Jacques Charles used this reaction to prepare hydrogen gas for his historic balloon flights: Fe(s) + H2SO4(aq) = FeSO4(aq) + H2(
g) He wants to prepare a small test balloon to check flight conditions before he lifts off in his giant balloon. He has 234 grams of iron and 382 mL of 12 M (mol / L) sulfuric acid available. What is the maximum size his test balloon can be in L? The temperature in Paris is a chilly 18 celsius and the atmospheric pressure if 770 mm Hg. (This is a limiting reactant problem).
1 answer:
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Answer:
The correct option is;
4 percent ionic, 96 percent covalent, 222 pm
Explanation:
The parameters given are;
Phosphorus:
Atomic radius = 109 pm
Covalent radius = 106 pm
Ionic radius = 212 pm
Electronegativity of phosphorus = 2.19
Selenium:
Atomic radius = 122 pm
Covalent radius = 116 pm
Ionic radius = 198 pm
Electronegativity of selenium= 2.55
The percentage ionic character of the chemical bond between phosphorus and selenium is given by the relation;
Using Pauling's alternative electronegativity difference method, we have;
Where:
Δ E.N. = Change in electronegativity = 2.55 - 2.19 = 0.36
Therefore;
Hence the percentage ionic character = 4.3% ≈ 4%
the percentage covalent character = (100 - 4.3)% = 95.7% ≈ 96%
The bond length for the covalent bond is found adding the covalent radii of both atoms as follows;
The bond length for the covalent bond = 106 pm + 116 pm = 222 pm.
The correct option is therefore, 4 percent ionic, 96 percent covalent, 222 pm.
Explanation:
- For path A, the calculation will be as follows.
As, for reversible isothermal expansion the formula is as follows.
W =
Since, we are not given the number of moles here. Therefore, we assume the number of moles, n = 1 mol.
As the given data is as follows.
R = 8.314 J/(K mol), T = 298 K ,
= 8.34 L,
= 2.67 L
Now, putting the given values into the above formula as follows.
W =
=
=
= -2818.68 J
Hence, work for path A is -2818.68 J.
- For path B, the calculation will be as follows.
Step 1: When there is no change in volume then W = 0
Hence, for step 1, W = 0
Step 2: As, the gas is allowed to expand against constant external pressure = 1.00 atm.
So, W =
Now, putting the given values into the above formula as follows.
W =
=
= -5.67 atm L
As we known that, 1 atm L = 101.33 J
Hence, work will be calculated as follows.
W =
= -574.54 J
Therefore, total work done by path B = 0 + (-574.54 J)
W = -574.54 J
Hence, work for path B is -574.54 J.