Attached is the MOT of NO molecule.
1. Bond order is calculated is
In present case, number of e- in bonding orbital = 6
number of e- in anti-bonding orbital = 1
∴ Bond-order =
= 2.5
2. From the attached figure, it can be seen that NO has one unpaired electron in π* orbital.
<span>a)
</span>First order
in A and zero order in B
<span>ln [A]
= (ln 0.1) (2) + ln Ao = ln 0.01 + ln Ao = ln 0.01 Ao = 1.0% of A will remain</span>
<span>b)
</span>First
order in A and first order in B
<span>1/[A] – 1/[A]0= kt where t+=1 and k=9</span>
[A]/[A]=1/19=0.053=5.3%
<span>c)
</span>Zero
order in both A and B
<span>[A]0-[A] = kt</span>
Then at 2
hours [A]=0 All of it has reacted.
<span> </span>
I think that the answer is B, but I may be wrong...
Here’s a simplified explanation.
The <em>protons</em> in the nucleus <em>repel each other</em>. The <em>neutrons provide the “glue”</em> that holds the nucleus together and prevents it from flying apart.
The “glue” is the strong nuclear force. It is strong but extremely short range. It falls off extremely rapidly as the p-n distance increases.
A <em>neon atom</em> has 10 protons. There are three stable isotopes, with 10, 11, and 12 neutrons.
With fewer than 10 protons, the glue is not strong enough to hold the nucleus together.
If there are more than 12 neutrons, the average p-n distance is great enough that the glue has again become too weak.
<em>Gold</em> has one stable isotope. It contains 79 protons and 118 neutrons.
If there are fewer than 118 neutrons, the proton repulsions will be too strong for the strong force. If there are more than 118 neutrons, the average p-n distance will be large enough that the glue will again be too weak to hold the nucleus toge
ther.
Answer:
165 of CO₂.
Explanation:
In the reaction:
Na2CO3 + 2HCl → 2NaCl + CO2 + H2O
2 moles of HCl reacts producing 1 mole o CO₂
If 7.5 moles of HCl reacts, moles of CO₂ produced are:
7.5 moles of HCl ₓ ( 1 mol CO₂ / 2 mol HCl) = 3.75 mol CO₂. As molar mass of CO₂ is 44g/mol, mass of CO₂ is:
3.75 mol CO₂ ₓ (44g / 1mol) = <em>165 of CO₂ </em>
