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elena-14-01-66 [18.8K]
2 years ago
5

Ammonia is a weak base and establishes the following equilibrium in water: NH3(aq) + H2O(l) ---> NH4+ (aq) + OH-(aq). In Part

III, the phenolphthalein indicator is used to monitor the equilibrium shifts of the ammonia/ammonium ion system. The phenolphthalein equilibrium established with water is Hph(aq)(colorless) + H2O(l) --> H3O+ (aq) + ph-(aq)(pink or red). You compared the color of the solutions in three test tubes that initially contained 3 mL of 0.1 M ammonium hydroxide and a few drops of phenolphthalein indicator. In a second test tube, you added 6 M NH4OH dropwise. Explain, in terms of the ammonia/ammonium ion equilibrium, why the phenolphthalein equilibrium shifted to the right.
Chemistry
1 answer:
CaHeK987 [17]2 years ago
4 0

Answer:

Explanation:

phenolphthalein indicator is an acid - alkaline organic dye that changes color over a particular pH range depending on the medium. As such , asides phenolphthalein, methyl orange, Alzarian yellow are other indicators that are used during avid - base titration experiment.

In this case, phenolphthalein in an acidic medium is colorless while in basic medium is pink. Also phenolphthalein been a weak acid reacting with the hydroxide thereby reducing the concentration, this will have an effect on the equilibrium position by favoring the forward reaction i.e product formation. hence more products will be formed.

In the case of ammonium chloride, the addition of more ammonium ions results in an increase in the concentration of the products and as such, if more ammonium ions are removed, the equilibrium position will shift to favor the backward reaction i.e reactant formation.

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vovangra [49]
This question is incomplete. Luckily, I found the same problem which is shown in the attached picture. To answer the question, we must know how the size and charge affect the lattice energy. The answer is: lattice energy increases with the increasing charge of the ions, and decreasing radius of the atoms. <em>Therefore, the ranking would be: A < B < C</em>.

4 0
2 years ago
During the discussion of gaseous diffusion for enriching uranium, it was claimed that 235UF6 diffuses 0.4% faster than 238UF6. S
Kay [80]

<u>Answer:</u> The below calculations proves that the rate of diffusion of ^{235}UF_6 is 0.4 % faster than the rate of diffusion of ^{238}UF_6

<u>Explanation:</u>

To calculate the rate of diffusion of gas, we use Graham's Law.

This law states that the rate of effusion or diffusion of gas is inversely proportional to the square root of the molar mass of the gas. The equation given by this law follows the equation:

\text{Rate of diffusion}\propto \frac{1}{\sqrt{\text{Molar mass of the gas}}}

We are given:

Molar mass of ^{235}UF_6=349.034348g/mol

Molar mass of ^{238}UF_6=352.041206g/mol

By taking their ratio, we get:

\frac{Rate_{(^{235}UF_6)}}{Rate_{(^{238}UF_6)}}=\sqrt{\frac{M_{(^{238}UF_6)}}{M_{(^{235}UF_6)}}}

\frac{Rate_{(^{235}UF_6)}}{Rate_{(^{238}UF_6)}}=\sqrt{\frac{352.041206}{349.034348}}\\\\\frac{Rate_{(^{235}UF_6)}}{Rate_{(^{238}UF_6)}}=\frac{1.00429816}{1}

From the above relation, it is clear that rate of effusion of ^{235}UF_6 is faster than ^{238}UF_6

Difference in the rate of both the gases, Rate_{(^{235}UF_6)}-Rate_{(^{238}UF_6)}=1.00429816-1=0.00429816

To calculate the percentage increase in the rate, we use the equation:

\%\text{ increase}=\frac{\Delta R}{Rate_{(^{235}UF_6)}}\times 100

Putting values in above equation, we get:

\%\text{ increase}=\frac{0.00429816}{1.00429816}\times 100\\\\\%\text{ increase}=0.4\%

The above calculations proves that the rate of diffusion of ^{235}UF_6 is 0.4 % faster than the rate of diffusion of ^{238}UF_6

3 0
2 years ago
Lars observes a substance to be a solid and to float in water at room temperature (23°C). Based on the given properties, which s
Triss [41]

Answer:

D. Sulfur Hexafluride

Explanation:

  • above it says the substance floats above water at room temperature and lists some substances and their density at room temp!
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I hope this helped !!

4 0
2 years ago
A sample of gas (1.3 mol) occupies __________ l at 22 °c and 2.5 atm.
Furkat [3]
To answer this question, you just need to put all the variable into <span>PV = nRT formula. In this case, the pressure(P) is 2.5 atm, the number of molecule(n) is 1.3 mol, the temperature(T) is 22 celcius= 295.15 kelvin.
Then the volume should be:

</span><span>PV = nRT
</span>2.5 * V = 1.3 * <span>0.08206* 295.15
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4 0
2 years ago
The van der waals constants a and b for benzene are 18.00 atm l2 mol−2 and 0.115 l mol−1, respectively. Calculate the critical c
Aneli [31]

There are three critical constants namely, critical temperature, critical pressure and critical volume.

Critical temperature is defined as temperature of gas below which the increase in pressure cause liquefaction of gas and above that liquefaction of gas do not take place.

Critical pressure is defined as pressure needed to liquefy a gas at critical temperature. Volume of 1 mol of gas at critical pressure and temperature is known as critical volume.

Critical temperature can be calculated as follows:

T_{C}=\frac{8a}{27Rb}

Putting the value,

T_{C}=\frac{8(18.00atm L^{2}mol^{-2})}{27(0.0821 atm L K^{-1} mol^{-1})(0.115 L mol^{-1})}=564.88 K

Thus, critical temperature is 564.88 K.

Critical pressure is calculated as follows:

P_{C}=\frac{a}{27b^{2}}

Putting the values,

P_{C}=\frac{18.00 atm L^{2}mol^{-2}}{27(0.115 L mol^{-1})^{2}}=50.41 atm

Therefore, critical pressure is 50.41 atm.

Now, calculate critical volume as follows:

V_{C}=3b

Putting the values,

V_{C}=3(0.115 L mol^{-1})=0.345 L mol^{-1}

Therefore, for 1 mol critical volume is 0.345 L.

5 0
2 years ago
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