Answer: The electron arrangement for gallium is:
Ar 3d10 4s2 4p1
Taking into account the scientific notation, the result of the subtraction is 6.5×10⁵.
<h3>Scientific notation</h3>
First, remember that scientific notation is a quick way to represent a number using powers of base ten.
The numbers are written as a product:
a×10ⁿ
where:
- a is a real number greater than or equal to 1 and less than 10, to which a decimal point is added after the first digit if it is a non-integer number.
- n is an integer, which is called an exponent or an order of magnitude. Represents the number of times the comma is shifted. It is always an integer, positive if it is shifted to the left, negative if it is shifted to the right.
<h3 /><h3>Subtraction in scientific notation</h3>
You want to subtract two numbers in scientific notation. It should be noted that when the numbers to be added do not have the same base 10 exponent, the base 10 power with the highest exponent must be found. In this case, the highest exponent is 5.
Then all the values are expressed as a function of the base 10 exponent with the highest exponent. In this case: 5.00×10⁴=0.500×10⁵
Taking the quantities to the same exponent, all you have to do is subtract what was previously called the number "a". In this case:
7.00×10⁵ - 0.500×10⁵= (7.00- 0.500)×10⁵= 6.5×10⁵
Finally, the result of the subtraction is 6.5×10⁵.
Learn more about scientific notation:
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Answer:
![[SO_3]=0.25M](https://tex.z-dn.net/?f=%5BSO_3%5D%3D0.25M)
Explanation:
Hello there!
In this case, since the integrated rate law for a second-order reaction is:
![[SO_3]=\frac{[SO_3]_0}{1+kt[SO_3]_0}](https://tex.z-dn.net/?f=%5BSO_3%5D%3D%5Cfrac%7B%5BSO_3%5D_0%7D%7B1%2Bkt%5BSO_3%5D_0%7D)
Thus, we plug in the initial concentration, rate constant and elapsed time to obtain:
![[SO_3]=\frac{1.44M}{1+14.1M^{-1}s^{-1}*0.240s*1.44M}\\\\](https://tex.z-dn.net/?f=%5BSO_3%5D%3D%5Cfrac%7B1.44M%7D%7B1%2B14.1M%5E%7B-1%7Ds%5E%7B-1%7D%2A0.240s%2A1.44M%7D%5C%5C%5C%5C)
![[SO_3]=0.25M](https://tex.z-dn.net/?f=%5BSO_3%5D%3D0.25M)
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Answer:
The three-step synthesis of trans-2-pentene from acetylene is as follows.
<u>Step -1:</u> Formation of higher order terminal alkyne on reaction with sodium acetylides with haloalkanes.
<u>Step -2:</u> Formation terminal alkyne to nonterminal alkynes.
<u>Step -3:</u> Formation of trans-pent - 2-pent-ene by reduction.
Explanation:
Synthesis of trans-pent-2-yne from ethyne takes place is mainly a three step synthesis which involves formation of higher order terminal alkyne on reaction with sodium acetylides with haloalkane. Second step involves the further alkylation of terminal alkynes to higher order nonterminal alkynes and the third step involves the formation of trans-2-ene by dissolving reduction method.
The chemical reaction of each step of chemical reactions is as follows.