Examples of fossil fuels are

Question 1 of 10

Stolb23 [73]

Answer:

C. The conclusions tell why the data support or reject the hypothesis.

8 0

3 years ago

How many moles of ammonia, NH,, are contained in<br><br> 6.21 x 1024 molecules of ammonia?

ankoles [38]

Answer:

10.32 moles of ammonia NH₃

Explanation:

From the question given above, the following data were obtained:

Number of molecules = 6.21×10²⁴ molecules

Number of mole of NH₃ =?

The number of mole of NH₃ can be obtained as follow:

From Avogadro's hypothesis,

6.02×10²³ molecules = 1 mole

Therefore,

6.21×10²⁴ molecules = 6.21×10²⁴ / 6.02×10²³

6.21×10²⁴ molecules = 10.32 moles

Thus, 6.21×10²⁴ molecules contains 10.32 moles of ammonia NH₃

6 0

3 years ago

What is the simplest type of pure substance

Eduardwww [97]

The answer is: An atom

6 0

3 years ago

What formula gives the ratio of elements?

USPshnik [31]

Answer:

In a chemical formula, the elements in a compound are represented by their chemical symbols, and the ratio of different elements is represented by subscripts.

Explanation:

6 0

3 years ago

Given the following reaction: 2K3PO4 + AL2(CO3)3 = 3K2CO3 + 2ALPO4 If I perform this reaction with 150 g of potassium phosphate

Novosadov [1.4K]

Answer : The theoretical yield of potassium carbonate is, 146.483 g

The percent yield of potassium carbonate is, 85.33 %

Solution : Given,

Mass of $K_3PO_4$ = 150 g

Mass of $Al_2(CO_3)_3$ = 90 g

Molar mass of $K_3PO_4$ = 212.27 g/mole

Molar mass of $Al_2(CO_3)_3$ = 233.99 g/mole

Molar mass of $K_2CO_3$ = 138.205 g/mole

First we have to calculate the moles of $K_3PO_4$ and $Al_2(CO_3)_3$

$\text{ Moles of }K_3PO_4=\frac{\text{ Mass of }K_3PO_4}{\text{ Molar mass of }K_3PO_4}=\frac{150g}{212.27g/mole}=0.7066moles$

$\text{ Moles of }Al_2(CO_3)_3=\frac{\text{ Mass of }Al_2(CO_3)_3}{\text{ Molar mass of }Al_2(CO_3)_3}=\frac{90g}{233.99g/mole}=0.3846moles$

The given balanced reaction is,

$2K_3PO_4+Al_2(CO_3)_3\rightarrow 3K_2CO_3+2AlPO_4$

From the given reaction, we conclude that

2 moles of $K_3PO_4$ react with 1 mole of $Al_2(CO_3)_3$

0.7066 moles of $K_3PO_4$ react with $\frac{1}{2}\times 0.7066=0.3533$ moles of $Al_2(CO_3)_3$

But the moles of $Al_2(CO_3)_3$ is, 0.3846 moles.

So, $Al_2(CO_3)_3$ is an excess reagent and $K_3PO_4$ is a limiting reagent.

Now we have to calculate the moles of $K_2CO_3$ .

As, 2 moles of $K_3PO_4$ react to give 3 moles of $K_2CO_3$

So, 0.7066 moles of $K_3PO_4$ react to give $\frac{3}{2}\times 0.7066=1.0599$ moles of $K_2CO_3$

Now we have to calculate the mass of $K_2CO_3$ .

$\text{ Mass of }K_2CO_3=\text{ Moles of }K_2CO_3\times \text{ Molar mass of }K_2CO_3$

$\text{ Mass of }K_2CO_3=(1.0599moles)\times (138.205g/mole)=146.483g$

The theoretical yield of potassium carbonate = 146.483 g

The experimental yield of potassium carbonate = 125 g

Now we have to calculate the % yield of potassium carbonate.

Formula for percent yield :

$\% yield=\frac{\text{ Theoretical yield}}{\text{ Experimental yield}}\times 100$

$\% \text{ yield of }K_2CO_3=\frac{125g}{146.483g}\times 100=85.33\%$

Therefore, the % yield of potassium carbonate is, 85.33%

6 0

3 years ago