<h3><u>Answer;</u></h3>
321.8 g CaF2
321.5 g Al2(CO3)3
<h3><u>Explanation;</u></h3>
The equation for the reaction is;
3 CaCO3 + 2 AlF3 → 3 CaF2 + Al2(CO3)3
Number of moles of CaCO3 will be;
=(412.5 g CaCO3) / (100.0875 g CaCO3/mol)
= 4.12139 mol CaCO3
Number of moles of AlF3 will be;
= (521.9 g AlF3) / ( 83.9767 g AlF3/mol)
= 6.21482 mol AlF3
But;
4.12139 moles of CaCO3 would react completely with 4.12139 x (2/3) = 2.74759 moles of AlF3.
Thus; there is more AlF3 present than that, so AlF3 is in excess, and CaCO3 is the limiting reactant.
Therefore;
Mass of CaF2 will be;
(4.12139 mol CaCO3) x (3/3) x (78.0752 g CaF2/mol) = 321.8 g CaF2
Mass of Al2(CO3)3 on the other hand will be;
(4.12139 mol CaCO3) x (1/3) x (233.9903 g Al2(CO3)3/mol) = 321.5 g Al2(CO3)3
Answer:
When designers require rugged, tough materials for their projects, steel and titanium are the first options that come to mind. These metals come in a wide assortment of alloys - base metals imbued with other metallic elements that produce a sum greater than its parts. There are dozens of titanium alloys and hundreds more steel alloys, so it can oftentimes be challenging to decide where to begin when considering these two metals. This article, through an examination of the physical, mechanical, and working properties of steel and titanium, can help designers choose which material is right for their job. Each metal will be briefly explored, and then a comparison of their differences will follow to show when to specify one over the other.
Explanation:
in the ph. above
Answer: Yttruim: 39 protons, 50 neutrons, 39 electrons
Ruthenium: 44 protons, 57 neutrons, 44 electrons
M(NaOH)=15 g
v=600 ml=0.6 l
n(NaOH)=m(NaOH)/M(NaOH)
C=n(NaOH)/v=m(NaOH)/[M(NaOH)v]
C=15g/[40g/mol*0.6 l]=0.625 mol/l
0.625M
[H⁺]=7.5*10⁻²
pOH=14-pH=14+lg[H⁺]
pOH=14+lg(7.5*10⁻²)=14-1.125=12.875
