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joja [24]
3 years ago
15

Lara’s chemistry teacher has a decorative metal cube on her desk. The underside of the cube indicates that is made of stainless

steel. However, Lara thinks it is too heavy to be this metal. She researches the density of stainless steel and finds it is 7.85g/cm³. She carries out this experiment to verify if the cube is actually made of stainless steel.
Chemistry
1 answer:
Yuri [45]3 years ago
7 0

Answer:

D: 4, 5

Explanation:

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Which of the following substances is formed deep under earth
EastWind [94]

Answer:

Oil

Explanation:

everything else is above the earth

6 0
2 years ago
What type of energy is this?
Firlakuza [10]
2 correct answers: Chemical and neither endothermic or exothermic
6 0
4 years ago
A silver coin with a mass of 14.1g has a density of 10.5 grams
Angelina_Jolie [31]

Answer:

1.34 cubic centimeter

Explanation:

v = m/d

5 0
3 years ago
3.10 mol of an ideal gas with CV,m=3R/2 is expanded adiabatically against a constant external pressure of 1.00 bar. The initial
Otrada [13]

Answer : The value of q, w, ΔU and ΔH for the process is, 0 J, -7.78 kJ, -7.78 kJ and -12.97 kJ respectively.

Explanation :

First we have to calculate the final temperature of the gas.

Using poission equation:

(\frac{P_1}{P_2})^{1-\gamma}=(\frac{T_2}{T_1})^{\gamma}

where,

P_1 = initial pressure = 14.7 bar

P_2 = final pressure = 1.00 bar

T_1 = initial temperature = 305 K

T_2 = final temperature = ?

\gamma=\frac{R}{C_v}+1=frac{R}{(\frac{3R}{2})}+1=\frac{5}{3}=1.67

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

(\frac{14.7}{1.00})^{1-1.67}=(\frac{T_2}{305})^{1.67}

T_2=103.7K

Now we have to calculate the q, w, ΔU and ΔH for the process.

As we know that, at adiabatic process the value of q=0. So,

dq=dU-w\\\\0=dU-w\\\\dU=w\\\\w=dU=n\times C_v\times (T_2-T_1)

w=dU=3.10mol\times \frac{3R}{2}\times (103.7-305)K

w=dU=3.10mol\times \frac{3\times 8.314J/mol.K}{2}\times (103.7-305)K

w=dU=-7782.278J=-7.78kJ

Now we have to calculate the value of C_p

C_p-C_v=R\\\\C_p=R+C_v\\\\C_p=R+\frac{3R}{2}\\\\C_p=\frac{5R}{2}

Now we have to calculate the value of ΔH.

dH=n\times C_p\times (T_2-T_1)

dH=3.10mol\times \frac{5R}{2}\times (103.7-305)K

dH=3.10mol\times \frac{5\times 8.314J/mol.K}{2}\times (103.7-305)K

dH=-12970.5J=-12.97kJ

Thus, the value of q, w, ΔU and ΔH for the process is, 0 J, -7.78 kJ, -7.78 kJ and -12.97 kJ respectively.

4 0
3 years ago
How are insects different from arachnids?
aalyn [17]

Arachnids lack antennae. The main differences between insects and arachnids are in their body structure and legs.

<h3>What are insects?</h3>

Both insects and arachnids are multi-legged creatures that tend to elicit fear, and while they have a lot in common, including that fear response, there are some distinct differences between the two.

The main differences between insects and arachnids are in their body structure and legs.

Arachnids have eight legs, whereas insects only have six. Arachnids have simple eyes, whereas insects have compound eyes. Arachnids lack antennae while insects do.

Therefore, Arachnids lack antennae. The main differences between insects and arachnids are in their body structure and legs.

To learn more about insects, refer to the link:

brainly.com/question/28174759

#SPJ1

8 0
2 years ago
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