To answer the problem given:
|0.53−4.0| / 4.0 * 100%
= 3.47 / 4.0 * 100%
= 87%
The maximum theoretical percent recovery from the
crystallization of 4.0 g of acetanilide from 100 ml of water is 87%. I
am hoping that this answer has satisfied your query and it will be able to help
you in your endeavor, and if you would like, feel free to ask another question.
Answer:
-) 3-bromoprop-1-ene
-) 2-bromoprop-1-ene
-) 1-bromoprop-1-ene
-) bromocyclopropane
Explanation:
In this question, we can start with the <u>I.D.H</u> (<em>hydrogen deficiency index</em>):

In the formula we have 3 carbons, 5 hydrogens, and 1 Br, so:

We have an I.D.H value of one. This indicates that we can have a cyclic structure or a double bond.
We can start with a linear structure with 3 carbon with a double bond in the first carbon and the Br atom also in the first carbon (<u>1-bromoprop-1-ene</u>). In the second structure, we can move the Br atom to the second carbon (<u>2-bromoprop-1-ene</u>), in the third structure we can move the Br to carbon 3 (<u>3-bromoprop-1-ene</u>). Finally, we can have a cyclic structure with a Br atom (<u>bromocyclopropane</u>).
See figure 1
I hope it helps!
Answer:
Explanation:
The following equation relates to Born-Haber cycle

Where
is enthalpy of formation
S is enthalpy of sublimation
B is bond enthalpy
is ionisation enthalpy of metal
is electron affinity of non metal atom
is lattice energy
Substituting the given values we have
-435.7 = 79.2 + 1/2 x 242.8 + 418.7 - 348 +U_L
= - 707 KJ / mol
First step is to get the mass of the mercury:
Pressure = mass/volume
mass = pressure x volume = 13.5 x 1.85 = 24.975 gm
Second step is to calculate the number of moles in 24.975 gm:
From the periodic table, the molar mass of mercury is 200.59
mass = number of moles x molar mass
number of moles = 24.975 / 200.59 = 0.1245 mole
Last step is to get the number of atoms:
Each mole contains 6.02 x 10^23 atoms
number of atoms = 0.1245 x 6.02 x 10^23 = 7.4949 x 10^22 atoms
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