Fire is actually a chemical reaction. It's an oxidation reaction to be specific. When wood gets hot enough (the part of the wood that is burning) the large hydrocarbons break down to charred solids and a gas. The gas is what reacts with oxygen in the atmosphere to produced light, CO2 and H2O.
The lattice energy of the compounds is distributed in the following decreasing order of magnitude: MgO > CaO > NaF > KCl.
<h3>KCl or NaF, which has a higher lattice energy?</h3>
The lattice energy increases with increasing charge and decreasing ion size.(Refer to Coulomb's Law.)MgF2 > MgO.Following that, we can examine NaF and KCl (both of which have 1+ and 1-charges), as well as atomic radii.NaF will have a larger LE than KCl since Na is smaller then K and F was smaller than Cl.
<h3>MgO or CaO, which has a larger lattice energy?</h3>
MGO is more difficult than CaO, hence.This is because "Mg" (two-plus) ions are smaller than "Ca" (two-plus) ions in size.MgO has higher lattice energy as a result.
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Answer:
P = 14.1 atm
Explanation:
Given data:
Mass of methane = 64 g
pressure exerted by water vapors = ?
Volume of engine = 24.0 L
Temperature = 515 K
Solution:
Chemical equation:
CH₄ + 2O₂ → CO₂ + 2H₂O + energy
Number of moles of methane:
Number of moles = mass / molar mass
number of moles = 64 g/ 16 g/mol
Number of moles = 4 mol
Now we will compare the moles of water vapors and methane.
CH₄ : H₂O
1 : 2
4 : 2/1×4 = 8 mol
Pressure of water vapors:
PV = nRT
R = general gas constant = 0.0821 atm.L/mol.K
P = 8 mol × 0.0821 atm.L/mol.K× 515 K / 24.0 L
P = 338.25 atm.L/ / 24.0 L
P = 14.1 atm
This question is testing to see how well you understand the "half-life" of radioactive elements, and how well you can manipulate and dance around them. This is not an easy question.
The idea is that the "half-life" is a certain amount of time. It's the time it takes for 'half' of the atoms in any sample of that particular unstable element to 'decay' ... their nuclei die, fall apart, and turn into nuclei of other elements.
Look over the table. There are 4,500 atoms of this radioactive substance when the time is 12,000 seconds, and there are 2,250 atoms of it left when the time is ' y ' seconds. Gosh ... 2,250 is exactly half of 4,500 ! So the length of time from 12,000 seconds until ' y ' is the half life of this substance ! But how can we find the length of the half-life ? ? ?
Maybe we can figure it out from other information in the table !
Here's what I found:
Do you see the time when there were 3,600 atoms of it ?
That's 20,000 seconds.
... After one half-life, there were 1,800 atoms left.
... After another half-life, there were 900 atoms left.
... After another half-life, there were 450 atoms left.
==> 450 is in the table ! That's at 95,000 seconds.
So the length of time from 20,000 seconds until 95,000 seconds
is three half-lifes.
The length of time is (95,000 - 20,000) = 75,000 sec
3 half lifes = 75,000 sec
Divide each side by 3 : 1 half life = 25,000 seconds
There it is ! THAT's the number we need. We can answer the question now.
==> 2,250 atoms is half of 4,500 atoms.
==> ' y ' is one half-life later than 12,000 seconds
==> ' y ' = 12,000 + 25,000
y = 37,000 seconds .
Check:
Look how nicely 37,000sec fits in between 20,000 and 60,000 in the table.
As I said earlier, this is not the simplest half-life problem I've seen.
You really have to know what you're doing on this one. You can't
bluff through it.
