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3 years ago

How many molecules are there in 223 grams of Na2SO4?

Chemistry

1 answer:

Harlamova29_29 [7]3 years ago

8 0

<h3><u>Answer;</u></h3>

= 9.45 × 10^23 molecules

<h3><u>Explanation; </u></h3>

The molar mass of Na2SO4 = 142.04 g/mol

Number of moles = mass/molar mass

= 223/142.04

= 1.57 moles

But;

1 mole = 6.02 × 10^23 molecules

Therefore;

1.57 moles = ?

= 1.57 × 6.02 × 10^23 molecules

<u>= 9.45 × 10^23 molecules </u>

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The rock in a particular iron ore deposit contains 82 % Fe2O3 by mass. How many kilograms of the rock must be processed to obtai

Zinaida [17]

Answer:

Mass of rock = 4533.4 kg

Explanation:

Percent composition is percentage by the mass of element present in the compound.

Given that the $Fe_2O_3$ is 82 % by mass.

It means that 82 kg of $Fe_2O_3$ is present in 100 kg of rock

To find the mass of rock which contains 2600 kg of iron

2600 kg = 2600*1000 g

2 moles of iron are present in 1 mole of $Fe_2O_3$

Molar mass of iron = 55.845 g/mol

Mass of 2 moles = 55.845 * 2 = 111.69 g/mol

Molar mass of $Fe_2O_3$ = 159.69 g/mol

It means,

111.69 g of iron are present in 159.69 g of $Fe_2O_3$

1 g of iron is present in 159.69/111.69 g of $Fe_2O_3$

2600*1000 g is present in (159.69/111.69)*2600*1000 g of $Fe_2O_3$

Mass of $Fe_2O_3$ = 3717378.47 g =3717.38 kg

Thus,

82 kg of $Fe_2O_3$ is present in 100 kg of rock

1 kg of $Fe_2O_3$ is present in 100/82 kg of rock

3717.38 kg of $Fe_2O_3$ is present in (100/82)*3717.38 kg of rock

Mass of rock = 4533.4 kg

5 0

3 years ago

Read 2 more answers

Question 30 (10 points) All unauthorized experiments are prohibited . O True False

denpristay [2]

Answer:

All unauthorized experiments are prohibited. You are allowed to enter the chemical preparation/storage area any time you need to get an item. Laboratory aprons should be worn during all lab activities. It's okay to pick up broken with our bare hand as long as the glass is placed in the trash.

Explanation:

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3 years ago

A material has an ASTM grain size number of 7. Determine the magnification, if the number of grains per square inch observed is:

Luden [163]

Answer:

A) M = 100X

B) M = 36X

C) M = 178.88X

Explanation:

Given data:

ASTM grain size number 7

a) total grain per inch^2 - 64 grain/inch^2

we know that number of grain per square inch is given as

$Nm = 2^{n-1} (\frac{100}{M})^2$

where M is magnification, n is grain size

therefore we have

$64 = 2^{7-1}(\frac{100}{M})^2$

solving for M we get

M = 100 X

B) total grain per inch^2 = 500 grain/inch^2

we know that number of grain per square inch is given as

$Nm = 2^{n-1} (\frac{100}{M})^2$

where M is magnification, n is grain size

therefore we have $500 = 2^{7-1}(\frac{100}{M})^2$

solving for M we get

M = 36 X

C) Total grain per inch^2 = 20 grain/inch^2

we know that number of grain per square inch is given as

$Nm = 2^{n-1} (\frac{100}{M})^2$

where M is magnification, n is grain size

therefore we have $20 = 2^{7-1}(\frac{100}{M})^2$

solving for M we get

M = 178.88 X

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3 years ago

You carry out a reaction that yields a major product Q (boiling point: 120°C) and side product R (boiling point: 150°C). The rat

d1i1m1o1n [39]

Answer:

Here's what I get

Explanation:

1. Requirements for a liquid/liquid extraction solvent

(a) Distribution Coefficient

The solute must be more soluble in the solvent (the extract phase) than in the raffinate phase (the other liquid).

(b) Insolubility

The solvent and the other liquid must be mutually immiscible. This means they will form two layers in the separatory funnel.

(c) Removability

It must be easy to separate the solvent from the solutes. This usually means the solvent must have a much lower boiling point than the solutes.

2. Extract layer

The density of ethyl acetate is 0.920 g/mL.

The density of water is 1.00 g g/mL.

The ethyl acetate layer will be on the top.

3. Separability of mixture

Two miscible liquids must have a boiling point difference of at least 40 °C to be separable by simple distillation.

4. Type of distillation

The boiling point difference is only 30°C, so you must use fractional distillation.

5. Apparatus

Your microscale kit must include an air condenser or a fractionating column.

Perhaps it looked like the fractionating head in the kit in Fig. 1.

The alternate distillation (simple distillation) would use a distillation head like that in Fig. 2.

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4 years ago

Identify the followings as a element or compound c12 h22o11

Anastaziya [24]

Answer:

Compound

Explanation:

The given specie is a compound.

A compound is made up of two or more kinds of atoms/elements joined together in a definite grouping.

The properties of a compound are distinct and different from those of the individual elements that are combined in its make-up.
For this given specie, the elements are carbon, hydrogen and oxygen.
So, this is a compound.

Elements are distinct substances that cannot be split - up into simpler substances. Such substances consist of only one kind of atom

6 0

3 years ago