Given that Oxygen has an oxidation state of 2 -, you can combine Mn 3+ with O 2- ions to form Mn2O3, and you can combine Mn 2+ with O 2- to form Mn2O2 which is MnO.
The other compounds imply oxidation states of N, Br and Cl that does not exist.
Therefore, the answer is the option D. MnO
Answer:
4.06 Hz
Explanation:
For simple harmonic motion, frequency is given by
where k is spring constant and m is the mass of the object.
Substituting 0.2 Kg for mass and 130 N/m for k then

A) d. 10T
When a charged particle moves at right angle to a uniform magnetic field, it experiences a force whose magnitude os given by

where q is the charge of the particle, v is the velocity, B is the strength of the magnetic field.
This force acts as a centripetal force, keeping the particle in a circular motion - so we can write

which can be rewritten as

The velocity can be rewritten as the ratio between the lenght of the circumference and the period of revolution (T):

So, we get:

We see that this the period of revolution is directly proportional to the mass of the particle: therefore, if the second particle is 10 times as massive, then its period will be 10 times longer.
B) 
The frequency of revolution of a particle in uniform circular motion is

where
f is the frequency
T is the period
We see that the frequency is inversely proportional to the period. Therefore, if the period of the more massive particle is 10 times that of the smaller particle:
T' = 10 T
Then its frequency of revolution will be:

Answer:
<em><u>a) </u></em><em><u> </u></em><em><u>Carbonic acid</u></em>
<em><u>b</u></em><em><u>)</u></em><em><u> </u></em><em><u>ammonium hydroxide</u></em>
<em><u>c</u></em><em><u>)</u></em><em><u> </u></em><em><u>Aluminum phosphate</u></em>
<em><u>d</u></em><em><u>)</u></em><em><u> </u></em><em><u>Sodium hydroxide</u></em>
<em><u>e</u></em><em><u>)</u></em><em><u> </u></em><em><u>Gold trichloride</u></em>
Explanation:
<em>I</em><em> </em><em>hope this</em><em> </em><em>will help</em><em> </em><em>you</em><em> </em><em>buddy</em><em> </em>