Answer:
The energy of photon is 33.28×10⁻¹⁴ J
Explanation:
Given data:
Frequency of photon = 5.02×10²⁰ Hz
Energy of photon = ?
Solution:
E = h.f
h = planc'ks constant = 6.63×10⁻³⁴ Js
by putting values,
E = 6.63×10⁻³⁴ Js × 5.02×10²⁰ Hz
Hz = s⁻¹
E = 33.28×10⁻¹⁴ J
The energy of photon is 33.28×10⁻¹⁴ J.
Answer:
The ¹³C-NMR Spectrum of <em>tert</em>-butyl alcohol will show only two signals.
(i) Signal at around 31 ppm:
This signal towards upfield is for the carbon atoms which are more shielded and are having rich electron surroundings. The height of peak at y-axis shows the number of carbon atoms as compared to other peaks. In this case it is three times the height of second signal hence, it shows that this peak corresponds to three carbon atoms.
(ii) Signal at around 70 ppm:
This signal towards downfield is for the carbon atom which is more deshielded and is having electron deficient surrounding. As compared to the second signal the height of this peaks corresponds to only one carbon. And the deshielded environment shows that this carbon is directly attached to an electronegative element.
Answer:
Explanation:
Flame test:
The metals ions can be detected through the flame test. Different ions gives different colors when heated on flame. Tom perform the flame test following steps should follow:
1. Dip a wire loop in the solution of compound which is going to be tested.
2. After dipping put the loop of wire on bunsen burner flame.
3. Observe the color of flame.
4. Record the flame color produce by compound
Color produce by metals:
Red = Lithium, zirconium, strontium, mercury, Rubidium (red violet)
Orange-red = calcium
Yellow = sodium, iron (brownish yellow)
Green = green
Blue = cesium. arsenic, copper, tantalum, indium, lead
Violet = potassium (lilac)
It is a covalent compound
Answer:
May be the answer is C. Coal is being used faster than it is being produced
