Answer:
Posible dimers structures: B) and D) (Figure 1)
Explanation:
The correct <u>dimer structures</u> are the ones that have a right <u>intermolecular bonding</u> interacction. In this case we will have hydrogen bonding. This type of interaction is due to the negative polarization in the oxygen and the positive polarization in the hydrogen. (Figure 2)
So, a correct dimer structure is posible if we have an interaction from the negative part (oxygen) to the positive part (hydrogen). The only structures that have this type of interactions are B and D.
The reaction between the lead iodide, PbI2 and sodium carbonate, Na2CO3, will form the chemical compounds, sodium iodide and lead carbonate as shown below.
PbI2 + Na2CO3 --> NaI + PbCO3
The white precipitate formed in the reaction is PbCO3.
The answers are a.) 0.03 mol KOH requires 0.03 mol HCl, b.) 2 mol NH3 requires 2 mol HCl and c.) 0.1 mol Ca(OH)2 requires 0.2 mol HCl.
Solution:
We need to write the balanced equations for each reactions to find out the stoichiometry for each reactants.
a.) HCl (aq) + KOH (aq) → KCl (aq) + H2O(ℓ)
From the balanced equations, we can see that 1 HCl reacts with 1 KOH, therefore if 0.03 mol KOH is reacted then 0.03 mol HCl must also be present.
b.) HCl(aq) + NH3(aq) ) → NH4Cl(aq)
If 2 moles of NH3 are reacted then 2 moles of HCl must also be present since 1 HCl reacts with 1 NH3 from the balanced reaction.
c.) 2HCl(aq) + Ca(OH)2(s) → CaCl2(aq) + 2H2O(ℓ)
We can see that 2 HCl react with 1 Ca(OH)2, hence if 0.1 mol of Ca(OH)2 is reacted then 0.2 mol HCl must also be present.
Answer:
The answer is Option a, that is "−9kJmole,5kJmole".
Explanation:
Please find the complete question in the attached file.
In the question, it uses the catalyst inside a process, which does not modify the process eigenvalues, however, it decreases the active energy with an enthalpy of -9kJmole, and also the power for activating decreases around 13 to 5 kJ mole, that's why the choice a is correct.
