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4 years ago

Which of the following is kept constant when using a bomb calorimeter?

1 answer:

6 0

Constant Volume Calorimetry, also know as bomb calorimetry, is used to measure the heat of a reaction while holding volume constant and resisting large amounts of pressure. Although these two aspects of bomb calorimetry make for accurate results, they also contribute to the difficulty of bomb calorimetry. In this module, the basic assembly of a bomb calorimeter will be addressed, as well as how bomb calorimetry relates to the heat of reaction and heat capacity and the calculations involved in regards to these two topics.

Introduction

Calorimetry is used to measure quantities of heat, and can be used to determine the heat of a reaction through experiments. Usually a coffee-cup calorimeter is used since it is simpler than a bomb calorimeter, but to measure the heat evolved in a combustion reaction, constant volume or bomb calorimetry is ideal. A constant volume calorimeter is also more accurate than a coffee-cup calorimeter, but it is more difficult to use since it requires a well-built reaction container that is able to withstand large amounts of pressure changes that happen in many chemical reactions.

Most serious calorimetry carried out in research laboratories involves the determination of heats of combustion ΔHcombustion" role="presentation" style="display: inline-table; font-style: normal; font-weight: normal; line-height: normal; font-size: 14.4px; text-indent: 0px; text-align: left; text-transform: none; letter-spacing: normal; word-spacing: normal; word-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; padding: 0px; margin: 0px; position: relative;">ΔHcombustionΔHcombustion, since these are essential to the determination of standard enthalpies of formation of the thousands of new compounds that are prepared and characterized each month. In a constant volume calorimeter, the system is sealed or isolated from its surroundings, and this accounts for why its volume is fixed and there is no volume-pressure work done. A bomb calorimeter structure consists of the following:

Steel bomb which contains the reactantsWater bath in which the bomb is submergedThermometerA motorized stirrerWire for ignition

is usually called a “bomb”, and the technique is known as bomb calorimetry

Another consequence of the constant-volume condition is that the heat released corresponds to qv , and thus to the internal energy change ΔUrather than to ΔH. The enthalpy change is calculated according to the formula

(1.1)ΔH=qv+ΔngRT" role="presentation" style="display: inline-table; font-style: normal; font-weight: normal; line-height: normal; font-size: 14.4px; text-indent: 0px; text-align: center; text-transform: none; letter-spacing: normal; word-spacing: normal; word-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; padding: 0px; margin: 0px; width: 10000em !important; position: relative;">ΔH=qv+ΔngRT(1.1)(1.1)ΔH=qv+ΔngRT

Δng" role="presentation" style="display: inline-table; font-style: normal; font-weight: normal; line-height: normal; font-size: 14.4px; text-indent: 0px; text-align: left; text-transform: none; letter-spacing: normal; word-spacing: normal; word-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; padding: 0px; margin: 0px; position: relative;">ΔngΔng is the change in the number of moles of gases in the reaction.

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Observations of the solar system support the nebular theory because

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Answer:

Because all the planets orbit the sun in the same direction, along a circular flattened disk.

Explanation:

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4 years ago

Read 2 more answers

A reaction vessel contains 10.0 g of CO and 10.0 g of O2. How many grams of CO2 could be produced according to the following rea

iren2701 [21]

Answer:

1. 15.71 g CO2

2. 38.19 % of efficiency

Explanation:

According to the balanced reaction (2 CO(g) + O2(g) → 2 CO2(g)), it is clear that the CO is the limitant reagent, because for every 2 moles of CO we are using only 1 mole of O2, so even if we have the same quantity for both reagents, not all of the O2 will be consumed. This means that we can just use the stoichiometric ratios of the CO and the CO2 to solve this question, and for that we need to convert the gram units into moles:

For CO:

C = 12.01 g/mol

O = 16 g/mol

CO = 28.01 g/mol

(10.0g CO) x (1 mol CO/28.01 g) = 0.3570 mol CO

For CO2:

C = 12.01 g/mol

O = 16 x 2 = 32 g/mol

CO2 = 44.01 g/mol

We now that for every 2 moles of CO we are going to get 2 moles of CO2, so we resolve as follows:

(0.3570 mol CO) x (2 mol CO2/2 mol CO) = 0.3570 moles CO2

We are obtaining 0.3570 moles of CO2 with the 10g of CO, now lets convert the CO2 moles into grams:

(0.3570 moles CO2) x (44.01 g/1 mol CO2) = 15.71 g CO2

Now for the efficiency question:

From the previous result, we know that if we produce 15.71 CO2 with all the 10g of CO used, we would have an efficiency of 100%. So to know what would that efficiency be if we would only produce 6g of CO2, we resolve as follows,

(6g / 15.71g) x 100 = 38.19 % of efficiency

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3 years ago

How many grams C3H7OH can be made by reacting with 7.3L of CO2 at STP

Komok [63]

Answer:

6.54g of C3H7OH

Explanation:

Step 1:

Determination of the number of mole of CO2 that occupy 7.3L at stp.

This can be obtained as follow:

1 mole of a gas occupy 22.4L at stp.

Therefore, Xmol of CO2 will occupy 7.3L at stp i.e

Xmol of CO2 = 7.3/22.4

Xmol of CO2 = 0.326 mole.

Therefore, 0.326 mole of CO2 was used in the reaction.

Step 2:

The balanced equation for the reaction. This is given below:

6CO2 + 8H2O —> 2C3H7OH + 9O2

Step 3:

Determination of the number of mole of C3H7OH produced from the reaction. This is illustrated below:

From the balanced equation above,

6 moles of CO2 reacted to produce 2 moles of C3H7OH.

Therefore, 0.326 mole of CO2 will react to produce = (0.326 x 2)/6 = 0.109 mole of C3H7OH.

Step 4:

Conversion of 0.109 mole of C3H7OH to grams. This is illustrated below:

Number of mole of C3H7OH = 0.109 mole.

Molar mass of C3H7OH = (12x3)+ (7x1) + 16 + 1 = 60g/mol

Mass of C3H7OH =..?

Mass = mole x molar mass

Mass of C3H7OH = 0.109 x 60

Mass of C3H7OH = 6.54g.

Therefore, 6.54g of C3H7OH is produced from the reaction.

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3 years ago

Girl does not see a puddle of liquid on the lab table and accidentally touches the liquid. What should hiro do next?

Shalnov [3]

Tell the teacher they have touched an unidetified liquid. This is because "hiro" can have a reaction with the chemical he touched that can harm his health.

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3 years ago

Resistance of a material being scratched in known as

Nastasia [14]

Friction is the answer

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3 years ago