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lora16 [44]
3 years ago
13

Easy Questions! Chemistry - Please help! I give thanks

Chemistry
2 answers:
Evgen [1.6K]3 years ago
8 0
This is not chemistry but

see it is a triangular prim on it's side
V=BH
are of base times height
the base is a triangle
height is 18.5

base=1/2bh
b=8.6
h=8.4
base=1/2(8.6)(8.4)
base=36.12

V=bh
V=36.12*18.5
V=668.22

round
V=668.2 ft^3
artcher [175]3 years ago
4 0
All sides have to be exact
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An unknown compound with a molar mass of 223.94 g/mol consists of 32.18% c, 4.50% h, and 63.32% cl. find the molecular formula f
Dima020 [189]

The actual number of atoms of each element present in the molecule of the compound is represented by the formula known as molecular formula.

Molar mass of the unknown compound = 223.94 g/mol (given)

Mass of each element present in the unknown compound is determined as:

  • Mass of carbon, C:

\frac{32.18}{100}\times 223.94 = 72.06 g

  • Mass of hydrogen, H:

\frac{4.5}{100}\times 223.94 = 10.08 g

  • Mass of chlorine, Cl:

\frac{63.32}{100}\times 223.94 = 141.79 g

Now, the number of each element in the unknown compound is determined by the formula:

number of moles = \frac{given mass}{molar mass}

  • Number of moles of C:

number of moles = \frac{72.06}{12} = 6.005 mole\simeq 6 mole

  • Number of moles of H:

number of moles = \frac{10.08}{1} = 10.08 mole\simeq 10 mole

  • Number of moles of Cl

number of moles = \frac{141.79}{35.5} = 3.99 mole\simeq 4 mole

Dividing each mole with the smallest number of mole, to determine the empirical formula:

C_{\frac{6}{4}}H_{\frac{10}{4}}Cl_{\frac{10}{4}}

C_{1.5}H_{2.5}Cl_{1}

Multiplying with 2 to convert the numbers in formula into a whole number:

So, the empirical formula is C_{3}H_{5}Cl_{2}.

Empirical mass = 12\times 3+1\times 5+2\times 35.5 = 112 g/mol

In order to determine the molecular formula:

n = \frac{molar mass}{empirical mass}

n = \frac{223.94}{112} = 1.99 \simeq 2

So, the molecular formula is:

2\times C_{3}H_{5}Cl_{2} =  C_{6}H_{10}Cl_{4}

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3 years ago
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