The combustion of an organic compound is mostly written as,
CaHbOc + O2 --> CO2 + H2O
where a, b, and c are supposed to be the subscripts of the elements C, H, and O in the compound. Determining the number of moles of C and H in the product which is the same as that in the compound,
(Carbon, C) : (561 mg) x (12/44) = 153 mg x (1 mmole/12 mg) = 12.75
(Hydrogen, H) : (306 mg) x (2/18) = 34 mg x (1 mmole/1 mg) = 34
Calculating for amount of O in the sample,
(oxygen, O) = 255 - 153 mg - 34 mg = 68 mg x (1mmole/16 mg) = 4.25
The empirical formula is therefore,
C(51/4)H34O17/4
C3H8O1
The molar mass of the empirical formula is 60. Therefore, the molecular formula of the compound is,
C9H24O3
Answer:
b is the anwer
Explanation:
the option is the explanation
As we know that
<span>V1/T1 = V2/T2
V1 = 9.10 L
T1 = 471 K
V2 = 2.50 L
T2 = 2.5 x 471 / 9.10 = 129.3 K
T2 = 129.3 - 273 =
-143.6 deg Celsiu
hope it helps</span>
The rate of reaction would increase because as pressure increases the molecules are more likely to bump into each other leading to a more likely hood of the molecules colliding properly to react leading to an increase in the reaction rate of the substance.
For this, we first calculate molecular weight of MgSiO₃:
Atomic masses:
Mg = 24
Si = 28
O = 16
Mr = 24 + 28 + 16 x 3
Mr = 100
moles = mass / Mr
moles = 237 / 100
moles = 2.37
