17.8 mL NaOH
<em>Step 1.</em> Write the chemical equation
Fe^(2+) + 2NaOH → Fe(OH)2 + 2Na^(+)
<em>Step 2.</em> Calculate the moles of Fe^(2+)
Moles of Fe^(2+) = 500 mL Fe^(2+) × [0.0230 mmol Fe^(2+)]/[1 mL Fe^(2+)]
= 11.50 mmol Fe^(2+)
<em>Step 3.</em> Calculate the moles of NaOH
Moles of NaOH = 11.50 mmol Fe^(2+) × [2 mmol NaOH]/[1 mmol Fe^(2+)]
= 23.00 mmol NaOH
<em>Step 4.</em> Calculate the volume of NaOH
Volume of NaOH = 23.00 mmol NaOH × (1 mL NaOH/1.29 mmol NaOH)
= 17.8 mL NaOH
All scientists try to base their conclusions on data and measurements.
1. No
2.No
I hope this helps:)
Explanation:
Mg(s) + Cr(C2H3O2)3 (aq)
Overall, balanced molecular equation
Mg(s) + Cr(C2H3O2)3(aq) --> Mg(C2H3O2)3(aq) + Cr(s)
To identify if an element has been reduced or oxidized, the oxidation number is observed in both the reactant and product phase.
An increase in oxidation number denotes that the element has been oxidized.
A decrease in oxidation number denotes that the element has been reduced.
Oxidation number of Mg:
Reactant - 0
Product - +3
Oxidation number of Cr:
Reactant - +3
Product - 0
Note: C2H3O2 is actually acetate ion; CH3COO- The oxidatioon number of C, H and O do not change.
Oxidized : Mg
Reduced : Cr
Answer:
59.8%
Explanation:
First find the Mr of manganese (III) nitrate.
Mr of Mn(NO₃)₃ = 54.9 + (14 × 3) + (16 × 3 × 3) = <u>240.9</u>
Since we have to find the percentage composition of oxygen, we need to find the Mr of oxygen in the compound, which is:
Mr of (O₃)₃ = (16 × 3) × 3 = <u>144</u>
Now we can find percentage composition / percentage by mass of oxygen.
% composition = 
× 100
% composition = 
× 100 = <u>59.776%</u>
∴ % compostion of oxygen in maganese(III)nitrate is 59.8% (to 3 significant figures).
