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Marina CMI [18]
2 years ago
12

What is the difference between a physical and chemical change? From what I know a chemical change modifies the atoms of an objec

t while a physical change modifies the molecules right?
Chemistry
1 answer:
blsea [12.9K]2 years ago
6 0
Physical change: a change in which no new substances are formed. the form of the substance is changed but not it's chemical composition (ice melting, bread toasting)

chemical change: any change that results in the formation of new chemical substances. this type of change modifies molecules and atoms by making and breaking the bonds between atoms! (iron rusting, gas burning)

so basically a physical change just changes the appearance of a substance, but a chemical change changes the makeup on a molecular level. i hope this helps you out!
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if a balloon had an initial volume of 100ml at pressure of 3.2 atm and the pressure was decreased to 1.9 atm, what would the new
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Answer:

168.4 mL

Explanation:

Data Given

initial volume V1 of gas in balloon = 100 mL

initial pressure P1 of gas in balloon = 3.2 atm

final pressure P2 of gas in balloon = 1.9 atm

final volume V2 of gas in balloon = ?

Solution:

This problem will be solved by using Boyle's law equation at constant Temperature.

The formula used

                  P1V1 = P2V2

As we have to find out Volume, so rearrange the above equation

                  V2 = P1V1 / P2

Put value from the data given

V2 = 100 mL x 3.2 atm / 1.9 atm

V2 = 168.4 mL

So the final Volume of gas in baloon =  168.4 mL

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Consider the reaction Mg(s) + I2 (s) → MgI2 (s) Identify the limiting reagent in each of the reaction mixtures below:
Lapatulllka [165]

Answer:

a) Nor Mg, neither I2 is the limiting reactant.

b) I2 is the limiting reactant

c) <u>Mg is the limiting reactant</u>

<u>d) Mg is the limiting reactant</u>

<u>e) Nor Mg, neither I2 is the limiting reactant.</u>

<u>f) I2 is the limiting reactant</u>

<u>g) Nor Mg, neither I2 is the limiting reactant.</u>

<u>h) I2 is the limiting reactant</u>

<u>i) Mg is the limiting reactant</u>

Explanation:

Step 1: The balanced equation:

Mg(s) + I2(s) → MgI2(s)

For 1 mol of Mg we need 1 mol of I2 to produce 1 mol of MgI2

a. 100 atoms of Mg and 100 molecules of I2

We'll have the following equation:

100 Mg(s) + 100 I2(s) → 100MgI2(s)

This is a stoichiometric mixture. <u>Nor Mg, neither I2 is the limiting reactant.</u>

b. 150 atoms of Mg and 100 molecules of I2

We'll have the following equation:

150 Mg(s) + 100 I2(s) → 100 MgI2(s)

<u>I2 is the limiting reactant</u>, and will be completely consumed. There will be consumed 100 Mg atoms. There will remain 50 Mg atoms.

There will be produced 100 MgI2 molecules.

c. 200 atoms of Mg and 300 molecules of I2

We'll have the following equation:

200 Mg(s) + 300 I2(s) →200 MgI2(s)

<u>Mg is the limiting reactant</u>, and will be completely consumed. There will be consumed 200 I2 molecules. There will remain 100 I2 molecules.

There will be produced 200 MgI2 molecules.

d. 0.16 mol Mg and 0.25 mol I2

We'll have the following equation:

Mg(s) + I2(s) → MgI2(s)

<u>Mg is the limiting reactant</u>, and will be completely consumed. There will be consumed 0.16 mol of I2. There will remain 0.09 mol of I2.

There will be produced 0.16 mol of MgI2.

e. 0.14 mol Mg and 0.14 mol I2

We'll have the following equation:

Mg(s) + I2(s) → MgI2(s)

This is a stoichiometric mixture. <u>Nor Mg, neither I2 is the limiting reactant.</u>

There will be consumed 0.14 mol of Mg and 0.14 mol of I2. there will be produced 0.14 mol of MgI2

f. 0.12 mol Mg and 0.08 mol I2

We'll have the following equation:

Mg(s) + I2(s) → MgI2(s)

<u>I2 is the limiting reactant</u>, and will be completely consumed. There will be consumed 0.08 moles of Mg. There will remain 0.04 moles of Mg.

There will be produced 0.08 moles of MgI2.

g. 6.078 g Mg and 63.455 g I2

We'll have the following equation:

Mg(s) + I2(s) → MgI2(s)

Number of moles of Mg = 6.078 grams / 24.31 g/mol = 0.250 moles

Number of moles I2 = 63.455 grams/ 253.8 g/mol = 0.250 moles

This is a stoichiometric mixture. <u>Nor Mg, neither I2 is the limiting reactant.</u>

There will be consumed 0.250 mol of Mg and 0.250 mol of I2. there will be produced 0.250 mol of MgI2

h. 1.00 g Mg and 2.00 g I2

We'll have the following equation:

Mg(s) + I2(s) → MgI2(s)

Number of moles of Mg = 1.00 grams / 24.31 g/mol = 0.0411 moles

Number of moles I2 = 2.00 grams/ 253.8 g/mol = 0.00788 moles

<u>I2 is the limiting reactant</u>, and will be completely consumed. There will be consumed 0.00788 moles of Mg. There will remain 0.03322 moles of Mg.

There will be produced 0.00788 moles of MgI2.

i. 1.00 g Mg and 2.00 g I2

We'll have the following equation:

Mg(s) + I2(s) → MgI2(s)

Number of moles of Mg = 1.00 grams / 24.31 g/mol = 0.0411 moles

Number of moles I2 = 20.00 grams/ 253.8 g/mol = 0.0788 moles

<u>Mg is the limiting reactant</u>, and will be completely consumed. There will be consumed 0.0411 moles of Mg. There will remain 0.0377 moles of I2.

There will be produced 0.0411 moles of MgI2.

4 0
3 years ago
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