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tekilochka [14]
3 years ago
13

If the heat of combustion for a specific compound is -1430.0 kj/mol and its molar mass is 59.25 g/mol, how many grams of this co

mpound must you burn to release 409.20 kj of heat?
Chemistry
1 answer:
Rina8888 [55]3 years ago
8 0
To be able to answer this item, one needs to use the proper dimensional analysis such that we arrive to the same unit of measures as given in the item. If we let x be the amount, in grams, of the certain substance then, we may use the equation,

                  409.20 kJ = (x g)(1 mol/59.25 g)(1430 kJ/mol)

The value of x from the equation is 16.95 g.

Answer: 16.95 g
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If 10 grams of potassium chlorate decompose to form potassium chloride and oxygen gas inside a 500 mL container at a temperature
Gekata [30.6K]

Answer:

10 atm

Explanation:

There's a lot to do here, but lets take it one step at a time. First, let's write a balanced equation for the decomposition of potassium chlorate into potassium chloride and oxgyen gas.

2 KClO3 → 2 KCl + 3 O2

Now let's find the moles of the KClO3 (molar mass 122.55 g/mol) that we have take 10 g/122.55 g/mol, grams will cancel and we are left with 0.0816 moles. lets divide that by two since we have a two in front of the KClO3 in the equation, and then multiply that number by 5 since it's the total moles of products, in summary, multiply by 5/2 to get 0.204 moles.

Now that we know the moles of our products, let's plug some stuff into the ideal gas law PV = nRT. We are looking for P so let's solve for that. P = (nRT)/V, now let's plug in our values. Make sure V is converted to liters so 0.5 L. And convert celcius to kelvin by adding 273

P = ((0.204 moles)(318 K)(0.08206 L atm mol^-1 K^-1))/0.5 L

A lot of units cancel, and we get about 10.65 atm, if you don't want the answer in atm, you can find a conversion equation. But let's round to sig figs for now, which will bring us to 10 atm.

3 0
2 years ago
A 1.775g sample mixture of potassium hydrogen carbonate is decomposed by heating. if the mass loss is 0.275g what is the percent
Marina86 [1]

A 1.775g sample mixture of KHCO₃ is decomposed by heating. if the mass loss is 0.275g, the mass percentage of KHCO₃ is 70.4%.

<h3>What is a decomposition reaction?</h3>

A decomposition reaction can be defined as a chemical reaction in which one reactant breaks down into two or more products.

  • Step 1: Write the balanced equation for the decomposition of KHCO₃.

2 KHCO₃(s) → K₂CO₃(s) + CO₂(g) + H₂O(l)

The mass loss of 0.275 g is due to the gaseous CO₂ that escapes the sample.

  • Step 2: Calculate the mass of KHCO₃ that formed 0.275 g of CO₂.

In the balanced equation, the mass ratio of KHCO₃ to CO₂ is 200.24:44.01.

0.275 g CO₂ × 200.24 g KHCO₃/44.01 g CO₂ = 1.25 g KHCO₃

  • Step 3: Calculate the mass percentage of KHCO₃ in the sample.

There are 1.25 g of KHCO₃ in the 1.775 g sample.

%KHCO₃ = 1.25 g/1.775 g × 100% = 70.4%

A 1.775g sample mixture of KHCO₃ is decomposed by heating. if the mass loss is 0.275g, the mass percentage of KHCO₃ is 70.4%.

Learn more about decomposition reactions here: brainly.com/question/14219426

7 0
2 years ago
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algol [13]
Magma is made from sediments 

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3 years ago
What is the molarity of a solution of 14.0 g NH4Br in enough H2O to make 150 mL of solution?
kiruha [24]
The molarity of a solution equals to the mole number of the solute/the volume of the solution. For NH4Br, we know that the mole mass is 98. So the molarity is (14/98) mol /0.15 L=0.95 mol/L.
8 0
3 years ago
Read 2 more answers
Write down five laboratory safety rules.
OLEGan [10]

Answer:

Five Laboratory Safety Rules:

1). Do not eat in the laboratory.

2). Do not touch any chemical or reagent unless you are told to do so.

3). Neither play in lab, nor sit on the table.

4). Don't remove labels on any reagent.

5). Don't taste anything in the laboratory, no matter how familiar it appears.

Hope it helps.

5 0
3 years ago
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