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dimaraw [331]
3 years ago
14

What is 1.23 x 10^-3 in standard notation

Chemistry
2 answers:
Sophie [7]3 years ago
7 0

Answer:

0.00123

Explanation:

Standard notation is the normal way of writing numbers. Examples include 1, 2, and 10. The number 1.23 x 10^-3 is written in scientific notation. The decimal goes after the first nonzero integer and it is multiplied by a power of 10. The power or exponent attached to the 10 tells you how many places over you need to move the decimal to get back into scientific notation. Examples include 1.00 x 10^2 (representing 100 in standard form because you would move the decimal two places to the right.), 2.0 x 10^1 (representing 20 in standard form because you would move the decimal one place to the right), and 3.0 x 10^-4 (representing 0.0003 in standard form because you would move the decimal four places to the left since it is a negative exponent).

The negative (-3) exponent in 1.23 x 10^-3 indicated to move the decimal three places to the left. If it was positive, you would move it three places to the right.

In 1.23 x 10^-3 move the decimal to the left 1 place to get:

0.123

two places to get:

0.0123

and a third place to get:

0.00123

The final answer is 0.00123

Andrew [12]3 years ago
6 0

Answer:

=0.00123

Explanation:

Look at the attachments below

Hope this helps (:

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Read 2 more answers
The heats of combustion of ethane (C2H6) and butane (C4H10) are 52 kJ/g and 49 kJ/g, respectively. We need to produce 1.000 x 10
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Answer :

(1) The number of grams needed of each fuel (C_2H_6)\text{ and }(C_4H_{10}) are 19.23 g and 20.41 g respectively.

(2) The number of moles of each fuel (C_2H_6)\text{ and }(C_4H_{10}) are 0.641 moles and 0.352 moles respectively.

(3) The balanced chemical equation for the combustion of the fuels.

C_2H_6+\frac{7}{2}O_2\rightarrow 2CO_2+3H_2O

C_4H_{10}+\frac{13}{2}O_2\rightarrow 4CO_2+5H_2O

(4) The number of moles of CO_2 produced by burning each fuel is 1.28 mole and 1.41 mole respectively.

The fuel that emitting least amount of CO_2 is C_2H_6

Explanation :

<u>Part 1 :</u>

First we have to calculate the number of grams needed of each fuel (C_2H_6)\text{ and }(C_4H_{10}).

As, 52 kJ energy required amount of C_2H_6 = 1 g

So, 1000 kJ energy required amount of C_2H_6 = \frac{1000}{52}=19.23g

and,

As, 49 kJ energy required amount of C_4H_{10} = 1 g

So, 1000 kJ energy required amount of C_4H_{10} = \frac{1000}{49}=20.41g

<u>Part 2 :</u>

Now we have to calculate the number of moles of each fuel (C_2H_6)\text{ and }(C_4H_{10}).

Molar mass of C_2H_6 = 30 g/mole

Molar mass of C_4H_{10} = 58 g/mole

\text{ Moles of }C_2H_6=\frac{\text{ Mass of }C_2H_6}{\text{ Molar mass of }C_2H_6}=\frac{19.23g}{30g/mole}=0.641moles

and,

\text{ Moles of }C_4H_{10}=\frac{\text{ Mass of }C_4H_{10}}{\text{ Molar mass of }C_4H_{10}}=\frac{20.41g}{58g/mole}=0.352moles

<u>Part 3 :</u>

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C_4H_{10}+\frac{13}{2}O_2\rightarrow 4CO_2+5H_2O

<u>Part 4 :</u>

Now we have to calculate the number of moles of CO_2 produced by burning each fuel to produce 1000 kJ.

C_2H_6+\frac{7}{2}O_2\rightarrow 2CO_2+3H_2O

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As, 0.641 mole of C_2H_6 react to produce 0.641\times 2=1.28 moles of CO_2

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C_4H_{10}+\frac{13}{2}O_2\rightarrow 4CO_2+5H_2O

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As, 1 mole of C_4H_{10} react to produce 4 moles of CO_2

As, 0.352 mole of C_4H_{10} react to produce 0.352\times 4=1.41 moles of CO_2

So, the fuel that emitting least amount of CO_2 is C_2H_6

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