Answer:
solute is that we disolve in solvent
solvent is in which we dissolve solute
Answer: Therefore, the volume of a 0.155 M potassium hydroxide solution is 56.0 ml
Explanation:
Molarity of a solution is defined as the number of moles of solute dissolved per Liter of the solution.
According to the neutralization law,
where,
= molarity of 
solution = 0.338 M
= volume of solution = 25.7 ml
= molarity of 
solution = 0.155 M
= volume of solution = ?
= valency of = 1
= valency of = 1
Therefore, the volume of a 0.155 M potassium hydroxide solution is 56.0 ml
Answer:
7.3 g (NH₄)₃PO₄
Explanation:
The balanced equation for the reaction is:
H₃PO₄ + 3 NH₃ ----> (NH₄)₃PO₄
To find the mass of ammonium phosphate ((NH₄)₃PO₄) produced, you need to (1) convert grams NH₃ to moles NH₃ (via the molar mass from the periodic table), then (2) convert moles NH₃ to moles (NH₄)₃PO₄ (via mole-to-mole ratio from balanced equation), and then (3) convert moles (NH₄)₃PO₄ to grams (NH₄)₃PO₄ (via molar mass from periodic table). Make sure to arrange the ratios/conversions in a way that allows for the cancellation of units. The final answer should have 2 sig figs because the given value (2.5 grams) has 2 sig figs.
Molar Mass (NH₃): 14.01 g/mol + 3(1.008 g/mol)
Molar Mass (NH₃): 17.034 g/mol
Molar Mass ((NH₄)₃PO₄):
3(14.01 g/mol) + 12(1.008 g/mol) + 30.97 g/mol + 4(16.00 g/mol)
Molar Mass ((NH₄)₃PO₄): 149.096 g/mol
2.5 g NH₃ 1 mole NH₃ 1 mole (NH₄)₃PO₄ 149.096 g
--------------- x -------------------- x --------------------------- x --------------------------
17.034 g 3 moles NH₃ 1 mole (NH₄)₃PO₄
= 7.3 g (NH₄)₃PO₄
Answer:
Explanation:
1. Calculate the initial moles of acid and base
2. Calculate the moles remaining after the reaction
OH⁻ + H₃O⁺ ⟶ 2H₂O
I/mol: 0.0053 0.005 00
C/mol: -0.00500 -0.005 00
E/mol: 0.0003 0
We have an excess of 0.0003 mol of base.
3. Calculate the concentration of OH⁻
Total volume = 53 mL + 25.0 mL = 78 mL = 0.078 L
![\text{[OH}^{-}] = \dfrac{\text{0.0003 mol}}{\text{0.078 L}} = \textbf{0.0038 mol/L}\\\\\text{The final concentration of OH$^{-}$ is $\large \boxed{\textbf{0.0038 mol/L}}$}](https://tex.z-dn.net/?f=%5Ctext%7B%5BOH%7D%5E%7B-%7D%5D%20%3D%20%5Cdfrac%7B%5Ctext%7B0.0003%20mol%7D%7D%7B%5Ctext%7B0.078%20L%7D%7D%20%3D%20%5Ctextbf%7B0.0038%20mol%2FL%7D%5C%5C%5C%5C%5Ctext%7BThe%20final%20concentration%20of%20OH%24%5E%7B-%7D%24%20is%20%24%5Clarge%20%5Cboxed%7B%5Ctextbf%7B0.0038%20mol%2FL%7D%7D%24%7D)
