of the enthalpy changes of the intermediate reactions.

The chemical reaction for the formation of PCl_5 follows:

2P(s)+5Cl_2(g)\rightarrow 2PCl_5(s)      \Delta H^o_{rxn}=?

The intermediate balanced chemical reaction are:

(1) PCl_5(s)\rightarrow PCl_3(g)+Cl_2(g)    \Delta H_1=87.9kJ   ( × 2)

(2) 2P(s)+3Cl_2(g)\rightarrow 2PCl_3(g)     \Delta H_2=-574kJ

The expression for enthalpy of the reaction follows:

\Delta H^o_{rxn}=[2\times (-\Delta H_1)]+[1\times \Delta H_2]

Putting values in above equation, we get:

\Delta H^o_{rxn}=[(2\times (-87.9))+(1\times (-574))=-749.8kJ

Hence, the \Delta H^o_{rxn} for the reaction is -749.8 kJ.

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What is the pOH of 0.50 molar H3BO3?
Crazy boy [7]

<u>Answer:</u>

<em>A. 10.25</em>

<em></em>

<u>Explanation:</u>

Pkb =4.77

So pka = 14 - pka = 9.23

Ka =10^{-pka}

H_3 BO_3 (aq)+ H_2 O(l) H_2 BO_3^- (aq)+H_3 O^+ (aq)

Initial                0.50M                                 0                                 0

Change                  -x                                 +x                               +x

Equilibrium    0.50M-x                               +x                               +x

Ka =\frac {((x)(x))}{(0.50M-x)}

5.88\times10^{-10}= \frac {x^2}{(0.50M-x)}

(-x is neglected) so we get

5.88\times10^{-10}\times0.50=x^2\\\\x^2=2.94\times10^{-10}

x=\sqrt{x^2}=1.72\times10^{-5} M=H^3 O^{+}

pH=-log[H^3 O^+]\\\\pH=-log[1.72\times10^{-5}]\\\\pH=4.76

pOH = 14 - pH

= 14 - 4.76

pOH = 9.24 is the answer

Option A - 10.25 is the answer which is close to 9.24

4 0
3 years ago
What oxidation state(s) can the alkaline earth metals exhibit?
iragen [17]

Answer:

+2

Explanation:

Alkaline earth metals are present in group 2 of periodic table. There are six elements in second group. Beryllium, magnesium, calcium, strontium, barium and radium.

All have two valance electrons.

Electronic configuration of Beryllium:

Be = [He] 2s²

Electronic configuration of magnesium.

Mg = [Ne] 3s²

Electronic configuration of calcium.

Ca = [Ar] 4s²

Electronic configuration of strontium.

Sr = [Kr] 5s²

Electronic configuration of barium.

Ba = [Xe] 6s²

Electronic configuration of radium.

Ra = [ Rn] 7s²

They are present in group two and have same number of valance electrons (two valance electrons) and show oxidation state +2 by loosing two valance electrons. They also show similar reactivity.

They react with oxygen and form oxide.

2Ba   +   O₂   →    2BaO

2Mg  +   O₂   →    2MgO

2Ca +   O₂   →    2CaO

this oxide form hydroxide when react with water,

BaO  + H₂O   →  Ba(OH)₂

MgO  + H₂O   →  Mg(OH)₂

CaO  + H₂O   →  Ca(OH)₂

With sulfur,

Mg + S   →  MgS

Ca + S   →  CaS

Ba + S   →  BaS

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C3Hg + 5 02 --&gt; 3 CO2 + 4 H20
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Answer:

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The answer is 197
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