Answer: 17.6×10^-3g of CO2
Explanation:
We first look at the stoichiometry of the balanced reaction equation. One mole of methane produces one mole of carbon dioxide. Hence 16g of methane yields 44g of carbon dioxide. If we now composed this with the given 6.40×10^-3g of methane as shown in the solution attached, we obtain the answer stated above.
No it does not have a coefficient.
Answer:
= 2.33 L
Explanation:
V1/T1 = V2/T2
V1 = 3.0 L
T1 = 78 + 273
= 351 K
At s.t.p the temperature is 273 K and pressure is 1 atm.
V2 = ?
T2 = 273 K
V2 = V1T2/T1
= (3.0 ×273)/351
= 2.33 L
Explanation:
ik it is confusing but that is what i got
If you have 100 g of compound, then based on the percentages given, there are 40.0 g C, 6.70 g H, and 53.3 g O. Convert those to moles (g / AW). (AW = atomic weight from the periodic table).
C: 40.0 g / 12.0 = 3.33 moles C
H: 6.70 g / 1.01 = 6.63 moles H
O: 53.3 / 16.0 = 3.33 moles O
Dividing by the smallest (3.33), we get a C:H:O mole ratio of 1:2:1. The empirical formula is CH2O. That formula has a molar mass of (12.0 + 2(1.0) + 16.0) = 30.0. How many times will that go into the actual molar mass of 150? 150/30 = 5, so multiply the emprical formula by 5.
5 x CH2O = C5H10O5, and that is the molecular formula.