The reaction between C2H2 and O2 is as follows:
2C2H2 + 5O2 = 4CO2 + 2H2O
After balancing the equation, the reaction ratio between C2H2 and O2 is 2:5.
The moles of O2 in this reaction is 84.0 mol. According to the above ratio, the moles of C2H2 needed to react completely with the O2 is 84.0mole *2/5 = 33.6 mole.
Answer:
2.64 M
Explanation:
To find the molarity, you need to (1) convert grams to moles (via molar mass), then (2) convert mL to L, and then (3) calculate the molarity (via molarity ratio). The final answer should have 3 sig figs to match the sigs figs of the given values.
(Step 1)
Molar Mass (NH₄NO₃): 2(14.007 g/mol) + 4(1.008 g/mol) + 3(15.998 g/mol)
Molar Mass (NH₄NO₃): 80.04 g/mol
66.5 grams NH₄NO₃ 1 mole
--------------------------------- x ---------------------- = 0.831 moles NH₄NO₃
80.04 grams
(Step 2)
1,000 mL = 1 L
315 mL 1 L
-------------- x ------------------ = 0.315 L
1,000 mL
(Step 3)
Molarity = moles / volume
Molarity = 0.831 moles / 0.315 L
Molarity = 2.64 M
Answer: -
IE 1 for X = 801
Here X is told to be in the third period.
So n = 3 for X.
For 1st ionization energy the expression is
IE1 = 13.6 x Z ^2 / n^2
Where Z =atomic number.
Thus Z =( n^2 x IE 1 / 13.6)^(1/2)
Z = ( 3^2 x 801 / 13.6 )^ (1/2)
= 23
Number of electrons = Z = 23
Nearest noble gas = Argon
Argon atomic number = 18
Number of extra electrons = 23 – 18 = 5
a) Electronic Configuration= [Ar] 3d34s2
We know that more the value of atomic radii, lower the force of attraction on the electrons by the nucleus and thus lower the first ionization energy.
So more the first ionization energy, less is the atomic radius.
X has more IE1 than Y.
b) So the atomic radius of X is lesser than that of Y.
c) After the first ionization, the atom is no longer electrically neutral. There is an extra proton in the atom.
Due to this the remaining electrons are more strongly pulled inside than before ionization. Hence after ionization, the radii of Y decreases.
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