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lakkis [162]
2 years ago
6

Given the two half-reactions, what must be done in the next step before the reaction can be balanced? First: upper A u superscri

pt 3 plus right arrow upper A u. Second: upper I superscript minus right arrow Upper I subscript 2. Show the number of electrons being lost or gained. Make sure that the charges are balanced. List the atoms that need to be balanced.
Mathematics
1 answer:
kenny6666 [7]2 years ago
8 0

In the first reaction, Gold (Au) gets 3 electrons and became Gold. In the second reaction, Iodine (I) loses 2 electrons and became Iodine.

<h3>What are cations and anions?</h3>

When the atom loses the election then it is called a cation and when it gains the electron is called an anion.

Two reactions are given.

First reaction;

Au³⁺ + 3e⁻ ⇒ Au

In the first reaction, Gold (Au) gets 3 electrons and became Gold.

Second reaction;

2I⁻ ⇒ I₂ + 2e⁻

In the second reaction, Iodine (I) loses 2 electrons and became Iodine.

More about the cation and the anion link is given below.

brainly.com/question/1659592

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Answer:

x=11

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3 years ago
HELP PLZ WILL GIVE BRAINLIEST! find the arrithmetic means in the given sequence<br> -3,?,?,?,93
pav-90 [236]

Answer:

Step-by-step explanation:

The standard form of an arithmetic sequence is

aₙ = a₁ + d(n - 1)

where aₙ is the number of the term in the sequence (in order from first term where n = 1, to second term where n = 2, to third term where n = 3, etc) a₁ is the the first term in the sequence, and d is the arithmetic difference or means.  This is what we are looking to solve for.  

In our sequence we have the first term, -3 (where n = 1) and the fifth term, 93 (where n = 5).  If we fill in what we have, the only unknown is d, our arithmetic difference (means) between each number in the sequence.

Because we have the fifth term, we can write our standard form to fit our needs:

a₅ = a₁ + d(n-1).  Therefore,

93 = -3 + d(5 - 1) and

93 = -3 + d(4) so

96 = 4d and

d = 24

Our arithmetic difference (means) is 24.  Let's test it on a few values of n.  Let's look for the second, third, and 4th terms, and then try it out for n = 5 to make sure the 5th term, using our arithmetic sequence with d = 24 works and we do, in fact, find the fifth term to be 93.

Testing n = 2

a₂ = -3 + 24(2 - 1) so

a₂ = -3 + 24(1)  and

a₂ = 21.  Second term is 21 (Notice that difference between -3 and 21 is 24)

Testing n = 3

a₃ = -3 + 24(3 - 1) so

a₃ = -3 + 24(2) and

a₃ = 48 - 3 and

a₃ = 45 (Notice the difference between 21 and 45 is 24)

Testing n = 4

a₄ = -3 + 24(4 - 1) so

a₄ = -3 + 24(3) and

a₄ = 72 - 3 and

a₄ = 69 (Notice the difference between 45 and 69 is 24)

Testing n = 5 (and it better come out as 93 or we did something wrong!)

a₅ = -3 + 24(5 - 1) and

a₅ = -3 + 24(4) so

a₅ = 96 - 3 so

a₅ = 93 (Phew!)  ; )

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Two integers, a and b have a product of 15. what is the least possible sum of a and b?
Neporo4naja [7]
If they're integers, that means they can't be fractions (such as 15/2 and 2).

So the only couples of integers that, when multiplied, result in 15 are:

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(there are no other possibilities as both 5 and 3 are not divisible further).
The smaller of the two sums (15+1; 5+3) is 5+3=8.

The least possible sum of a and b is 8.
7 0
3 years ago
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