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pochemuha
3 years ago
14

How many primary tectonic plates make up the surface of Earth?

Chemistry
2 answers:
dimaraw [331]3 years ago
7 0


There are about 7 or 8 primary tectonic plates. The lithosphere is made of about a dozen plates. The plates fit together like the pieces of a jigsaw puzzle. However, the plates can move. This causes earthquakes, volcanoes, and mountains to form.

Hope it helps

deff fn [24]3 years ago
4 0
There are 15 to 20 primary tectonics that make up the surface of Earth. They are called Lithosphere. 
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Write each element and elements of atoms of the compound Na2PO3F
Rudik [331]

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)

7 0
3 years ago
13. What is the charge of an aluminum ion that has 13 protons and 10 electrons?
Ksenya-84 [330]

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!

7 0
3 years ago
Read 2 more answers
The anwser to this problem
Ilya [14]
Answer: 500

Explanation: Since there are fifty tens, you must multiply 50 by 10, in which will get you 500 as your final answer.
8 0
4 years ago
12. A heterogeneous mixture is a
TEA [102]
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.
3 0
3 years ago
Three positive charges lie on the x axis: q1 = 1 × 10−8 C at x1 = 1 cm, q2 = 2 × 10−8 C at x2 = 2 cm, and q3 = 3 × 10−8 C at x3
NARA [144]

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}}]

As it is given that q_{1} = 1 \times 10^{-8} C, q_{2} = 2 \times 10^{-8} C, and q_{3} = 3 \times 10^{-8} C.

        Distance between the charges = 1 cm = 1 \times 10^{-2} m  (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}}]

            = 9 \times 10^{9} [\frac{1 \times 10^{-8} \times 2 \times 10^{-8}}{10^{-2}} + \frac{2 \times 10^{-8} \times 3 \times 10^{-8}}{10^{-2}} + \frac{3 \times 10^{-8} \times 1 \times 10^{-8}}{10^{-2}}]    

            = 9 \times 10^{9} [2 + 6 + 1.5]

            = 85.5 \times 10^{-5} 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.              

8 0
3 years ago
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