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3 years ago

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1 answer:

7 0

How would you calculate the change in entropy when the pressure of a perfect gas is changed isothermally from P_i to P_f

Answer:

ΔS = nRIn(P_i/P_f)

Explanation:

Formula for entropy change is;

ΔS = nC_p•In(T_f/T_i) + nRIn(P_i/P_f)

Since the pressure of the perfect gas is changed isothermally, it means that temperature remains the same and as such T_f = T_i

Thus, we now have;

ΔS = nC_p•In(1) + nRIn(P_i/P_f)

In(1) = 0

Thus;

ΔS = nRIn(P_i/P_f)

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Suppose a 1.0 L solution contains 0.020 M in Cu(NO3)2, then 0.40 moles of NH3 are added. Assuming no change in volume, what is t

Rasek [7]

Answer:

$4.6*10^{-14}$ M

Explanation:

Concentration of $Cu^{2+} = [Cu(NO_3)_2]$ = 0.020 M

Constructing an ICE table;we have:

$Cu^{2+}+4NH_3_{aq} \rightleftharpoons [Cu(NH_3)_4]^{2+}_{(aq)}$

Initial (M) 0.020 0.40 0

Change (M) - x - 4 x x

Equilibrium (M) 0.020 -x 0.40 - 4 x x

Given that: $K_f =1.7*10^{13}$

$K_f } = \frac{[Cu(NH_3)_4]^{2+}}{[Cu^{2+}][NH_3]^4}$

$1.7*10^{13} = \frac{x}{(0.020-x)(0.40-4x)^4}$

Since x is so small; 0.40 -4x = 0.40

Then:

$1.7*10^{13} = \frac{x}{(0.020-x)(0.0256)}$

$1.7*10^{13} = \frac{x}{(5.12*10^{-4}-0.0256x)}$

$1.7*10^{13}(5.12*10^{-4} - 0.0256x) = x$

$8.704*10^9-4.352*10^{11}x =x$

$8.704*10^9 = 4.352*10^{11}x$

$x = \frac{8.704*10^9}{4.352*10^{11}}$

$x = 0.0199999999999540$

$Cu^{2+}= 0.020 - 0.019999999999954$

$Cu^{2+} = 4.6*10^{-14}$ M

8 0

3 years ago

In a combustion reaction, pentane (molar mass (MM) = 72.17 g/mol) burns, in the presence of oxygen (MM = 32.00 g/mol), to produc

ladessa [460]

It is C bc I’m right anyits a free country

3 0

3 years ago

¿como afecta la temperatura el volumen de un gas?

Oxana [17]

Lo afect porque cuando la temperature aumenta, el volumen aumentará, luego, cuando we mantiene la presión, es constante. Calentar el gas aumenta la emergía cinética we law partículas, lo que have que el gas se expanda.

Espero que esto ayude :)

8 0

3 years ago

What element in the second period has the largest atomic radius?

Allushta [10]

Answer:

Na is the element from the second period that has the largest atomic radius

Explanation:

The atomic radius is a chemist property from the periodic table. It is decreased when we move in a period from the periodic, so the element in the second period that has the largest radius is Na, and the shortest, the Ar

The atomic radius indicates the distance between the nucleus and the outermost valence layer. In the periods it decreases with increasing Atomic Number, to the right, due to the attraction that the nucleus exerts on the electrons of the outermost orbitals, thus decreasing the core-electron distance.

3 0

3 years ago

Balance this nuclear reaction by supplying the missing nucleus: 249/98 Cf +__ --> 263/106 Sg +4 1/0 n

Sholpan [36]

The total atomic number must be the same on each side. The total mass number must be the same on both side.

On the RHS, for the mass number, we have 257 + 4 = 261 (the 4 comes from the 4 neutrons). That means the mass number of the missing piece on the LHS is 261 - 247 = 14. 

One the RHS, for the atomic number we have a total of 104 since the 4 neutrons are all neutral. On the LHS, we have this: 104 - 98 = 6. 

The missing piece is a nucleus of carbon 14. Done in your style, it is 14/6C

5 0

4 years ago

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