The second level has s and p orbitals, the s orbital holds two electrons and all three p orbitals also hold two each for a total of eight electrons in the second energy level.
Answer:
The student should weigh out 61.2g of ethanolamine [6.12 * 10]
Explanation:
In this question, we are expected to calculate the mass of ethanolamine needed to make 60.0ml of it given that the density of the ethanolamine in question is 1.02g/cm^3
Mathematically, it has been shown that mass = density * volume
Hence, by multiplying the density by the volume, we get the mass.
Now, from the question we can see that we have the values for the density and the volume. We now need to get the mass.
Since cm^3 is same as ml, we need not perform any conversion.
Hence, the needed mass is:
60 * 1.02 = 61.2g
Answer:
16:1
Explanation:
Atoms of element X weigh 32 times more than atoms of element Y. We can write this in a symbolic way.
mX = 32 mY [1]
where,
- mX and mY are the masses of X and Y, respectively
A compound has the formula: XY₂, that is, in 1 molecule of XY₂ there is 1 atom of X and 2 atoms of Y. The ratio of the mass of X to the mass of Y in this compound equals:
mX/2 mY [2]
If we substitute [1] in [2], we get:
mX/2 mY = 32 mY/2 mY = 16 = 16:1
Answer:
2 L is the new volume
Explanation:
We can apply the Ideal Gases Law to solve the problem.
At STP, we kwow that 1 mol of gas occupy a volume of 22.4 L
Then, how many moles do we have in 1 L?
Let's do it by a rule of three:
(1L . 1 mol) / 22.4L = 0.0446 moles
These moles are at 1 atm and 273 K of temperature, so let's change our conditions.
P . V = n . R .T
1 atm . V = 0.0446 mol . 0.082 L.atm/mol K . 546 K
V = 2 L
If we pay attention, we can notice that, if we double temperature, we double the volume.
