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:a line or a circle but i think circle
Explanation:
Answer:
Heat transfer = 3564 Jolues
The same value
Explanation:
The heat of combustion is the heat released per 1 mole of substunce experimenting the combustion at standard conditions of pressure and temperature ( 1 atm, 298 K):
Qtransfer = - mol x ΔHºc Qtransfer
So look up in appropiate reference table ΔHºc and solve the problem:
ΔHºc = - 891 kJ/mol
Qtransfer = - (4 x 10³ mol x -891 kJ/mol ) = 3564 J
if the combustion were achieved with 100 % excess air, the result will still be the same. As long as the standard conditions are maintained, the heat of combustion remains constant. In fact in many cases the combustion is performed under excess oxygen to ensure complete combustion.
Answer:
The colliding molecules need to possess certain energy which is greater than the activation energy Ea and proper orientation.
<h3>Not all collisions result in a chemical reaction because not all collisions have the have sufficient amount of energy or have the appropriate activation energy. Nor do they all have this correct orientation. Molecules need to collide in such a way that they're oriented, so the correct bonds could break in. The correct bonds conform. And then, of course, they need to have sufficient activation energy in order to initiate the breaking of the bonds.</h3>
<h3>Hope this helps please let me know If I'm wrong.</h3>
Physicist Ernest Rutherford<span> established the nuclear theory of the atom with his </span>gold-foil experiment<span>. When he shot a beam of alpha particles at a sheet of </span>gold foil<span>, a few of the particles were deflected. He concluded that a tiny, dense nucleus was causing the deflections.</span>
