18. <span>Answer is </span>
A<span>
<span>Since the enthalpy of reaction is positive, the
forward reaction is<span> an endothermic reaction which means the energy
is gained from the surrounding to happen the reaction. If the temperature
decreases, according to the </span></span>Le Chatelier's principle, the system tries to become equilibrium
by increasing temperature. Since forward reaction is endothermic (because of
the bond breaking), the backward reaction is exothermic (because of the bond
making) which releases the energy to the surroundings. This makes the increase
of temperature. So if the backward reaction is promoted because of the decrease
of temperature, then the concentration of H</span><span>₂ will decrease.</span>
<span>
</span>
19. Answer is A.
The reactant side
has 2 moles/molecules of reactants and the product side has 4 moles/molecules
of products which come from 1 N₂(g) and 3 H₂<span>(g). If the pressure is reduced in the system, according to the Le Chatelier's principle, the
system tries to increase the pressure. </span><span>Hence, forward
reaction is promoted because of the higher number
of molecules in product side. If the forward reaction is promoted, the
concentration of NH</span>₃(g) will decreased.
<span>20. </span>Answer is C.
If the concentration
of reactant is increased in the
system, according to the Le Chatelier's principle, the system tries
to reduce the concentration of that reactant. So if NH₃(g) concentration
is increased, then to be equilibrium, the forward reaction will be promoted.
Then the concentration of N₂<span>(g) will increase.</span>
<span> </span>
Answer:
![n=0.286mol](https://tex.z-dn.net/?f=n%3D0.286mol)
Explanation:
Hello,
In this case, we consider oxygen as an ideal gas, for that reason, we use yhe ideal gas equation to compute the moles based on:
![PV=nRT\\\\n=\frac{PV}{RT}](https://tex.z-dn.net/?f=PV%3DnRT%5C%5C%5C%5Cn%3D%5Cfrac%7BPV%7D%7BRT%7D)
Hence, at 3.50 atm and 25 °C for a volume of 2.00 L we compute the moles considering absolute temperature in Kelvins:
![n=\frac{3.50atm*2.00L}{0.082\frac{atm*L}{mol*K}(25+273)K} \\\\n=0.286mol](https://tex.z-dn.net/?f=n%3D%5Cfrac%7B3.50atm%2A2.00L%7D%7B0.082%5Cfrac%7Batm%2AL%7D%7Bmol%2AK%7D%2825%2B273%29K%7D%20%5C%5C%5C%5Cn%3D0.286mol)
Best regards.
Nitrogen has five valence electrons
Answer:
A positive and negative ion.
Explanation:
Ionic bond:
It is the bond which is formed by the transfer of electron from one atom to the atom of another element.
Both bonded atoms have very large electronegativity difference.
The atom with large electronegativity value accept the electron from other with smaller value of electronegativity.
For example:
Sodium chloride is ionic compound. The electronegativity of chlorine is 3.16 and for sodium is 0.93. There is large difference is present. That's why electron from sodium is transfer to the chlorine. Sodium becomes positive and chlorine becomes negative ion. Sodium have one valance electron while chlorine have 7 valance electrons. In order to complete the octet chlorine require one electron while sodium need to lose its one electrons. That's why when both atom combine sodium lose its electron and becomes positive ion i.e cation while chlorine accept its electron and becomes negative ion called anion and bond between them is ionic bond.
Answer: 1.
moles
2. 90 mg
Explanation:
![O_3(g)+2NaI(aq)+H_2O(l) \rightarrow O_2(g)+I_2(s)+2NaOH(aq)](https://tex.z-dn.net/?f=O_3%28g%29%2B2NaI%28aq%29%2BH_2O%28l%29%20%5Crightarrow%20O_2%28g%29%2BI_2%28s%29%2B2NaOH%28aq%29)
According to stoichiometry:
1 mole of ozone is removed by 2 moles of sodium iodide.
Thus
moles of ozone is removed by =
moles of sodium iodide.
Thus
moles of sodium iodide are needed to remove
moles of ![O_3](https://tex.z-dn.net/?f=O_3)
2. ![\text{Number of moles of ozone}=\frac{0.01331g}{48g/mol}=0.0003moles](https://tex.z-dn.net/?f=%5Ctext%7BNumber%20of%20moles%20of%20ozone%7D%3D%5Cfrac%7B0.01331g%7D%7B48g%2Fmol%7D%3D0.0003moles)
According to stoichiometry:
1 mole of ozone is removed by 2 moles of sodium iodide.
Thus 0.0003 moles of ozone is removed by =
moles of sodium iodide.
Mass of sodium iodide=
(1g=1000mg)
Thus 90 mg of sodium iodide are needed to remove 13.31 mg of
.