The answer is both B and C
positive chromic acid test is indicated by disappearance of orange colour from chromic ions and appearance of blue-green color from Chromium (iii) ion (reduction of chromium ion from CrO4 - to Cr3+)
positive chromic test indicated functional groups that can be oxidized.
cyclohexanol can be oxidized to become cyclohexanone
and pentan-3-ol can be oxidized to become pentan-3-one
hence both B and C will show positive chromic acid test
A) tert butanol although contains alcohol functional group, cannot be further oxidized as it is a tertiary alcohol
Pure substances are further broken down into elements and compounds. Mixture are physically combined structures that can be separated into their original components. A chemical substance is composed of one type of atom or molecule.
Answer:
um pretty sure this is it
Explanation:
The fission process also releases extra neutrons, which can then split additional atoms, resulting in a chain reaction that releases a lot of energy.
Answer:
12. is the pressure equilibrium constant for the decomposition of ammonia at the final temperature of the mixture.
Explanation:
initially
3.0 atm 0 0
At equilibrium
(3.0-2p) p 3p
Equilibrium partial pressure of nitrogen gas = p = 0.90 atm
The expression of a pressure equilibrium constant will be given by :
12. is the pressure equilibrium constant for the decomposition of ammonia at the final temperature of the mixture.
Answer:
The final temperature of the solution is 44.8 °C
Explanation:
assuming no heat loss to the surroundings, all the heat of solution (due to the dissolving process) is absorbed by the same solution and therefore:
Q dis + Q sol = 0
Using tables , can be found that the heat of solution of CaCl2 at 25°C (≈24.7 °C) is q dis= -83.3 KJ/mol . And the molecular weight is
M = 1*40 g/mol + 2* 35.45 g/mol = 110.9 g/mol
Q dis = q dis * n = q dis * m/M = -83.3 KJ/mol * 13.1 g/110.9 gr/mol = -9.84 KJ
Qdis= -9.84 KJ
Also Qsol = ms * Cs * (T - Ti)
therefore
ms * Cs * (T - Ti) + Qdis = 0
T= Ti - Qdis * (ms * Cs )^-1 =24.7 °C - (-9.84 KJ/mol)/[(104 g + 13.1 g)* 4.18 J/g°C] *1000 J/KJ
T= 44.8 °C
