After 4 half-lives have elapsed, the amount of a radioactive sample which has not decayed is

When octane (C8H18) is burned in the presence of oxygen, the yield of products (carbon dioxide and water) is 87%. What mass of c

ahrayia [7]

Answer:

14.5g of CO₂ are produced

Explanation:

The reaction of octane with oxygen is:

C₈H₁₈ + 25/2O₂ → 8CO₂ + 9H₂O

<em>Where 1 mole of octane (Molar mass: 114.23g/mol) reacts with 25/2 moles of O₂ (Molar mass 32g/mol) to produce 8 moles of CO₂ and 9 moles of water.</em>

When 21.0 g of octane is burned with 19.0 g of oxygen gas you need to find <em>limiting reactant </em>to find how many moles of products are formed:

Octane: 21.0g ₓ (1mol / 114.23g) = 0.184 moles octane

Oxygen: 19.0g ₓ (1 mol / 32g) = 0.594 moles oxygen

For a complete reaction of 0.184 moles of octane you will need:

0.184 moles C₈H₁₈ ₓ (25/2 moles O₂ / 1 mole C₈H₁₈) = <em>2.3 moles of oxygen</em>

As you have just 0.594 moles of oxygen, <em>Oxygen is limiting reactant.</em>

Based on chemical equation, 25/2 of O₂ produce 8 moles of CO₂, that means theoretical yield of CO₂ with 0.594 moles of O₂ is:

0.594 moles O₂ ₓ (8 moles CO₂ / 25/2 moles O₂) = 0.380 moles of CO₂

But, as yield of products is 87%, moles produced of CO₂ are:

0.380 moles of CO₂ ₓ 87% = 0.331 moles CO₂ are produced.

As molar mass of CO₂ is 44g/mol, mass of CO₂ in 0.331 moles is:

0.331 moles CO₂ ₓ (44g / mol) =

<h3>14.5g of CO₂ are produced</h3>

6 0

3 years ago

A 99.8 mL sample of a solution that is 12.0% KI by mass (d: 1.093 g/mL) is added to 96.7 mL of another solution that is 14.0% Pb

andre [41]

Answer:

$m_{PbI_2}=18.2gPbI_2$

Explanation:

Hello,

In this case, we write the reaction again:

$Pb(NO_3)_2(aq) + 2 KI(aq)\rightarrow PbI_2(s) + 2 KNO_3(aq)$

In such a way, the first thing we do is to compute the reacting moles of lead (II) nitrate and potassium iodide, by using the concentration, volumes, densities and molar masses, 331.2 g/mol and 166.0 g/mol respectively:

$n_{Pb(NO_3)_2}=\frac{0.14gPb(NO_3)_2}{1g\ sln}*\frac{1molPb(NO_3)_2}{331.2gPb(NO_3)_2} *\frac{1.134g\ sln}{1mL\ sln} *96.7mL\ sln\\\\n_{Pb(NO_3)_2}=0.04635molPb(NO_3)_2\\\\n_{KI}=\frac{0.12gKI}{1g\ sln}*\frac{1molKI}{166.0gKI} *\frac{1.093g\ sln}{1mL\ sln} *99.8mL\ sln\\\\n_{KI}=0.07885molKI$

Next, as lead (II) nitrate and potassium iodide are in a 1:2 molar ratio, 0.04635 mol of lead (II) nitrate will completely react with the following moles of potassium nitrate:

$0.04635molPb(NO_3)_2*\frac{2molKI}{1molPb(NO_3)_2} =0.0927molKI$

But we only have 0.07885 moles, for that reason KI is the limiting reactant, so we compute the yielded grams of lead (II) iodide, whose molar mass is 461.01 g/mol, by using their 2:1 molar ratio:

$m_{PbI_2}=0.07885molKI*\frac{1molPbI_2}{2molKI} *\frac{461.01gPbI_2}{1molPbI_2} \\\\m_{PbI_2}=18.2gPbI_2$

Best regards.

5 0

3 years ago

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At sea level, a kilogram weighs approximately:

Marysya12 [62]

The answer would be b, or 2

7 0

3 years ago

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Indicate how you might prepare acetic acid from ethene

ruslelena [56]

Acetic acid can be prepared from ethane by oxidizing it. Acetic acid is generally a carboxylic acid. These acids can be made by the oxidation of alcohols using potassium dichromate(VI) solution with sulfuric acid. Ethane is made into acetic acid in two steps. First it is formed into an aldehyde and then to acetic acid.

8 0

3 years ago

How does water affect the growth of a plant?

irina1246 [14]

Answer:

Water transports important nutrients throughout the plant.

Explanation:

Much like blood in an animal, water is necessary for a plants survival, as it transports important nutrients from the soil all through the plant's stems and leaves.

3 0

3 years ago

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