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Alenkinab [10]
3 years ago
6

In a coffe cup calorimeter, 50.0mL of 0.100M of AgNO3 and 50mL of 0.100M HCl are mixed to yield the following reaction:

Chemistry
1 answer:
Jet001 [13]3 years ago
7 0

Answer:

The enthalpy change of the reaction is -66.88 kJ/mol.

Explanation:

Mass of the solution = m = 100 g

Heat capacity of the solution = c = 4.18 J/g°C

Initial temperature of the solutions before mixing = T_1=22.60^oC

Final temperature of the solution after mixing = T_2=23.40^oC

Heat gained by the solution due to heat released by reaction between HCl and silver nitrate = Q

Q=m\times c\times (T_2-T_1)

Q=100 g\times 4.18 J/g^oC\times (23.40^oC-22.60^oC)=334.4 J

Heat released due to reaction = Q' =-Q = -334.4 J

Moles of silver nitrate = n

Molarity of silver nitrate solution = 0.100 M

Volume of the silver nitrate solution = 50.0 mL = 0.050 L ( 1 mL = 0.001 L)

Moles =Molarity\times Volume (L)

n=0.100 M\times 0.050 L=0.005 mol

Enthalpy change of the reaction = \Delta H

=\Delta H=\frac{-334.4 J}{0.005 mol}=-66,880 J/mol=-66.88 kJ/mol

1 J = 0.001 kJ

The enthalpy change of the reaction is -66.88 kJ/mol.

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Andru [333]

Answer: Please see below for answers

Explanation: Matching appropriate labels , we have

1)3/4 of the way to second equivalence point of a diprotic acid/strong base titration-- pH=pka₂

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equivalence point of a strong acid/strong base titration= pH=7

equivalence point of a weak acid/strong base titration=pH>7

half-way to equivalence point of a weak acid/strong base titration pH =pka

where

pH gives the measure of the amount of concentration of hydrogen ions in an aqueous solution.

pKa  is known as acid dissociation constant which explains the equilibrum at which a chemical species can give out or receive proton

pka₂ is the acid dissociation constant for the second ionization energy.

5 0
2 years ago
When an atom becomes an anion it becomes Smaller,Larger,neutral,positive
liraira [26]

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3 0
2 years ago
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emmasim [6.3K]
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Hope this helps!
7 0
2 years ago
1. Using the balanced equation, answer the following questions:
nalin [4]

Answer:

                     a)  2.53 × 10²³ molecules of O₂

                     b)  31.90 g of KCl

Explanation:

                  The balance chemical equation for given decomposition reaction is as follow;

                                   2 KClO₃ → 2 KCl + 3 O₂

<h3>Part 1:</h3>

Step 1: <u>Calculate Moles for given amount of KClO₃;</u>

                    Moles  =  Mass / M.Mass

                    Moles  =  34.35 g / 122.55 g/mol

                    Moles  =  0.280 moles of KClO₃

Step 2: <u>Find out  moles of O₂ produced;</u>

According to eq,

                  2 moles of KClO₃ produces  =  3 moles of O₂

So,

            0.280 moles of KClO₃ will produce  =  X moles of O₂

Solving for X,

                    X  =  0.280 mol × 3 mol / 2 mol

                     X =  0.42 moles of O₂

Step 3: <u>Calculate No. of Molecules of O₂ as,</u>

No. of Molecules  =  Moles × 6.022 × 10²³

No. of Molecules  =  0.42 mol × 6.022 × 10²³ molecules/mol

No. of Molecules  =  2.53 × 10²³ molecules of O₂

<h3>Part 2:</h3>

Step 1: <u>Calculate Moles for given amount of KClO₃;</u>

                    Moles  =  Mass / M.Mass

                    Moles  =  52.53 g / 122.55 g/mol

                    Moles  =  0.428 moles of KClO₃

Step 2: <u>Find out  moles of KCl produced;</u>

According to eq,

                  2 moles of KClO₃ produces  =  2 moles of KCl

So,

            0.428 moles of KClO₃ will produce  =  X moles of KCl

Solving for X,

                    X  =  0.428 mol × 2 mol / 2 mol

                     X =  0.428 moles of KCl

Step 3: <u>Calculate Mass of KCl as;</u>

                         Mass  =  Moles × M.Mass

                         Mass  =  0.428 mol × 74.55 g/mol

                         Mass  =  31.90 g of KCl

6 0
3 years ago
Is the following reaction spontaneous at 298 K? Answer by calculating ΔG. H2O(g) + C(s) → CO(g) + H2(g) ΔH = 131.3 kJ/mole ΔS =
SpyIntel [72]

Is the following reaction spontaneous at 298 K? Answer by calculating ΔG. H2O(g) + C(s) → CO(g) + H2(g) ΔH = 131.3 kJ/mole ΔS = 134 J/mole˙K


No

7 0
3 years ago
Read 2 more answers
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