We will use Ideal gas equation.
PV=nRT
where
P= pressure(in atm)=5.82atm
V=Volume=?
n= moles of gases=0.682
R = gas constant= 0.0821L atm / mol K
T = temperature= 68.2 C= 68.2 + 273.15 = 341.35 K
V= nRT/ P= 0.682 X 0.0821 X 341.35/ 5.82
V= 3.284 L.
Hello!
The set of compounds that have the same empirical formula is b) N₂O₄ and NO₂.
The Empirical Formula is the most simple representation of the atom ratio in a chemical compound. Of the listed sets of compounds, the only in which the atom ratio is the same for both compounds is the pair N₂O₄ and NO₂ in which the atom ratio N:O is 1:2. The Empirical Formula for this pair is NO₂.
N₂O₄ has the same atom ratio as NO₂ but this formula has each atom multiplied by two. However, its Empirical Formula is the same.
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<em>The cathode is Y</em>
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Further explanation</em></h3>
Electrolysis uses electrical energy to carry out redox reactions that are not spontaneous.
The ions in the solution flowing electrically will move towards to opposite charge of the electrode
The electrolysis material is an electrolyte which can be a solution or a melt.
In positive pole electrolysis cells - the anode is the site of the oxidation reaction, while the negative pole - the cathode is the reduction reaction site.
The result of the reaction in the anode is based on a substance that easily oxidized while the reaction in the cathode is based on a substance that easily reduced.
Electrons (electricity) enter an electrolysis cell through the negative pole (cathode)
The negative ion from the solution will move towards the positive electrode and release the electrons around the positive electrode (oxidation) and the electrons flow to the negative pole
Whereas around the negative electrode, there is electron binding and a reduction reaction occurs
So if we see the picture the cathode is Y
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Learn more</em></h3><h3><em>
reaction related to electrochemistry brainly.com/question/3461108</em></h3>
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Answer details </em></h3>
Grade: Senior High School
Subject: Chemistry
Chapter: Electrochemistry
Keywords: cathode, anode, oxidation, reduction, negative pole, electrode
Answer:
1.10 g H2
Explanation:
I took the test and got the answer correct
Answer:
159.609 g/mol
Explanation:
According to the CuSO4.5H2O (k) heat CuSO4 (k) + 5 H2O (g) equation, the crystal water amount of copper sulfate and its rough formula will be calculated.
Weight of copper sulfate containing crystal water = m1 = 249.62… g
Weight of copper sulfate without crystal water weighed = m2 = 159.62 g
Accordingly, calculate the x and y values in the molecular formula of copper sulfate (xCuSO4.yH2O).
