Answer:
The amount of energy released from the combustion of 2 moles of methae is 1,605.08 kJ/mol
Explanation:
The chemical reaction of the combustion of methane is given as follows;
CH₄ (g) + 2O₂ (g) → CO₂ (g) + 2H₂O (g)
Hence, 1 mole of methane combines with 2 moles of oxygen gas to form 1 mole of carbon dioxide and 2 moles of water vapor
Where:
CH₄ (g): Hf = -74.6 kJ/mol
CO₂ (g): Hf = -393.5 kJ/mol
H₂O (g): Hf = -241.82 kJ/mol
Therefore, the combustion of 1 mole of methane releases;
-393.5 kJ/mol × 1 + 241.82 kJ/mol × 2 + 74.6 kJ/mol = -802.54 kJ/mol
Hence the combustion of 2 moles of methae will rellease;
2 × -802.54 kJ/mol or 1,605.08 kJ/mol.
Molarity=Moles of solute/Volume of solution in L
So
- 0.56M=moles/2.5L
- moles=0.56(2.5)
- moles of Iodine=1.4mol
Mads of Iodine
- Moles(Molar mass)
- 1.4(126.9)
- 177.66g
Answer:
B) Symmetrical and nonpolar
Step-by-step explanation:
The formula is H-C≡C-H.
Each C atom has <em>two</em> electron regions, so VSEPR theory predicts a <em>linear molecular geometry</em> (see image below).
The molecule is symmetrical, because the green line divides the molecule into two halves that are mirror images of each other.
The C-H bonds are slightly polar, because C is more electronegative than H (µ ≈ 0.4 D).
The C atoms are partially negative (red), while the H atoms are partially positive (blue).
However, the two C-H bond dipoles point in <em>opposite directions</em>, so they cancel each other. The molecule has <em>no net dipole moment.</em>
Acetylene is nonpolar.
Hey there!
For SN1 mechanism; the activation barrier is the C-I bond energy which is broken in the first step of the reaction.
The activation barrier is : 56 kcal/ mol = 5.6 kcal/ mole ( nearest 0.1)
Answer:
a) 0.100 M
b) 0.395 M
Explanation:
a) Calculate the molarity of a solution that contains 0.200 moles of NaOH (solute) in 2.00 L of solution
We will use the following expression for molarity.
[NaOH] = moles of solute / liters of solution
[NaOH] = 0.200 mol/2.00 L = 0.100 M
b) Calculate the molarity for a solution that contains 15.5 g of NaCl (solute, 58.44 g/mol) in 671 mL of solution
We will use the following expression for molarity.
[NaCl] = mass of solute / molar mass of solute × liters of solution
[NaCl] = 15.5 g / 58.44 g/mol × 0.671 L = 0.395 M
