Answer:
86.3 g of N₂ are in the room
Explanation:
First of all we need the pressure from the N₂ in order to apply the Ideal Gases Law and determine, the moles of gas that are contained in the room.
We apply the mole fraction:
Mole fraction N₂ = N₂ pressure / Total pressure
0.78 . 1 atm = 0.78 atm → N₂ pressure
Room temperature → 20°C → 20°C + 273 = 293K
Let's replace data: 0.78 atm . 95L = n . 0.082 . 293K
(0.78 atm . 95L) /0.082 . 293K = n
3.08 moles = n
Let's convert the moles to mass → 3.08 mol . 28g /1mol = 86.3 g
Answer:
537.68 torr.
Explanation:
- We can use the general law of ideal gas:<em> PV = nRT.</em>
where, P is the pressure of the gas in atm.
V is the volume of the gas in L.
n is the no. of moles of the gas in mol.
R is the general gas constant,
T is the temperature of the gas in K.
- If n and V are constant, and have different values of P and T:
<em>(P₁T₂) = (P₂T₁).</em>
P₁ = 485 torr, T₁ = 40°C + 273 = 313 K,
P₂ = ??? torr, T₂ = 74°C + 273 = 347 K.
∴ P₂ = (P₁T₂)/(P₁) = (485 torr)(347 K)/(313 K) = 537.68 torr.
Answer:<span>d. 145 minutes
</span>
Half-life is the time needed for a radioactive to decay half of its weight. The formula to find the half-life would be:
Nt= N0 (1/2)^ t/h
Nt= the final mass
N0= the initial mass
t= time passed
h= half-life
If 25.0% of the compound decomposes that means the final mass would be 75% of initial mass. Then the half-live for the compound would be:
Nt= N0 (1/2)^ t/h
75%= 100% * (1/2)^ (60min/h)
3/4= 1/2^(60min/h)
log2 3/4 = log2 1/2^(60min/h)
0.41503749928 = -60min/h
h= -60 min / 0.41503749928= 144.6min
Answer:
252.68 K or -20.46 °C
Explanation:
According to Gay-Lussac's Law, "Pressure and Temperature at given volume are directly proportional to each other".
Mathematically,
P₁ / T₁ = P₂ / T₂ ---- (1)
Data Given:
P₁ = 30.7 kPa
T₁ = 0.00 °C = 273.15 K
P₂ = 28.4 kPa
T₂ = <u>???</u>
Solving equation for T₂,
T₂ = P₂ T₁ / P₁
Putting values,
T₂ = 28.4 kPa × 273.15 K / 30.7 kPa
T₂ = 252.68 K or -20.46 °C
Answer:
Your question is complex, because I think you wrote it wrong.
Although in front of this what I can help you is that the carbons are associated between a single, double or triple union.
This depends on whether they are attached to more or less carbons or hydrogens, the carbons have the possibility of joining 4 radicals, both other carbons and hydrogens.
Simple junctions talks about compound organisms called ALKANS.
The double unions, in organic these compounds are called as ALQUENOS.
And as for the tertiary unions, the organic chemistry names them as ALQUINOS.
These compounds that we write, a simple union, the less energy, the less this union, that is why the triple bond is the one that contains the most energy when breaking or destroying it in a reaction.
Explanation:
In a chemical compound the change of these unions if we modified them we would generate changes even in the classifications naming them as well as different compounds and not only that until they change their properties
