Okay so, 24.00 mL of a 0.25 M NaOH solution is titrated.
You can use grease or oil
The other dude is rude wrong
The correct answer is Dmitri Mendeleev
equalize the coefficients:
Al(NO3)3+3NaCl=3NaNO3+AlCl3
according to the equation 1 mol Al(NO3)3 reacted with 3 moles NaCl. hence, 4 moles Al(NO3)3 should react with 12 moles NaCl, and 9 moles reacted NaCl. So NaCl is limiting, we decide further on this reagent.
\frac{9}{3}=\frac{x}{3}
3
9
=
3
x
x=9
9 moles NaNO3
Answer:
286.825 or 287
Explanation:
To do this, you need to convert 38.5% to a decimal (0.385) and multiply it by 745.
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