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:
58.9mL
Explanation:
Given parameters:
Initial volume = 34.3mL = 0.0343dm³
Initial concentration = 1.72mM = 1.72 x 10⁻³moldm⁻³
Final concentration = 1.00mM = 1 x 10⁻³ moldm⁻³
Unknown:
Final volume =?
Solution:
Often times, the concentration of a standard solution may have to be diluted to a lower one by adding distilled water. To find the find the final volume, we must recognize that the number of moles of the substance in initial and final solutions are the same.
Therefore;
C₁V₁ = C₂V₂
where C and V are concentration and 1 and 2 are initial and final states.
now input the variables;
1.72 x 10⁻³ x 0.0343 = 1 x 10⁻³ x V₂
V₂ = 0.0589dm³ = 58.9mL
Answer:
<h2>All Group 1 metals form halides that are white solids at room temperature. The melting point is correlated to the strength of intermolecular</h2>
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