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2 years ago

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

Chemistry

1 answer:

anyanavicka [17]2 years 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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How many grams of the excess reagent are left over when 6.00g of CS2 gas react with 10.0g of Cl2 gas in the following reaction:

WARRIOR [948]

Answer:

CS₂ = 2.43 g

Explanation:

Data Given:

CS₂ gas = 6.00g

Cl₂ = 10.0g

Reaction Given:

CS₂(g) + 3Cl₂(g) --------> CCl₄(l) + S₂Cl₂(l)

Solution

Limiting Reagent :

The reactant which is less in amount control the amount of product and know as limiting reagent.

Excess Reagent:

The amount of reactant which are in excess and leftover at the end of reaction and product formed.

Now we have to find the reactant that is in excess

For this we will look at the Reaction

CS₂ + 3Cl₂ --------> CCl₄ + S₂Cl₂

1 mol 3 mol 1 mol 1 mol

we come to know from the above reaction that

1 mole of CS₂ react with 3 mole of Cl₂ to produce 1 mole of CCl₄ and 1 mole of S₂Cl₂

Now to convert mole to mass we required molar masses

molar mass of CS₂ = 12 + 2(32)

molar mass of CS₂ = 76 g/mol

molar mass of Cl₂ = 2(35.5)

molar mass of Cl₂ = 71 g/mol

if we represent mole in grams then

CS₂ + 3Cl₂ --------> CCl₄ + S₂Cl₂

1 mol (76 g/mol) 3 mol (71 g/mol)

CS₂ + 3Cl₂ --------> CCl₄ + S₂Cl₂

76 g 213 g

So,

we come to know that 76 g of CS₂ will combine with 213 g of Cl₂ to form product.

So now look for the ratio of both reactant

CS₂ : Cl₂

76 g : 213 g

1 g : 2.8 g

So, the for every one gram of CS₂ required 2.8g Cl₂

So from this details

we apply unity formula

2.8 g of Cl₂ ≅ 1 g of CS₂

10 g of Cl₂ ≅ ? g of CS₂

by doing cross multiplication

g of CS₂ = 10 x 1 / 2.8

g of CS₂ = 3.6 g

So,

10.0g of Cl₂ used 3.57 g of CS₂

It showed that 3.57 g of CS₂ used and the remaining amount of CS₂ is

Remaining amount of CS₂ = 6 - 3.57 = 2.43 g

8 0

2 years ago

What is the name of Br6F10 ?

astraxan [27]

Answer:

idfk

Explanation:

3 0

2 years ago

Rhett is solving the quadratic equation 0= x2 – 2x – 3 using the quadratic formula. Which shows the correct substitution of the

bekas [8.4K]

The correct values I believe would be a=1 b=-2 and c=-3.

8 0

2 years ago

In this problem, you will use Lenz's law to explore what happens when an electromagnet is activated a short distance from a wire

Vsevolod [243]

Answer:

The correct answer is Option A (There is no magnetic flux through the wire loop.)

Explanation:

Magnetic flux measures the entire magnetic field that passes through the wire loop.

The right hand rule can be demonstrated on how magnetic flux is generated through the moving current in the wire loop. The magnetic flux through the wire loop will decrease as it moves upward through the magnetic field region.

If the direction of the vector area of the wire loop is to the right, and the switch is closed, it will push the magnetic flux to the right which will now be increased due to an equal increase in the current in the wire loop. But, when the switch is open, this will halt the movement of current through the wire loop thus affecting the generation of magnetic field. This would make the magnetic flux to be zero.

4 0

2 years ago

To convert m-ethylaniline to m-ethylfluorobenzene, it should be treated with nitrous acid followed by _____________.

bezimeni [28]

Answer:

$NaBF_4$

Explanation:

In this reaction, we must exchange the amino group ( $NH_2$ ) for a fluorine atom ( $F$ ). Also, the first step in this reaction is the addition of nitrous acid.

We must remember that the amino group in the presence of nitrous acid produces a diazonium salt. The $N_2$ group is a very good leaving group and many benzene derivatives can be produced from this intermediate (see figure 1).

If what we want is to bond a fluorine atom we must use $NaBF_4$ to be able to produce m-ethylfluorobenzene (see figure 2).

I hope it helps!

6 0

2 years ago