Define element and compound. explain the difference between an element and a compound.

In a closed system, If a gas is transported to a container with double the volume of the previous container, the gas was held in

Maslowich

If it goes from a volume of x, then when the volume is doubled it will go to a volume of 2x. If you meant what happens to the pressure, then the pressure halves as volume doubles due to Boyle's law.

7 0

3 years ago

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Show your work. Write a balanced net ionic equation for the following reaction. H3PO4(aq) + Ca(OH)2(aq) Ca3(PO4)2(aq) + H2O(l)

viktelen [127]

To do a balancing to result to a balanced equation, we do elemental balance from each side of the equation. Hence we do separate balances for H, P, O and Ca. In this case, the final balanced equation after balancing elementally is <span>3 Ca(OH)2(aq) + 2 H3Po4(aq) = 6 H2O(l) + Ca3(Po4)2(s)</span>

6 0

4 years ago

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How many moles of sulfur are present in 5 moles of h2so4?.

dusya [7]

Answer:

One mole of sulfuric acid consists of one mole of sulphur.

Explanation:

This is because in the molecule of sulfuric acid there is only one atom of sulphur that is present. As a result of this since in case of one mole of sulfuric acid there should be only one 1 mole of sulphur.

According to the molar concept one mole of sulphur means 6.023 x10^ 23 atoms of sulphur is present in one complete mole of sulfuric acid. in addition to this the molecule also consists of 2 moles of hydrogen and 4 moles of oxygen in 1 mole of sulfuric acid

to know more about sulfuric acid,

. Calculate the mass (in g) of 0.500 mol of sulfuric acid (H2SO4 ...

4 0

2 years ago

How many grams of al(oh)3 (molar mass = 78.0 g/mol) can be produced from the reaction of 48.6 ml of .15 m koh with excess al2(so

Fudgin [204]

Taking into account the reaction stoichiometry, 0.18954 grams of Al(OH)₃ are formed from the reaction of 48.6 mL (0.0486 L) of 0.15 M KOH with excess Al₂(SO₄)₃.

<h3>Reaction stoichiometry</h3>

In first place, the balanced reaction is:

Al₂(SO₄)₃ + 6 KOH → 2 Al(OH)₃ + 3 K₂SO₄

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

Al₂(SO₄)₃: 1 mole
KOH: 6 moles
Al(OH)₃: 2 moles
K₂SO₄: 3 moles

The molar mass of the compounds is:

Al₂(SO₄)₃: 342 g/mole
KOH: 56.1 g/mole
Al(OH)₃: 78 g/mole
K₂SO₄: 174.2 g/mole

Then, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:

Al₂(SO₄)₃: 1 mole ×342 g/mole= 342 grams
KOH: 6 moles ×56.1 g/mole= 336.6 grams
Al(OH)₃: 2 moles ×78 g/mole= 156 grams
K₂SO₄: 3 moles ×174.2 g/mole= 522.6 grams

<h3>Definition of molarity</h3>

Molarity is a measure of the concentration of a solute in a solution and indicates the number of moles of solute that are dissolved in a given volume:

Molarity= number of moles÷ volume

<h3>Mass of Al(OH)₃ formed</h3>

In firts place, you know that 48.6 mL (0.0486 L) of 0.15 M KOH react with excess Al₂(SO₄)₃.

Replacing in the definition, you can calculate the amount of moles of KOH that react:

0.15 M= number of moles÷ 0.0486 L

Solving:

0.15 M× 0.0486 L= number of moles

<u><em>0.00729 moles= number of moles</em></u>

Then, 0.00729 moles of KOH react with excess Al₂(SO₄)₃. So, following rule of three can be applied: if by reaction stoichiometry 6 moles of KOH form 156 grams of Al(OH)₃, 0.00729 moles of KOH form how much mass of Na₂SO₄?

$mass of Al(OH)_{3} =\frac{0.00729 moles of KOHx156 grams of Al(OH)_{3}}{6 moles of KOH}$