Answer:
CHCL3= 11.73 mL
CHBr3= 8.268 mL
Explanation:
Let x be the mL of CHCl3 and y be the mL of CHBr3, then we have:
y+x = 20 mL
1.492x+2.89y=2.07* 20.0 mL
Oxidation state number can be used to determine the unknown element in these two compounds. They are used to determine how many electrons are given, taken or shared to form compounds.
Recall the elementary rules of oxidation numbers.
1. The sum of all oxidation numbers in a neutral compound is zero.
2. Chlorine, bromine, iodine have oxidation number of -1 ( except compounds with fluorine and oxygen)
Let oxidation number of element M be x.
Check rule 2. Chlorine has -1 oxidation number.
Now we write an equation of MCl₂ (neutral compound)
x + (2 * -1)= 0 ⇒ x₁= +2
For MCl₃
x + (3 * -1)= 0 ⇒ x₂= +3
So element has 2 different oxidation number in compounds, +2 and +3.
The element is iron (Fe) since it has +2 and +3 oxidation numbers in the compounds.
You need to learn it by hard. Unfortunately there is not an easier way to work out with these oxidation numbers.
The answer is iron (Fe).
Answer:
Ratios in order of increasing value ; The ratio of the mass ratio of Y to X in XY2 to the mass ratio of Y to X in XY, The ratio of the mass ratio of Y to X in XY3 to the mass ratio of Y to X in XY, The ratio of the mass ratio of Y to X in XY4 to the mass ratio of Y to X in XY
1) Mass ratio = 3
2) Mass ratio = 2
3) Mass ratio = 4
Explanation:
The detailed and step by step calculation is shown in the attachment.
Answer is: the boiling point of the resulting solution of sucrose is 100.42°C.
m(H₂<span>O) = 15.2 g ÷ 1000 g/kg = 0.0152 kg.
</span>m(C₁₂H₂₂O₁₁<span>) = 4.27 g.
n</span>(C₁₂H₂₂O₁₁) = m(C₁₂H₂₂O₁₁) ÷ M(C₁₂H₂₂O₁₁).
n(C₁₂H₂₂O₁₁) = 4.27 g ÷ 342.3 g/mol.
n(C₁₂H₂₂O₁₁) = 0.0125 mol.
b(solution) = n(C₁₂H₂₂O₁₁) ÷ m(H₂O).
b(solution) = 0.0125 mol ÷ 0.0152 kg.
b(solution) = 0.82 m.
ΔT = b(solution) · Kb(H₂O).
ΔT = 0.82 m · 0.512°C/m.
ΔT = 0.42°C.
Tb = 100°C + 0.42°C = 100.42°C.
Answer:
324.18 g/mol
Explanation:
Let the molecular mass of the antimalarial drug, Quinine is x g/mol
According to question,
Nitrogen present in the drug is 8.63% of x
So, mass of nitrogen = 
Also, according to the question,
2 atoms are present in 1 molecule of the drug.
Mass of nitrogen = 14.01 amu = 14.01 g/mol (grams for 1 mole)
So, mass of nitrogen = 14.01×2 = 28.02
These 2 must be equal so,
solving for x, we get:
<u>x = 324.18 g/mol</u>
