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1 year ago

How many molecules are in a 189 g sample of carbon tetrabromide, CBr4 ?

Chemistry

1 answer:

anyanavicka [17]1 year ago

5 0

Answer:

3.4 × 10²³ molecules of CBr₄

Explanation:

Given data:

Mass of CBr₄ = 189 g

Number of molecules = ?

Solution:

First of all we will calculate the number of moles.

Number of moles = mass / molar mass

Number of moles = 189 g/ 331.63 g/mol

Number of moles = 0.6 mol

Now the given problem will solve by using Avogadro number.

It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance.

The number 6.022 × 10²³ is called Avogadro number.

For example,

18 g of water = 1 mole = 6.022 × 10²³ molecules of water

Foe 0.6 moles of CBr₄:

0.6 mol × 6.022 × 10²³ molecules of CBr₄ / 1 mol

3.4 × 10²³ molecules of CBr₄

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1 year ago

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2 years ago

Read 2 more answers

Consider the chemical reaction: N2 3H2 yields 2NH3. If the concentration of the reactant H2 was increased from 1.0 x 10-2 M to 2

Brilliant_brown [7]

The equilibrium constant of a reaction is defined as:

"The ratio between equilibrium concentrations of products powered to their reaction quotient and equilibrium concentration of reactants powered to thier reaction quotient".

The reaction quotient, Q, has the same algebraic expressions but use the actual concentrations of reactants.

To solve this question we need this additional information:

<em>For this reaction, K = 6.0x10⁻² and the initial concentrations of the reactants are:</em>

<em />

Thus, for the reaction:

N₂ + 3H₂ ⇄ 2NH₃

The equilibrium constant, K, of this reaction, is defined as:

$K = 6.0x10^{-2} = \frac{[NH_3]^2}{[N_2][H_2]^3}$

And Q, is:

$Q = \frac{[NH_3]^2}{[N_2][H_2]^3}$

Where actual concentrations are:

[NH₃] = 1.0x10⁻⁴M

[N₂] = 4.0M

[H₂] = 2.5x10⁻¹M

Replacing:

$Q = \frac{[1.0x10^{-4}]^2}{[4.0][2.5x10^{-1}]^3}$

As Q < K,

<h3>The chemical system will shift to the right in order to produce more NH₃</h3>

Learn more about chemical equililbrium in:

brainly.com/question/24301138

8 0

1 year ago

How many molecules of carbon dioxide are in 243.6 g of carbon dioxide?

german

Hey there !

Molar mass carbon dioxide:

CO2 = 44.01 g/mol

1) number of moles :

1 mole CO2 ------------- 44.01 g
(moles CO2) ------------ 243.6 g

moles CO2 = 243.6 * 1 / 44.01

moles CO2 = 243.6 / 44.01

=> 5.535 moles of CO2

Therefore:

1 mole -------------------- 6.02x10²³ molecules
5.535 moles ------------ ( molecules CO2)

molecules CO2 = 5.535 * ( 6.02x10²³) / 1

=> 3.33x10²⁴ molecules of CO2

3 0

1 year ago