I think the answer is C. Runoff hopefully this helps
Answer:
cinnamic acid - 150 mg
cis-stilbene - 100 μL
trans- stilbene - 100 mg
pyridinium tribromide - 200-385 mg
For this data:
moles of cinnamic acid = 0.150 g/148.16 g/mol = 0.001 mols
Theoretical mass of dibromoproduct formed = 0.001 mol x 307.97 g/mol = 0.312 g
cis-stilbene (100 ul = 0.1 ml)
moles of cis-stilbene = 0.1 ml x 1.01 g/mol/180.25 g/mol = 0.00056 mols
Theoretical mass of dibromoproduct formed = 0.00056 mol x 340.05 g/mol = 0.19 g
trans-stilbene
moles of tran-stilbene = 0.1 g/180.25 g/mol = 0.00055 mols
Theoretical mass of dibromoproduct formed = 0.00055 mol x 340.05 g/mol = 0.19 g
Explanation:
A is correct. water is incompressible, so it will always have the same density, regardless of volume
Answer:
259.497 mg, 58.84%
Explanation:
BaSO₄ → Ba²⁺ + SO₄²⁻
to calculate the mole of BaSO₄
mole BaSO₄ = mass given / molar mass = 403 mg / 233.38 g/mol = 1.7268 mol
comparing the mole ratio
1.7268 mol of BaSO₄ yields 1.7268 mol of Ba²⁺
403 mg BaSO₄ yields ( 1.7268 × 137.327 ) where 137.327 is the molar mass of Barium mol of Ba²⁺
441 mg BaSO₄ will yield ( 1.7268 × 137.327 × 441 mg ) / 403 mg = 259 .497 mg
mas percentage of the Barium compound = 259 .497 mg / 441 mg × 100 = 58.84%
