Answer is: volume of carbon dioxide is 1,84·10⁸ l.
Chemical reaction: C + O₂ → CO₂.
m(C) = 100 t · 1000 kg/t = 100000 kg
m(C) = 100000 kg · 1000 g/kg = 10⁸ g.
n(C) = m(C) ÷ M(C).
n(C) = 10⁸ g ÷ 12 g/mol.
n(C) = 8,33·10⁶ mol.
From chemical reaction: n(C) . n(CO₂) = 1 : 1.
n(CO₂) = 8,33·10⁶ mol.
m(CO₂) = 8,33·10⁶ mol · 44 g/mol.
m(CO₂) = 3,66·10⁸ = 3,66·10⁵ kg.
V(CO₂) = 3,66·10⁵ kg ÷ 1,98 kg/m³ = 1,84·10⁵ m³.
V(CO₂) = 1,84·10⁵ m³ · 1000 l/m³ = 1,84·10⁸ l.
Answer:
5.56 × 10^23 molecules
Explanation:
The number of molecules in a molecule can be calculated by multiplying the number of moles in that molecule by Avagadro's number (6.02 × 10^23)
Using mole = mass/molar mass
Molar mass of N2O4 = 14(2) + 16(4)
= 28 + 64
= 92g/mol
mole = 85.0/92
= 0.9239
= 0.924mol
number of molecules of N2O4 (nA) = 0.924 × 6.02 × 10^23
= 5.56 × 10^23 molecules
Carbons starting from the left end:
- sp²
- sp²
- sp²
- sp
- sp
Refer to the sketch attached.
<h3>Explanation</h3>
The hybridization of a carbon atom depends on the number of electron domains that it has.
Each chemical bond counts as one single electron domain. This is the case for all chemical bonds: single, double, or triple. Each lone pair also counts as one electron domain. However, lone pairs are seldom seen on carbon atoms.
Each carbon atom has four valence electrons. It can form up to four chemical bonds. As a result, a carbon atom can have up to four electron domains. It has a minimum of two electron domains, with either two double bonds or one single bond and one triple bond.
- A carbon atom with four electron domains is sp³ hybridized;
- A carbon atom with three electron domains is sp² hybridized;
- A carbon atom with two electron domains is sp hybridized.
Starting from the left end (H₂C=CH-) of the molecule:
- The first carbon has three electron domains: two C-H single bonds and one C=C double bond; It is sp² hybridized.
- The second carbon has three electron domains: one C-H single bond, one C-C single bond, and one C=C double bond; it is sp² hybridized.
- The third carbon has three electron domains: two C-C single bonds and one C=O double bond; it is sp² hybridized.
- The fourth carbon has two electron domains: one C-C single bond and one C≡C triple bond; it is sp hybridized.
- The fifth carbon has two electron domains: one C-H single bond and one C≡C triple bond; it is sp hybridized.
Answer:
During nuclear fission and fusion matter that seems to disappear but is actually converted into energy. The amount of energy (E) produced in such a reaction can be calculated using Einstein's formula for the equivalence of mass and energy: E = mc^2.
Explanation:
Answer:
0.1 M
Explanation:
The following data were obtained from the question:
Initial volume (V1) = 100 mL
Initial concentration (C1) = 0.5 M
Final volume (V2) = 500 mL
Final concentration (C2) =?
Using the dilution formula C1V1 = C2V2, the new concentration of the solution can be obtained as follow:
C1V1 = C2V2
0.5 × 100 = C2 × 500
50 = C2 × 500
Divide both side by 500
C2 = 50/500
C2 = 0.1 M
Therefore, the new concentration of the solution is 0.1 M