Answer:
9.72 grams.
Explanation:
From the equation, 4 moles of NH₃ produce 6 moles of water.
Therefore the reaction to product ratio of NH₃ to H₂O is 4:6
and 2:3 into its simplest form.
The number of moles of NH₃ in 6.12 g is:
Number of moles=mass/ RMM
=6.12 g/17 G/mol
=0.36 moles.
Therefore the number of moles of H₂O produced is calculated as follows.
(0.36 Moles×3)2 = 0.54 moles
Mass= Number of moles × RMM
=0.54 moles×18g/mol
=9.72 grams.
Answer:
a) 48KJ
b) -48KJ
Explanation:
Given that;
ln(K2/K1) = ΔH°/R(1/T2 - 1/T1)
K2= equilibrium constant at T2
K1 = equilibrum constant at T1
R = gas constant
T1 = initial temperature
T2 = final temperature
When we double the equilibrium constant K1; K2 = 2K1
T1 = 310 K
T2 = 310 + 15 = 325 K
ln(2K1/K1) =- ΔH°/R(1/T2 - 1/T1)
ln2 = -ΔH°/8.314(1/325 - 1/310)
0.693 = -ΔH°/8.314(3.08 * 10^-3 - 3.2 * 10^-3)
0.693 = -ΔH°/8.314 (-0.00012)
0.693 = 0.00012ΔH°/8.314
0.693 * 8.314 = 0.00012ΔH°
ΔH° = 0.693 * 8.314/0.00012
ΔH° = 48KJ
b) K2 =K1/2
ln(K1/2/K1) =- ΔH°/R(1/T2 - 1/T1)
ln (1/2) = -ΔH°/8.314 (1/325 - 1/310)
-0.693 = -ΔH°/8.314 (-0.00012)
-0.693 = 0.00012ΔH°/8.314
-0.693 * 8.314 = 0.00012ΔH°
ΔH°= -0.693 * 8.314/0.00012
ΔH°= -48KJ
Answer:
c...............
Explanation:.............................yuuuuuuuuuuuuuuuuuuuuuwoopuuuuuuuu
The effect of an insoluble impurity, such as sand, on the observed melting point of a compound would be none. It will not depress or elevate the melting point of the compound. Instead, it would affect the reading if you are trying to determine the melting point of the compound. This is because you might be missing the actual melting point of the compound since you will be waiting for the whole sample to liquify. You would not be able to determine exactly that temperature because of the insoluble impurity would have a different melting point than that of the compound.
Answer:
Explanation:
Remark
Interesting que8stion. You have to figure out how many mols are present in each reactant. Since all periodic tables are different, I'm going to use rounded numbers. If it is too close, I will go further.
NaCl
Na = 23
Cl = 35.5
1 mol = 58.5 grams
given = 50.0 grams
Mols for the reaction = 50/58.5 = 0.855
H2SO4
H2 = 2*1 2
S = 1 * 32 32
O4 = 4*16 64
1 mol = 98 grams
mols present = 50/98 = 0.510
MnO2
Mn = 1 * 55 = 55
O2 = 2*16 = 32
1 mol = 87 grams
mols available = 50/87 = 0.5747
Discussion
Na Cl and H2SO4 both require 2 moles for every mol of Cl2 produces.
H2SO4 has 0.51 mols available for a reaction
NaCl has 0.855 moles available for a reaction
MnO2 has 0.575 moles available for a reaction.
Given those numbers 0.510 mols of H2SO4 will only produce 0.255 mols of chlorine and the rest will be reduced in a similar manner. H2SO4 is the limiting reagent (reactant).
In other words only 0.510 moles of NaCl will be used and 0.855 - 0.510 moles will be left over on the reactants side.
only 0.575 moles of MnO2 will be used and 0.065 moles will be left over.
The oddity in the result shows up because the balance numbers in the equation give a ratio of 2 to 1 for H2SO4 and NaCl The 2 belongs to the reactants and the 1 for the chlorine.
