The lattice energy of the compounds is distributed in the following decreasing order of magnitude: MgO > CaO > NaF > KCl.
<h3>KCl or NaF, which has a higher lattice energy?</h3>
The lattice energy increases with increasing charge and decreasing ion size.(Refer to Coulomb's Law.)MgF2 > MgO.Following that, we can examine NaF and KCl (both of which have 1+ and 1-charges), as well as atomic radii.NaF will have a larger LE than KCl since Na is smaller then K and F was smaller than Cl.
<h3>MgO or CaO, which has a larger lattice energy?</h3>
MGO is more difficult than CaO, hence.This is because "Mg" (two-plus) ions are smaller than "Ca" (two-plus) ions in size.MgO has higher lattice energy as a result.
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The answer is 1/16.
Half-life is the time required for the amount of a sample to half its value.
To calculate this, we will use the following formulas:
1.

,
where:
<span>n - a number of half-lives
</span>x - a remained fraction of a sample
2.

where:
<span>

- half-life
</span>t - <span>total time elapsed
</span><span>n - a number of half-lives
</span>
So, we know:
t = 10 min
<span>

= 2.5 min
We need:
n = ?
x = ?
</span>
We could first use the second equation to calculate n:
<span>If:

,
</span>Then:

⇒

⇒

<span>
</span>
Now we can use the first equation to calculate the remained fraction of the sample.
<span>

</span>⇒

<span>⇒

</span>
pH is the measure of the hydrogen ion concentration while pOH is of hydroxide ion concentration in the solution. The pH is 0.939 and pOH is 13.061 pOH.
pH is the concentration of the hydrogen ion released or gained by the species in the solution that depicts the acidity and basicity of the solution.
pOH is the concentration of the hydroxide ion in the solution and is dependent on the pH as an increase in pH decreases the pOH and vice versa.
Both HCl and HBr are strong acids and gets ionized 100 % in the solution. If we let 1 L of solution for the acids then the concentration of the hydrogen ion will be 0.100 M.
Since both completely dissociate we would just add the molarities of each of the H+ ions together and then calculate the PH and POH from that :
HCL(0.040M)----> H+(0.040M) +CL-(0.040M)
HBr(0.075M)----> H+(0.075M) +Br-(0.075M)
so 0.040M (H+ from HCL) + 0.075M (H+ from HBr) = 0.115M H+ in total.
pH is calculated as:
pH = -log[H+]
Substituting values in the equation:
log(0.115M)= 0.939 pH
pOH is calculated as:
14 - pH = pOH
Substituting values in the equation above:
14 - 0.939= 13.061 pOH
Therefore, pH is 0.939 and pOH is 13.061.
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Answer:
See explaination
Explanation:
1)
we know that
half cell with higher reduction potential is cathode
so
cathode :
N20 + 2H+ + 2e- ---> N2 + H20
anode :
Cr(s) ---> Cr+3 + 3e-
so
overall reaction is
3 N20 + 6H+ + 2 Cr ---> 3N2 + 3H20 + 2Cr+3
now
Eo cell = Eo cathode - Eo anode
so
EO cell = 1.77 + 0.74
Eo cell = 2.51 V
now
in this case
oxidizing agents are N20 and Cr+3
reducing agents are Cr and N2
higher the reduction potential , stronger the oxidizing agent
lower the reduction potential , stronger the reducing agent
so
oxidzing agents
N20 > Cr+3
reducing agents
Cr > N2
2)
cathode :
Au+ + e- --> Au
anode :
Cr ---> Cr+3 + 3e-
overall reaction
3Au+ + Cr ---> 3Au + Cr+3
Eo cell = 1.69 + 0.74
Eo cell = 2.43
now
oxidizing agents :
Au+ > Cr+3
reducing agents :
Cr > Au
3)
cathode :
N20 + 2H+ + 2e- ---> N2 + H20
andoe :
Au ---> Au+ + e-
overall
2 Au + N20 + 2H+ --> 2 Au+ + N2 + H20
Eo cell = 1.77 - 1.69
Eo cell = 0.08
oxidizing agents
N20 > Au+
reducing agents
Au > N2