According to sources, the most probable answer to this query is that when solutions reaches equilibrium, the amount of concentration of two or more matter combined in this solution becomes equal.
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Answer:
643g of methane will there be in the room
Explanation:
To solve this question we must, as first, find the volume of methane after 1h = 3600s. With the volume we can find the moles of methane using PV = nRT -<em>Assuming STP-</em>. With the moles and the molar mass of methane (16g/mol) we can find the mass of methane gas after 1 hour as follows:
<em>Volume Methane:</em>
3600s * (0.25L / s) = 900L Methane
<em>Moles methane:</em>
PV = nRT; PV / RT = n
<em>Where P = 1atm at STP, V is volume = 900L; R is gas constant = 0.082atmL/molK; T is absolute temperature = 273.15K at sTP</em>
Replacing:
PV / RT = n
1atm*900L / 0.082atmL/molK*273.15 = n
n = 40.18mol methane
<em>Mass methane:</em>
40.18 moles * (16g/mol) =
<h3>643g of methane will there be in the room</h3>
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Answer:
d) A - 70% B - 30%
Explanation:
If x is the abundance of A, and 1−x is the abundance of B, then:
x (32.0) + (1−x) (33.0) = 32.3
32x + 33 − 33x = 32.3
33 − x = 32.3
x = 0.7
The abundance of A is 70%, and the abundance of B is 30%.
