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3 years ago

Provide a balanced molecular equation, total ionic, and net ionic equation for sodium phosphate and zinc acetate.

Chemistry

1 answer:

vivado [14]3 years ago

5 0

Answer: Balanced molecular equation :

$2Na_3PO_4(aq)+3(CH_3COO)_2Zn(aq)\rightarrow 6CH_3COONa(aq)+Zn_3(PO_4)_2(s)$

Total ionic equation:

$6Na^+(aq)+3PO_4^{2-}(aq)+6CH_3COO^-(aq)+3Zn^{2+}(aq)\rightarrow 6CH_3COO^-(aq)+6Na^+(aq)+Zn_3(PO_4)_2(s)$

The net ionic equation:

$2PO_4^{3-}(aq)+3Zn^{2+}(aq)\rightarrow Zn_3(PO_4)_2(s)$

Explanation:

Complete ionic equation : In complete ionic equation, all the substances that are strong electrolyte are present in an aqueous state as ions.

Net ionic equation : In the net ionic equations, we are not include the spectator ions in the equations.

Spectator ions : The ions present on reactant and product side which do not participate in a reactions. The same ions present on both the sides.

When sodium phosphate and zinc acetate then it gives zinc phosphate and sodium acetate as product.

The balanced molecular equation will be,

$2Na_3PO_4(aq)+3(CH_3COO)_2Zn(aq)\rightarrow 6CH_3COONa(aq)+Zn_3(PO_4)_2(s)$

The total ionic equation in separated aqueous solution will be,

$6Na^+(aq)+2PO_4^{3-}(aq)+6CH_3COO^-(aq)+3Zn^{2+}(aq)\rightarrow 6CH_3COO^-(aq)+6Na^+(aq)+Zn_3(PO_4)_2(s)$

In this equation, and are the spectator ions.

By removing the spectator ions from the balanced ionic equation, we get the net ionic equation.

The net ionic equation will be,

$2PO_4^{3-}(aq)+3Zn^{2+}(aq)\rightarrow Zn_3(PO_4)_2(s)$

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3 years ago

The indicator propyl red has a Ka of 3.3 X 10-6 . It is red at low pH and yellow at higher pH. What is the approximate pH range

Alik [6]

Answer:

4.48 - 6.48

Explanation:

A pH indicator works in a better way in a range of pH = pKa ± 1. That means we need to determine the pKa of the indicator propyl red to find the range over which it change its color. That is:

pKa = -log Ka

pKa = -log 3.3x10⁻⁶

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That means the range over propyl red will change from yellow to red or vice versa is:

4.48 - 6.48

5 0

2 years ago

Hydrogen peroxide decomposes to water and oxygen at constant pressure by the following reaction: 2H2O2(l) → 2H2O(l) + O2(g) ΔH =

Mashcka [7]

<u>Answer:</u> The amount of heat released is -7.203 kJ

<u>Explanation:</u>

The given chemical equation follows:

$2H_2O_2(l)\rightarrow 2H_2O(l )+O_2(g);\Delta H=-196kJ$

To calculate the enthalpy change for 1 mole of the hydrogen peroxide, we use unitary method:

When 2 moles of hydrogen peroxide is reacted, the enthalpy of the reaction is -196 kJ

So, when 1 mole of hydrogen peroxide will react, the enthalpy of the reaction will be $\frac{-196}{2}\times 1=-98kJ$

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Given mass of hydrogen peroxide = 2.50 g

Molar mass of hydrogen peroxide = 34 g/mol

Putting values in above equation, we get:

$\text{Moles of hydrogen peroxide}=\frac{2.50g}{34g/mol}=0.0735mol$

To calculate the heat of the reaction, we use the equation:

$\Delta H_{rxn}=\frac{q}{n}$

where,

$q$ = amount of heat released

n = number of moles = 0.0735 moles

$\Delta H_{rxn}$ = enthalpy change of the reaction = -98 kJ/mol

Putting values in above equation, we get:

$-98kJ/mol=\frac{q}{0.0735mol}\\\\q=(-98kJ/mol\times 0.0735mol)=-7.203kJ$

Hence, the amount of heat released is -7.203 kJ

8 0

3 years ago