Answer:
The total pressure is 25.7 atm
Explanation:
The total pressure is the sum of the pressures of the three gases in the flask.
total pressure=(11.4+5.6+8.7) atm
=25.7 atm
Answer: D. Made of only one kind of atom
Explanation:A compound forms whenever two or more atoms form chemical bonds with each other.
69.9%
Explanation:
To find the mass percentage of iron in the compound in Fe₂O₃, we would go ahead to express the given molar mass of the iron to that of the compound.
Mass percentage =
x 100
Molar mass of Fe = 55.85g/mol
Molar mass of O = 16g/mol
Molar mass of Fe₂O₃ = 2(55.85) + 3(16) = 159.7g/mol
Mass percentage =
= 69.94% = 69.9%
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Answer: a. two substances present; two phases present : Heterogeneous mixture
b. two substances present; one phase present
: Homogeneous mixture
c. one substance present; one phase present
: pure substance.
d. one substance present; two phases present: Heterogeneous mixture
Explanation:
A pure substance is a substance which contains definite composition of only one type of component. Hence, it cannot be separated by physical means.
Mixture is a substance which contains two or more than two types of components and they can be separated by physical means as well.
Homogeneous mixtures: It is a mixture that has uniform composition throughout the solution and the particle size or shapes are not different. There is no physical boundary between the dispersed phase and dispersion medium.
Heterogeneous mixtures: It is a mixture that has non-uniform composition throughout the solution and the particle size or shapes are also different. There is a physical boundary between the dispersed phase and dispersion medium.
The given equation from the problem above is already balance,
N2O5 ---> 2NO2 + 0.5O2
Since, in every mole of N2O5 consumed, 2 moles of NO2 are formed, we can answer the problem by multiplying the given rate, 7.81 mol/L.s with the ratio.
(7.81 mol/L.s) x (2 moles NO2 formed/ 1 mole of N2O5 consumed)
= 15.62 mol/L.s
The answer is the rate of formation of NO2 is approximately 15.62 mol/L.s.