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IRINA_888 [86]
3 years ago
11

Write the formula unit and the net-ionic equations to describe the reaction between potassium hydrogen phthalate (khp) and cesiu

m hydroxide.

Chemistry
2 answers:
dybincka [34]3 years ago
8 0
C₆H₄(COOH)(COOK) + CsOH = C₆H₄(COOCs)(COOK) + H₂O

C₆H₄(COOH)(COO)⁻ + K⁺ + Cs⁺ + OH⁻ = C₆H₄(COO)₂²⁻ + K⁺ + Cs⁺ + H₂O

C₆H₄(COOH)(COO)⁻ + OH⁻ = C₆H₄(COO)₂²⁻ + H₂O

Alborosie3 years ago
6 0

Explanation:

The formula unit pf potassium phthalate is C_6H_4(COOH)(COOK)

The reaction between  potassium hydrogen phthalate and cesium hydroxide.

C_6H_4(COOH)(COOK)(aq)+CsOH(aq)\rightarrow C_6H_4(COOCs)(COOK)(aq)+H_2O(l)

In the above reaction, proton will get combine with hydroxide ion from cesium hydroxide to form a water molecule.

C_6H_4(COOH)(COO^-)(aq)+K^+(aq)+Cs^+(aq)+OH^-(aq)\rightarrow C_6H_4(COO^-)_2^{2-}(aq)+H_2O(l)+K^+(aq)+Cs^+(aq)

Removing common ions from both sides :

The net ionic equation is :

C_6H_4(COOH)(COO^-)(aq)+OH^-(aq)\rightarrow C_6H_4(COO^-)_2^{2-}(aq)+ H_2O(l)

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How many moles of gas would occupy 22.4 L at 273K in one arm?
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Answer:

1 mol

Explanation:

Using the general gas law equation as follows:

PV = nRT

Where;

P = pressure (atm)

V = volume (L)

n = number of moles (mol)

R = gas law constant (0.0821 Latm/molK)

T = temperature (K)

According to the provided information in the question;

V = 22.4L

T = 273K

P = 1 atm

R = 0.0821 Latm/molK

n = ?

Using PV = nRT

n = PV/RT

n = (1 × 22.4) ÷ (0.0821 × 273)

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Automobile airbags contain solid sodium azide, NaN3, that reacts to produce nitrogen gas when heated, thus inflating the bag. 2N
Vitek1552 [10]

Answer : The value of work done for the system is 1144.69 J

Explanation :

First we have to calculate the moles of NaN_3

\text{Moles of }NaN_3=\frac{\text{Mass of }NaN_3}{\text{Molar mass of }NaN_3}

Molar mass of NaN_3 = 65.01 g/mole

\text{Moles of }NaN_3=\frac{20.2g}{65.01g/mole}=0.311mole

Now we have to calculate the moles of nitrogen gas.

The balanced chemical reaction is,

2NaN_3(s)\rightarrow 2Na(s)+3N_2(g)

From the balanced reaction we conclude that

As, 2 mole of NaN_3 react to give 3 mole of N_2

So, 0.311 moles of NaN_3 react to give \frac{0.311}{2}\times 3=0.466 moles of N_2

Now we have to calculate the volume of nitrogen gas.

Using ideal gas equation:

PV=nRT

where,

P = Pressure of N_2 gas = 1.00 atm

V = Volume of N_2 gas = ?

n = number of moles N_2 = 0.466 mole

R = Gas constant = 0.0821L.atm/mol.K

T = Temperature of N_2 gas = 22^oC=273+22=295K

Putting values in above equation, we get:

1.00atm\times V=0.466mole\times (0.0821L.atm/mol.K)\times 295K

V=11.3L

As initially no nitrogen was present. So,

Volume expanded = Volume of nitrogen evolved

Thus,

Expansion work = Pressure × Volume

Expansion work = 1.00 atm × 11.3 L

Expansion work = 11.3 L.atm

Conversion used : (1 L.atm = 101.3 J)

Expansion work = 11.3 × 101.3 = 1144.69 J

Therefore, the value of work done for the system is 1144.69 J

5 0
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