The volume of 0.555M KNO3 solution would contain 12.5 g of solute iss 223 mL.
<h3>What is the relationship between mass of solute and concentration of solution?</h3>
The mass of solute in a given volume of solution is related by the formula below:
- Molarity = mass/(molar mass * volume)
Therefore, volume of solution is given by:
Volume = Mass /molarity * molar mass
Molar mass of KNO₃ = 101 g/mol
Volume = 12.5/(0.555 * 101)
Volume = 0.223 L or 223 mL
In conclusion, the volume of the solution is obtained from the molarity of solution as well as mass and molar mass of solute.
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Answer:
Explanation:
Hello there!
In this case, according to the given information, it turns out firstly necessary for us to set the equation for the calculation of density and mass divided by volume:
Thus, we can find the mass of the unknown by subtracting the total mass of the liquid to the mass of the flask and the liquid:
So that we are now able to calculate the density in g/mL first:
Now, we proceed to the conversion to lb/in³ by using the following setup:
Regards!
Answer:
<em> ionic equation : </em>3Fe(2+)(aq) + 3SO4(2-)(aq)+ 6Na(+)(aq) + 2PO4 (3-) (aq) → Fe3(PO4)2(s)+ 6Na(+) + 3SO4(2-)(aq)
<em> net ionic equation: </em>3Fe(2+)(aq) + 2PO4 (3-)(aq) → Fe3(PO4)2(s)
Explanation:
The balanced equation is
3FeSO4(aq)+ 2Na3PO4(aq) → Fe3(PO4)2(s)+ 3Na2SO4(aq)
<em>Ionic equations: </em>Start with a balanced molecular equation. Break all soluble strong electrolytes (compounds with (aq) beside them) into their ions
. Indicate the correct formula and charge of each ion. Indicate the correct number of each ion
. Write (aq) after each ion
.Bring down all compounds with (s), (l), or (g) unchanged. The coefficents are given by the number of moles in the original equation
3Fe(2+)(aq) + 3SO4(2-)(aq)+ 6Na(+)(aq) + 2PO4 (3-) (aq) → Fe3(PO4)2(s)+ 6Na(+) + 3SO4(2-)(aq)
<em>Net ionic equations: </em>Write the balanced molecular equation. Write the balanced complete ionic equation. Cross out the spectator ions, it means the repeated ions that are present. Write the "leftovers" as the net ionic equation.
3Fe(2+)(aq) + 2PO4 (3-)(aq) → Fe3(PO4)2(s)
