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mixer [17]
3 years ago
15

For a solar eclipse to occur which of the following alignments is necessary? A. The moon is located along a straight line betwee

n the sun and earth B. the is located along a straight line between the sun and moon C. the moon is located from a line between the sun and earth D. the earth is located from a line 90 degrees between the sun and moon
Chemistry
1 answer:
Aloiza [94]3 years ago
8 0

For a solar eclipse t occur;

C. The moon is located from a line between the sun and earth

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The brown Haze that forms over Sunny cities like Los Angeles is called
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3 years ago
What is the composition, in atom percent, of an alloy that consists of a) 5.5 wt% Pb and b) 94.5 wt% of Sn? Assume that the atom
Anastaziya [24]

Answer : The percent composition of Pb and Sn in atom is, 3.21 % and 96.8 % respectively.

Explanation :

First we have to calculate the number of atoms in 5.5 wt% Pb and 94.5 wt% of Sn.

As, 207.2 g of lead contains 6.022\times 10^{23} atoms

So, 5.5 g of lead contains \frac{5.5}{207.2}\times 6.022\times 10^{23}=1.59\times 10^{22} atoms

and,

As, 118.71 g of lead contains 6.022\times 10^{23} atoms

So, 94.5 g of lead contains \frac{94.5}{118.71}\times 6.022\times 10^{23}=4.79\times 10^{23} atoms

Now we have to calculate the percent composition of Pb and Sn in atom.

\% \text{Composition of Pb}=\frac{\text{Atoms of Pb}}{\text{Atoms of Pb}+\text{Atoms of Sn}}\times 100

\% \text{Composition of Pb}=\frac{1.59\times 10^{22}}{(1.59\times 10^{22})+(4.79\times 10^{23})}\times 100=3.21\%

and,

\% \text{Composition of Sn}=\frac{\text{Atoms of Sn}}{\text{Atoms of Pb}+\text{Atoms of Sn}}\times 100

\% \text{Composition of Sn}=\frac{4.79\times 10^{23}}{(1.59\times 10^{22})+(4.79\times 10^{23})}\times 100=96.8\%

Thus, the percent composition of Pb and Sn in atom is, 3.21 % and 96.8 % respectively.

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3 years ago
10 moles of carbon dioxide has a mass of 440 g. What is the relative formula mass of carbon dioxide?
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Moles =  Mass / Molar Mass or Formular Mass.

Base on this question; Moles = 10, Mass = 440g, and Formular Mass = ?

Making 'Formular Mass', subject of the formular; we thus have;

Formular mass = Mass / Moles = 440/ 10 = 44g

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