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Nata [24]
2 years ago
15

A typical virus is 5x10-6 cm in diameter. if avogadro's number of these virus particles were laid in a row, how many kilometers

long would the line be?
Chemistry
2 answers:
Debora [2.8K]2 years ago
8 0
Answer is: line be long 3,011·10¹³ kilometers.
diametar of virus = 5·10⁻⁶ cm ÷ 100000 = 5·10⁻¹¹ km.
line lenght = 5·10⁻¹¹ km · 6,023·10²³.
line lenght = 3,011·10¹³ km.
Avogadro number = 6,023·10²³.
1 cm = 10⁻² m = 10⁻⁵ km.
ad-work [718]2 years ago
6 0

Answer:

3.01x10^{12} Km

Explanation:

The avogrado's number is A = 6.02x10^{23}, which means that in one mol of a substance, there is this number of molecules, or atoms, for example.

So, to know how long would be the line (L), we must multiply the numbers of diameter and avogrado:

L = 5x10^{-6} x 6.02x10^{23}

To do it, we must multiply the numbers, repeat the basis 10 and add the exponents, then:

L = 3.01x10^{18} cm 1 cm ------ ------------------10^{-6} Km3.01 x 10^{18} cm --------xBy simple direct rule: x = 3.01x10^{12} Km

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How much maleic anhydride would you need to react 178 mg of anthracene? Assume 1:1 ratio from maleic anhydride to anthracene.
Licemer1 [7]

Answer:

(1) 0.10      (2) 17.8 g

Explanation:

Since the reaction ratio is 1:1 what we need is to convert the given masses to moles and you will have the answer:

MW anthracene = 178.23 g/mol

MW maleic anhydride = 98.06 g/mol

a) mass anthracene = 178 mg x 1 g/ 1000 mg = 0.178 g anthracene

Moles anthracene = 0.178 g anthracene/ 178.23 g/mol

= 0.001 mol anthracene

0.001 mol anthracene x 1 mol maleic acid/mol anthracene

= 0.001 mol maleic anhydride

mass maleic anhydride  = 0.001 mol x 98.06 g/mol =  0.10 g

b) moles maleic anhydride = 9.8 g/ 98.06 g/mol = 0.099 moles

0.099 moles maleic anhydride x 1 mol anthracene/mol  maleic anhydride =

0.099 mol anthracene

g anthracene = 0.10mol x 178 g/mol = 17.8 g

8 0
2 years ago
in collecting the precipitate, why would it be inappropriate to heat the reacted mixture and evaporate off the water?
djverab [1.8K]
In collecting the precipitate, it is inappropriate to heat <span>the reacted mixture and evaporate off the water because it is possible that the mixture contains other substances that precipitates as well when the mixture is being heated so you will not be able to collect what you want.</span>
6 0
2 years ago
A gas sample occupies 3.50 liters of volume at 20.°c. what volume will this gas occupy at 100.°c (reported to three significant
VLD [36.1K]

Explanation:

According to Charle's law, at constant pressure the volume of an ideal gas is directly proportional to the temperature.

That is,             Volume \propto Temperature

Hence, it is given that V_{1} is 3.50 liters, T_{1} is 20 degree celsius, and T_{2} is 100 degree celsius.

Therefore, calculate V_{2} as follows.

                           \frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}}

                           \frac{3.50 liter}{20^{o}C} = \frac{V_{2}}{100^{o}C}

                                V_{2} = 17.5 liter

Thus, we can conclude that volume of gas required at 100 degree celsius is 17.5 liter.

6 0
2 years ago
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