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const2013 [10]
3 years ago
6

Which of the following is not equal to 0.536 kilometers? 53.6 hectometers (hm) 536 meters (m) 536,000 millimeters (mm) 53,600 ce

ntimeters (cm)
Chemistry
2 answers:
garri49 [273]3 years ago
8 0
I believe it would be 53.6 hectometers, but I'm not 100% sure 
Tom [10]3 years ago
4 0

Answer:

53.6 hectometers

Explanation:

In this case, we have to write out of given options which is not equivalent to 0.536 kilometers.

1 kilometers = 1000 m

So, 0.536 kilometers = 536 meters

1 kilometers = 1000000 millimeters

So, 0.536 kilometers = 536000 millimeters

1 kilometers = 100000 centimeters

0.536 kilometers = 53600 centimeters

And 1 kilometers = 10 hectometers

So, 0.536 kilometers = 5.36 hectometers

Hence, the correct option is (a) " 53.6 hectometers "

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The question is incomplete, here is the complete question:

A chemistry student weighs out 0.104 g of sulfurous acid, a diprotic acid, into a 250.0 mL volumetric flask and dilutes to the mark with distilled water. He plans to titrate the acid with 0.0700 M NaOH solution. Calculate the volume of NaOH solution the student will need to add to reach the final equivalence point. Be sure your answer has the correct number of significant digits.

<u>Answer:</u> The volume of NaOH needed is 36.2 mL

<u>Explanation:</u>

To calculate the molarity of solution, we use the equation:

\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}

Given mass of sulfurous acid = 0.104 g

Molar mass of sulfurous acid = 82 g/mol

Volume of solution = 250 mL

Putting values in above equation, we get:

\text{Molarity of sulfurous acid}=\frac{0.104\times 1000}{82\times 250}\\\\\text{Molarity of sulfurous acid}=5.07\times 10^{-3}M

To calculate the volume of base, we use the equation given by neutralization reaction:

n_1M_1V_1=n_2M_2V_2

where,

n_1,M_1\text{ and }V_1 are the n-factor, molarity and volume of acid which is H_2SO_3

n_2,M_2\text{ and }V_2 are the n-factor, molarity and volume of base which is NaOH.

We are given:

n_1=2\\M_1=5.07\times 10^{-3}M\\V_1=250mL\\n_2=1\\M_2=0.0700M\\V_2=?mL

Putting values in above equation, we get:

2\times 5.07\times 10^{-3}\times 250.0=1\times 0.0700\times V_2\\\\V_2=\frac{2\times 5.07\times 10^{-3}\times 250}{1\times 0.0700}=36.2mL

Hence, the volume of NaOH needed is 36.2 mL

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