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julsineya [31]
2 years ago
12

A sample of hexane (C6H14) has a mass of 0.580 g. The sample is burned in a bomb calorimeter that has a mass of 1.900 kg and a s

pecific heat of 3.21 J/giK. What amount of heat is produced during the combustion of hexane if the temperature of the calorimeter increases by 4.542 K?
Chemistry
2 answers:
Mama L [17]2 years ago
7 0
Use the formula, Q= mcT

Q= heat
m= mass= 1.900Kg= 1.900 x 10^3 grams
c= specific heat= 3.21 
T= 4.542 K

Q= (1.900 x10^3g)(3.21)(4.542K)= 14.6 Joules.
OleMash [197]2 years ago
6 0

Answer : The amount of heat produced during the combustion of hexane per mole is 4109.79 kJ/mole.

Explanation :

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

Formula used :

Q=m\times c\times \Delta T

where,

Q = heat absorb = ?

m = mass of calorimeter = 1.900 kg  = 1900 g

c = specific heat of calorimeter = 3.21J/g.K

\Delta T = change in temperature = 4.542 K

Now put all the given value in the above formula, we get:

Q=1900g\times 3.21J/g.K\times 4.542K

Q=27701.658J=27.70kJ

Now we have to calculate the moles of hexane.

\text{Moles of }C_6H_{14}=\frac{Mass of }C_6H_{14}}{Molar mass of }C_6H_{14}=\frac{0.580g}{86.18g/mole}=0.00674mole

Now we have to calculate the heat produced during the combustion of hexane per mole.

\text{Heat produced}=\frac{27.70kJ}{0.00674mole}=4109.79kJ/mole

Therefore, the amount of heat produced during the combustion of hexane per mole is 4109.79 kJ/mole.

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Aliun [14]

Answer:

A) 3.59 cm

Explanation:

Given that :-

The density of the gold ingot = 19.31\ g/cm^3

Given that:- Mass = 5.50 lbs

Also, considering the conversion of lbs to g as shown below:-

1 lb = 453.592 g

Thus,

Mass = 5.50\times 453.592\ g = 2494.756 g

The volume = Length*Breadth*Height

Given that:- Length = 12.0 cm , Breadth = 3.00 cm

Considering the expression for density as:-

Density=\frac{Mass}{Volume}

19.31=\frac{2494.756}{12.0\times 3.00\times Height}

Solving for height, we get that:-

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san4es73 [151]

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Albus Dumbledore provides his students with a sample of 19.3 g of sodium sulfate. How many oxygen atoms are in this sample
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Answer:

<em>3.27·10²³ atoms of O</em>

Explanation:

To figure out the amount of oxygen atoms in this sample, we must first evaluate the sample.

The chemical formula for sodium sulfate is <em>Na₂SO₄, </em>and its molar mass is approximately 142.05\frac{g}{mol}.

We will use stoichiometry to convert from our mass of <em>Na₂SO₄ </em>to moles of <em>Na₂SO₄</em>, and then from moles of <em>Na₂SO₄ </em>to moles of <em>O </em>using the mole ratio; then finally, we will convert from moles of <em>O </em>to atoms of <em>O </em>using Avogadro's constant.

19.3g <em>Na₂SO₄</em> · \frac{1 mol Na^2SO^4}{142.05g Na^2SO^4} · \frac{4 mol O}{1 mol Na^2SO^4} ·\frac{6.022x10^2^3}{1 mol O}

After doing the math for this dimensional analysis, you should get a quantity of approximately <em>3.27·10²³ atoms of O</em>.

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Answer:

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From the above reaction, we have 2 moles of KOH combining with 1 mole of H₂SO₄ to produce 1 mole of K₂SO₄  and 2 moles of H₂O.

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