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nekit [7.7K]
4 years ago
5

What is kelvin and why must we use it.

Chemistry
1 answer:
Yakvenalex [24]4 years ago
8 0
Kelvin is the standard unit for temperature, just how like grams is for weight. We need to use it because it’s based on the idea of absolute zero, or 0K, which is a point where time itself freeezes
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Answer:

The correct answer is c add more HCO3- by adding NaHCO3.

Explanation:

The reaction mentioned in the question is carried out by bicarbonate buffer system of our body to maintain the normal acid  base balance.

         Now concentration of the reactant (H2C03) is decreased then NaHCO3 should be added which undergo breakdown to release bicarbonate ions(HCO3-).

        The released bicarbonate ions then reacts with H+ to form Carbonic acid(H2CO3).Thus homeostasis of H2CO3 is maintained.

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When wax is heated and changes to liquid how are the molecule affected
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A system is composed of 7.14 grams of Ne gas at 298 K and 1 atm. When 2025 joules of heat are added to the system at constant pr
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Answer:

(a) 1 atm, 8.72 L and 298 K respectively.

(b) 1 atm, 16.72 L and 573.35 K respectively.

(c) \Delta U=1215J

Explanation:

Hello,

(a) In this case, considering that neon could be considered as an ideal gas, we can compute its volume as follows:

PV=nRT\\\\V=\frac{nRT}{P}=\frac{7.14g*\frac{1mol}{20g}0.082\frac{atm*L}{mol*K}*298K}{1atm}\\  \\V=8.72L

Thus, initial conditions of pressure volume and temperature are 1 atm, 8.72 L and 298 K respectively.

(b) Since the process was carried out at constant pressure, the work is defined as:

W=P(V_2-V_1)

Thus, the final volume is:

V_2=\frac{W}{P} +V_1=\frac{810Pa*m^3}{1atm*\frac{101325Pa}{1atm} } *\frac{1000L}{1m^3}+8.72L\\ \\V_2=16.72L

And the final temperature is computed by considering the pressure-constant specific heat of neon (1.03 J/g*K) and the added heat:

Q=mCp(T_2-T_1)\\\\T_2=\frac{Q}{mCp}+T_1 =\frac{2025J}{7.14g*1.03J/(g*K)}+298K\\ \\T_2=573.35K

Therefore, final volume is 16.72 L, final pressure is also 1 atm and final temperature is 573.35 K for this expansion process.

(c) Finally, the change in the internal energy is computed via the first law of thermodynamics:

Q-W=\Delta U\\\\\Delta U=2025J-810J\\\\\Delta U=1215J

Best regards.

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