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4 years ago

11

How many moles of CO2 are produced when 5.00 moles of C2H5OH are burned? Please include the steps.

1 answer:

Burka [1]4 years ago

6 0

Just construct the chemical equation.Knowing that a molecule of ethanol has 2 carbon atoms, 2 molecules of carbon dioxide are produced.So the mol ratio of ethanol and CO2 is 1:2 ,so 10 mols of CO2 is produced.Of course this is under the assumption that it is burnt completely.If It is not burnt completely then thats a different story.

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Read 2 more answers

The density of potassium, which has the BCC structure, is 0.855 g/cm3. The atomic weight of potassium is 39.09 g/mol. Calculate

Delicious77 [7]

<u>Answer:</u>

<u>For a:</u> The edge length of the crystal is 533.5 pm

<u>For b:</u> The atomic radius of potassium is 231.01 pm

<u>Explanation:</u>

<u>For a:</u>

To calculate the lattice parameter or edge length of the crystal, we use the equation:

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

where,

$\rho$ = density = $0.855g/cm^3$

Z = number of atom in unit cell = 2 (BCC)

M = atomic mass of metal = 39.09 g/mol

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

a = edge length of unit cell = ?

Putting values in above equation, we get:

$0.855=\frac{2\times 39.09}{6.022\times 10^{23}\times (a)^3}\\\\\a=5.335\times 10^{-8}cm=533.5pm$

<u>Conversion factor:</u> $1cm=10^{10}pm$

Hence, the edge length of the crystal is 533.5 pm

<u>For b:</u>

To calculate the edge length, we use the relation between the radius and edge length for BCC lattice:

$R=\frac{\sqrt{3}a}{4}$

where,

R = radius of the lattice = ?

a = edge length = 533.5 pm

Putting values in above equation, we get:

$R=\frac{\sqrt{3}\times 533.5}{4}=231.01pm$

Hence, the atomic radius of potassium is 231.01 pm

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3 years ago