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2 years ago

1. The density of solid Cu is 8.96g/cm3 . How many atoms are present per cubic centimeter(cm3) of Cu ?

Chemistry

1 answer:

Leni [432]2 years ago

4 0

Answer:

8.5 × 10²² atom / cm³, 2.125 × 10²² unit cells / cm³, 4.7 × 10⁻²³ cm³ / cell unit, 3.61 × 10⁻⁸ cm /cell

Explanation:

The density of the solid Cu = 8.96 g/cm³

a) number of atoms present

mole = mass / molar mass = (8.96 g/cm³) / 63.546 = 0.141 mol

number of atoms = mole × avogadro's constant = 0.141 × 6.02 × 10²³ = 8.5 × 10²² atom / cm³

b) a face-centered cubic unit cell contains 4 atoms

8.5 × 10²² atom / cm³ / 4 = 2.125 × 10²² unit cells / cm³

c) the volume of a unit cell of the metal = reciprocal of 2.125 × 10²² unit cells / cm³ = 1 ÷ ( 2.125 × 10²² unit cells / cm³ ) = 4.7 × 10⁻²³ cm³ / cell unit

the edge length of a cell of cu = ∛(4.7 × 10⁻²³ cm³ / cell unit) = 3.61 × 10⁻⁸ cm /cell

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2. Consider the reaction 2 Cg H18 (4) +250â (9) âº 16 co, (g) + 18 HâO(g) la How many moles of H20co) are produced, when |--16:1

Sladkaya [172]

Answer :

(a) The moles of water produced are 145.35 moles.

(b) The mass of oxygen needed are 3080.8 grams.

<u>Solution for part (a) : Given,</u>

Moles of $C_8H_{18}$ = 16.15 moles

First we have to calculate the moles of $H_2O$

The balanced chemical reaction is,

$2C_8H_{18}+25O_2\rightarrow 16CO_2+18H_2O$

From the balanced reaction we conclude that

As, 2 moles of $C_8H_{18}$ react to give 18 moles of $H_2O$

So, 16.15 moles of $C_8H_{18}$ react to give $\frac{16.15}{2}\times 18=145.35$ moles of $H_2O$

The moles of water produced are 145.35 moles.

<u>Solution for part (b) : Given,</u>

Mass of $C_8H_{18}$ = 878 g

Molar mass of $C_8H_{18}$ = 114 g/mole

Molar mass of $O_2$ = 32 g/mole

First we have to calculate the moles of $C_8H_{18}$ .

$\text{ Moles of }C_8H_{18}=\frac{\text{ Mass of }C_8H_{18}}{\text{ Molar mass of }C_8H_{18}}=\frac{878g}{114g/mole}=7.702moles$

Now we have to calculate the moles of $O_2$

The balanced chemical reaction is,

$2C_8H_{18}+25O_2\rightarrow 16CO_2+18H_2O$

From the balanced reaction we conclude that

As, 2 moles of $C_8H_{18}$ react with 25 moles of $O_2$

So, 7.702 moles of $C_8H_{18}$ react with $\frac{7.702}{2}\times 25=96.275$ moles of $O_2$

Now we have to calculate the mass of $O_2$ .

$\text{ Mass of }O_2=\text{ Moles of }O_2\times \text{ Molar mass of }O_2$

$\text{ Mass of }O_2=(96.275moles)\times (32g/mole)=3080.8g$

The mass of oxygen needed are 3080.8 grams.

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2 years ago

A student needs to convert a volume of 3 Tbsp (tablespoons) to mL. The student finds out that 1 tablespoon is equivalent to 14.7

garri49 [273]

Answer:

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Explanation:

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77julia77 [94]

Answer:

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Explanation:

c + 273.15 = k

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kap26 [50]

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tia_tia [17]

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