CHEMISTRY COLLEGE

Ammonium nitrate dissociates in water according to the following equation:43() = 4+()+03−()

When a student mixes 5.00 g of NH4NO3 with 50.0 mL of water in a coffee-cup calorimeter, the temperature of the resultant solution decreases from 22.0 °C to 16.5 °C. Assume the density of water is 1.00 g/ml and the specific heat capacity of the resultant solution is 4.18 J/g·°C.

1) Calculate q for the reaction. You must show your work.

2) Calculate the number of moles of NH4NO3(s) which reacted. You must show your work.

3) Calculate ΔH for the reaction in kJ/mol. You must show your work.

Answers

Answer 1

Answer:

Answer:

Explanation:

NH₄NO₃ = NH₄⁺ +NO₃⁻

heat released by water = msΔ T

m is mass , s is specific heat and ΔT is fall in temperature

= 50 x 4.18 x ( 22 - 16.5 ) ( mass of 50 mL is 50 g )

= 1149.5 J .

This heat will be absorbed by the reaction above .

q for the reaction = + 1149.5 J

2 )

molecular weight of NH₄NO₃ = 80

No of moles reacted = 5/80 = 1 / 16 moles.

3 )

5 g absorbs 1149.5 J

80 g absorbs 1149.5 x 16 J

= 18392 J

= 18.392 kJ.

= + 18.392 kJ

ΔH = 18.392 kJ / mol

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2 SO2 (g) + O2 (g) 2 SO3 (g)If the TEMPERATURE on the equilibrium system is suddenly increased :The value of Kc A. IncreasesB. DecreasesC. Remains the sameThe value of Qc A. Is greater than KcB. Is equal to KcC. Is less than KcThe reaction must: A. Run in the forward direction to restablish equilibrium.B. Run in the reverse direction to restablish equilibrium.C. Remain the same. Already at equilibrium.The concentration of O2 will: A. Increase.B. Decrease.C. Remain the same.

Answers

Answer:

Part one: B. Kc decreases

Part two: B. Is equal to Kc

Part three: B. Run in the reverse direction to reestablish equilibrium

Part four: A. Increase

Explanation:

Part one: Sulfur dioxide combines with oxygen to form sulphur trioxide in an exothermic reaction. If the temperature is suddenly is increased, while the reaction is at equilibrium, the backward reaction (the endothermic one) is favored to "sweep up the excess heat". An increase in reactants means a decrease in Kc since the denominator(reactants) is becoming bigger while the numerator (products) become smaller.

Part two: Qc is a varying version of Kc. For this set of circumstances, it will be equal to Kc since Kc varies with temperature

Part three: The reaction must run in the reverse to reestablish the equilibrium.

Part four: The concentration of of oxygen will increase as more of the reactants are formed

Final answer:

The increase in temperature for this exothermic reaction will cause the value of Kc to decrease, the value of Qc to be greater than Kc, the reaction to run in the reverse direction, and the concentration of O2 to increase.

Explanation:

The given chemical reaction represents a type of equilibrium reaction, specifically an exothermic reaction, as it produces sulfur trioxide (SO3), which releases heat. According to Le Chatelier's principle, to maintain equilibrium, if a system is disturbed by an external factor, the system will adjust accordingly.

Here are my answers to the specific questions:

When the temperature is increased in an exothermic reaction, the system tries to consume the excess heat by moving in the endothermic direction, which is the reverse reaction in this case. Therefore, the value of Kc decreases (B).
Since the equilibrium has been disturbed, the value of Qc will not be equal to Kc. Considering more products are formed, Qc will be greater than Kc (A).
As a response to the increase in temperature, to re-establish equilibrium, the reaction will run in the reverse direction (B).
As the reaction goes in reverse to establish a new equilibrium, the concentration of reactants increases. Thus, the concentration of O2 will increase (A).

Learn more about Chemical Equilibrium here:

brainly.com/question/3920294

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Write the equation for the alpha decay of C-11. Write the equation for the beta decay of Be-10
HELP HELP HELP

Answers

Answer:

We know that in the decay process the sum of molecular number as well as molecular weight should be constant.

The following three reaction are as follows

1 .

Explanation:

Which TWO properties are characteristic of iconic compounds?brittleness
ductility
high melting point
low boiling point
malleability

Answers

Answer : The correct options are, brittleness and high melting point

Explanation :

Ionic compound : Ionic compounds are the compounds which are formed when a metal cation bonded with non-metal anion. The metal cation and non-metal anion bonded with an electrostatic force of attraction.

The properties of ionic compounds are :

Ionic compounds are brittle and hard. They breaks easily into small pieces.

They have high melting point and boiling point.

They conduct electricity in liquid state not in solid state.

Hence, the brittleness and high melting point properties are the characteristic of ionic compounds.

Some characteristics of Ionic compounds by Mimiwhatsup: brittle, high melting point, conducts electricity when molten or dissolved in water.

What pressure is exerted on the bottom of a 0.500-m-wide by 0.900-m-long gas tank that can hold 50.0 kg of gasoline by the weight of the gasoline in it when it is full?

Answers

Answer:

1088.89 Pa

Explanation:

According to the Newton's second law of motion:-

$Force=Mass* Acceleration$

Mass = 50.0 kg

Acceleration = g = 9.81 m/s²

So,

$Force=50.0* 9.8\ kgm/s^2$

Force = 490 N

Area of the base = $length* breath$ = $0.500* 0.900$ m² = 0.45 m²

Pressure = Force/Area = $(490\ N)/(0.45\ m^2)$ = 1088.89 Pa

The enthalpy of vaporization (ΔH°vap) of benzene is 30.7 kJ/mol at its normal boiling point of 353.3 K. What is ΔS°vap at this temperature? a. 383 J/(mol·K) b. 0.0115 J/(mol·K) c. 86.9 J/(mol·K) d. 0.087 J/(mol·K) e. 11.5 J/(mol·K)

Answers

Answer: The correct answer is Option c.

Explanation:

Vaporization is defined as the physical process in which liquid particles get converted to gaseous particles.

$Liquid\rightleftharpoons Gas$

The value of standard Gibbs free energy is 0 for equilibrium reactions.

To calculate $\Delta S^o_(vap)$ for the reaction, we use the equation:

$\Delta S^o_(vap)=(\Delta H^o_(vap))/(T)$

where,

$\Delta S^o_(vap)$ = standard entropy change of vaporization

$\Delta H^o_(vap)$ = standard enthalpy change of vaporization = 30.7 kJ/mol = 30700 J/mol (Conversion factor: 1 kJ = 1000 J)

T = temperature of the reaction = 353.3 K

Putting values in above equation, we get:

$\Delta S^o_(vap)=(30700J/mol)/(353.3K)=86.9J/(mol.K)$

Hence, the correct answer is Option c.

Final answer:

The change in entropy (ΔS°vap) of benzene at its normal boiling point can be calculated using the enthalpy of vaporization (ΔH°vap) and the temperature (T). The ΔS°vap for benzene at 353.3 K is 0.087 J/(mol·K).

Explanation:

The enthalpy of vaporization (ΔH°vap) of a substance is related to the change in entropy (ΔS°vap) by the equation ΔS°vap = ΔH°vap / T, where T is the temperature in Kelvin. We are given that the enthalpy of vaporization of benzene at its normal boiling point (T = 353.3 K) is 30.7 kJ/mol. Plugging these values into the equation, we can calculate ΔS°vap to be 30.7 kJ/mol / 353.3 K = 0.087 J/(mol·K).