Can someone solve the equations for these? 45 points

Chemistry

1 answer:

Volgvan3 years ago

8 0

2)

formula equation: Pb(NO₃)₂(aq) + 2KI(aq) ----> PbI₂(s) + 2KNO₃(aq)
total ionic equation: Pb⁺² + 2NO₃⁻¹ + 2K⁺¹ + 2I⁻¹----> PbI₂(s) + 2K⁺¹ + 2NO₃⁻¹
net ionic equation: Pb⁺² + 2 I⁻¹----> PbI₂(s)

3)

formula equation: Zn(NO₃)₂(aq) + K₂CO₃(aq) -----> ZnCO₃(s) + 2KNO₃(aq)
total ionic equation: Zn⁺² + 2NO₃⁻¹ + 2K⁺¹ + CO₃⁻² ---> ZnCO₃ (s) + 2K⁺¹ + 2NO₃⁻¹
net ionic equation: Zn⁺² + CO₃⁻² ----> ZnCO₃ (s)

note: if I did not specify the state of the molecule in the reaction, you can assume they are aqueous unless state otherwise.

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Calculate the molarity of a solution of monoprotic KHP prepared by mixing a 0.5237 g KHP in 50.0 mL of water .

Sergio039 [100]

Answer:

$M=0.0513M$

Explanation:

Hello!

In this case, since the molarity of a solution is computed by dividing the moles of solute by the volume of solution, we notice we first need the moles of KHP as shown below:

$n=0.5237g*\frac{1mol}{204.22 g} =0.00256mol$

Next, the volume in liters:

$V=50.0mL*\frac{1L}{1000mL}\\\\V=0.0500L$

Thus, the molarity turns out to be:

$M=\frac{n}{V}\\\\M=\frac{0.00256mol}{0.05000L} \\\\M=0.0513M$

Best regards!

3 0

2 years ago

A 3.0 L solution contains 73.5 g of H2SO4. Calculate the molar concentration of the solution.

kozerog [31]

$C_{m}=\frac{m}{MV}=\frac{73,5g}{98\frac{g}{mol}*3L}=0,25M\\\\
C_{1}V_{1}=C_{2}V_{2}\\\\
0,25M*3L=0,18M*V_{2}\\\\
V_{2}=\frac{0,25M*3L}{0,18M}\approx 4,17L\\\\
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