Aldehydes are usually more reactive<span> toward nucleophilic substitutions </span>than<span> ketones because of both steric and electronic effects. In aldehydes, the relatively small hydrogen atom is attached to one side of the carbonyl group, while a larger R group is affixed to the other side.</span>
Answer:
We need 17.2 L of Ca(OH)2
Explanation:
Step 1: Data given
Concentration of Ca(OH)2 = 1.45 M
Moles of H2SO4 = 25.0 moles
Step 2: The balanced equation
Ca(OH)2 + H2SO4 ⟶2H2O + CaSO4
Step 3: Calculate moles Ca(OH)2
For 1 mol Ca(OH)2 we need 1 mol H2SO4 to produce 2 moles H2O and 1 mol CaSO4
For 25.0 moles H2SO4 we'll need 25.0 moles Ca(OH)2 to produce 50 moles H2O and 25.0 moles CaSO4
Step 4: Calculate volume of Ca(OH)2
Volume Ca(OH)2 = moles Ca(OH)2 / concentration Ca(OH)2
Volume Ca(OH)2 = 25.0 moles / 1.45 M
Volume Ca(OH)2 = 17.2 L
We need 17.2 L of Ca(OH)2
Answer:
Explanation:
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In this case, since the net ionic equation of a chemical reaction shows up the ionic species that result from the simplification of the spectator ions, which are those at both reactants and products sides, we take into account that aqueous species ionize into ions whereas liquid, solid and gas species remain unionized. In such a way, for the reaction of cesium phosphate and silver nitrate we can write the complete molecular equation:
Whereas the three aqueous salts are ionized in order to write the following complete ionic equation:
In such a way, since the cesium and nitrate ions are the spectator ions because of the aforementioned, the net ionic equation turns out:
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The balanced chemical reaction is written as :
Na2CO3<span> + 2HCl === 2NaCl + H2O + CO2
</span>
We are given the amount of NaCl to be produced from the reaction. This will be the starting point for the calculations. We do as follows:
120 g NaCl ( 1 mol / 58.44 g) ( 1 mol Na2CO3 / 2 mol NaCl)( 105.99 g / 1 mol ) = 1108.82 g Na2CO3 needed
Answer:
Check the explanation
Explanation:
When,
pH = -log[H+] = 3.30
[H+] =
=
=
When,
pH = -log[H+] = 10.15
[H+] =
Ka1 = 1 ; Ka2 = 0.0316 ; Ka3 = 0.01 ; Ka4 = 0.002 ; Ka5 = ; Ka6 =
=
=