Answer:
X= Be
Y= B
Z=O
Explanation:
From the description of the compound XCl2, among the options listed only beryllium can form such compound with three lone pairs in the two chlorine atoms and no lone pair on the central atom X.
From the description of YCl3, only Boron among the options listed can form such a compound with no lone pair on the central atom and three lone pairs on each of the chlorine atoms.
From the description of ZCl2, only oxygen forms the compound OCl2 among the elements listed where oxygen possesses two lone pairs and each chlorine atom possesses three lone pairs each.
Answer:
90.3 L
Explanation:
Given data:
Volume of water produced = 77.4 L
Volume of oxygen required = ?
Solution:
Chemical equation:
2C₂H₆ + 7O₂ → 4CO₂ + 6H₂O
It is known that,
1 mole = 22.414 L
There are 7 moles of oxygen = 7×22.414 = 156.9 L
There are 6 moles of water = 6×22.414 = 134.5 L
Now we will compare:
H₂O : O₂
134.5 : 156.9
77.4 : 156.9/134.5×77.4 =90.3 L
So for the production of 77.4 L water 90.3 L oxygen is required.
Explanation:
As it is given that solubility of water in diethyl ether is 1.468 %. This means that in 100 ml saturated solution water present is 1.468 ml.
Hence, amount of diethyl ether present will be calculated as follows.
(100ml - 1.468 ml)
= 98.532 ml
So, it means that 98.532 ml of diethyl ether can dissolve 1.468 ml of water.
Hence, 23 ml of diethyl ether can dissolve the amount of water will be calculated as follows.
Amount of water = 
= 0.3427 ml
Now, when magnesium dissolves in water then the reaction will be as follows.
Molar mass of Mg = 24.305 g
Molar mass of 
= 18 g
Therefore, amount of magnesium present in 0.3427 ml of water is calculated as follows.
Amount of Mg = 
= 0.462 g
Answer is: <span>yield of a reaction is 56,4%.
</span>Chemical reaction: PCl₃ + 3H₂O → 3HCl + H₃PO₃.
m(PCl₃) = 200 g.
m(HCl) = 91,0 g.
n(PCl₃) = m(PCl₃) ÷ M(PCl₃).
n(PCl₃) = 200 g ÷ 137,33 g/mol.
n(PCl₃) = 1,46 mol.
n(HCl) = m(HCl) ÷ M(HCl).
n(HCl) = 91 g ÷ 36,45 g/mol.
n(HCl) = 2,47 mol.
From reaction: n(PCl₃) : n(HCl) = 1 : 3.
n(HCl) = 1,46 mol · 3 = 4,38 mol.
Yield of reaction: 2,47 mol ÷ 4,38 mol · 100% = 56,4%.
Answer:
fH = - 3,255.7 kJ/mol
Explanation:
Because the bomb calorimeter is adiabatic (q =0), there'is no heat inside or outside it, so the heat flow from the combustion plus the heat flow of the system (bomb, water, and the contents) must be 0.
Qsystem + Qcombustion = 0
Qsystem = heat capacity*ΔT
10000*(25.000 - 20.826) + Qc = 0
Qcombustion = - 41,740 J = - 41.74 kJ
So, the enthaply of formation of benzene (fH) at 298.15 K (25.000 ºC) is the heat of the combustion, divided by the number of moles of it. The molar mass od benzene is: 6x12 g/mol of C + 6x1 g/mol of H = 78 g/mol, and:
n = mass/molar mass = 1/ 78
n = 0.01282 mol
fH = -41.74/0.01282
fH = - 3,255.7 kJ/mol
