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professor190 [17]
3 years ago
15

In two or more complete sentences explain how to balance the chemical equation, KClO3 ⟶ KCl + O2 and include all steps. Write yo

ur response in the essay box below.
Chemistry
1 answer:
Serggg [28]3 years ago
5 0

Answer:

2KClO3 -------> 2KCl + 3O2

Explanation:

First, in balancing a chemical reaction such as the one given in the question, you should understand that for an equation to be balanced, the number of atoms and ions on both sides of the equation that is the right and left side must be equal. This follows the law of conservation of mass which tells us that matter can neither be created nor destroyed but can be changed into another form.

Next is to begin balancing the equation by identifying and writing down the substances given:

KCl03 ---------> KCl + O2

Next is to count he number of the individual atoms on each side and find out if they are the same on both sides and if not you must follow the next step.

Add a corresponding number and use it to multiply the atoms involved

KClO3 ---------> KCl + O2

Oxygen is 3 on the left side and two on the other side, so we multiply the left hand side by 2 and the right hand side by 3

2KClO3 -----> KCl + 3O2

The potassium and Chlorine are no longer balanced, so you multiply the right had=nd side of KCl by 2.

2KClO3 -----> 2KCl + 3O2

The reactionis herefore balanced as both sides have equal number of atoms and ions.

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Assume you mix 100.0 mL of 200 M CsOH with 50.0 mL 0f 0.400 M HCl in a coffee cup calorimeter. A reaction occurs. The temperatur
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Answer:

The enthalpy change for the reaction per mole of CsOH is -56.1 kJ/mol

Explanation:

Step 1: Data given

Volume of a CsOH solution = 100.0 mL

Molarity of a CsOH solution = 0.200 M

Volume of HCl solution = 50.0 mL

Molarity of HCl solution = 0.400M

The temperature of both solutions before mixing was 22.50 °C, and it rises to 24.28 °C after the acid-base reaction.

Density = 1.00 g/mL

Specific heat = 4.2 J/gK = 4.2 J/g°C

Step 2: The balanced equation

CsOH + HCl → CsCl + H2O

Step 3: Calculate the energy

Q = m*c*ΔT

⇒with Q = the heat transfer = TO BE DETERMINED

⇒with m = the mass of the solution = (100+ 50 mL) * 1.00 g/mL = 150 grams

⇒with c = the specific heat of the solution = 4.2 J/g°C

⇒with ΔT = The change of temperature = T2 - T1  = 24.28 °C - 22.50 °C = 1.78 °C

Q = 150 grams * 4.2 J/g°C * 1.78 °C

Q = 1121.4 J

Step 4: Calculate moles CsOH

Moles CsOH = molarity CsOH * volume CsOH

Moles CsOH = 0.200M * 0.100 L

Moles CsOH = 0.0200 moles

Step 5: Calculate  the enthalpy change for the reaction per mole of CsOH

ΔH is negative since this is an exothermic reaction

ΔH = -Q/moles

ΔH = -1121.4 J / 0.0200 moles

ΔH = -56070 J/mol = -56.1 kJ/mol

The enthalpy change for the reaction per mole of CsOH is -56.1 kJ/mol

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The second form is dependent on how the heating is taking place. If the bottle is out in sunlight, the form of heat transfer is radiation from the sun's rays. If heat is directly being applied to it, then the form is conduction, which occurs in solids and through direct contact.
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