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docker41 [41]
3 years ago
10

Ac sample of octane that has a mass of 0.750 g is burned in a bone calorimeter. As a result, the temperature if the calorimeter

increases from 21.0 °C to 41°C . The specific heatv of the calorimeter is 1.50 J/(g*°C), and it's nass is 1.0 kg. How much heat is released during the combustion
Chemistry
2 answers:
Vinvika [58]3 years ago
7 0
<span>How much heat is released during the combustion?

30.0 kJ

Hope this helps. 
</span>
Daniel [21]3 years ago
6 0
<span>From the specific heat of the calorimete which is 1.50 J/(g*°C), and it's mass is 1.0 kg, then the amount of energy  that is released during the combustion is </span>30.0 kJ. 
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When discussing discordant and harmonious sound waves, which statement is false?
notka56 [123]
When discussing discordant and harmonious sound waves, the statement tat is false is : if the original waves combine to form irregular displacement of air, the sound will be discordant. The fact is, irregular displacement of air does not guarantee sound discordance

hope this helps
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3 years ago
How much heat energy is needed to raise the temperature of 59.7g of cadmium from 25°C to 100°C? The specific heat of cadmium is
labwork [276]
     Using the Fundamental Equation of Calorimetry, we have:

Q=mc\Delta T \\ Q=59.7.0.231.(100-25) \\ \boxed {Q=1034.3025J}      

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7 0
3 years ago
51. The radius of gold is 144 pm and the density is 19.32 g/cm3. Does elemental gold have a face-centered cubic structure or a b
Serjik [45]

Answer:

Elemental gold to have a Face-centered cubic structure.

Explanation:

From the information given:

Radius of gold = 144 pm

Its density = 19.32 g/cm³

Assuming the structure is a face-centered cubic structure, we can determine the density of the crystal by using the following:

a = \sqrt{8} r

a = \sqrt{8} \times 144 pm

a = 407 pm

In a unit cell, Volume (V) = a³

V = (407 pm)³

V = 6.74 × 10⁷ pm³

V = 6.74 × 10⁻²³ cm³

Recall that:

Net no. of an atom in an FCC unit cell = 4

Thus;

density = \dfrac{mass}{volume}

density = \dfrac{ 4 \ atm ( 196.97 \ g/mol) (\dfrac{1 \ mol }{6.022 \times 10^{23} \ atoms})}{6.74 \times 10^{-23} \ cm^3}

density d = 19.41 g/cm³

Similarly; For a  body-centered cubic structure

r = \dfrac{\sqrt{3}}{4}a

where;

r = 144

144 = \dfrac{\sqrt{3}}{4}a

a = \dfrac{144 \times 4}{\sqrt{3}}

a = 332.56 pm

In a unit cell, Volume V = a³

V = (332.56 pm)³

V = 3.68 × 10⁷ pm³

V  3.68 × 10⁻²³   cm³

Recall that:

Net no. of atoms in BCC cell = 2

∴

density = \dfrac{mass}{volume}

density = \dfrac{ 2 \ atm ( 196.97 \ g/mol) (\dfrac{1 \ mol }{6.022 \times 10^{23} \ atoms})}{3.68 \times 10^{-23} \ cm^3}

density =17.78 g/cm³

From the two calculate densities, we will realize that the density in the face-centered cubic structure is closer to the given density.

This makes the elemental gold to have a Face-centered cubic structure.

3 0
3 years ago
PLEASE HELP WITH THIS QUESITION WILL GIVE BRAINLIST 2 BEST ANSWER.
Lera25 [3.4K]

Answer: show all

Explanation:

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3 years ago
Sulfur dioxide and oxygen react to form sulfur trioxide during one of the key steps in sulfuric acid synthesis. An industrial ch
Inessa [10]

Answer:

The answer is "\bold{4.97 \times 10^{-2}}"

Explanation:

Please find the complete question in the attached file.

Equation:

2SO_2+O_2  \leftrightharpoons 2SO_3

at t=0 3.3   \ \ \ \ \ \ \ \ \ \ 0.79

at equilibrium 3.3-p \ \ \ \ \ \ \ \ \ \ 0.79 - \frac{P}{2} \ \ \ \ \ \ \ \ \ \ \ \ P

p= 0.47 \ \ atm\\\\SO_2=3.3-0.47 = 2.83 \ \ atm\\\\O_2= 0.74 -\frac{0.47}{2}=0.74-0.235=0.555 \ atm\\\\K_P=\frac{[PSO_3]^2}{[PSO_2]^2[PO_2]}\\\\

     =\frac{0.47^2}{2.83^2\times 0.555}\\\\=4.97 \times 10^{-2}

8 0
3 years ago
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