Answer is: volume of helium is 244.72 liters.
m(He) = 43.7 g.
n(He) = m(He) ÷ M(He).
n(He) = 43.7 g ÷ 4 g/mol.
n(He) = 10.925 mol.
V(He) = n(He) · n(He).
V(He) = 10.925 mol · 22.4 L/mol.
V(He) = 244.72 L.
Vm - molar volume at STP.
n - amount of substance.
Answer:
C. 0.04 moles per cubic decimeter.
Explanation:
The molar mass of the Iodine is 253.809 grams per mole and a cubic decimeter equals 1000 cubic centimeters. The concentration of Iodine (
), measured in moles per cubic decimeter, can be determined by the following formula:
(1)
Where:
- Mass of iodine, measured in grams.
- Molar mass of iodine, measured in grams per mol.
- Volume of solution, measured in cubic decimeters.
If we know that
,
and
, then the concentration of iodine in a solution is:


Hence, the correct answer is C.
Answer:
The possible structures are ketone and aldehyde.
Explanation:
Number of double bonds of the given compound is calculated using the below formula.

=Number of double bonds
= Number of carbon atoms
= Number of hydrogen atoms
= Number of nitrogen atoms
The number of double bonds in the given formula - 

The number of double bonds in the compound is one.
Therefore, probable structures is as follows.
(In attachment)
The structures I and III are ruled out from the probable structures because the signal in 13C-NMR appears at greater than 160 ppm.
alkene compounds I and II shows signal less than 140 ppm.
Hence, the probable structures III and IV are given as follows.
The carbonyl of structure I appear at 202 and ketone group of IV appears at 208 in 13C, which are greater than 160.
Hence, the molecular formula of the compound
having possible structure in which the signal appears at greater than 160 ppm are shown aw follows.
Complete Question
The complete question is shown on the first uploaded image
Answer:
The correct option is 
Explanation:
From the question we are told that
the cell voltage for AD is 
From the data give we can see that

i.e 
In the same way we can say that

=> 

2AgNO3+K2CrO4⇒Ag2CrO4(s)+2KNO3
Hence by mixing 0.0024M AgNO3 and 0.004M
K2CrO4, we will have Ag2CrO4 which is precipitated out and leave us with
0.0024M KN03 which is mixed with (0.004-0.0024/2)M, it can be 0.0028M, of K2Cr04