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lbvjy [14]
2 years ago
5

14. Would you expect each of the following atoms to gain or lose electrons when forming ions ? what ion is the most likely in ea

ch case ? a. Ra b. Br c. In d. Te e. Ca f. As g. Al h. Rb
Chemistry
1 answer:
yanalaym [24]2 years ago
7 0

Answer:

The answer to your question is below

Explanation:

Metals lose electrons and Nonmetals gain them.

a. Ra is in group IIA it will lose electrons, the ion will be Ra⁺²

b. Br is in group VIIA it will gain electrons, the ion will be Br⁻¹

c. In is in group IIIA it will lose electrons, the ion will be In⁺³

d. Te is in group VIA it will gain electrons, the ion will be Te⁻²

e. Ca is in group IIA it will lose electrons, the ion will be Ca⁺²

f. As is in group VA it will gain electrons, the ion will be As⁻³

g. Al is in group IIlA it will lose electrons, the ion will be Al⁺³

h. Rb is in group IA it will lose electrons, the ion will be Rb⁺¹

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<u>Answer:</u> The rate of heat flow is 3.038 kW and the rate of lost work is 1.038 kW.

<u>Explanation:</u>

We are given:

C_p=\frac{7}{2}R\\\\T=475K\\P_1=100kPa\\P_2=50kPa

Rate of flow of ideal gas , n = 4 kmol/hr = \frac{4\times 1000mol}{3600s}=1.11mol/s    (Conversion factors used:  1 kmol = 1000 mol; 1 hr = 3600 s)

Power produced = 2000 W = 2 kW     (Conversion factor:  1 kW = 1000 W)

We know that:

\Delta U=0   (For isothermal process)

So, by applying first law of thermodynamics:

\Delta U=\Delta q-\Delta W

\Delta q=\Delta W      .......(1)

Now, calculating the work done for isothermal process, we use the equation:

\Delta W=nRT\ln (\frac{P_1}{P_2})

where,

\Delta W = change in work done

n = number of moles = 1.11 mol/s

R = Gas constant = 8.314 J/mol.K

T = temperature = 475 K

P_1 = initial pressure = 100 kPa

P_2 = final pressure = 50 kPa

Putting values in above equation, we get:

\Delta W=1.11mol/s\times 8.314J\times 475K\times \ln (\frac{100}{50})\\\\\Delta W=3038.45J/s=3.038kJ/s=3.038kW

Calculating the heat flow, we use equation 1, we get:

[ex]\Delta q=3.038kW[/tex]

Now, calculating the rate of lost work, we use the equation:

\text{Rate of lost work}=\Delta W-\text{Power produced}\\\\\text{Rate of lost work}=(3.038-2)kW\\\text{Rate of lost work}=1.038kW

Hence, the rate of heat flow is 3.038 kW and the rate of lost work is 1.038 kW.

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2 years ago
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