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Leto [7]
3 years ago
5

Given the balanced equation:

Chemistry
1 answer:
Liula [17]3 years ago
8 0

Answer:

In our case, the equation  2KI + F2 → 2 KI + I2 is a single-replacement (also called single- displacement) reaction.

Explanation:

<u>Step 1:</u> The balanced equation:

2KI + F2 → 2 KI + I2

<u>Step 2</u>: What kind of reaction

Since we start with 2 reactants and end with 2 products, we can say it won't be a synthesis or decomposition reaction.

In our case, It looks like it's a replacement reaction.

- A single-replacement reaction is a reaction in which one element replaces a similar element in a compound. The general form of a single-replacement reaction is:  A+BC→AC+B

where element  A  is a metal and replaces element  B , also a metal, in the compound.

When the element that is doing the replacing is a nonmetal, it must replace another nonmetal in a compound.

- A double-replacement reaction is a reaction in which the positive and negative ions of two ionic compounds exchange places to form two new compounds. The general form of a double-replacement reaction is:  AB+CD→AD+BC

In our case, the equation  2KI + F2 → 2 KI + I2 is a <u><em>single-replacement</em></u> (also called single- displacement) reaction.

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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.

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What two structures would provide a positive identification of a plant cell under a microscope? a)cell wall, mitochondria b)plas
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A gas at constant temperature has a pressure of 404.6 kPa with a volume of 12 ml. If the volume changes to 43ml, what is the new
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Answer:

The answer is

<h2>112.912 kPa</h2>

Explanation:

The new pressure can be found by using the formula for Boyle's law which is

P_1V_1 = P_2V_2

Since we are finding the new pressure

P_2 =  \frac{P_1V_1}{V_2}  \\

404.6 kPa = 404600 Pa

From the question we have

P_2 =  \frac{404600 \times 12}{43}  =  \frac{4855200}{43}  \\  = 112911.6279... \\  = 112912

We have the final answer as

<h3>112.912 kPa</h3>

Hope this helps you

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