How many valence electrons does p3- have
1 answer:
You might be interested in
(1) • Balanced Equation : 2HgO (s) ---> 2 Hg (l) + O₂ (g)
• Line a is represented by activation energy, or the smallest amount of energy needed for molecules to undergo a chemical reaction.
• Activated complex is the intermediate state during the conversion of reactants into products. In this case it is represented by Hg----O-----O----Hg. The dashed lines indicate that the bond between Hg and O are breaking and their length is increasing and the dashed line between oxygen atoms indicate that the bond is decreasing and bond formation is taking place.
• This reaction is endothermic as the energy of reactants is smaller than the energy of products.
(2) • Balanced Equation : CH₃Br + OH⁻ ---> CH₃OH + Br⁻
• The hydroxide ion reacts with the bromomethane, resulting in a double replacement reaction. The 'CH₃' in 'CH₃Br' joins with the hydroxide ion to form CH₃OH, while the Br breaks off as an ion.
• Take a look at the energy diagram below.
Answer:
(B) 0.160 M
Explanation:
Considering:
Or,
Given :
For :
Molarity = 0.200 M
Volume = 20.0 mL
The conversion of mL to L is shown below:
1 mL = 10⁻³ L
Thus, volume = 20.0×10⁻³ L
Thus, moles of :
Moles of = 0.004 moles
For :
Molarity = 0.400 M
Volume = 30.0 mL
The conversion of mL to L is shown below:
1 mL = 10⁻³ L
Thus, volume =30.0×10⁻³ L
Thus, moles of :
Moles of = 0.012 moles
According to the given reaction:
1 mole of potassium carbonate react with 1 mole of barium nitrate
0.004 moles potassium carbonate react with 0.004 mole of barium nitrate
Moles of barium nitrate = 0.004 moles
Available moles of barium nitrate = 0.012 moles
Limiting reagent is the one which is present in small amount. Thus, potassium carbonate is limiting reagent.
The formation of the product is governed by the limiting reagent. So,
1 mole of potassium carbonate gives 1 mole of barium carbonate
Also,
0.004 mole of potassium carbonate gives 0.004 mole of barium carbonate
Mole of barium carbonate = 0.004 moles
Also, consumed barium nitrate = 0.004 moles (barium ions precipitate with carbonate ions)
Left over moles = 0.012 - 0.004 moles = 0.008 moles
Total volume = 20.0 + 30.0 mL = 50.0 mL = 0.05 L
So, Concentration = 0.008/0.05 M = 0.160 M
<u>(B) is correct.</u>