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julia-pushkina [17]
3 years ago
12

What is the solubility of NH4Cl at 50°C?

Chemistry
2 answers:
NISA [10]3 years ago
8 0

Answer : Option E) 50 grams.

Explanation : According to the solubility curves the compound NH_{4}Cl to dissolve at 50 °C in 100 mL of water will need 50 grams of the compound. It is clearly indicated in the graph which is marked with red that at 50°C approximately 50.4 grams of the compound NH_{4}Cl will be needed to dissolved in 100 mL of water to form a solution.


timurjin [86]3 years ago
6 0
Hi there

Got it, the solubility is 50 grams, i found a solubility chart online for the compound that you are looking for
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a solution is an acid or base and it doesn't react with metal. is it pH more likely to be 4 or 9? explain your answer
coldgirl [10]
It is more likely 9. pH 4 is acidic and pH 9 is basic, and as the pH of a substance gets closer to 0 or 14, the substance becomes more corrosive or reactive. As 4 is closer to 0 than 9 is to 14, there is a much higher chance the solution has a pH of 9, because pH 4 is less neutral and therefore more corrosive/reactive than pH 9.
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3 years ago
Rotation about a carbon-carbon double bond does not readily occur because: __________.1) the overlap of the p orbitals of the ca
Advocard [28]

Answer:

1) The overlap of the p orbitals of the carbon-carbon π bond would be lost

Explanation:

Unlike simple bonds, a double bond can not rotate, since it is not possible to twist the ends of the molecule without breaking the π bond.

In the structure of but-2-ene present in the attachment, we can see the two isomers, <em>cis</em> and<em> trans</em>. These isomers cannot be interconverted by rotation around the carbon-carbon double bond without breaking the π bond.

5 0
3 years ago
a) Whatis the composition in mole fractions of a solution of benzene and toluene that has a vapor pressure of 35 torr at 20 °C?
iogann1982 [59]

Answer:

molar composition for liquid

xb= 0.24

xt=0.76

molar composition for vapor

yb=0.51

yt=0.49

Explanation:

For an ideal solution we can use the Raoult law.

Raoult law: in an ideal liquid solution, the vapor pressure for every component in the solution (partial pressure) is equal to the vapor pressure of every pure component multiple by its molar fraction.  

For toluene and benzene would be:

P_{B}=x_{B}*P_{B}^{o}

P_{T}=x_{T}*P_{T}^{o}

Where:

P_{B} is partial pressure for benzene in the liquid  

x_{B} is benzene molar fraction in the liquid  

P_{B}^{o} vapor pressure for pure benzene.  

The total pressure in the solution is:

P= P_{T}+ P_{B}

And  

1=x_{B}+x_{T}

Working on the equation for total pressure we have:

P=x_{B}*P_{B}^{o} + x_{T}*P_{T}^{o}

Since x_{T}=1-x_{B}

P=x_{B}*P_{B}^{o} + (1-x_{B})*P_{T}^{o}

We know P and both vapor pressures so we can clear x_{B} from the equation.

x_{B}=\frac{P- P_{T}^{o}}{ P_{B}^{o} - P_{T}^{o}}

x_{B}=\frac{35- 22}{75-22} = 0.24

So  

x_{T}=1-0.24 = 0.76

To get the mole fraction for the vapor we know that in the equilibrium:

P_{B}=y_{B}*P

y_{T}=1-y_{B}

So  

y_{B} =\frac{P_{B}}{P}=\frac{ x_{B}*P_{B}^{o}}{P}

y_{B}=\frac{0.24*75}{35}=0.51

y_{T}=1-0.51=0.49

Something that we can see in these compositions is that the liquid is richer in the less volatile compound (toluene) and the vapor in the more volatile compound (benzene). If we take away this vapor from the solution, the solution is going to reach a new state of equilibrium, where more vapor will be produced. This vapor will have a higher molar fraction of the more volatile compound. If we do this a lot of times, we can get a vapor that is almost pure in the more volatile compound. This is principle used in the fractional distillation.

7 0
3 years ago
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