Answer:
C. Cs, because it has the lower ionization energy and more easily gives up its valence electrons to participate in a reaction.
Explanation:
The Cs will produce the most hydrogen gas because it has a lower ionization energy and more easily gives up its valence electrons to participate in the reaction.
- Ionization energy is the energy required to remove the most loosely held electrons in an atom.
- Atoms with lower ionization energy readily goes into reactions.
- This is because they will present little barrier for the reaction to occur.
Answer:
Al + MnO4- + 2H2O → Al(OH)4- + MnO2
Explanation:
First of all, we out down the skeleton equation;
Al + MnO4- → MnO2 + Al(OH)4-
Secondly, we write the oxidation and reduction equation in basic medium;
Oxidation half equation:Al + 4H2O + 4OH- → Al(OH)4- + 4H2O + 3e-
Reduction half equation:MnO4- + 4H2O + 3e- → MnO2 + 2H2O + 4OH-
Thirdly, we add the two half reactions together to obtain:
Al + MnO4- + 8H2O + 4OH- + 3e- → Al(OH)4- + MnO2 + 6H2O + 3e- + 4OH-
Lastly, cancel out species that occur on both sides of the reaction equation;
Al + MnO4- + 8H2O→ Al(OH)4- + MnO2 + 6H2O
The simplified equation now becomes;
Al + MnO4- + 2H2O → Al(OH)4- + MnO2
Answer:
2Fe(s) + 3Cl2(g) → 2FeCl3(s)
Explanation:
Step 1: Data given
iron = Fe = solid = Fe(s)
chlorine = Cl2 = gas = Cl2(g)
iron(III) chloride = FeCl3 = solid = FeCl3(s)
Step 2: The unbalanced equation
Fe(s) + Cl2(g) → FeCl3(s)
Step 3: Balancing the equation
Fe(s) + Cl2(g) → FeCl3(s)
On the left we have 2x Cl (in Cl2) and on the right side we have 3x Cl (in FeCl3). To balance the amount of Cl we have to multiply Cl2 (on the left) by 3 and FeCl3 by 2.
Fe(s) + 3Cl2(g) → 2FeCl3(s)
On the left side we have 1x Fe and on the right side we have 2x Fe (in 2FeCl3). To balance the amount of Fe, we have to multiply Fe on the left side by 2. Now the equation is balanced.
2Fe(s) + 3Cl2(g) → 2FeCl3(s)
Find the number of moles of sodium you have:
<span>n = m/M where m is your 20g of sodium and M is 22.99 g/mol. </span>
<span>Look at the stoichiometry of the equation - it's 2:2 when you are producing NaOH. So if you took 1 mole of Na, it'd produce 1 mole of NaOH (as the ratio is equal). </span>
<span>That means that your moles of sodium is equal to the moles of NaOH produced. Use the molar mass of NaOH - which is 39.998 g/mol along with your calculated number of moles to get the mass (the formula rearranges to m = nM). </span>
<span>This figure is the theoretical yield - what you would get if every last mole of sodium was converted into NaOH. </span>
<span>What you get in practice is the experminetal yield, and the percentage yield is the experimental yield divided by the theoretical yield - and then multiplied by 100%.</span>
Answer: I think the answer is Cesium (Cs)
Explanation:
A half-filled 6s subshell would be 6s^1
