Answer:
m(Na3AlF6)=57.6537 kg. Leftover mass will be 52.8644 kg
Explanation:
Al2O3(s)+6NaOH(l)+12HF(g)⟶2Na3AlF6+9H2O(g)
n(Al2O3)=m(Al2O3)/M(Al2O3)=14000/101.96=137.31 mol
n(NaOH)=m(NaOH)/M(NaOH)=59400/40=1485 mol
n(HF)=m(HF)/M(HF)=59400/20.01=2968 mol
To know what is in excess we divide moles by coefficients and compare numbers.
n(Al2O3) = 137.31 mol
n(NaOH)/6=247.5 mol
n(HF)/12=247.33 mol
so NaOH and HF are in excess
So
m(Na3AlF6)=n(Na3AlF6)*M(Na3AlF6)/1000=n(Al2O3)*2*M(Na3AlF6)/1000=57.6537 kg.
Leftover mass will be (n(NaOH)-n(Al2O3)*6)*40 + (n(HF)-n(Al2O3)*12)*20.01=26445.6+26418.8=52864.4 g= 52.8644 kg
It’s a convergent boundary which means they are sliding
Stratosphere :) hope this helped
The buoyancy of an object is dictated by its density. So let us calculate for density, where:density = mass / volume
Calculate the volume first of a solid cube:volume = (6 cm)^3 = 216 cm^3 = 216 mL
Therefore density is:density = 270 g / 216 mLdensity = 1.25 g / mL
Therefore this object will float in the layer in which the density is more than 1.25 g / mL.