The titration reaction can be represented as mgcl2(aq) 2 naoh(aq) → mg(oh)2(s) 2 nacl(aq) the solubility equilibrium is written as mg(oh)2(s) ⇌ mg2 (aq) 2 oh-(aq) kemmi is trying to teach noke how to calculate the concentration of the mg2 cation at the titration endpoint. she explains that the initial mgcl2 concentration must be multiplied by the volume of mgcl2 pipetted and divided by the total volume at the endpoint. (note: the calculation page in the lab manual has a similar equation for the lead cation.) noke do her calculation. for the first titration in the set the initial concentration of mgcl2(aq) was 0.0050 m, the volume measured was 100.00 ml, and the total volume of solution at the endpoint was 104.12 ml. calculate the concentration of mg2 at the endpoint. (answer with 2 significant digits and units of m)

The concentration of Mg²⁺ cation at end point: M₂ = 0.0048 M

Explanation:

The titration reaction involved: MgCl₂(aq) + 2 NaOH(aq) → Mg(OH)₂(s) + 2 NaCl(aq)

and the solubility equilibrium for the product: Mg(OH)₂(s) ⇌ Mg²⁺ (aq) + 2 OH⁻(aq)

Given: Concentration of MgCl₂: M₁ = 0.005 M, volume of MgCl₂: V₁ = 100 mL, Total volume of solution at end point = 104.12 mL

Concentration of Mg²⁺ cation at end point: M₂ = ?M

Now, to calculate the concentration of Mg²⁺ cation at end point, we use the equation:    M₁ × V₁ = M₂ × V₂

⇒ M₂ = M₁ × V₁ ÷ V₂

⇒ M₂ = (0.005 M × 100 mL) ÷ 104.12 mL

⇒ M₂ = 0.0048 M (2 significant digits)

Therefore, the concentration of Mg²⁺ cation at end point: M₂ = 0.0048 M

c. because step up transformer steps up the input voltage i.e increases the input voltage  

the number of turns are greater on the secondary side of step up transformer  

transformer does not handle dc current

answer: the value of enthalpy change will be -152.80 kj.

solution:

pressure = 1atm , temperature = = 298 k

given that pressure and temperature are same in both gases , the number of moles of both gases can be determined by the ideal gas equation.

according to reaction 3 moles no are reacting with 1 mol of ,then 0.49 mol of no will react with moles of ozone that is 0.16 moles of ozone. since, no is limiting reagent and is an excessive reagent.

enthalpy change for the reaction between the given volumes of the gases is :