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2 years ago

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2.2 Chromium has four naturally-occurring isotopes: 4.34% of 50Cr, with an atomic weight of 49.9460 amu; 83.79% of 52Cr, with an

atomic weight of 51.9405 amu; 9.50% of 53Cr, with an atomic weight of 52.9407 amu; and 2.37% of 54Cr, with an atomic weight of 53.9389 amu. On the basis of these data, confirm that the average atomic weight of Cr is 51.9963 amu.

1 answer:

shutvik [7]2 years ago

4 0

Answer:

Average atomic mass = 51.9963 amu

Explanation:

Given data:

Abundance of Cr⁵⁰ with atomic mass= 4.34% , 49.9460 amu

Abundance of Cr⁵² with atomic mass = 83.79%, 51.9405 amu

Abundance of Cr⁵³ with atomic mass =9.50%, 52.9407 amu

Abundance of Cr⁵⁴ with atomic mass = 2.37%, 53.9389 amu

Average atomic mass = 51.9963 amu

Solution:

Average atomic mass = (abundance of 1st isotope × its atomic mass) +(abundance of 2nd isotope × its atomic mass +....n) / 100

Average atomic mass = (4.34×49.9460)+(83.79×51.9405) +(9.50×52.9407)+ (2.37×53.9389) / 100

Average atomic mass = 216.7656 + 4352.0945 + 502.9367 +127.8352 / 100

Average atomic mass = 5199.632 / 100

Average atomic mass = 51.9963 amu

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Solid calcium carbonate breaks down into carbon dioxide gas, oxygen gas, and solid calcium, write an equation for described reac

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Answer:

CaCO₃ ------> CaO + CO₂

<em>Note: The breakdown products of solid calcium carbonate is not carbon dioxide gas, oxygen gas, and solid calcium. Rather, it is calcium oxide and carbon dioxide as shown in the reaction equation above.</em>

Explanation:

Calcium carbonate is a mineral which occurs in nature in rocks as calcite and limestone. It is also the main component of eggshells, snail shells, seashells and pearls. It has the molecular formula CaCO₃.

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CaCO₃ ------> CaO + CO₂

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A 27 kg iron block initially at 375 C is quenched in an insulated tank that contains 130kg of water at 26 C. Assume the water th

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Solution :

a). Applying the energy balance,

$\Delta E_{sys}=E_{in}-E_{out}$

$0=\Delta U$

$0=(\Delta U)_{iron} + (\Delta U)_{water}$

$0=[mc(T_f-T_i)_{iron}] + [mc(T_f-T_i)_{water}]$

$0 = 27 \times 0.45 \times (T_f - 375) + 130 \times 4.18 \times (T_f-26)$

$t_f=33.63^\circ C$

b). The entropy change of iron.

$\Delta s_{iron} = mc \ln\left(\frac{T_f}{T_i} \right)$

$= 27 \times 0.45\ \ln\left(\frac{33.63 + 273}{375 + 273} \right)$

= -9.09 kJ-K

Entropy change of water :

$\Delta s_{water} = mc \ \ln\left(\frac{T_f}{T_i} \right)$

$= 130 \times 4.18\ \ln\left(\frac{33.63 + 273}{26 + 273} \right)$

= 10.76 kJ-K

So, the total entropy change during the process is :

$\Delta s_{tot} = \Delta s_{iron} + \Delta s_{water}$

= -9.09 + 10.76

= 1.67 kJ-K

c). Exergy of the combined system at initial state,

$X=(U-U_{0}) - T_0(S-S_0)+P_0(V-V_0)$

$X=mc (T-T_0) - T_0 \ mc \ \ln \left(\frac{T}{T_0} \right)+0$

$X=mc\left((T-T_0)-T_0 \ ln \left(\frac{T}{T_0} \right)\right)$

$X_{iron, i} = 27 \times 0.45\left(((375+273)-(12+273))-(12+273) \ln \frac{375+273}{12+273}\right)$

$X_{iron, i} =63.94 \ kJ$

$X_{water, i} = 130 \times 4.18\left(((26+273)-(12+273))-(12+273) \ln \frac{26+273}{12+273}\right)$

$X_{water, i} =-13.22 \ kJ$

Therefore, energy of the combined system at the initial state is

$X_{initial}=X_{iron,i} +X_{water, i}$

= 63.94 -13.22

= 50.72 kJ

Similarly, Exergy of the combined system at initial state,

$X=(U_f-U_{0}) - T_0(S_f-S_0)+P_0(V_f-V_0)$

$X=mc\left((T_f-T_0)-T_0 \ ln \left(\frac{T_f}{T_0} \right)\right)$

$X_{iron, f} = 27 \times 0.45\left(((33.63+273)-(12+273))-(12+273) \ln \frac{33.63+273}{12+273}\right)$

$X_{iron, f} = 216.39 \ kJ$

$X_{water, f} = 130 \times 4.18\left(((33.63+273)-(12+273))-(12+273) \ln \frac{33.63+273}{12+273}\right)$

$X_{water, f} =-9677.95\ kJ$

Thus, energy or the combined system at the final state is :

$X_{final}=X_{iron,f} +X_{water, f$

= 216.39 - 9677.95

= -9461.56 kJ

d). The wasted work

$X_{in} - X_{out}-X_{destroyed} = \Delta X_{sys}$

$0-X_{destroyed} =$

$X_{destroyed} = X_{initial} - X_{final}$

= 50.72 + 9461.56

= 9512.22 kJ

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Amiraneli [1.4K]

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</span>Intermolecular Forces present in CH₄:
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