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Kryger [21]
3 years ago
9

2.2 liters at 20 degrees Celsius if the pressure doesn’t change what is the new temperature at 2.6 liters

Chemistry
1 answer:
Mamont248 [21]3 years ago
7 0
Based on the question, it is evident that this question rests on the premise of Charle's Law, which essentially states that temperature is proportional to volume once pressure and mass remain constant.

Thus by Charle's Equation:   \frac{ V_{1} }{ T_{1} }  =  \frac{ V_{2} }{ T_{2} }

Since the initial volume = 2.2 L; the initial temperature = 20 C; and the final volume = 2.6 L,
                           then V₁ = 2.2 L; T₁ = 20 C; V₂ = 2.6 L and what we would be
                           finding is T₂ (the final/new temperature)

Now,   T_{2}  =  \frac{(V_{2})   (T_{1}) }{V_{1}} 

     ⇒   T_{2}  =  \frac{(2.6 L)   (20  ^{o} C) }{2.2 L}

     ⇒   T_{2} = 23.64   ^{o}C


        ∴  the new temperature is ≈  23.6 °C
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Phosphate buffers are commonly used in biological research. If a small amount of strong acid is added to a buffer solution that
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A) [H3PO4] will increase, [KH2PO4] will decrease, and pH will slightly decrease.

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3 years ago
Calculate the standard entropy change for the following reactions at 25°C.
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Answer:

(a) ΔSº = 216.10 J/K

(b) ΔSº = - 56.4 J/K

(c) ΔSº = 273.8 J/K

Explanation:

We know the standard entropy change for a given reaction is given by the sum of the entropies of the products minus the entropies of reactants.

First we need to find in an appropiate reference table the standard  molar entropies entropies, and then do the calculations.

(a)        C2H5OH(l)          +        3 O2(g)         ⇒        2 CO2(g)     +    3 H2O(g)

Sº            159.9                          205.2                         213.8                  188.8

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ΔSº = [ 2(213.8) + 3(188.8) ]   - [ 159.9  + 3(205.) ]  J/K

ΔSº = 216.10 J/K

(b)         CS2(l)               +         3 O2(g)               ⇒      CO2(g)      +      2 SO2(g)

Sº          151.0                              205.2                         213.8                 248.2

(J/Kmol)

ΔSº  = [ 213.8 + 2(248.2) ] - [ 151.0 + 3(205.2) ] J/K = - 56.4 J/K

(c)        2 C6H6(l)           +        15 O2(g)                     12 CO2(g)     +     6 H2O(g)

Sº           173.3                           205.2                           213.8                    188.8

(J/Kmol)  

ΔSº  = [ 12(213.8) + 6(188.8) ] - [ 2(173.3) + 15( 205.2) ] = 273.8 J/K

Whenever possible we should always verify if our answer makes sense. Note that the signs for the entropy change agree with the change in mol gas. For example in reaction (b) we are going from 4  total mol gas reactants to 3, so the entropy change will be negative.

Note we need to multiply the entropies of each substance by  its coefficient in the balanced chemical equation.

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