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Ammonia, NH3, is used as a refrigerant. At its boiling point of –33 oC, the standard enthalpy of vaporization of ammonia is 23.3 kJ/mol. How much heat is released when 50.0 g of ammonia is condensed at –33 oC?–0.466 kJ–7.94 kJ–36.6 kJ–68.4 kJ–1.17 x 103 kJ

Answers

Answer 1

Answer:

Answer:

-68.4 kJ

Explanation:

The standard enthalpy of vaporization = 23.3 kJ/mol

which means the energy required to vaporize 1 mole of ammonia at its boiling point (-33 °C).

To calculate heat released when 50.0 g of ammonia is condensed at -33 °C.

This is the opposite of enthalpy of vaporization which means that same magnitude of heat is released.

Thus, Q = -23.3 kJ/mol

Where negative sign signifies release of heat

Given: mass of 50.0 g

Molar mass of ammonia = 17.034 g/mol

Moles of ammonia = 50.0 /17.034 moles = 2.9353 moles

Also,

1 mole of ammonia when condenses at -33 °C releases 23.3 kJ

2.9412 moles of ammonia when condenses at -33 °C releases 23.3×2.9353 kJ

Thus, amount of heat released when 50 g of ammonia condensed at -33 °C= -68.4 kJ, where negative sign signifies release of heat.

Answer 2

Answer:

Final answer:

The heat released when 50.0 g of ammonia condenses at its boiling point is -68.4 kJ. This is calculated by multiplying the moles of ammonia by the enthalpy of vaporization and recognizing that heat is released in condensation.

Explanation:

To solve this problem, we need to understand the concept of enthalpy of vaporization, which is the heat needed to convert 1 mole of a substance from a liquid to a gas at constant pressure and temperature. For ammonia (NH3), which boils at -33 °C, the enthalpy of vaporization is 23.3 kJ/mol. However, we want the heat released when 50.0 g (around 2.94 moles) of ammonia condenses, which is the reverse process of vaporization. Thus, the energy would be released rather than absorbed.

Now, let's calculate this value. We multiply the number of moles of ammonia by the enthalpy of vaporization:

2.94 moles x 23.3 kJ/mol = 68.4 kJ

Since this is the reverse of the process of vaporization, heat is released, so the enthalpy change is negative (-68.4 kJ). Therefore, the correct answer is -68.4 kJ.

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Most wine is prepared by the fermentation of the glucose in grape juice by yeast: C6H12O6(aq) --> 2C2H5OH(aq) + 2CO2(g) How many grams of glucose should there be in grape juice to produce 725 mLs of wine that is 11.0% ethyl alcohol, C2H5OH (d=0.789 g/cm3), by volume?

Answers

Answer:

123.41 g

Explanation:

Given that the ethyl alcohol produced is 11.0 % by volume.

It means that 1000 mL contains 110 mL of ethyl alcohol

Given that the volume is:- 725 mL

So,

Volume of ethyl alcohol = $(110)/(1000)* 725\ mL$ = 79.75 mL

Given that:- Density = 0.789 g/cm³ = 0.789 g/mL

So, Mass = Density*Volume = $0.789* 79.75\ g$ = 62.92 g

Calculation of the moles of ethyl alcohol as:-

Molar mass of ethyl alcohol = 46.07 g/mol

The formula for the calculation of moles is shown below:

$moles = (Mass\ taken)/(Molar\ mass)$

Thus,

$Moles= (62.92\ g)/(46.07\ g/mol)$

$Moles=1.37\ mol$

According to the reaction:-

$C_6H_(12)O_6_((aq))\rightarrow 2C_2H_5OH_((aq)) +2CO_2_((g))$

2 moles of ethyl alcohol is produced when 1 mole of glucose reacts

Also,

1.37 moles of ethyl alcohol is produced when $(1)/(2)* 1.37$ mole of glucose reacts

Moles of glucose = 0.685 Moles

Molar mass of glucose = 180.156 g/mol

Mass = Moles*Molar mass = $0.685* 180.156\ g$ = 123.41 g

What volume will 12 g of oxygen gas (O2) occupy at 25 °C and a pressure of 53 kPa?

