Answer:
[Ne] 2s2 2p3
Explanation:
Phosphorus will most likely have an ion that will be 3- because it wants to have a full outer shell. Thus, the elctron configuration is: 1s2 2s2 2p6 3s2 3p3.
The question incomplete , the complete question is:
A student dissolves of 18.0 g urea in 200.0 mL of a solvent with a density of 0.95 g/mL . The student notices that the volume of the solvent does not change when the urea dissolves in it. Calculate the molarity and molality of the student's solution. Round both of your answers to significant digits.
Answer:
The molarity and molality of the student's solution is 1.50 Molar and 1.58 molal.
Explanation:
Moles of urea = 
Volume of the solution = 200.0 mL = 0.2 L (1 mL = 0.001 L)
Molarity of the urea solution ;
Mass of solvent = m
Volume of solvent = V = 200.0 mL
Density of the urea = d = 0.95 g/mL
(1 g = 0.001 kg)
Molality of the urea solution ;
The molarity and molality of the student's solution is 1.50 Molar and 1.58 molal.
Balance Chemical Equation,
2 CO + O₂ → 2CO₂
Acc. to this reaction,
88 g (2 mole) of CO₂ was produced when = 56 g (2 mole)of CO was reacted
So,
24.7 g of CO₂ will be produced by reacting = X g of CO
Solving for X,
X = (56 g × 24.7 g) ÷ 88 g
X = 2.26 g ÷ 88 g
X = 0.0257 g of CO
Result:
0.0257 g of CO is required to be reacted with excess of O₂ to produce 24.7 g of CO₂.
Answer:
V = 134.5 L
Explanation:
Given data:
Number of moles of KClO₃ = 4 mol
Litters of oxygen produced at STP = ?
Solution:
Chemical equation:
2KClO₃ → 2KCl + 3O₂
Now we will compare the moles of KClO₃ with oxygen.
KClO₃ : O₂
2 : 3
4 ; 3/2×4 = 6 mol
Litters of oxygen at STP:
PV = nRT
V = nRT/P
V = 6 mol × 0.0821 atm.L/mol.K × 273 K / 1atm
V = 134.5 L / 1
V = 134.5 L
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(I hope this helps you!)
