Answer:Two more hydrogen atoms will be required
Explanation:
A carbon atom has 4 valence electrons in its outermost shell,thus it can form 4 covalent bonds.Two pairs of electrons are shared in a double bond between C-C atom.two more electrons are left which is shared with two hydrogen atoms.
Answer:
CH₅N
Explanation:
In the combustion, all of the C in the compound was used to produce CO₂ in a 1:1 ratio. Thus, the moles of CO₂ (MW 44.01 g/mol) produced equals the moles of C in the compound:
(44.0 g)(mol/44.01g) = 0.99977 mol CO₂ = 0.99977... mol C
Similarly, all of the H in the compound was used to produce H₂O in a ratio of 2H:1H₂O. The moles of H₂O (MW 18.02 g/mol) produced was:
(45.0 g)(mol/18.02g) = 2.497...mol H₂O
Moles of H is found using the molar ratio of 2H:1H₂O:
(2.497...mol H₂O)(2H/1H₂O) = 4.994...mol H
The ratio of H to C in the compound is:
(4.994...mol H)/(0.99977... mol C) = 5 H:C
Some NO₂ was produced from the N in the compound. Assuming a 1:1 ratio of C:N, the simplest empirical formula is: CH₅N.
Answer:
There are 2 double bond units and 1 lone pair, which will try to get as far apart as possible - taking up a trigonal planar arrangement. Because the lone pair isn't counted when you describe the shape, SO2 is described as bent or V-shaped.
Explanation:
There are 2 double bond units and 1 lone pair, which will try to get as far apart as possible - taking up a trigonal planar arrangement. Because the lone pair isn't counted when you describe the shape, SO2 is described as bent or V-shaped.
The volume in liters of Hydrogen gas = 49.28 L
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Further explanation
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The reaction equation is the chemical formula of reagents and product substances
A reaction coefficient is a number in the chemical formula of a substance involved in the reaction equation. The reaction coefficient is useful for equalizing reagents and products
Reaction
2 Al (s) + 6 HCl (aq) → 2 AlCl₃ (aq) + 3 H₂ (g)
mol Al :
mol H₂ :
At STP, 1 mol =22.4 L
so the volume of hydrogen :
Answer:
So the volume will be 2.33 L
Explanation:
The reaction for the combustion is:
2 C₄H₁₀ (g) + 13 O₂ (g) → 8 CO₂ (g) + 10 H₂O (l)
mass of butane to moles (mass / molar mass)
1.4 g / 58 g/mol
= 0.024 moles
2 moles of butane can produce 8 moles of carbon dioxide
0.024 moles of butane must produce (0.024 × 8) /2
= 0.096 moles of CO₂
Now we apply the Ideal Gases Law to find out the volume formed.
P . V = n . R . T
p = 1atm
n = 0.096 mol
R = 0.082 L.atm/mol.K
T = 273 + 23 = 296K
V = ?
1atm × V = 0.096 mol × 0.082 L.atm/mol.K × 296K
V = 0.096 mol × 0.082 L.atm/mol.K × 296K / 1atm
= 2.33 L
So the volume will be 2.33 L
