All categories

3 years ago

How can we separate the components of a compound?

Chemistry

1 answer:

Alinara [238K]3 years ago

6 0

Answer:

A mixture is a physical combination of substances thus it only requires physical processes to separate. However, a compound is chemically combined and can only be separated by chemical processes. Therefore, it is much harder to separate a compound than a mixture.

You might be interested in

What is the chemical formula for ammonium carbonate?

xenn [34]

(NH4),CO

(NH₄)₂ CO₃

Explanation:

The chemical formula for ammonium carbonate is (NH4),CO

Ammonium carbonate is made up of:

Ammonium ion

Carbonate ion

Ammonium ion is represented by NH₄⁺

Carbonate ion is represented by CO₃²⁻

Now to write the formula, we are going to use their combining powers;

NH₄⁺ CO₃²⁻

Combining

power 1 2

Exchange

of power 2 1

Formula of compound (NH₄)₂ CO₃

learn more:

Chemical compound brainly.com/question/10585691

#learnwithBrainly

3 0

3 years ago

How many grams of chlorine gas are present in a 150. liter cylinder of chlorine held at a pressure of 1.00 atm and 0. °C? Group

OlgaM077 [116]

Answer:

474 grams of chlorine gas are present in a 150 liter cylinder of chlorine held at a pressure of 1.00 atm and 0 °C

Explanation:

An ideal gas is a theoretical gas that is considered to be composed of randomly moving point particles that do not interact with each other. Gases in general are ideal when they are at high temperatures and low pressures.

The pressure, P, the temperature, T, and the volume, V, of an ideal gas, are related by a simple formula called the ideal gas law:

P*V = n*R*T

where P is the gas pressure, V is the volume that occupies, T is its temperature, R is the ideal gas constant, and n is the number of moles of the gas.

In this case:

P= 1.00 atm
V= 150 L
n= ?
R= 0.082 $\frac{atm*L}{mol*K}$
T= 0 C= 273 K

Replacing:

1.00 atm* 150 L= n*0.08206 $\frac{atm*L}{mol*K}$ *273 K

Solving:

$n=\frac{1.00 atm* 150 L}{0.08206 \frac{atm*L}{mol*K}*273 K}$

n= 6.69 moles

Being Cl= 35.45 g/mole, the molar mass of chlorine gas is:

Cl₂=2*35.45 g/mole= 70.9 g/mole

So if 1 mole has 70.9 grams, 6.69 moles of the gas, how much mass does it have?

$mass=\frac{6.69 moles*70.9 grams}{1 mole}$

mass= 474.321 grams ≅ 474 grams

474 grams of chlorine gas are present in a 150 liter cylinder of chlorine held at a pressure of 1.00 atm and 0 °C

4 0

3 years ago

A powder contains FeSO4 · 7 H2O (molar mass = 278.01 g/mol), among other components. A 3.055 g sample of the powder was dissolve

solong [7]

Answer:

The mass of FeSO4*7H2O is 1.023 grams

Explanation:

Step 1: Data given

Molar mass of FeSO4 * 7H2O = 278.01 g/mol

Mass of Fe2O3 = 0.294 grams

Mass of the sample = 3.055 grams

Step 2: The balanced equation

4 Fe + 3O2 → 2Fe2O3

Step 3: Calculate moles Fe2O3

Moles Fe2O3 = mass Fe2O3 / molar mass Fe2O3

Moles Fe2O3 = 0.294 grams / 159.59 g/mol

Moles Fe2O3 = 0.00184 moles

Step 3: Calculate moles Fe

In 2 moles Fe2O3 we have 4 moles Fe

For 0.00184 moles we'll have 2*0.00184 = 0.00368 moles

Step 4: The balanced equation

Fe + H2SO4 + 7H2O → FeSO4*7H20 + H2

Step 5: Calculate moles FeSO4* 7H2O

For 1 mol Fe we have 1 mol FeSO4*7H2O

For 0.00368 moles Fe we have 0.00368 moles FeSO4*7H2O

Step 6: Calculate the mass of FeSO4*7H2O

Mass FeSO4*7H2O = moles * molar mass

Mass FeSO4*7H2O = 0.00368 moles * 278.01 g/mol

Mass FeSO4*7H2O = 1.023 grams

The mass of FeSO4*7H2O is 1.023 grams

6 0

3 years ago

When naming compounds part of the second element's name is dropped and what is added in its place?

FromTheMoon [43]

Answer:

ide is the correct answer

Explanation:

example compound of sodium and hydrogen in sodium hydroxide

and compound of hydrogen and chlorine is hydrogen chloride

3 0

3 years ago

Calculate the [OH-] [H+]=0.000025

Harman [31]

Explanation:

There are several ways to define acids and bases, but pH and pOH refer to hydrogen ion concentration and hydroxide ion concentration, respectively. The "p" in pH and pOH stands for "negative logarithm of" and is used to make it easier to work with extremely large or small values. pH and pOH are only meaningful when applied to aqueous (water-based) solutions. When water dissociates it yields a hydrogen ion and a hydroxide.

3 0

3 years ago