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Rasek [7]
4 years ago
11

What is 45,000 written in scientific notation?

Chemistry
2 answers:
MAVERICK [17]4 years ago
7 0

Your answer will be (A) - 4.5 x 104

You always want to follow a x 10^b (you are doing calculating).

so A is going to be 45,000, but you need to turn it into a decimal so you will make it -> 4.5000.

The next step would be able to find be. So you currently have --> 4.5000, you will count how many numbers are right of the decimal so you have 4.

The next step is writing it out, but you are going to build of the information that you have provided off the information given to you. So to start the process you are using the a (your 45,000 that you turned into a decimal 4.5000) x 10 ^ B ( and B is going to be the number of place from the decimal which is 4 -> B = 4).

So then you will right the whole thing out 4.5 (you remove the 0's and you would in any standard equation to solve a problem), x 10^4 - > 4.5 x 10^4.

I always recommend checking your work to make sure you are correct, and in this case you can check your work by reversing the information that you got so...

10^4 x 4.5 = 45,000

aev [14]4 years ago
6 0

45,000 written in scientific notation would be 4.5×10⁴.

The number first has to be written so that there's only one number to the left of the decimal point.

4.5000

The decimal point would have to be moved 4 places to the right for the decimal to become 45,000 again. Therefore, the exponent on the 10 would be positive 4.

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<u>Answer:</u> The entropy change of the ethyl acetate is 133. J/K

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

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Given mass of ethyl acetate = 398 g

Molar mass of ethyl acetate = 88.11 g/mol

Putting values in above equation, we get:

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To calculate the entropy change for different phase at same temperature, we use the equation:

\Delta S=n\times \frac{\Delta H_{fusion}}{T}

where,  

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n = moles of ethyl acetate = 4.52 moles

\Delta H_{fusion} = enthalpy of fusion = 10.5 kJ/mol = 10500 J/mol   (Conversion factor:  1 kJ = 1000 J)

T = temperature of the system = 84.0^oC=[84+273]K=357K

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After one half-life, half (50 %) of the original amount will remain.  

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We can construct a table as follows:  

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