What happens to the amount of carbon dioxide as the population increases?

Identify which best describes the energy used to pluck guitar strings to make sound. A) electrity b) electromagnetic c) heat d)

DochEvi [55]

The capability to do work is said to be energy. It can neither be created nor be destroyed according to law of conservation of energy. Energy exist in different forms depends upon the work to be done such as potential energy, kinetic energy, mechanical energy, heat energy etc.

Now, when the guitarist plucks the guitar strings to make sound, he uses mechanical energy and when guitarist plucks the guitar strings guitar produces sound energy also.

Thus, option (d) is the correct answer.

6 0

4 years ago

Read 2 more answers

Reactions that absorb thermal energy are called blank reactions

Valentin [98]

Endothermic reactions

7 0

3 years ago

How is data not actually obtained from the experiment represented in a line graph?

vfiekz [6]

Answer:

with a double line

8 0

3 years ago

Please HELP! Use the following Equations to answer the problem: CH3OH + O2 —> CO2 + H2O

snow_lady [41]

Answer:

The answer to your question is below

Explanation:

1)

Balanced chemical reaction

2CH₃OH + 3O₂ ⇒ 2 CO₂ + 4H₂O

Reactant Element Product

2 C 2

8 H 8

8 O 8

Molar mass of CH₃OH = 2[12 + 16 + 4]

= 2[32]

= 64 g

Molar mass of O₂ = 3[16 x 2] = 96 g

Theoretical proportion CH₃OH/O₂ = 64 g/96g = 0.67

Experimental proportion CH₃OH/O₂ = 60/48 = 1.25

Conclusion

The limiting reactant is O₂ because the Experimental proportion was higher than the theoretical proportion

2)

Balanced chemical reaction

S₈ + 12O₂ ⇒ 8SO₃

Reactant Elements Products

8 S 8

24 O 24

Molar mass of S₈ = 32 x 8 = 256 g

Molar mass of O₂ = 12 x 32 = 384 g

Theoretical proportion S₈ / O₂ = 256 / 384

= 0.67

Experimental proportion S₈ / O₂ = 40 / 35

= 1.14

Conclusion

The limiting reactant is O₂ because the experimental proportion was lower than the theoretical proportion.

6 0

4 years ago

Nitrogen (N2) enters a well-insulated diffuser operating at steady state at 0.656 bar, 300 K with a velocity of 282 m/s. The inl

Tasya [4]

Answer:

The exit temperature of the Nitrogen would be 331.4 K.
The area at the exit of the diffuser would be $7*10^{-3} m^2$ .
The rate of entropy production would be 0.

Explanation:

First it is assumed that the diffuser works as a isentropic device. A isentropic device is such that the entropy at the inlet is equal that the entropy T the exit.
It will be used the subscript 1 for the inlet conditions of the nitrogen, and the subscript 2 for the exit conditions of the nitrogen.
It will be called: v the velocity of the nitrogen stream, T the nitrogen temperature, V the volumetric flow of the specific stream, A the area at the inlet or exit of the diffuser and, P the pressure of the nitrogen flow.
It is known that for a fluid flowing, its volumetric flow is obtain as: $V=v*A$ ,
Then for the inlet of the diffuser: $V_1=v_1*A_1=282\frac{m}{s}*4.8*10^{-3}m^2=1.35\frac{m^3}{s}$
For an ideal gas working in an isentropic process, it follows that: $\frac{T_{2} }{T_1}=(\frac{P_2}{P_1})^k$ where each variable is defined according with what was presented in step 2 and 3, and k is the heat values relationship, 1.4 for nitrogen.
Then solving for $T_2$ , the temperature of the nitrogen at the exit conditions: $T_2=T_1(\frac{P_2}{P_1})^k$ then, $T_2=300 K (\frac{0.9 bar}{0.656 bar})^{(\frac{1.4-1}{1.4})}=331.4 K$
Also, for an ideal gas working in an isentropic process, it follows that: $\frac{P_2}{P_1}= (\frac{V_1}{V_2})^k$ , where each variable is defined according with what was presented in step 2 and 3, and k is the heat values relationship, 1.4 for nitrogen.
Then solving for $V_2$ the volumetric flow at the exit of the diffuser: $V_2=V_1*\frac{1}{\sqrt[k]{\frac{P_2}{P_1}}}=\frac{1.35\frac{m^3}{s}}{\sqrt[1.4]{\frac{0.9bar}{0.656bar} }}=1.080\frac{m^3}{s}$ .
Knowing that $V_2=1.080\frac{m^3}{s}$ , it is possible to calculate the area at the exit of the diffuser, using the relationship presented in step 4, and solving for the required parameter: $A_2=\frac{V_2}{v_2}=\frac{1.08\frac{m^3}{s} }{140\frac{m}{s}}=7.71*10^{-3}m^2$ .
To determine the rate of entropy production in the diffuser, it is required to do a second law balance (entropy balance) in the control volume of the device. This balance is: $S_1+S_{gen}-S_2=\Delta S_{system}$ , where: $S_1$ and $S_2$ are the entropy of the stream entering and leaving the control volume respectively, $S_{gen}$ is the rate of entropy production and, $\Delta S_{system}$ is the change of entropy of the system.
If the diffuser is operating at stable state is assumed then $\Delta S_{system}=0$ . Applying the entropy balance and solving the rate of entropy generation: $S_{gen}=S_2-S_1$ .
Finally, it was assume that the process is isentropic, it is: $S_1=S_2$ , then $S_{gen}=0$ .

6 0

3 years ago