3. What volume of \( 0.905 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4} \) will react with 26.7 mL of 0.554 M NaOH ? \( \underline{H}_{2} \mathrm{SO}_{4}+2 \mathrm{NaOH}^{-} \mathrm{Na}_{2} \mathrm{SO}_{4}+2 \mathrm{H}_{2} \mathrm{O} \) 4. What volume of \( 1.000 \mathrm{M} \mathrm{Na}_{2} \mathrm{CO}_{3} \) will react with 342 mL of \( 0.733 \mathrm{M} \mathrm{H}_{3} \mathrm{PO}_{4} \) ? \( 3 \mathrm{Na}_{2} \mathrm{CO}_{3}+2 \mathrm{H}_{3} \mathrm{PO}_{4} \rightarrow 2 \mathrm{Na}_{3} \mathrm{PO}_{4}+3 \mathrm{H}_{2} \mathrm{O}+ \) \( 3 \mathrm{CO}_{2} \) 5. It takes 23.77 mL of 0.1505 M HCl to titrate with 15.00 mL of \( \mathrm{Ca}(\mathrm{OH}) 2 \). What is the concentration of \( \mathrm{Ca}(\mathrm{OH})_{2} \) ? You will need
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- Use the formula: \(\text{moles} = \text{molarity} \times \text{volume (L)}\). - Molarity of NaOH = 0.554 M, Volume = 26.7 mL = 0.0267 L. - Moles of NaOH = \(0.554 \, \text{M} \times 0.0267 \, \text{L} = 0.0148 \, \text{mol}\). Show more…
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