Answer:
Explanation:
a ) Total mixture = 4.656 g
Sand recovered = 2.775 g
percent composition of sand in the mixture
= (2.775 g / 4.656 g ) x 100
= 59.6 % .
b )
Total of sand and salt recovered = 2.775 g + .852 g = 3.627 g .
Total mixture = 4.656 g
percent recovery = (3.627 / 4.656 ) x 100
= 77.9 % .
Answer:
i think so
Ba(OH)2 + H2SO4 ------> BaSO4 + 2H2O
1) Moles of Ba(OH)2 = moles of H2SO4 = 0.025L x 2)0.02M = 5.0 x 10^-4M
Concn of Ba(OH)2 in g/L = 5.0 x 10^-4M x 171.33g/mol = 0.086g/mol
the compounds in which phosphorous posses the highest possible oxidation have to mention here.
The species in which phosphorous have the highest oxidation state are: H₃PO₄, P₂O₅, PCl₅
The possible oxidation state of phosphorous is III and V. The highest oxidation state is V. There are several compounds in which phosphorous posses the +5 oxidation state. Like- Phosphoric acid (H₃PO₄), phosphorous pentoxide (P₂O₅), Phosphorous chloride (PCl₅) etc.
The oxidation state of an element depends upon the valence electron the valence shell of phosphorous is 3s² 3p³. Thus there are 5 electrons, as it has vacant 3d orbital thus it can easily form compound having +5 oxidation state.
Answer:
T₂ = 721 k
Explanation:
Given data:
Initial volume = 285 mL
Initial pressure = 1.88 atm
Initial temperature = 355 K
Final temperature = ?
Final volume = 435 mL
Final pressure = 2.50 atm
Formula:
P₁V₁/T₁ = P₂V₂/T₂
P₁ = Initial pressure
V₁ = Initial volume
T₁ = Initial temperature
P₂ = Final pressure
V₂ = Final volume
T₂ = Final temperature
Solution:
P₁V₁/T₁ = P₂V₂/T₂
T₂ = P₂V₂ T₁ / P₁V₁
T₂ = 2.50 atm × 435 mL × 355 K / 1.88 atm × 285 mL
T₂ = 386062.5 atm. mL. K /535.8 atm. mL
T₂ = 721 k
Answer is: B₂O₃ + Mg → B + MgO.
B₂O₃ + 3Mg → 2B + 3MgO, balanced.
B₂O₃ - diboron trioxide. White, glassy solid compound.
B - boron. Metalloid, crystalline boron is and amorphous<span> boron is a brown powder.
Mg - magnesium. S</span><span>hiny, gray and solid metal.
MgO - magnesium oxide. W</span><span>hite, </span>hygroscopic<span> solid powder.</span>