The molar mass of the unknown compound is calculated as follows
let the unknown gas be represented by letter Y
Rate of C2F4/ rate of Y = sqrt of molar mass of gas Y/ molar mass of C2F4
= (4.6 x10^-6/ 5.8 x10^-6) = sqrt of Y/ 100
remove the square root sign by squaring in both side
(4.6 x 10^-6 / 5.8 x10^-6)^2 = Y/100
= 0.629 =Y/100
multiply both side by 100
Y= 62.9 is the molar mass of unknown gas
Answer:
A. percentage mass of iron = 5.17%
percentage mass of sand = 8.62%
percentage mass of water = 86.205%
B. (Iron + sand + water) -------> ( iron + sand) ------> sand
C. The step of separation of iron and sand
Explanation:
A. Percentage mass of the mixtures:
Total mass of mixture = (15.0 + 25.0 + 250.0) g =290.0 g
percentage mass of iron = 15/290 * 100% = 5.17%
percentage mass of sand = 25/290 * 100% = 8.62%
percentage mass of water = 250/290 * 100% = 86.205%
B. Flow chart of separation procedure
(Iron + sand + water) -------> separation by filtration using filter paper and funnel to remove water --------> ( iron + sand) -----------> separation using magnet to remove iron ------> sand
C. The step of separation of iron and sand by magnetization of iron will have the highest amount of error because during the process, some iron particles may not readily be attracted to the magnet as they may have become interlaced in-between sand grains. Also, some sand particle may also be attracted to the magnet as they are are borne on iron particles.
Answer:
1000 g
Explanation:
d = m/v
We are given d: 10g/cm3
and v: 100cm3
Plug them into the equation to get 10 = m/100
Then, cross multiply 10x100 to get mass which is: 1000g
Answer:
0 g.
Explanation:
Hello,
In this case, since the reaction between methane and oxygen is:
If 0.963 g of methane react with 7.5 g of oxygen the first step is to identify the limiting reactant for which we compute the available moles of methane and the moles of methane consumed by the 7.5 g of oxygen:
Thus, since oxygen theoretically consumes more methane than the available, we conclude the methane is the limiting reactant, for which it will be completely consumed, therefore, no remaining methane will be left over.
Regards.
