A CH compound is combusted to produce CO2 and H2O
CnHm + O2 -----> CO2 + H2O
Mass of CO2 = 23.1g
Mass of H2O = 10.6g
Calculate by mass of the compounds
For Carbon C, divide by molecular weight of CO2 and multiply with Carbon
molecular weight. So C in grams = 23.1 x (12.01 / 44.01) = 6.3 g C
For Hydrogen H, divide by molecular weight of H2O and multiply with Hydrogen molecular weight. So H in grams = 10.6 x (2.01 / 18.01) = 0.53 g C
= 1.18 of H
Calculate the moles for C and H
6.3 grams of C x (1 mole/12.01 g C) = 0.524 moles of C
1.18 grams of H x (1 mole/1.008 g H) = 1.17 moles of H
Divides by both mole entities with smallest
C = 0.524 / 0.524 = 1 x 4 = 4
H = 1.17 / 0.524 = 2.23 x 4 = 10
The empirical formula is C4H10.
Answer:
C3 H6 O2
Explanation:
first divide their mass by their respective molar mass, we get:
30.4 moles of C
61.2 moles of H
20.25 moles of O
now divide everyone by the smallest one of them then we get
C= 1.5
H= 3
O= 1
since our answer of C is not near to any whole number so we will multiply all of them by 2
so,
C3 H6 O2 is our answer
Answer: Yes we agree with the student's claim.
Explanation:
When the molecules are present in smaller size, more reactants can react as decreasing the size increases the surface area of the reactants which will enhance the contact of molecules.Hence, more products will form leading to increased rate of reaction.
On increasing the temperature will make more reactant molecules will have sufficient energies to cross the energy barrier and thus the number of effective collisions increases, thus leading to more products and increased rate of reaction.
When the solution is stirred , the molecule's kinetic energy and thus the rate of reaction increases.
Thus smaller size, stirring and increase of temperature will make the solution quickly.
Answer: 0.52V
Explanation:
Ecell = Ecell(standard) - [(0.0592 logQ)/n]
Q = product of the quotient
n = no of electrons transferred = 2
Ecell = 0.63 - [(0.0592*Log(1 / 2.0 * 10-4) / 2]
Ecell = 0.63 - 0.0194
Ecell = 0.5205V
Answer:
[Ar] 3d10 4s2 4p3 is shorthand, 1s22s22p63s23p63d104s24p3 long hang
Explanation:
The shorthand is made using the lowest & closest noble gas, and picking up where it leaves off as follows, and longhand is made from a followed pattern you can easily find
