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Vitek1552 [10]
3 years ago
5

The chain on a bicycle rusts faster when the bicycle is left outside in damp conditions. Which of the following factors affect t

he rate at which the bicycle chain rusts?
a) Surface area
b) concentration of oxygen atoms in air
c) temperature
d) all of the above
e) b and c
f) a and c
g) a and b
Chemistry
1 answer:
andreyandreev [35.5K]3 years ago
6 0
D
Larger surface area speeds up reactions 
Rust is an oxide so more oxygen should speed up rate of rust
Temperature also speeds up reactions

+ The humidity affects it aswell

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Answer : Option C) 2.5 units

Explanation : For calculating the strength of a unknown magnetic field in comparison with the given magnetic field we can use the magnetic formula as;

\frac{ S_{1}}{S_{2}} =  \frac{ (D_{1}^{2}) }{(D_{2}^{2})} .

On substituting the values we get; 

\frac{10}{x}  =  \frac{4^{2}}{2^{2}} ;

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This material has a density of 1 g/cm3.
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How many moles of carbon, hydrogen, and oxygen are present in a 100-g sample of ascorbic acid?
Y_Kistochka [10]

There are:

3.41 moles of C

4.54 moles of H

3.40 moles of O.

Why?

To solve the problem, the first thing that we need to do is to write the chemical formula of the ascorbic acid.

C_{6}H_{8}O_{6}

Now, we know that there are 100 grams of the compound, so, the masses of each element will represent the percent in the compound.

We have that:

C_{6}=12.0107g*6=72.08g\\\\H_{8}=1.008g*8=8.064g\\\\O_{6}=15.999g*6=95.994g\\\\C_{6}H_{8}O_{6}=72.08g+8.064g+95.994g=176.138g

To know the percent of each element, we need to to the following:

C=\frac{72.08g}{176.138g}*100=0.409*100=40.92(percent)\\\\H=\frac{8.064g}{176.138g}*100=4.58(percent)\\\\O=\frac{95.994}{176.138g}*100=54.49(percent)

So, we know that for the 100 grams of the compound, there are:

40.92 grams of C

4.58 grams of H

54.49 grams of O

We know the molecular masses of each element:

C=12.0107\frac{g}{mol}\\\\H=1.008\frac{g}{mol}\\\\O=15.999\frac{g}{mol}{mol}

Now, to calculate the number of moles of each element, we need to divide the mass of each element by the molecular mass of each element:

C=\frac{40.92g}{12.010\frac{g}{mol}}=3.41mol\\\\H=\frac{4.58g}{1.008\frac{g}{mol}}=4.54mol\\\\O=\frac{54.49g}{15.999\frac{g}{mol}}=3.40mol

Hence, we have that there are 3.41 moles of C, 4.54 moles of H, and 3.40 moles of O.

Have a nice day!

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