Which of the following is and example of a chemical change

What is the theoretical yield of Calcium if we begin with 12.6 grams of Aluminium?

AlladinOne [14]

The question is incomplete, here is the complete question:

The given chemical reaction is:

$2Al+3Cu(NO_3)_2\rightarrow 3Cu+2Al(NO_3)_3$

What is the theoretical yield of Calcium if we begin with 12.6 grams of Aluminium?

<u>Answer:</u> The theoretical yield of copper is 44.48 grams

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

Given mass of aluminium = 12.6 g

Molar mass of aluminium = 27 g/mol

Putting values in equation 1, we get:

$\text{Moles of aluminium}=\frac{12.6g}{27g/mol}=0.467mol$

The given chemical equation follows:

$2Al+3Cu(NO_3)_2\rightarrow 3Cu+2Al(NO_3)_3$

By Stoichiometry of the reaction:

2 moles of aluminium produces 3 moles of copper

So, 0.467 moles of aluminium will produce = $\frac{3}{2}\times 0.467=0.7005mol$ of copper

Now, calculating the mass of copper from equation 1, we get:

Molar mass of copper = 63.5 g/mol

Moles of copper = 0.7005 moles

Putting values in equation 1, we get:

$0.7005mol=\frac{\text{Mass of copper}}{63.5g/mol}\\\\\text{Mass of copper}=(0.7005mol\times 63.5g/mol)=44.48g$

Hence, the theoretical yield of copper is 44.48 grams

4 0

3 years ago

Which statement describes climate and not weather?

sasho [114]

I would answer B!! But your guess is as good as mine!!

3 0

3 years ago

Read 2 more answers

In an experiment, a student needs 250.0 mL of a 0.100 M copper (II) chloride solution. A stock solution of 2.00 M copper (II) ch

LekaFEV [45]

Answer : The volume of stock solution needed are, 12.5 mL

Explanation :

Formula used :

$M_1V_1=M_2V_2$

where,

$M_1\text{ and }V_1$ are the initial molarity and volume of copper (II) chloride.

$M_2\text{ and }V_2$ are the final molarity and volume of stock solution of copper (II) chloride.

We are given:

$M_1=0.100M\\V_1=250.0mL\\M_2=2.00M\\V_2=?$

Putting values in above equation, we get:

$0.100M\times 250.0mL=2.00M\times V_2\\\\V_2=12.5mL$

Hence, the volume of stock solution needed are, 12.5 mL

8 0

3 years ago

Vanillin (used to flavor vanilla ice cream and other foods) is the substance whose aroma the human nose detects in the smallest

balu736 [363]

Answer:

Cost to supply enough vanillin is $3.2\$$

Explanation:

Threshold limit of vanillin in air is $2.0\times 10^{-11}g$ per litre means there should be $2.0\times 10^{-11}g$ of vanillin in 1L of air to detect aroma of vanillin.

$1ft^{3}=28.32L$

So, $5.0\times 10^{7}ft^{3}=(5.0\times 10^{7}\times 28.32)L$

So amount of vanillin should be present to detect = $(2.0\times 10^{-11}\times 5.0\times 10^{7}\times 28.32)g$

As cost of 50 g vanillin is $112\$$ therefore cost of $(2.0\times 10^{-11}\times 5.0\times 10^{7}\times 28.32)g$ vanillin = $(2.0\times 10^{-11}\times 5.0\times 10^{7}\times 28.32\times 112)\$ = 3.2\$$

5 0

3 years ago

The OAC simulation showed that warm air over a cold current:

vladimir2022 [97]

Answer:

Option B

Transfers energy to the water

Explanation:

Warm air transfers energy to the water when it flows over cold currents. This means that the warm air loses heat energy to the cold currents thus, raising its temperature.

Whenever there is a temperature difference between two bodies in contact with each other, the Fouriers law explains that there is always a transfer of heat from the hotter body to the colder body until they become the same temperature.

Thus, following this, heat will flow from the warm air to the cold currents.

4 0

2 years ago