Answer:
The frequency of photon is 0.75×10¹⁵ s⁻¹.
Explanation:
Given data:
Energy of photon = 5×10⁻¹⁹ J
Frequency of photon = ?
Solution:
Formula;
E = hf
h = planck's constant = 6.63×10⁻³⁴ Js
5×10⁻¹⁹ J = 6.63×10⁻³⁴ Js ×f
f = 5×10⁻¹⁹ J / 6.63×10⁻³⁴ Js
f = 0.75×10¹⁵ s⁻¹
The frequency of photon is 0.75×10¹⁵ s⁻¹.
Answer:
a) increases
b) decreases
c) does not change
d) increases
Explanation:
The vapour pressure of a liquid is dependent on;
I) the magnitude of intermolecular forces
II) the temperature of the liquid
Hence, when any of these increases, the vapour pressure increases likewise.
Similarly, the boiling point of a liquid depends on the magnitude of intermolecular forces present because as intermolecular forces increases, more energy is required to break intermolecular bonds.
Lastly, increase in surface area of a liquid does not really affect it's vapour pressure.
Your answer is (4) 4, because the alpha particle is a helium nucleus (either an ion or an ejected particle from alpha decay), and it has a mass number of 4, being consisted of four nucleons, (2 protons and 2 neutrons).
Answer:
2. All the naturally occurring isotopes of Mg.
Explanation:
You want to know the atomic mass of the magnesium you use in the lab. That’s “natural” magnesium. So, you must use the weighted average of all the naturally occurring isotopes in natural Mg.
1. and 3. are <em>wrong</em>. You won’t get the correct mass for natural Mg if you use only the artificial isotopes for your calculation.
4. is <em>wrong</em>. You must use all the naturally occurring isotopes. The two most abundant isotopes of Mg account for only 90 % of the atoms. If you ignore the other 10 %, your calculation will be wrong.
Answer:
d)Cells 1 and 2
Explanation:
In a voltaic cell, oxidation occurs at the anode and reduction occurs at the cathode. The half cell that function as anode or cathode in a voltaic cell depends strictly on the reduction potential of the metal ion/metal system in that half cell.
Examining the reduction potentials of the various metal ion/metal systems in the three half cells;
Cu= +0.34 V
Ni= -0.25 V
Zn= -0.76 V
Fe(Fe2+)= -0.44 V
Hence only Zn2+ has a more negative reduction potential than Fe2+. The more negative the reduction potential, the greater the tendency of the system to function as the anode. Thus iron half cell will function as anode in cells 1&2 as explained in the argument above.
