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3 years ago

10

Phosphorus crystallizes in several different forms, one of which is a simple cube with an edge length of 238 pm. What is the den

sity of phosphorus in the simple cubic crystal form?

1 answer:

algol133 years ago

6 0

<u>Answer:</u> The density of phosphorus is $3.81g/cm^3$

<u>Explanation:</u>

To calculate the density of phosphorus, we use the equation:

$\rho=\frac{Z\times M}{N_{A}\times a^{3}}$

where,

$\rho$ = density

Z = number of atom in unit cell = 1 (CCP)

M = atomic mass of phosphorus = 31 g/mol

$N_{A}$ = Avogadro's number = $6.022\times 10^{23}$

a = edge length of unit cell = $238pm=238\times 10^{-10}cm$ (Conversion factor: $1cm=10^{10}pm$ )

Putting values in above equation, we get:

$\rho=\frac{1\times 31}{6.022\times 10^{23}\times (238\times 10^{-10})^3}\\\\\rho=3.81g/cm^3$

Hence, the density of phosphorus is $3.81g/cm^3$

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A certain alcohol contains only three elements, carbon, hydrogen, and oxygen. Combustion of a 60.00 gram sample of the alcohol p

marin [14]

Answer:

$C_2H_6O$

Explanation:

The first step is the <u>calculation of the moles</u> of $H_2O$ and $CO_2$ , so:

$114.6~g~CO_2\frac{1~mol~CO_2}{44~g~CO_2}=2.6~mol~of~CO_2$

$70.44~g~H_2O\frac{1~mol~H_2O}{18~g~H_2O}=~3.9~mol~H_2O$

Now, in 1 mol of CO2 we have 1 mol of C and in 1 mol of $H_2O$ we have 1 mol of H. Additionally, if we want to calculate the moles of oxygen we need to <u>calculate the grams of C and O</u> and then do the <u>substraction</u> form the initial amount, so:

$2.6~mol~CO_2\frac{1~mol~C}{1~mol~CO_2}\frac{12~g~C}{1~mol~C}=31.25~g~of~C$

$3.9~mol~H_2O\frac{2~mol~H}{1~mol~H_2O}\frac{1~g~H}{1~mol~H}=7.82~g~of~H$

$Total~grams=~31.25~+~7.82=39.08~g$

$grams~of~O=60.00~g-~39.08~g=20.92~g~of~O$

Now we can <u>convert the grams</u> of O to moles, so:

$20.92~g~of~O\frac{1~mol~O}{16~g~O}=1.30~mol~O$

The next step is to divide all the mol values by the <u>smallest one</u>:

$O=\frac{1.30~mol~O}{1.30~mol~O}=~1$

$C=\frac{2.6~mol~C}{1.30~mol~O}=~2$

$H=\frac{7.82~mol~H}{1.30~mol~O}=6$

Therefore the formula is $C_2H_6O$

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4 years ago

In which time period did alchemists attempt to change less valuable metals into gold? 8000 BC – 1000 BC 1000 BC – 450 BC 300 BC

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Time Period 3 (300 BC to 1400 AD)

The alchemists were in pursuit of 2 ideals. First, alchemy came about from the belief that cheap metals can be transmuted to gold using the legendary substance <em>Philosopher's Stone</em> for the conversion. These views were greatly influenced by Aristotle. Thus, from 300 BC to 1400 AD, alchemists worked to make the conversion of cheap metals to gold a reality. However, it never succeeded.

After this, the second ideal that the alchemists were in pursuit of was the concoction of an elixir of life to enable people to live longer and cure all ailments. This took place in the 1500s up to the end of the 1600s. Just like the pursuit of the first ideal, efforts to succeed in the second one all led to failure.

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3 years ago

Rank the nonmetals in each set from most reactive (1) to least reactive (3). Neon: Selenium: Fluorine

expeople1 [14]

The periodic trend of reactivity depends on whether the elements are metals or nonmetals. For metals, reactivity increases down a group and from right to left across a period. Nonmetals do the opposite. For nonmetals, reactivity increases up a group and from left to right across a period. Therefore, the rank is as follows:

Fluorine

Selenium

Neon

Hope this answers the question. Have a nice day.

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Answer:

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