Answer:
Carbonyl
Explanation:
While the diagram is slightly unclear, the molecule most likely being shown is a carbonyl. A molecule is a carbonyl when there is a carbon double-bonded to an oxygen.
Answer:
Molecular weight of the compound = 372.13 g/mol
Explanation:
Depression in freezing point is related with molality of the solution as:
Where,
= Depression in freezing point
= Molal depression constant
m = Molality
m = 0.26
Molality =
Mass of solvent (toluene) = 15.0 g = 0.015 kg
Moles of compound = 0.015 × 0.26 = 0.00389 mol
Mass of the compound = 1.450 g
Molecular weight =
Reactant are those that combine or reacts to give products !!
so in combustion of ethane ; ethane and o2 are reactants so ... your answer is B !!
Answer:
IO₂
Explanation:
We have been given the mass percentages of the elements that makes up the compound:
Mass percentage given are:
Iodine = 79.86%
Oxygen = 20.14%
To calculate the empirical formula which is the simplest formula of the compound, we follow these steps:
> Express the mass percentages as the mass of the elements of the compound.
> Find the number of moles by dividing through by the atomic masses
> Divide by the smallest and either approximate to nearest whole number or multiply through by a factor.
> The ratio is the empirical formula of the compound.
Solution:
I O
% of elements 79.86 20.14
Mass (in g) 79.86 20.14
Moles(divide by
Atomic mass) 79.86/127 20.14/16
Moles 0.634 1.259
Dividing by
Smallest 0.634/0.634 1.259/0.634
1 2
The empirical formula is IO₂