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Vlada [557]
3 years ago
9

A sample of nitrogen gas is collected over water at temperature of 20.0˚C. What is the pressure of the nitrogen gas if atmospher

ic pressure is 1.01atm? (vapor pressure of water at 20.0˚C is 17.5 torr)
Chemistry
1 answer:
Levart [38]3 years ago
4 0

Answer:

P_N=0.987atm

Explanation:

Hello there!

In this case, for these problems about collecting a gas over water, we must keep in mind that once the gas has been collected, the total pressure of the system is given by the atmospheric pressure, in this case 1.01 atm. Next, since we also have water in the mixture, we can write the following equation:

P_T=P_w+P_N

Thus, by solving for the pressure of nitrogen and using consistent units, we obtain:

1.01atm=17.5torr*\frac{1atm}{760torr} +P_N\\\\P_N=1.01atm-0.023atm\\\\P_N=0.987atm

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a single-replacement reaction replaces one element for another in a compound. A double-replacement reaction exchanges the cations, or the anions, of two ionic compounds.

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adam wants to investigate the strengths of different acids and alkalis. He has 2 different acid solutions and 2 different alkali
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Answer:

The strength of an acid or alkali depends on the degree of dissociation of the acid or alkali in water. The degree of dissociation measures the percentage of acid molecules that ionise when dissolved in water. He could use universal indicators or litmus paper for this.

Explanation:

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Activity C: Hydrolysis of Carbohydrates As discussed earlier, disaccharides are composed of two monosaccharides linked together.
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Sucrose: glucose and fructose

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<em>What monosaccharides will result from the hydrolysis of sucrose?</em>

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4 0
3 years ago
The protein lysozyme unfolds at a transition temperature of 75.5°C, and the standard enthalpy of transition is 509 kJ mol-1. Cal
spin [16.1K]

Answer:

0.4774 KJ/K.mol

Explanation:

We are told that the transition at 25.0°C occurs in three steps. Steps i, ii and iii.

Thus;

the entropy of unfolding of lysozyme = ΔS_i + ΔS_ii + ΔS_iii

Now,

C_p,m(unfolded protein) = C_p,m(folded protein) + 6.28 kJ/K.mol

Now, for the first process, ΔS_i is given as;

ΔS_i = C_p,m × In(T2/T1)

We are given;

T1 = 25°C = 25 + 273.15K = 298.15 K

T2 = 75.5°C = 75.5 + 273.15 K=348.65 K

Thus;

ΔS_i = C_p,m × In(348.65/298.15)

Now, for the third process, ΔS_iii is given as;

ΔS_iii = (C_p,m + 6.28 kJ/K.mol) × In(T1/T2)

Thus;

ΔS_iii = (C_p,m + 6.28 kJ/K.mol) × In(298.15/348.65)

Now, we don't know C_pm. So, we have to find a way to eliminate it. We will do it by rewriting In(298.15/348.65) in such a way that when ΔS_iii is added to ΔS_i, C_p,m will cancel out. Thus;

In(298.15/348.65) can also be written as;

In(348.65/298.15)^(-1) or

- In(348.65/298.15)

Thus;

ΔS_iii = - [(C_p,m + 6.28 kJ/K.mol) × In(298.15/348.65)]

Now, let's add ΔS_iii to ΔS_i to get;

ΔS_i + ΔS_iii = [C_p,m × In(348.65/298.15)] + [(-C_p,m - 6.28 kJ/K.mol) × In(348.65/298.15)]

ΔS_i + ΔS_iii = [C_p,m × In(348.65/298.15)] - [C_p,m × In(348.65/298.15)] - [6.28In(348.65/298.15)]

First 2 terms will cancel out to give;

ΔS_i + ΔS_iii = -6.28In(348.65/298.15)

ΔS_i + ΔS_iii = -0.9826 KJ/K.mol

Now,for process ii;

ΔS_ii = standard enthalpy of transition/Transition Temperature

Thus;

ΔS_ii = (509 KJ/K.mol)/348.65

ΔS_ii = 1.46 KJ/K.mol

Thus;

the entropy of unfolding of lysozyme = ΔS_i + ΔS_ii + ΔS_iii = -0.9826 + 1.46 = 0.4774 KJ/K.mol

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