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vladimir2022 [97]
3 years ago
15

Argon occupies a volume of 28.3 L at a pressure of 0.36 atm. Find the volume of argon when the pressure is increased to 0.85 atm

.
Chemistry
1 answer:
FromTheMoon [43]3 years ago
3 0

Answer:

12

Explanation:

Boyles law

P1 (.36)

V1(28.3)

P2(.85)

V2(?)

P1*V1=P2*V2

plug it in, and you get 12

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Kinetics is a study of the rate at which __ disappear and __ appear.​
Korvikt [17]

Answer:

Reactants disappear while the Products appear

Explanation:

8 0
3 years ago
Describe the ways in which ionic compound nomenclature differs from covalent compound nomenclature
Igoryamba

Answer:

Ionic compound are when electrons are given to another element, making one atom positive and the other negative, so they attract. Covalent compound is when both atoms share electrons with each other.

7 0
3 years ago
For the reaction N2(g) + 3H2(g) 2NH3(g), what will happen if hydrogen gas was removed from the reaction mixture? There will be a
Hoochie [10]
This uses something called <span>Le Chatelier's principle. It states essentially that any stress put upon a system will be corrected.

In more simple terms, it means that in an equilibrium, such as the equation N2(g) + 3H2(g) <=> 2NH3(g), removing a reactant will cause the system to create more of said reactant to compensate for its loss, or adding excess reactant will cause the system to remove some of the added reactant. For future reference, the same principle applies to products in an equilibrium as well. 

In this case, hydrogen gas is a reactant, and hydrogen is being removed. According to </span><span>Le Chatelier's principle, the system will shift to create more hydrogen gas. In essence, it will shift in the direction of the hydrogen gas, so there will be a shift toward the reactants.

To clear something up, Keq will not change, as it is a constant value with constant conditions (such as temperature, pressure, etc.).</span>
7 0
3 years ago
During the chemical reaction given below 21.71 grams of each reagent were allowed to react. Determine how many grams of the exce
swat32

Answer: 16.32 g of O_2 as excess reagent are left.

Explanation:

To calculate the moles :

\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}    

\text{Moles of} SO_2=\frac{21.71g}{64g/mol}=0.34mol

\text{Moles of} O_2=\frac{21.71g}{32g/mol}=0.68mol

2SO_2(g)+O_2(g)\rightarrow 2SO_3(g)  

According to stoichiometry :

2 moles of SO_2 require = 1 mole of O_2

Thus 0.34 moles of SO_2 will require=\frac{1}{2}\times 0.34=0.17moles  of O_2

Thus SO_2 is the limiting reagent as it limits the formation of product and O_2 is the excess reagent.

Moles of O_2 left = (0.68-0.17) mol = 0.51 mol

Mass of O_2=moles\times {\text {Molar mass}}=0.51moles\times 32g/mol=16.32g

Thus 16.32 g of O_2 as excess reagent are left.

3 0
2 years ago
In one of his experiments, Lavoisier placed 10.0 grams of mercury (II) oxide into a sealed container and heated it. The mercury
inysia [295]

Oxygen gas produced : 0.7 g

<h3>Further explanation</h3>

Given

10.0 grams HgO

9.3 grams Hg

Required

Oxygen gas produced

Solution

Reaction⇒Decomposition

2HgO(s)⇒2Hg(l)+O₂(g)

Conservation of mass applies to a closed system, where the masses before and after the reaction are the same

mass of reactants = mass of products

mass  HgO = mass Hg + mass O₂

10 g = 9.3 g + mass O₂

mass O₂ = 0.7 g

4 0
3 years ago
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