The resulting pressure of the gas after decreasing the initial volume from 2 L to 1 L is 3 atm.
<h3>What is
Boyle's Law?</h3>
According to the Boyle's Law at constant temperature, pressure of the gas is inversely proportional to the volume of that gas.
For the given question we use the below equation is:
P₁V₁ = P₂V₂, where
P₁ = initial pressure of gas = 1.5 atm
V₁ = initial volume of gas = 2 L
P₂ = final pressure of gas = ?
V₂ = final volume of gas = 1 L
On putting all these values on the above equation, we get
P₂ = (1.5atm)(2L) / (1L) = 3 atm
Hence required pressure of the gas is 3 atm.
To know more about Boyle's Law, visit the below link:
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43.8 kJ
<h3>
Explanation</h3>
There are two electrodes in a voltaic cell. Which one is the anode?
The lithium atom used to have no oxygen atoms when it was on the reactant side. It gains two oxygen atoms after the reaction. It has gained more oxygen atoms than the manganese atom. Gaining oxygen is oxidation. As a result, lithium is being oxidized.
Oxidation takes place at the anode of a cell. Therefore, the anode of this cell is made of lithium.
Lithium has an atomic mass of 6.94. Each gram of Li would contain 1/6.94 = 0.144 moles of Li atoms. Each Li atom loses one electron in this cell. Therefore, the number of electron transferred, <em>n</em>, equals 0.144 moles for each gram of the anode.
Let 
represents the electrical energy produced.
, where
- <em>n</em> is the <em>number of moles</em> electrons transferred,
- <em>F</em> is the Faraday's constant,
- <em>E</em>
is the cell potential,
<em>n </em>= 0.144 mol, as shown above, and
<em>F </em>= 96.486 kJ / (
).
Therefore,
.
Answer:
sample A
Explanation:
the first one because of the ppm value
Answer:
The equilibrium shifts to the left, and the concentration of Ba2+(aq) decreases
Explanation:
Whenever a solution of an ionic substance comes into contact with another ionic compound with which it shares a common ion, the solubility of the ionic substance in solution decreases significantly.
In this case, both BaSO4 and Na2SO4 both possess the SO4^2- anion. Hence SO4^2- anion is the common ion. Given the equilibrium;
BaSO4(s) <—> Ba2+ (aq) + SO4 2- (aq), addition of Na2SO4 will decrease the solubility of BaSO4 due to the presence of a common SO4^2- anion compared to pure water.
This implies that the equilibrium will shift to the left, (more undissoctiated BaSO4) hence decreasing the Ba^2+(aq) concentration.
