To find the mass you you need multiply its volume by its density.
6 cm^3 • 2.76 g/cm^3 = 16.56 g
Answer:
8.88 grams of aluminum nitrate should be weighted.
Explanation:
Hello there!
In this case, according to the given information, it turns out possible for us to use the definition of molarity to calculate the moles of aluminum nitrate as follows:
Now, since the molar mass of aluminum nitrate is 212.996 g/mol, we obtain the following mass:
Therefore, 8.88 grams of aluminum nitrate should be weighted.
Regards!
Explanation:
It is known that the maximum value of ml is equal to the vale of l. But the minimum value of n is as follows.
n = l + 1
where, n = principle quantum number
l = azimuthal quantum number
Values of n can be 1, 2, 3, 4 and so on. Whereas the values of l is 0 for s, 1 for p, 2 for d, 3 for f, and so on.
Also, "m" is known as magnetic quantum number whose values can be equal to -l and +l.
- Electronic configuration of Rb is . So here, n = 5, l = 0, m = 0 and s = ± .
- Electronic configuration of is . So here, n = 3, l = 1, m = -1, 0, +1, and s = ± .
- Electronic configuration of is . So here, n = 4, l = 2, m = -2, -1, 0, +1, +2 and s = ± .
- Electronic configuration of F is . So here, n = 2, l = 1, m = -1, 0, +1, and s = ± .
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