Answer:
Cells need to maintain homeostasis in order for our body to be healthy and fully functional. If the internal conditions of our cells are not stable, then the normal cell functions would be disrupted.
Explanation:
Answer:
C₂H₂ + 3H₂ ⟶ 2CH₄
Explanation:
The initial concentrations are:
[CH₄] = 6.30 ÷ 6.00 = 1.05 mol·L⁻¹
[C₂H₂] = 4.20 ÷ 6.00 = 0.700 mol·L⁻¹
[H₂] = 11.15 ÷ 6.00 = 1.858 mol·L⁻¹
2CH₄ ⇌ C₂H₂ + 3H₂
I/mol·L⁻¹: 1.05 0.700 1.858
![Q = \dfrac{\text{[C$_{2}$H$_{2}$][H$_{2}$]}^{3}}{\text{[CH$_{4}$]}^{2}} = \dfrac{ 0.700\times 1.858^{3}}{1.05^{2}}= 4.07](https://tex.z-dn.net/?f=Q%20%3D%20%5Cdfrac%7B%5Ctext%7B%5BC%24_%7B2%7D%24H%24_%7B2%7D%24%5D%5BH%24_%7B2%7D%24%5D%7D%5E%7B3%7D%7D%7B%5Ctext%7B%5BCH%24_%7B4%7D%24%5D%7D%5E%7B2%7D%7D%20%3D%20%5Cdfrac%7B%200.700%5Ctimes%201.858%5E%7B3%7D%7D%7B1.05%5E%7B2%7D%7D%3D%204.07)
Q > K
That means we have too many products.
The reaction will go to the left to get rid of the excess products.
C₂H₂ + 3H₂ ⟶ 2CH₄
Isolate the lab immediately mercury can be easily tracked throughout a lab by people walking on spill. laboratory staff can safely clean up small spill . don't stamp on small pieces . carefully inspect the bench tops and floors before sitting. to avoid hurting .
You need to use the % information to determine the empirical formula of the compound first.
The empirical formula is the simplest ratio of atoms in the molecule.
Then use the rest of the data to determine moles of gas, and use this to determine molar mass of gas...
Empirical formula calculations:
Assume you have 100 g, calculate the moles of each atom in the 100 g
moles = mass / molar mass
molar mass C = 12.01 g/mol
molar mass H = 1.008 g/mol
molar mass O = 16.00 g/mol
C = 64.9 % = 64.6 g
H = 13.5 % = 13.5 g
O = 21.6 % = 21.6 g
moles C = 64.6 g / 12.01 g/mol = 5.38 mol
moles H = 13.5 g / 1.008 g/mol = 13.39 mol
moles O = 21.6 g / 16.00 g/mol = 1.35 mol
So ratio of C : H : O
is 5.38 mol : 13.39 mol : 1.35 mol
Divide each number in the ratio by the lowest number to get the simplest whole number ratio
(5.38 / 1.35) : (13.39 / 1.35) : (1.35 / 1.35)
4 : 10 : 1
empirical formula is
C4H10O
Finding moles and molar mass calcs
Now, you know that at 120 deg C and 750 mmHg that 1.00L compound weighs 2.30 g.
We can use this information to determine the molar mass of the gas after first working out how many moles the are in the 1.00 L
PV = nRT
P = pressure = 750 mmHg
V = volume = 1.00 L
n = moles (unknown)
T = temp in Kelvin (120 deg C = (273.15 + 120) Kelvin)
- T = 393.15 Kelvin
R = gas constant, which is 62.363 mmHg L K^-1 mol^-1 (when your P is in mmHg and volume is in L)
n = PV / RT
n = (750 mmHg x 1.00 L) / (62.363mmHg L K^-1 mol^-1 x 393.15 K)
n = 0.03059 moles of gas
We know moles = 0.03509 and mass = 2.30 g
So we can work out molar mass of the gas
moles = mass / molar mass
Therefore molar mass = mass / moles
molar mass = 2.30 g / 0.03059 mol
= 75.19 g/mol
Determine molecular formula
So empirical formula is C4H10O
molar mass = 75.19 g/mol
To find the molecular formula you divide the molar mass by the formula weight of the empirical formula...
This tells you how many times the empirical formula fits into the molecular formula. Tou then multiply every atom in the empirical formula by this number
formula weight C4H10O = 74.12 g/mol
Divide molar mass by formula weight empirical
75.15 g/mol / 74.12 g/mol
= 1
(It doesn't matter that the number don't quite match, they rarely do in this type of calc (although I could have made a slight error somewhere) but the numbers are very close, so we can say 1.)
The empirical formula only fits into the molar mass once,
molecular formula thus = empirical formula
<span>
C4H10O
Therefore, the </span>molecular formula of the compound is <span>C4H10O.
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