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2 years ago

What is the guiding formula for combustion reactions

Chemistry

1 answer:

IgorLugansk [536]2 years ago

8 0

Answer:

Explanation: so a combustion occurs when we react a substance with oxygen .you commonly call this "burning'' .therefore combustion will always include oxygen in the equation and the product will include carbon dioxide and give off water vapours

Therefore the general equation for a complete combustion reaction would be

Fuel + O2 ------ CO2 + H2O

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How many moles are equal to 1.4 x 10^24 atoms of aluminum

Svetlanka [38]

Answer: 2.3 moles

Explanation:

Recall that based on Avogadro's law, 1 mole of any substance has 6.02 x 10^23 atoms

So if 1 mole of Aluminum = 6.02 x 10^23 atoms

Then, Z moles = 1.4 x 10^24 atoms

To get the value of Z, we cross multiply:

1 mole x 1.4 x 10^24 atoms = Z x (6.02 x 10^23 atoms)

1.4 x 10^24 atoms = Z x (6.02 x 10^23)

Hence, Z = (1.4 x 10^24 atoms) ➗ (6.02 x 10^23 atoms)

Z =2.3 moles

Thus, there are 2.3 moles in 1.4 x 10^24 atoms of aluminum.

4 0

3 years ago

Find the molarity of the following solutions: 0.5 moles of sodium chloride is dissolved to make 0.05 liters of solution.

vredina [299]

M(molarity)=moles/volume
M=.5mol/.05L=10M

4 0

2 years ago

What is the pOH of a solution of HNO3 that has [OH-] = 9.50 10-9 M?

bekas [8.4K]

Answer:

The pOH of HNO₃ solution that ha OH⁻ concentration 9.50 ×10⁻⁹M is 8.

Explanation:

Given data:

[OH⁻] = 9.50 ×10⁻⁹M

pOH = ?

Solution:

pOH = -log[OH⁻]

Now we will put the value of OH⁻ concentration.

pOH = -log[9.50 ×10⁻⁹M]

pOH = 8

Thus the pOH of HNO₃ solution that ha OH⁻ concentration 9.50 ×10⁻⁹M is 8.

6 0

2 years ago

Read 2 more answers

During the chemical reaction given below 21.71 grams of each reagent were allowed to react. Determine how many grams of the exce

swat32

Answer: 16.32 g of $O_2$ as excess reagent are left.

Explanation:

To calculate the moles :

$\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}$

$\text{Moles of} SO_2=\frac{21.71g}{64g/mol}=0.34mol$

$\text{Moles of} O_2=\frac{21.71g}{32g/mol}=0.68mol$

$2SO_2(g)+O_2(g)\rightarrow 2SO_3(g)$

According to stoichiometry :

2 moles of $SO_2$ require = 1 mole of $O_2$

Thus 0.34 moles of $SO_2$ will require= $\frac{1}{2}\times 0.34=0.17moles$ of $O_2$

Thus $SO_2$ is the limiting reagent as it limits the formation of product and $O_2$ is the excess reagent.

Moles of $O_2$ left = (0.68-0.17) mol = 0.51 mol

Mass of $O_2=moles\times {\text {Molar mass}}=0.51moles\times 32g/mol=16.32g$

Thus 16.32 g of $O_2$ as excess reagent are left.

3 0

2 years ago

density of a piece of ebony wood is 1.174 g/ml. what is the volume in quarts of a 2.1549 lb piece of this ebony wood?

lapo4ka [179]

0.88qt

Explanation:

Given parameters:

Density of ebony wood = 1.174g/ml

Mass of wood = 2.1549lb

Unknown:

Volume of the wood in quarts = ?

Solution:

Density is defined as the mass per unit volume of a substance.

Mass is the quantity of matter the wood contains.

Volume is the space it occupies.

To solve this problem, convert the mass form pounds into grams and plug into the density equation to find volume.

Then convert the volume from mL to quarts.

Density = $\frac{mass}{volume}$

Volume = $\frac{mass}{density}$

1 pound = 453.592g

2.1549 lb = 2.1549 x 453.592 = 977.5g

Volume = $\frac{977.5}{1.174}$ = 832.6mL

1 quart = 946.353mL

Therefore 832.6mL x $\frac{1qt}{946.353}$ = 0.88qt

learn more:

Density brainly.com/question/8441651

#learnwithBrainly

7 0

3 years ago

What is the guiding formula for combustion reactions​

What is the guiding formula for combustion reactions