Mass of SiC = 2 g
<h3>Further explanation</h3>
Given
Reaction
SiO₂(s) + 3C(s) → SiC(s) + 2CO(g)
3.00 g of SiO₂
4.50 g of C
Required
mass of SiC
Solution
mol SiO₂ (MW=60,08 g/mol) :
= 3 g : 60.08 g/mol
= 0.0499
mol C(Ar = 12 g/mol) :
= 4.5 g : 12 g/mol
= 0.375
mol : coefficient of reactants =
SiO₂ : 0.0499/1 = 0.0499
C : 0.375/3 = 0.125
SiO₂ as a limiting reactant(smaller ratio)
Mol SiC based on mol SiO₂ = 0.0499
Mass SiC :
= mol x MW
= 0.0499 x 40,11 g/mol
= 2 g
Answer:
C2H6O + 3O2 —> 2CO2 + 3H2O
Explanation:
Answer:
5.96 g/cm^3
Explanation:
Corner atom = 1/8
Atoms in center = 1
Atoms in face of the cube= 1/2
Molar mass of V = 50.94 g/mol <em>(from period table)</em>
1 mole = 6.02x10^23
<em>In BCC unit cell:</em>
(8 x 1/8)+ 1=2 per 1 unit cell
<em>Mass: </em>2(50.94g)/6.02x10^23 = 1.69x10^-22 g/unit cell
305pm=(305x10^-12m÷10^-2m) x (1mL÷1cm^3)
= 2.837 x 10^-23 mL
<em>1pm=10^-12m</em>
<em>1cm=10^-2m</em>
<em>1mL=1cm^3</em>
<em></em>
density=mass/volume
density of V = 1.69x10^-22g÷2.837x10^-23mL
=5.957g/mL
=5.96g/cm^3
Answer:
D. Air pressure lowers the temperature of the liquid molecules.
