D.
Misinterpretation of scientific knowledge for the influence of public opinion has been a repeating problem over the past, and usually creates a rift between the scientific community and the general public.
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.
Answer:
K = 10
Explanation:
Using Hess's law, it is possible to obtain the equilibrium constant, K, of a reaction using K of similar reactions. For example:
<em> If A ⇄ B K = X</em>
B ⇄ A K = 1/X
2A ⇄ 2B K = X².
Thus, if A(g) ⇄ 2B(g) K = 0.010
2B(g) ⇄ A(g) K = 1 / 0.010; K = 100
B(g) ⇄ A(g) K = √100 = 10
<h3>K = 10</h3>
In a titration, for an acid to neutralize a base, at the equivalence point, there should be an equal number of moles of H+ and OH-.
Moles of OH- can be found by multiplying the concentration of the base by the volume. (You will need to keep in mind the stoichimetric coefficients if the strong base is Ca(OH)₂, Ba(OH)₂, or Sr(OH)₂.
Moles of OH- = moles of H+
(0.253 M) * 0.005 L = 0.01000 L * c
c = 0.1265 M
The concentration of HBr is 0.127 M.
