Answer:
The compound which are boh soluble in water and hexane is
B. Ethanol and 1-propanol
Explanation:
The compounds ethanol and 1-propanol are soluble in both hexane and water.
It is soluble in water as both consists of polar end due to hydrogen bonding present in the -OH functional group.
and both are soluble in hexane as they contain a non polar end and the alliphatic hydrocarbon chain in them.
The solubility of alcohols varies in increasing order as the hydrocarbon chain increases. And becaue of this it becomes more non polar.
Non polar properties decreases for branched molecules.
so, the correct option is ethanol and 1-propanol.
Answer:
1
Explanation:
Using the Rydberg formula as:

where,
λ is wavelength of photon
R = Rydberg's constant (1.097 × 10⁷ m⁻¹)
Z = atomic number of atom
n₁ is the initial final level and n₂ is the final energy level
For Hydrogen atom, Z= 1
n₂ = 2
Wavelength = 410.1 nm
Also,
1 nm = 10⁻⁹ m
So,
Wavelength = 410.1 × 10⁻⁹ m
Applying in the formula as:

Solving for n₁ , we get
n₁ ≅ 1
Answer:
A. Chipping ice to flakes
Explanation:
You can't reverse the ice becoming flakes. They will stay like that until they melt
Answer:
the answer should be A. the sum of all forces acting on an object
Explanation:
Answer:
(a) Rate at which
is formed is 0.050 M/s
(b) Rate at which
is consumed is 0.0250 M/s.
Explanation:
The given reaction is:-

The expression for rate can be written as:-
![-\frac{1}{2}\frac{d[NO]}{dt}=-\frac{d[O_2]}{dt}=\frac{1}{2}\frac{d[NO_2]}{dt}](https://tex.z-dn.net/?f=-%5Cfrac%7B1%7D%7B2%7D%5Cfrac%7Bd%5BNO%5D%7D%7Bdt%7D%3D-%5Cfrac%7Bd%5BO_2%5D%7D%7Bdt%7D%3D%5Cfrac%7B1%7D%7B2%7D%5Cfrac%7Bd%5BNO_2%5D%7D%7Bdt%7D)
Given that:-
(Negative sign shows consumption)
![-\frac{1}{2}\frac{d[NO]}{dt}=\frac{1}{2}\frac{d[NO_2]}{dt}](https://tex.z-dn.net/?f=-%5Cfrac%7B1%7D%7B2%7D%5Cfrac%7Bd%5BNO%5D%7D%7Bdt%7D%3D%5Cfrac%7B1%7D%7B2%7D%5Cfrac%7Bd%5BNO_2%5D%7D%7Bdt%7D)
![-\frac{d[NO]}{dt}=\frac{d[NO_2]}{dt}](https://tex.z-dn.net/?f=-%5Cfrac%7Bd%5BNO%5D%7D%7Bdt%7D%3D%5Cfrac%7Bd%5BNO_2%5D%7D%7Bdt%7D)
![-(-0.050\ M/s)=\frac{d[NO_2]}{dt}](https://tex.z-dn.net/?f=-%28-0.050%5C%20M%2Fs%29%3D%5Cfrac%7Bd%5BNO_2%5D%7D%7Bdt%7D)
![\frac{d[NO_2]}{dt}=0.050\ M/s](https://tex.z-dn.net/?f=%5Cfrac%7Bd%5BNO_2%5D%7D%7Bdt%7D%3D0.050%5C%20M%2Fs)
(a) Rate at which
is formed is 0.050 M/s
![-\frac{1}{2}\frac{d[NO]}{dt}=-\frac{d[O_2]}{dt}](https://tex.z-dn.net/?f=-%5Cfrac%7B1%7D%7B2%7D%5Cfrac%7Bd%5BNO%5D%7D%7Bdt%7D%3D-%5Cfrac%7Bd%5BO_2%5D%7D%7Bdt%7D)
![-\frac{1}{2}\times -0.050\ M/s=-\frac{d[O_2]}{dt}](https://tex.z-dn.net/?f=-%5Cfrac%7B1%7D%7B2%7D%5Ctimes%20-0.050%5C%20M%2Fs%3D-%5Cfrac%7Bd%5BO_2%5D%7D%7Bdt%7D)
![\frac{d[O_2]}{dt}=0.0250\ M/s](https://tex.z-dn.net/?f=%5Cfrac%7Bd%5BO_2%5D%7D%7Bdt%7D%3D0.0250%5C%20M%2Fs)
(b) Rate at which
is consumed is 0.0250 M/s.