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juin [17]
3 years ago
11

The chart shows the solubility of different substances.

Chemistry
2 answers:
Annette [7]3 years ago
8 0
Answer: 1) Temperature can change the solubility of a solute.

Explanation:

The chart is missing so there is no way to tell what does the graph show.

Yet, I can help you because I can explain the status of each statement of the choices. As you will see there is only one possibility..

<span>1) Temperature can change the solubility of a solute.

Yes, temperature definetly can, and mostly do, modify the solubility of a solute.

You can search any chart of solubility and will find that.

I can give you two examples:

a) Sodium chloride: dissolve some spoons of salt in a cold water  until you can not dissolve more. Then, heat the water, you will find that more salt will get dissolved, proving that the temperature of the solution increases the solubility of sodium chloride.

b) Carbon dioxide gas: the soft drinks have CO₂ molecules dissolved in it.
 
The higher the temperature of the soft drink the less the amount of CO₂(g) that can be dissolved. That is why the soda bottling plants cool the beverage before adding the CO₂(g).

2) </span><span>Temperature has no affect on the solubility of a solute.

Since this is the opposite to the first statement and the first is true, this is false.

3) Salt has a greater solubility than sugar.

False.

This is an empirical result, which you cannot predict theoretically. So you need to see at the data either in a table or in a chart. Else you can test it at home. After the empirical data are shown it results that more grams of sugar can be dissolved in water compared to salt.

That is something you ca see in a chart or you can prove by yourself.

4) Nitrite salt has a greater solubility than sugar. </span>

False.

Looking at some data you can find that sodium nitrite solutiliby is aroun  70 - 100 g/10 g while sugar (sucrose) solutiblity is around 180 - 235 g/ 100 g.

Gnoma [55]3 years ago
4 0

Answer:

Temperature can change the solubility of a solute

Explanation:

i know this because i am big brained

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A balloon, whose volume at 23°C is 535 mL, is heated to 46°C. Assuming the pressure and amount of gas remain constant, what is t
NikAS [45]
This problem requires a certain equation.  That equation is V1/T1=V2/T2, where V1 is your initial volume (535 mL in this case), T1 is your initial temperature in Kelvin(23 degrees C = 296 K), V2 is your final volume (unknown), and T2 is your final temperature (46 degrees C = 319 K). By plugging in these values, the equation looks like this: 535/296=V2/319.  Now multiply both sides of the equation by 319, and your final answer is V2= 576.6 mL
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Answer:

Q = 114349.5 J

Explanation:

Hello there!

In this case, since this a problem in which we need to calculate the total heat of the described process, it turns out convenient to calculate it in three steps; the first one, associated to the heating of the liquid water from 40 °C to 100 °C, next the vaporization of liquid water to steam at constant 100 °C and finally the heating of steam from 100 °C to 115 °C. In such a way, we calculate each heat as shown below:

Q_1=45g*4.18\frac{J}{g\°C}*(100\°C-40\°C)=11286J\\\\Q_2=45g* 2260 \frac{J}{g} =101700J\\\\Q_3=45*2.02\frac{J}{g\°C}*(115\°C-100\°C)=1363.5J

Thus, the total energy turns out to be:

Q_T=11286J+101700J+1363.5J\\\\Q_T=114349.5J

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