Answer: The percentage by mass of sulphur in 
is 9.36%
Explanation:
Mass percent of an element is the ratio of mass of that element by the total mass expressed in terms of percentage.
Given: mass of sulphur = 32 g/mol
mass of = 342 g/mol
Putting in the values we get:
The percentage by mass of sulphur in is 9.36%
To get the theoretical yield of ammonia NH3:
first, we should have the balanced equation of the reaction:
3H2(g) + N2(g) → 2NH3(g)
Second, we start to convert mass to moles
moles of N2 = N2 mass / N2 molar mass
= 200 / 28 = 7.14 moles
third, we start to compare the molar ratio from the balanced equation between N2 & NH3 we will find that N2: NH3 = 1:2 so when we use every mole of N2 we will get 2 times of that mole of NH3 so,
moles of NH3 = 7.14 * 2 = 14.28 moles
finally, we convert the moles of NH3 to mass again to get the mass of ammonia:
mass of NH3 = no.moles * molar mass of ammonia
= 14.28 * 17 = 242.76 g
Answer: Too much base was added
i guessed
Explanation:
(missing part of your question):
when we have K = 1 x 10^-2 and [A] = 2 M & [B] = 3M & m= 2 & i = 1
So when the rate = K[A]^m [B]^i
and when we have m + i = 3 so the order of this reaction is 3 So the unit of K is L^2.mol^-2S^-1
So by substitution:
∴ the rate = (1x 10 ^-2 L^-2.mol^-2S^-1)*(2 mol.L^-1)^2*(3mol.L^-1)
= 0.12 mol.L^-1.S^-1
For an element whose third shell contains six electrons, the appropriate electron configuration is; 1s2 2s2 2p6 3s2 3p4.
The electron configuration shows the distribution of electrons in the shells of an atom and in orbitals.
We have been told that the six electrons are found in the third shell. This shell has n=3 and the configuration of this shell must ns2 np4.
The only electron configuration that meets this standard is 1s2 2s2 2p6 3s2 3p4.
Learn more: brainly.com/question/18704022
