The symbol of Radon??

What would the name of a compound composed of nickel, sulfur, and oxygen be? A. Nickel sulfur oxide B. Nickel sulfate C. Nickel

dexar [7]

Answer:

The answer to your question is: B. Nickel Sulfate

Explanation:

A. Nickel sulfur oxide this compound do not exist, this option is incorrect.

B. Nickel sulfate This compund exist and the formula is NiSO₄

C. Nickel oxygen sulfide This compound do not exist, the order of the names is inocrrect.

D. None of these This option is incorrect because because letter B is correct.

3 0

4 years ago

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both sides in the war used military observers in hydrogen-filled balloons to watch the enemy's movements

lesya692 [45]

Answer:

Yes.

Explanation:

Hydrogen-filled balloons were widely used by the militaries during World War I (1914–1918). The main purpose of these hydrogen-filled balloons to detect movements of enemy troops and to provide direction to the artillery fire. Balloons were the targets of opposing aircraft because they knew the purpose of these balloons so they hit it whenever seen by the enemies so we can say that both sides used hydrogen-filled balloons as military observer to watch the enemy's movements.

7 0

3 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?

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