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aalyn [17]
3 years ago
6

A chemical equation is shown:

Mathematics
1 answer:
ra1l [238]3 years ago
5 0

Answer:

Option C is correct.

Atoms of oxygen exist in the products of this reaction is 6.

Step-by-step explanation:

Given a chemical equation:

C_2H_4 +O_2 \rightarrow CO_2 + H_2O

Law of conservation of mass states that it is observed in a balanced chemical equation, which is a chemical equation that shows all mass is conserved throughout the reaction.

To balance the given equation:

First start off by balancing the elements in the complex molecules first and then the single element molecules.

First, we will start with balancing the Carbons.

On the left we have 2 but on the right only 1, so we multiply the CO_2 on the right by 2.

Next we look at the Hydrogen.

We have 4 on the left to the given equation, but only 2 on the right.

To balance them;

so, multiply H_2O by 2.

then, equation become;

 C_2H_4 +O_2 \rightarrow 2 CO_2 +2H_2O         .....[1]

Now, looking at the right hand side we have oxygen, 4 from CO_2 and 2 from H_2O, then we have total number of oxygen on right hand side is 6.

In order to have 6 on left side we balanced the equation [1] multiply O_2 by 3.

then the balanced given equation is:

C_2H_4 +3O_2 \rightarrow 2 CO_2 + 2H_2O

Number of oxygen atom exist = 2 \times 2 +2 \times 1 = 4 + 2 =6

Therefore, number of  atoms of oxygen exist in the products of this reaction is 6.

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In the expansion of ( x^3 - 2/x^2 ) ^10 , find the coefficient of 1/x^5​
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Answer:

240

Step-by-step explanation:

We need to find the coeffeicent of the binomial expansion of

( {x}^{3}  - 2 {x}^{ - 2} ) {}^{10}

Note that

- 2 {x}^{ - 2}  = -  \frac{2}{ {x}^{2} }

The binomial theorem states that

(x + y) {}^{n}  = x {}^{n} y {}^{0}  +  \binom{n}{1} x {}^{n - 1} y +  \binom{n}{2} x {}^{n - 2} y {}^{2} ....... + x {}^{0} y {}^{n} ( \binom{n}{n} )

Using this, we let expand our series

( {x}^{3}   - 2 {x}^{ - 2} ) {}^{10}  = x {}^{30}  +  \binom{10}{1} ( {x}^{27}     2 {x}^{ - 2} ) +  \binom{10}{2}  {x}^{24} 2x {}^{ - 4}  +  \binom{10}{3}  {x}^{21} 2x {}^{ - 6}  +  \binom{10}{4}  {x}^{18} 2x { }^{ - 8}  +  \binom{10}{5} x {}^{15} 2x {}^{ - 10}  +  \binom{10}{6} x {}^{12}2 x {}^{ - 12}  +  \binom{10}{7} x {}^{ 9} 2x {}^{ - 14}  +  \binom{10}{8} x {}^{ 6} 2x {}^{ - 16}  +  \binom{10}{9} ( {x}^{3} )2x {}^{ - 18}  + 2x {}^{ - 20}

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So what term in the series eqaul x^-5.

That term is the 10 choose 7 term.

\binom{10}{7}  {x}^{9} 2x {}^{ - 14}

Because

=  \binom{10}{7} 2x {}^{ - 14}  {x}^{9}  =  \binom{10}{7} 2 {x}^{ - 5}

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120(2) {x}^{ - 5}

240 {x}^{ - 5}

So the coeffceint u

is 240

3 0
2 years ago
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