<span>Calculate the mass of 1 L of solution. Mass of solution=1000mL soln ×1.19 g soln1mL soln =1190 g soln (3 significant figures + 1 guard digit)Calculate the mass of HCl . Mass of HCl=1190g soln ×37.7g HCl100g soln =448.6 g HCl.Calculate the moles of HCl . ...Calculate the molarity of the HCl.</span>
A compound has to be chemically bonded, however, air is not chemically bonded.
This can be proven by freezing air. By freezing air, it yields different liquids at different temperature. Liquid nitrogen has a different boiling point than liquid oxygen.
If air was a compound, they would all have a single boiling point and a single freezing point.
Hope this helps :)
Answer: The average atomic mass of the element = 88.242amu
Explanation:
The abundance of the first isotope is =35.5%
Atomic mass of first isotope = 68.9257
The average atomic mass of the first isotope =86.95amu X 35.5% =86.95amu X 0.355 =30.8725 amu
The abundance of the second isotope =64.5%
Atomic mass of the second isotope =88.95amu
The average atomic mass of second isotope =88.95amu x 64.5% = 88.95amu x 0.645= 57.37275 amu
Now the average atomic mass =30.8725 +57.37275 = 88.242amu
OR using the formulae
Average atomic mass = [mass of isotope× its abundance] + [mass of isotope× its abundance] +...[ ] / 100
{(86.95amu X 35.5 )+(88.95amu x 64.5)}/100
8,824/100
=88.24amu
The effect of an insoluble impurity, such as sand, on the observed melting point of a compound would be none. It will not depress or elevate the melting point of the compound. Instead, it would affect the reading if you are trying to determine the melting point of the compound. This is because you might be missing the actual melting point of the compound since you will be waiting for the whole sample to liquify. You would not be able to determine exactly that temperature because of the insoluble impurity would have a different melting point than that of the compound.
