I don’t know let me go call and ask NASA real quick hold up
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
0,15 moles of NaOH-------in------------1000ml
x moles of NaOH------------in--------100ml
x = 0,015 moles of NaOH
final volume = 150ml
0,015 moles of NaOH---in-------150ml
x moles of NaOH--------------in-----1000ml
x = 0,1 moles of NaOH
answer: 0,1mol/dm³ (molarity)
<span>c. Real gases and fusion reactions </span>
1 mole of nitric acid produce 1 mole of ammonium nitrate.
moles in 5000 kg of ammonium nitrate :
( molecular mass of ammonium nitrate is 80 gm/mol )
So, number of moles of nitric acid required are also 62500 moles.
Mass of 62500 moles of nitric acid :
Hence, this is the required solution.
