How many moles of iron (Fe) can be obtained from 3.80 g iron (III) oxide (Fe2O3) reacting with excess carbon monoxide
1 answer:
You might be interested in
Answer:
The concentration of methyl isonitrile will become 15% of the initial value after 10.31 hrs.
Explanation:
As the data the rate constant is not given in this description, However from observing the complete question the rate constant is given as a rate constant of 5.11x10-5s-1 at 472k .
Now the ratio of two concentrations is given as
Here C/C_0 is the ratio of concentration which is given as 15% or 0.15.
k is the rate constant which is given as
So time t is given as
So the concentration will become 15% of the initial value after 10.31 hrs.
Answer: polar molecule.
Explanation:
The boiling point is the temperature at which the vapor pressure of a liquid equals the external pressure surrounding the liquid. The boiling point is dependent on the type of forces present.
Iodine monochloride (ICl) is a polar molecule due to the difference in electronegativities of iodine and chlorine. Thus the molecules are bonded by strong dipole dipole forces. Thus a higher temperature is needed to generate enough vapor pressure.
Bromine is a non polar molecule as there is no electronegativity difference between two bromine atoms. The molecules are bonded by weak vanderwaal forces and thus has low boiling point.
The question in incomplete, complete question is;
Determine the theoretical yield:
Excess aqueous copper(II) nitrate reacts with aqueous sodium sulfide to produce aqueous sodium nitrate and copper(II) sulfide as a precipitate. In this reaction 469 grams of copper(II) nitrate were combined with 156 grams of sodium sulfide to produce 272 grams of sodium nitrate.
Answer:
The theoretical yield of sodium nitrate is 340 grams.
Explanation:
Moles of copper(II) nitrate =
Moles of sodium sulfide =
According to reaction, 1 mole of copper (II) nitrate reacts with 1 mole of sodium sulfide.
Then 2 moles of sodium sulfide will react with:
of copper (II) nitrate
As we can see from this sodium sulfide is present in limiting amount, so the amount of sodium nitrate will depend upon moles of sodium sulfide.
According to reaction, 1 mole of sodium sulfide gives 2 mole of sodium nitrate, then 2 mole of sodium sulfide will give:
sodium nitrate
Mass of 4 moles of sodium nitrate :
85 g/mol × 4 mol = 340 g
Theoretical yield of sodium nitrate = 340 g
The theoretical yield of sodium nitrate is 340 grams.