Answer:
2 mol of CO₂
Solution:
The reaction is as follow,
H₂CO + O₂ → CO₂ + H₂O
According to this equation,
1 mole of H₂CO produces = 1 mole of CO₂
So,
2 moles of H₂CO will produce = X moles of CO₂
Solving for X,
X = (2 mol × 1 mol) ÷ 1 mol
X = 2 mol of CO₂
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.
The molarity of a hydrochloric acid solution : 0.32 M
<h3>Further explanation </h3>
Titration is a procedure for determining the concentration of a solution by reacting with another solution which is known to be concentrated (usually a standard solution).
Titrations can be distinguished including acid-base titration, depositional titration, and redox titration. An acid-base titration is the principle of neutralization of acids and bases is used.
Acid-base titration formula
Ma. Va. na = Mb. Vb. nb
Ma, Mb = acid base concentration
Va, Vb = acid base volume
na, nb = acid base valence
1 ⇒HCl (valence=1, HCl ⇒H⁺+Cl⁻, one H⁺)
2⇒Ca(OH)₂(valence=2, Ca(OH)₂⇒Ca²⁺+2OH⁻, two OH⁻)
M₂=0.1 M
V₂=48 ml=0.048 L
V₁=30 ml=0.03 L
<span>Hydrogen bonds are
approximately 5% of the bond strength of covalent bonds, for example (C-C or C-H
bonds).
Hydrogen bonds strength in water is approximately 20
kJ/mol, strenght of carbon-carbon bond is approximately 350 kJ/mol
and strengh of carbon-hydrogen bond is approximately 340 kJ/mol.
20 kJ/350 kJ = 0,057 = 5,7 %.</span>
