The common neutralization reaction that involve NaOH reacting with HNO3 produces
NaNO3 and H2O
The equation for reaction is as folows
NaOH + HNO3 = NaNO3 + H2O
that is 1 mole of NaOH reacted with 1 mole of HNO3 to form 1 mole of NaNO3 and 1 mole of H2O
Answer:
44.91% of Oxygen in Iron (III) hydroxide
Explanation:
To solve this question we must find the molar mass of Fe(OH)3 and the molar mass of the oxygen in this molecule. Percent composition will be:
<em>Molar mass Oxygen / molar mass Fe(OH)3 * 100</em>
<em />
<em>Molar mass Fe(OH)3 and oxygen:</em>
1Fe = 55.845g/mol*1 = 55.845
3O = 16.00g/mol*3 = 48.00 - Molar mass of Oxygen
3H = 1.008g/mol*3 = 3.024
55.845 + 48.00 + 3.024 =
106.869g/mol is molar mass of Fe(OH)3
% Composition of oxygen is:
48.00g/mol / 106.869g/mol * 100 =
<h3>44.91% of Oxygen in Iron (III) hydroxide</h3>
Answer:
The answer to your question is 88.7 ml
Explanation:
Data
Volume = ?
Concentration of NaOH = 0.142 M
Volume of H₂C₄H₄O₆ = 21.4 ml
Concentration of H₂C₄H₄O₆ = 0.294 M
Balanced chemical reaction
2 NaOH + H₂C₄H₄O₆ ⇒ Na₂C₄H₄O₆ + 2H₂O
1.- Calculate the moles of H₂C₄H₄O₆
Molarity = moles/volume
Solve for moles
moles = Molarity x volume
Substitution
moles = 0.294 x 21.4/1000
Result
moles = 0.0063
2.- Use proportions to calculate the moles of NaOH
2 moles of NaOH ------------------ 1 moles of H₂C₄H₄O₆
x ------------------ 0.0063 moles
x = (0.0063 x 2) / 1
x = 0.0126 moles of NaOH
3.- Calculate the volume of NaOH
Molarity = moles / volume
Solve for volume
Volume = moles/Molarity
Substitution
Volume = 0.0126/0.142
Result
Volume = 0.088 L or 88.7 ml
Answer:
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Explanation:
Ca + NaCl ----> CaCl2 + Na
