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
Yo sup??
1 lmole of O2 contains 22.4 litre
therefore 10.2 miles will contain
=10.2*22.4
=228.48 litre
Hence the correct answer is option D
Hope this helps
Answer:
Explanation:
Hello there!
In this case, since the radioactive decay is considered as first-order kinetic model, whereby the remaining mass of the radioactive material involves the initial one, the rate constant and elapsed time:
As we were not initially given the rate constant, we can use the half-life to calculate it as follows:
Thus, we can calculated the elapsed time for the given conditions to obtain:
Regards!
We know that C has a formal charge of +4 while each O element
has a formal charge of -2, therefore the formal charge of X is:
X + 2 – 2 * 3 = 0
X + 2 – 6 = 0
X = 4
<span>So X has a group number of 4.</span>
Answer:
Molar Mass (moles
⇌
grams)
Stoichiometric Coefficients (moles )
Avagadro's number
Explanation: