V ( HCl ) = 45.00 mL in liters : 45.00 / 1000 => 0.045 L
M ( HCl ) = ?
V ( NaOH ) = 25.00 / 1000 => 0.025 L
M ( NaOH) = 0.2000 M
number of moles NaOH :
n = M x V = 0.2000 x 0.025 => 0.005 moles of NaOH
Mole ratio:
HCl + NaOH = NaCl + H2O
1 mole HCl ---------- 1 mole NaOH
? mole HCl ---------- 0.005 moles NaOH
moles HCl = 0.005 x 1 / 1
= 0.005 moles of HCl :
M ( HCl ) = n / V
M ( HCl ) = 0.005 / 0.045
= 0.1111 M
hope this helps!
To communicate the results in an organized report
Answer:
Universe, galaxy, solar system, star, planet, moon and asteroid.
Explanation:
You're welcome!
F. <em>None of the above
</em>
<em>No O atoms are present</em> as reacting substances, only O_2 and H_2O molecules.
O_2 + 2H_2O + 2e^(-) → 4OH^(-)
We must use <em>oxidation numbers</em> to decide whether oxygen or water is the substance reduced.
The oxidation number of O changes from 0 in O_2 to -2 in OH^(-).
A decrease in oxidation number is <em>reduction</em>, so O_2 is the substance reduced.
The oxidation number of O is -2 in both H_2O and OH^(-), so water is <em>neither oxidized nor reduced</em>.
