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3 years ago

HELPPPP I NEED THIS ANSWERED ASAPP!!!

1 answer:

4 0

This is because the animals with the favorable traits would survive. Therefore they will be the only ones left because they were able to adapt to the environment. And when they breed, the offspring will gain those favorable and the less favorable traits will continue to not be used which is why they “fade away” overtime.

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You are investigating an organic compound. you discover the following: it contains no hydrogen atoms. what property do you expec

Juli2301 [7.4K]

the answer is C because it has no water there will be nothing to boil and odor has nothing to do with hydrogen

I hope this helps! :)

-Ayden

5 0

3 years ago

Read 2 more answers

A 0.600 g sample of diprotic acid H2X is dissolved in water, and titrated to the equivalence point with 40.0 mL of 0.200 M NaOH.

9966 [12]

Answer:

150 g/mol

Explanation:

Let's consider the complete neutralization of a diprotic acid H₂X with NaOH.

H₂X + 2 NaOH → Na₂X + 2 H₂O

40.0 mL of 0.200 M NaOH. were required to reach the endpoint. The reacting moles of NaOH are:

0.0400 L × 0.200 mol/L = 8.00 × 10⁻³ mol

The molar ratio of H₂X to NaOH is 1:2. The reacting moles of H₂X are 1/2 × 8.00 × 10⁻³ mol = 4.00 × 10⁻³ mol.

4.00 × 10⁻³ moles of H₂X have a mass of 0.600 g. The molar mass of H₂X is:

0.600 g/4.00 × 10⁻³ mol = 150 g/mol

8 0

3 years ago

A car with constant speed travelso 150 km in 7200 seconds what is the speed of the car

liberstina [14]

7200sec÷60sec in a minute÷60minutes in an hour=2hours. 150km÷ 2 hours= 75 km/hour

4 0

3 years ago

Burning 10g of propane produces twice as much carbon dioxide as burning 5g of propane what law does it apply?

Marysya12 [62]

This corresponds to the law of multiple proportions which states that when a certain amount of compound reacts to form another compound, the ratio of the amounts between the reactants and the products are a fraction of simple or whole numbers.

7 0

3 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