The subscripts tell you <em>how many atoms</em> of an element are in one formula unit of a compound.
If there is no subscript, there is only one atom of the element.
In one formula unit of Na₂PO₃F, there are
Two atoms of sodium (Na)
One atom of phosphorus (P)
Three atoms of oxygen (O)
One atom of fluorine (F)
Answer:
+ 3
Ex planation:
Al3+ indicates an ion of aluminum having a charge of + 3. I.e., since an aluminum atom normally has 13 protons and 13 electrons, this ion has 10 electrons (-10 charge) and 13 protons (+ 13 charge) giving it a charge of + 3 (-10 + 13 = +3).
hope this helps!
Answer: 500
Explanation: Since there are fifty tens, you must multiply 50 by 10, in which will get you 500 as your final answer.
D. mixture in which its components retain their identity
In a heterogeneous mixture, multiple substances are mixed, but they don't chemically react and they remain chemically the same.
Answer: Option (4) is the correct answer.
Explanation:
Relation between potential energy and charge is as follows.
U = ![\frac{1}{4 \pi \epsilon_{o}}[\frac{q_{1}q_{2}}{r_{12}} + \frac{q_{2}q_{3}}{r_{23}} + \frac{q_{3}q_{1}}{r_{31}}]](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B4%20%5Cpi%20%5Cepsilon_%7Bo%7D%7D%5B%5Cfrac%7Bq_%7B1%7Dq_%7B2%7D%7D%7Br_%7B12%7D%7D%20%2B%20%5Cfrac%7Bq_%7B2%7Dq_%7B3%7D%7D%7Br_%7B23%7D%7D%20%2B%20%5Cfrac%7Bq_%7B3%7Dq_%7B1%7D%7D%7Br_%7B31%7D%7D%5D)
As it is given that
,
, and
.
Distance between the charges = 1 cm =
(as 1 cm = 0.01 m)
Hence, putting these given values into the above formula as follows.
U = ![\frac{1}{4 \pi \epsilon_{o}}[\frac{q_{1}q_{2}}{r_{12}} + \frac{q_{2}q_{3}}{r_{23}} + \frac{q_{3}q_{1}}{r_{31}}]](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B4%20%5Cpi%20%5Cepsilon_%7Bo%7D%7D%5B%5Cfrac%7Bq_%7B1%7Dq_%7B2%7D%7D%7Br_%7B12%7D%7D%20%2B%20%5Cfrac%7Bq_%7B2%7Dq_%7B3%7D%7D%7Br_%7B23%7D%7D%20%2B%20%5Cfrac%7Bq_%7B3%7Dq_%7B1%7D%7D%7Br_%7B31%7D%7D%5D)
=
= ![9 \times 10^{9} [2 + 6 + 1.5]](https://tex.z-dn.net/?f=9%20%5Ctimes%2010%5E%7B9%7D%20%5B2%20%2B%206%20%2B%201.5%5D)
=
J
= 0.00085 J
Thus, we can conclude that the potential energy of this arrangement, relative to the potential energy for infinite separation, is about 0.00085 J.