There are 2 moles of O stones present in 88 grams of CO2. Why? Well, we can find the amount of moles present in 88 grams of CO2 by dividing the mass by the molar mass. The mass of CO2 comes out to be 88 grams. The molar mass of CO2 comes out to be 44 grams. Because 88 is the mass of CO2 and 44 is the molar mass of CO2, we can divide 88 by 44 to identify that there are 2.0 moles of O atoms present in 88 grams of CO2.
Your final answer: There are 2.0 moles of O atoms present in 88 grams of CO2. Your final answer to this question is D, or 2.0 moles. If you need to better understand, let me know and I will gladly assist you.
The molecular formula of methylpropan-1-ol is C4H10O, so the complete combustion equation is: C4H10O + 6O2 --> 4CO2 + 5H2O. This mean to completely combust 1.0mol of methylpropan-1-ol, 6 mol of O2 is required. Molar mass of O2 is 32 g/mol, so 32g/mol x 6mol = 192 g of O2 is required. At room temperature and pressure, the density of O2 is 1.3315 g/L (this can be obtained by density of gas = P/RT). So the volume of O2 = mass/density = 192g/1.3315(g/L) = 144 L = 144 dm3. The answer is B.
739 degrees K equals 465.85 degrees Celcius.
Answer:
See explanation below
Explanation:
To get this, we need to apply the general expression for half life decay:
N = N₀e(-λt) (1)
Where:
N and N₀ would be the final and innitial quantities, in this case, masses.
t: time required to decay
λ: factor related to half life
From the above expression we need λ and t. To get λ we use the following expression:
λ = t₁₂/ln2 (2)
And we have the value of half life, so, replacing we have:
λ = 8.04 / ln2 = 11.6
Now, we can replace in (1) and then, solve for t:
0.75 = 40 exp(-11.6t)
0.75 / 40 = exp(-11.6t)
ln(0.01875) = -11.6t
-3.9766 = -11.6t
t = -3.9766 / -11.6
<h2>
t = 0.34 days</h2><h2>
</h2>
4Na +2e- ---->2Na2
O2 +2e- ------>2O
