The middle carbon is 4-degree since it is attached to 4 carbons. All other carbons are 1-degree since they are attached to only 1 carbon.
Hydrogens attached with 1-degree carbon are all same. Hydrogen are often refereed to as protons. No carbon is attached to 4-degree carbon. So all hydrogens in this structure are same.
This structure is called
NeoPentane
Answer:
Energy would be absorbed.
Explanation:
Lattice energy is defined as the energy required to break apart an ionic solid and convert its component atoms into gaseous ions. That is what you're doing in:
KCl (s) → K⁺(g) + Cl⁻(g)
The energy you require to obtain this reaction is 701 kJ/mol. As the value is positive, <em>energy would be absorbed.</em>
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I hope it helps!
Answer:
The true statement is option A.
Explanation:
Using ideal gas equation:
PV = nRT
where,
P = Pressure of gas = 1 atm
V = Volume of gas = ?
n = number of moles of gas = 1 mol
R = Gas constant = 0.0821 L.atm/mol.K
T = Temperature of gas = 273.15 K
V = 22.42 L
This means that 1 mole of an ideal gas at STP occupies 22.42 liters of volume.
So, 1 mole of helium gas and 1 mole of oxygen gas will have same value of volume in their respective balloons at STP.
Answer:
0.74 grams of methane
Explanation:
The balanced equation of the combustion reaction of methane with oxygen is:
it is clear that 1 mol of CH₄ reacts with 2 mol of O₂.
firstly, we need to calculate the number of moles of both
for CH₄:
number of moles = mass / molar mass = (3.00 g) / (16.00 g/mol) = 0.1875 mol.
for O₂:
number of moles = mass / molar mass = (9.00 g) / (32.00 g/mol) = 0.2812 mol.
- it is clear that O₂ is the limiting reactant and methane will leftover.
using cross multiplication
1 mol of CH₄ needs → 2 mol of O₂
??? mol of CH₄ needs → 0.2812 mol of O₂
∴ the number of mol of CH₄ needed = (0.2812 * 1) / 2 = 0.1406 mol
so 0.14 mol will react and the remaining CH₄
mol of CH₄ left over = 0.1875 -0.1406 = 0.0469 mol
now we convert moles into grams
mass of CH₄ left over = no. of mol of CH₄ left over * molar mass
= 0.0469 mol * 16 g/mol = 0.7504 g
So, the right choice is 0.74 grams of methane
55.9 kPa; Variables given = volume (V), moles (n), temperature (T)
We must calculate <em>p</em> from <em>V, n</em>, and <em>T</em>, so we use <em>the Ideal Gas Law</em>:
<em>pV = nRT</em>
Solve for <em>p</em>: <em>p = nRT/V</em>
R = 8.314 kPa.L.K^(-1).mol^(-1)
<em>T</em> = (265 + 273.15) K = 538.15 K
<em>V</em> = 500.0 mL = 0.5000 L
∴ <em>p</em> = [6.25 x 10^(-3) mol x 8.314 kPa·L·K^(-1)·mol^(-1) x 538.15 K]/(0.5000 L) = 55.9 kPa
