Answer:
V CH4(g) = 190.6 L
Explanation:
assuming ideal gas:
∴ STP: T =298 K and P = 1 atm
∴ R = 0.082 atm.L/K.mol
∴ moles (n) = 7.80 mol CH4(g)
∴ Volume CH4(g) = ?
⇒ V = RTn/P
⇒ V CH4(g) = ((0.082 atm.L/K.mol)×(298 K)×(7.80 mol)) / (1 atm)
⇒ V CH4(g) = 190.6 L
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the answer is 43.129310000000004
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Answer:
Explanation:
Hello there!
In this case, according to the given chemical reaction whereas the sodium chloride is in a 2:1 mole ratio with chlorine, the required moles of the later are computed as shown below:
So we cancel out the moles of NaCl to obtain:
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