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3 years ago

What would be an appropriate way to display the results of the experiment outlined?

Chemistry

2 answers:

miv72 [106K]3 years ago

6 0

The best way (in my opinion) is to put it in a table or spreadsheet.

Elodia [21]3 years ago

3 0

Answer:

Use a table or spreasheet

Explanation:

This will depend of what kind of experiment you are doing, but in general terms, you can use a table or spreadsheet to display the results of any experiment.

Let's do an example, with a simple experiment. Let's suppose we are doing an experiment to determine the density of some objects, For this, we are going to use a rock, a ball and a little box.

The experiment states that in order to determine the density, you need to weight the objects, then calculate the volume, and finally calculate the density with the expression:

d = m/V

Now, let's suppose that you weight the 3 objects, and the values are:

rock: 150 g

ball: 300 g

box: 200 g

The volume for them are:

rock: 35 cm³

Ball: 45 cm³

Box: 64 cm³

Finally the density:

rock: 4.3 g/cm³

ball: 6.7 g/cm³

box: 3.1 g/cm³

These data, will be better understandable if you write a table like the one in picture 1. In that way, you show the data you have and the results of that experiment in a better way:

Download xlsx

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Answer:

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i think it's right

hope it helps :)

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A photon has a wavelength of 6.2 meters. Calculate the energy of the photon in joules. (Planck's constant is 6.626 × 10-34 joule

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3 years ago

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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 $S_{gen}=0$ .

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3 years ago

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