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3 years ago

How much energy, in joules, is released when 17 kilograms of granite is cooled from 45 C to 21 C?

1 answer:

6 0

The final answer is -322, 320 Joules. The solution for the problem is:

The equation that must be used in this problem is:

U = mCp(Tf-Ti)

where:
U = energy released or absorbed in Joules
m = mass in kg
Cp = specific heat of material in J/kg-C
Tf = final temperature, C
Ti = initial temperature, C

Looking up the Cp of granite gives a value of 790 J/kg-C.

U = 17kg (790 J/kg-C) (21-45)
U = - 322, 320 Joules (negative means heat is released)

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Rolls of foil are 304 mmmm wide and 0.014 mmmm thick. (The density of foil is 2.7 g/cm3g/cm3 .) What maximum length of foil can

ratelena [41]

Answer:

92.2 m

Explanation:

Given that:=

The breadth = 304 mm

Height = 0.014 mm

Let Length = x mm

Volume = $Length\times breadth\times height$

Thus,

Volume = $304\times 0.014\times x\ mm^3=4.256x\ mm^3$

Also, 1 mm³ = 0.001 cm³

So, volume = 0.004256 cm³

Given that density = 2.7 g/cm³

Mass = 1.06 kg = 1060 g

So,

$Volume=\frac{Mass}{Density}=\frac{1060}{2.7}\ cm^3=392.59\ cm^3$

So,

0.004256*x = 392.59

x = 92243.89 mm

Length of foil = 92243.89 mm = 92.2 m

3 0

3 years ago

In the laboratory a general chemistry student finds that when 2.84 g of KClO4(s) are dissolved in 107.70 g of water, the tempera

vredina [299]

Answer : The enthalpy change of dissolution of $KClO_4$ is 54.3 kJ/mole

Explanation :

Heat released by the reaction = Heat absorbed by the calorimeter + Heat absorbed by the water

$q=[q_1+q_2]$

$q=[c_1\times \Delta T+m_2\times c_2\times \Delta T]$

where,

q = heat released by the reaction

$q_1$ = heat absorbed by the calorimeter

$q_2$ = heat absorbed by the water

$c_1$ = specific heat of calorimeter = $1.55J/^oC$

$c_2$ = specific heat of water = $4.184J/g^oC$

$m_2$ = mass of water = 107.70 g

$\Delta T$ = change in temperature = $T_2-T_1=(22.80-20.34)=2.46^oC$

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

$q=[(1.55J/^oC\times 2.46^oC)+(107.70g\times 4.184J/g^oC\times 2.46^oC)]$

$q=1112.3J=1.1123kJ$

Now we have to calculate the enthalpy change of dissolution of $KClO_4$

$\Delta H=\frac{q}{n}$

where,

$\Delta H$ = enthalpy change = ?

q = heat released = 1.1123 kJ

m = mass of $KClO_4$ = 2.84 g

Molar mass of $KClO_4$ = 138.55 g/mol

$\text{Moles of }KClO_4=\frac{\text{Mass of }KClO_4}{\text{Molar mass of }KClO_4}=\frac{2.84g}{138.55g/mole}=0.0205mole$

$\Delta H=\frac{1.1123kJ}{0.0205mole}=54.3kJ/mole$

Therefore, the enthalpy change of dissolution of $KClO_4$ is 54.3 kJ/mole

5 0

4 years ago

In the citric acid cycle, atp molecules are produced by _____.

IgorLugansk [536]

A glucose molecule is completely broken down to carbon dioxide and water in glycolysis and the citric acid cycle, but together these two<span> processes yield only a few molecules of ATP.</span>

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3 years ago

Fill in the blanks. A person would weight more on Neptune than on planet blank<br> because<br> blank

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A person would weigh more on Neptune than on planet earth because the gravitational field strength is stronger on Neptune.

4 0

3 years ago

A 3 mL sample of a liquid has a mass of 3 g. What is the density of this liquid?

ad-work [718]

Hey there!

Mass = 3 g

Volume = 3 mL

Density = mass / volume

D = 3 / 3

D = 1,0 g/mL

5 0

3 years ago

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