Answer:
Percentage mass of copper in the sample = 32%
Explanation:
Equation of the reaction producing Cu(NO₃) is given below:
Cu(s)+ 4HNO₃(aq) ---> Cu(NO₃)(aq) + 2NO₂(g) + 2H₂O(l)
From the equation of reaction, 1 mole of Cu(NO₃) is produced from 1 mole of copper. Therefore, 0.010 moles of Cu(NO₃) will be produced from 0.010 mole of copper.
Molar mass of copper = 64 g/mol
mass of copper = number of moles * molar mass
mass of copper = 0.01 mol * 64 g/mol = 0.64 g
Percentage by mass of copper in the 2.00 g sample = (0.64/2.00) * 100%
Percentage mass of copper in the sample = 32%
2.47 moles are produce if 85g of N2 react with 180g of Mg
I'm guessing water vapour. i can't think of any other gas limited to only hydrogen and oxygen
hope it helps :)
Answer:
The answer to this can be arrived at by clculating the mole fraction of atoms higher than the activation energy of 10.0 kJ by pluging in the values given into the Arrhenius equation. The answer to this is 20.22 moles of Argon have energy equal to or greater than 10.0 kJ
Explanation:
From Arrhenius equation showing the temperature dependence of reaction rates.
where
k = rate constant
A = Frequency or pre-exponential factor
Ea = energy of activation
R = The universal gas constant
T = Kelvin absolute temperature
we have

Where
f = fraction of collision with energy higher than the activation energy
Ea = activation energy = 10.0kJ = 10000J
R = universal gas constant = 8.31 J/mol.K
T = Absolute temperature in Kelvin = 400K
In the Arrhenius equation k = Ae^(-Ea/RT), the factor A is the frequency factor and the component e^(-Ea/RT) is the portion of possible collisions with high enough energy for a reaction to occur at the a specified temperature
Plugging in the values into the equation relating f to activation energy we get
or f =
= 20.22 moles of argon have an energy of 10.0 kJ or greater