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PilotLPTM [1.2K]
3 years ago
8

A sample of octane (c8h18 that has a mass of 0.750 g is burned in a bomb calorimeter. as a result, the temperature of the calori

meter increases from 21.0°c to 41.0°c. the specific heat of the calorimeter is 1.50 j/(g °c, and its mass is 1.00 kg. how much heat is released during the combustion of this sample? use .
Chemistry
2 answers:
timurjin [86]3 years ago
7 0
In a closed system, heat should be conserved which means that the heat produced in the calorimeter is equal to the heat released by the combustion reaction. We calculate as follows:

Heat of the combustion reaction = mC(T2-T1)
                                                          = 1 (1.50) (41-21)
                                                          = 30 kJ
GenaCL600 [577]3 years ago
3 0

Answer:

30 kJ is released during the combustion of this sample.

Explanation:

Heat released during combustion = Heat gained by calorimeter

Mass of calorimeter= m = 1.00 kg=1000 g

Change in temperature of the calorimeter = ΔT = 41.0 °C - 21.0 °C =20.0 °C

Specific heat if calorimeter = c = 1.50 J/g °C

Heat released during combustion = Q

Q=mc\Delta T

Q=1000.00 g\times 1.50 J/g ^oC\times 20.0 ^oC=30,000 =30 kJ

30 kJ is released during the combustion of this sample.

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Ammonium hydroxide is neutralized by sulfuric acid to produce ammonium sulfate and water. It will make ___ mol ammonium sulfate
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Answer:

1) Ammonium hydroxide is neutralized by sulfuric acid to produce ammonium sulfate and water. It will make 0.157 mol ammonium sulfate when you neutralize 11.00 g ammonium hydroxide.

2) 2NH₄OH + H₂SO₄ → (NH₄)₂SO₄ + 2H₂O.

Explanation:

  • Firstly, we should balance the equation of heptane combustion.
  • We can balance the equation by applying the conservation of mass to the equation.
  • The balanced equation is: <em>2NH₄OH + H₂SO₄ → (NH₄)₂SO₄ + 2H₂O.</em>
  • This means that every 2.0 moles of ammonium hydroxide (NH₄OH) will produce 1.0 mole of ammonium sulfate (NH₄)₂SO₄ when it is neutralized by sulfuric acid.
  • We need to calculate the no. of moles in 11.0 g of ammonium hydroxide that is neutralized using the relation: <em>n = mass/molar mass. </em>

n of 11.0 g of ammonium hydroxide (NH₄OH) = mass/molar mass = (11.0 g)/(35.04 g/mol) = 0.314 mol.

<u><em>Using cross multiplication: </em></u>

2.0 moles of NH₄OH make → 1.0 mole of (NH₄)₂SO₄.

0.314 mol of NH₄OH make → ??? moles of (NH₄)₂SO₄.

∴ The no. of moles of (NH₄)₂SO₄ that will be made from neutralizing (11.0 g) of NH₄OH = (0.314 mol)(1.0 mol)/(2.0 mol) = 0.157 mol.

<em>∴ Ammonium hydroxide is neutralized by sulfuric acid to produce ammonium sulfate and water. It will make </em><em>0.157</em><em> mol ammonium sulfate when you neutralize 11.00 g ammonium hydroxide.</em>

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At 741 torr and 44°C, 7.10 g of a gas occupy a volume of 5.40 L. What is the molar mass
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The molar mass of the gas is 35. glmol.

the molar mass of a chemical compound is described as the mass of a sample of that compound separated by the amount of substance in that sample, measured in moles.

The molar mass is a bulk, not molecular, effects of a substance.

<h3>What is molar mass and how is it calculated?</h3>

The molar mass is the mass of one mole of a sampling. To find the molar mass, count the atomic masses (atomic weights) of all of the atoms in the molecule. Find the atomic mass for each element by using the mass shown in the Periodic Table or table of atomic weights.

<h3>What is molar ?</h3>

Molar refers to the unit of concentration molarity, which is equivalent to the number of moles per liter of a solution. In chemistry, the word most often guides to molar concentration of a solute in a solution. Molar attention has the units mol/L or M.

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Write a balanced net ionic equation showing only the particles involved in the reaction and balancing the charges.Pb(s) AgNO3aq)
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Answer:

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Explanation:

Step 1: The unbalanced equation

Pb(s) + AgNO3aq) → Ag(s) + Pb(NO3)2(aq)

Step 2: Balancing the equation

Pb(s) + AgNO3aq) → Ag(s) + Pb(NO3)2(aq)

On the right side we have 2x NO3, on the left side we have 1x NO3.

To balance the amount of NO3 on both sides, we have to multiply AgNO3 (on the left side) by 2.

Pb(s) + 2AgNO3aq) → Ag(s) + Pb(NO3)2(aq)

On the left side we have 2x Ag, on the right side we have 1x Ag.

To balance the amount of Ag on both sides, we have to multiply Ag (on the right side) by 2. Now the equation is balanced.

Pb(s) + 2AgNO3(aq) → 2Ag(s) + Pb(NO3)2(aq)

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