Answer: Thus 0.724 mol of 
are needed to obtain 18.6 g of 
Explanation:
To calculate the moles :
According to stoichiometry :
2 moles of are produced by = 1 mole of
Thus 1.09 moles of will be produced by =
of
But as yield of reaction is 75.6 %, the amount of needed is =
Thus 0.724 mol of are needed to obtain 18.6 g of
The glass opposite to the negative electrode started to glow. Hence, option B is correct.
<h3>What is a cathode ray tube?</h3>
A cathode-ray tube (CRT) is a specialized vacuum tube in which images are produced when an electron beam strikes a phosphorescent surface.
J.J. Thomson, through his famous Cathode ray experiment, proved that all atoms contain small negatively charged particles known as electrons. In the experiment, he applied electric voltage across a cathode ray tube. a fluorescent material coating was done on the positive side. When the voltage was applied, the positive side has glowing dots.
Hence, option B is correct.
Learn more about the cathode ray tube here:
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Sucrose, a sweet, white crystalline substance, C12 H22 O11, OBTAINED CHIEFLY FROM THE JUICE OF THE SUGAR CANE AND SUGAR BEET, BUT ALSO PRESENT IN SORGHUM, THE sugar maple, some palms, and various other plants, and having extensive nutritional, pharmaceutical, and industrial uses; any of the class of carbohydrates to which this substance belongs, as glucose, levulose, and lactose.
Answer:
3.64g
Explanation:
Given parameters:
Mass of NH₃ = 18.1g
Mass of Cu₂O = 90.4g
Unknown:
Limiting reactant = ?
Mass of N₂ formed = ?
Solution:
The reaction equation is given as:
Cu₂O + 2NH₃ → 6Cu + N₂ + 3H₂O
The limiting reactant is the one in short supply in the reaction. Let us find the number of moles of the given species;
Number of moles = 
Molar mass of Cu₂O = 2(63.6) + 16 = 143.2g/mol
Molar mass of NH₃ = 14 + 3(1) = 17g/mol
Number of moles of Cu₂O = 
= 0.13moles
Number of moles of NH₃ = 
= 5.32moles
From this reaction;
1 mole of Cu₂O combines with 2 mole of NH₃
So 0.13moles of Cu₂O will combine with 0.13 x 2 mole of NH₃
= 0.26moles of NH₃
Therefore, Cu₂O is the limiting reactant. Ammonia is in excess;
Mass of N₂;
Mass = number of moles x molar mass
1 mole of Cu₂O will produce 1 mole of N₂
0.13 mole of Cu₂O will produce 0.13 mole of N₂
Mass = 0.13 x (2 x 14) = 3.64g
Given:
P1 = 13.0 atm
T1 = 20 °C
T2 = 102 °C
Required:
P2 of oxygen
Solution:
At constant volume,
we can apply Gay-Lussac’s law of pressure and temperature relationship
P1/T1=P2/T2
(13.0 atm) / (20 °C)
= P2 / (102 °C)
P2 = 66.3 atm
The answer is not in the choices given.