Answers

ANSWER

The volume of the oxygen gas is 17.5 L

EXPLANATION

Given that;

The mass of oxygen gas is 12 grams

The temperature of the gas is 25 degrees Celcius

The pressure of the gas is 53 kPa

To find the volume of the oxygen gas, follow the steps below

Step 1; Assume the gas behaves like an ideal gas

Therefore, apply the ideal gas equation to find the volume of the gas

$\text{ PV }=\text{ nRT}$

Where

P is the pressure of the gas

V is the volume of the gas

n is number of moles of the gas

R is the universal gas constant

T is the temperature of the gas

Step 2: Find the number of moles of the oxygen gas using the below formula

$\text{ mole }=\text{ }\frac{\text{ mass}}{\text{ molar mass}}$

Recall, that the molar mass of the oxygen gas is 32 g/mol

$\begin{gathered} \text{ mole }=\text{ }\frac{12}{\text{ 32}} \n \text{ mole }=\text{ 0.375 mol} \end{gathered}$

Step 3; Convert the temperature to degree Kelvin

$\begin{gathered} \text{ T }=\text{ t }+\text{ 273.15} \n \text{ t }=\text{ 25}\degree C \n \text{ T }=25\text{ }+\text{ 273.15} \n \text{ T }=\text{ 298.15K} \end{gathered}$

Step 4; Substitute the given data into the formula in step 1

Recall, that R is 8.314 L kPa K^-1 mol^-1

$\begin{gathered} \text{ 53 }*\text{ V }=\text{ 0.375}*\text{ 8.314}*\text{ 298.15} \n \text{ 53V }=\text{ 929.557} \n \text{ Divide both sides by 53} \n \text{ }\frac{\cancel{53}V}{\cancel{53}}\text{ }=\text{ }(929.557)/(53) \n \text{ V }=\text{ }(929.557)/(93) \n \text{ V }=\text{ 17.5 L} \end{gathered}$

Hence, the volume of the oxygen gas is 17.5 L

If the heat of combustion of carbon monoxide (CO) is −283.0 kJmole, how many grams of carbon monoxide must combust in order to release 2,500.kJ of energy?

Answers

Answer:

247.4 g

Explanation:

Let's consider the thermochemical equation for the combustion of carbon monoxide.

CO(g) + 0.5 O₂(g) ⇒ CO₂(g) ΔH°c = -283.0 kJ/mol

The moles of carbon monoxide required to release 2500 kJ (-2500 kJ) are:

-2500 kJ × (1 mol CO/-283.0 kJ) = 8.834 mol CO

The molar mass of CO is 28.01 g/mol. The mass corresponding to 8.834 moles of CO is:

8.834 mol × 28.01 g/mol = 247.4 g

Calculate the mass of a sample of lead (cPb = 0.16 J/g℃) when it loses 200 J cooling from 75.0℃ to 42.0℃.

Answers

looses 299 please collins 75.0

The correct electron configuration for magnesium is: 1s 22s 22p 63s 3 True False

Answers

Answer:

False

Explanation:

Magnesium is the element of second group and third period. The electronic configuration of magnesium is - 2, 8, 2 or $1s^22s^22p^63s^2$

There are 2 valence electrons of magnesium.

Only the valence electrons are shown by dots in the Lewis structure.

As, stated above, there are only two valence electrons of magnesium, so in the Lewis structure, two dots are made around the magnesium symbol.

Given that the electronic configuration is:- $1s^22s^22p^63s^3$ .

Orbital s cannot accommodate 3 electrons and also in magnesium it has $3s^2$ . Hence, the statement is false.

A rock has a mass of 15 grams and a volume of 5 cm3 what is the density of the object

Answers

Answer:

The answer is

3.0 g/cm³

Explanation:

To find the density of a substance when given the mass and volume we use the formula

$density = (mass)/(volume)$

From the question

mass = 15 g

volume of rock = 5 cm³

The density of the substance is

$density = (15)/(5)$

We have the final answer as

3.0 g/cm³

Hope this helps you

Final answer:

The density of an object is calculated by dividing its mass by its volume. In this case, the rock's density is 3 g/cm3.

Explanation:

The process to determine the density of an object involves dividing its mass by its volume. Here, the rock has a mass of 15 grams and a volume of 5 cm3. Thus, the density can be calculated by the formula:

Density = Mass / Volume.

Plugging the given numbers into this formula results in:

Density = 15 grams / 5 cm3.

Therefore, the density of the rock is 3 g/cm3.

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