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yan [13]
3 years ago
13

Rita wants to make some toast for breakfast. She puts a slice of bread in the toaster. After 3 min, Rita notices that the sides

of the bread are black. What has happened?
A. The bread has undergone a change of state.

B. The bread has undergone a change in density.

C. Some of the matter in the bread was destroyed.

D. New substances have formed as the result of a chemical change.
Chemistry
2 answers:
fredd [130]3 years ago
6 0
The answer is D


Because the bread cannot go back to its previous state, and because a new substance was formed. :)
goldfiish [28.3K]3 years ago
5 0
I believe the answer is c.

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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
What ratio of nitrogen and hydrogen molecules would result in no left-over reactants? Explain your answer.
madam [21]

A ration, Nitrogen to Hydrogen, of 1 : 3 would result in no left-over reactants

Explanation:

This is because Nitrogen and Hydrogen would react to form ammonia as shown in the following reaction below;

N₂ (g) + 3H₂ (g) → 2NH₃ (g)

For every mole of Nitrogen gas, you would require 3 moles of Hydrogen gas. This way they would react to form Ammonia with no leftover reactants.

Learn More:

For more on balanced gas reaction equations check out;

brainly.com/question/3407131

brainly.com/question/13607761

brainly.com/question/3458848

brainly.com/question/9557732

brainly.com/question/11575604

#LearnWithBrainly

6 0
2 years ago
Please help. Thank you:)
bearhunter [10]
These are guesses because there is no word bank or anything

A. Water
B. Housing
C. Plants
D. Animals
E. Warmth
5 0
3 years ago
How many milliners of hydrogen gas ar produced by the reaction 0.020 moles of magnesium with excess of hydrochloride acid at sto
jeyben [28]

Answer:- 448 mL of hydrogen gas are formed.

Solution:- It asks to calculate the volume of hydrogen gas formed in milliliters at STP when 0.020 moles of magnesium reacts with excess HCl acid. The balanced equation is:

Mg+2HCl\rightarrow MgCl_2+H_2(g)

There is 1:1 mol ratio between Mg and hydrogen gas. So, the moles of hydrogen gas is also equals to the moles of Mg reacted.

moles of Hydrogen gas formed = 0.020 mol

At STP, volume of 1 mol of the gas is 22.4 L. We need to calculate the volume of 0.02 moles of hydrogen gas.

0.02mol(\frac{22.4L}{1mol})

= 0.448 L

They want answer in mL. So, let's convert L to mL using the conversion formula, 1L = 1000mL

0.448L(\frac{1000mL}{1L})

= 448 mL

So, 0.020 moles of magnesium would produce 448 mL of hydrogen gas at STP on reacting with excess of HCl acid.

5 0
3 years ago
Which of the following is a unit of length? O A. A liter O B. A kilogram C. A meter O D. A degree​
sergeinik [125]

Answer:

C.) A meter

Explanation:

5 0
2 years ago
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