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anygoal [31]
3 years ago
13

What is the answer to this question

Chemistry
2 answers:
kumpel [21]3 years ago
8 0
Second one only the sons
Korolek [52]3 years ago
3 0

2

Explanation:

50/50 chance of getting correct

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Which could describe the motion of an object? Select 3 correct answers. distance = 10 meters and displacement = 1 meter distance
Anit [1.1K]

Answer:

The three correct statements which could describe the motion of an object are:

<u><em /></u>

  • <u><em>distance = 10 meters and displacement = 1 meter </em></u>
  • <u><em>distance = 10 meters and displacement = 0 </em></u>
  • <u><em>distance = 10 meters and displacement = 10 meters</em></u>

Explanation:

<em>Motion</em> is the change of position of an object along any path. As such, the path can be rectilinear or along a curve whose direction is variable.

Two important characteristics used to describe a motion are the length of the path traveled and the final distance from the starting point.

The length of the path traveled is what distance describes. The final distance from the starting point to the final point is what displacement describes.

The maginitude of the displacement is the difference between the final position and the starting position.

When the path traveled is rectilinear, the distance and the magnitude of the displacement are the same.

When the path is not rectilinear, the distance (the length of the path) is always greater than the magnitude of the displacement.

Thus, as for the correct statements, you have:

  • <u>distance = 10 meters and displacement = 1 meter </u>

        This means that the object traveled along a 10 meters path, but since it was not rectilinear, the difference between the final position and the starting position is just 1 meter.

  • <u>distance = 10 meters and displacement = 0 </u>

<u />

       This means that the object traveled 10 meters but it ended back in the very same place from which it departed (a complete turn around a table, for example).

  • <u>distance = 10 meters and displacement = 10 meters</u>

        This means that the travel was in the trip was alogn a straight line, and the object traveled 10 meters and ended 10 meters away from the starting place.

As for the two incorrect statements:

  • <u>distance = 0 and displacement = 10 meters </u>

  • <u>distance = 1 meter and displacement = 10 meters</u>

It is not possible that the object ended up 10 meters away from where it was (the displacement), by traveling just 1 meter less than that.

8 0
3 years ago
Read 2 more answers
Sodium (NA) and potassium (K) are in the same group in the periodic table. Sodium is in the 11th position. How many valence elec
garri49 [273]
C= 1 Valence electron
3 0
3 years ago
Read 2 more answers
1. The Yerkes-Dodson Law says
goblinko [34]

Answer:

The Yerkes-Dodson Law suggests that there is a relationship between performance and arousal. Increased arousal can help improve performance, but only up to a certain point.

Explanation:

4 0
3 years ago
the combustion of octane is given in this reaction 2C8H18+2502-----16CO2+18H2 O, identify the limit reagent for this reaction wi
Archy [21]

Answer:

Liming reagent in the given reaction is oxygen.

Explanation:

Mass of octane = 12.85 g

Mass of oxygen = 7.46 g

Molar mass of octane = 114.23 g/mol

Molar mass of oxygen = 32

Mole=\frac{Mass\;in\;g}{Molar\;mass}

No.\;of\;moles\;of\;octane=\frac{12.85}{114.23} = 0.1125\;mol

No.\;of\;moles\;of\;oxygen=\frac{7.46}{32} = 0.2331\;mol

2C_8H_{18} + 25O_2 \rightarrow 16CO_2 +18H_2O

According to the reaction, 2 moles of octane requires 25 moles of O2

so, 0.1125 moles of octane requires 0.1125\times \frac{25}{2}  = 1.40625\;mol of oxygen.

but only 0.2331 mol of oxygen is present.

Hence, oxygen is the limiting reagent

4 0
3 years ago
At a certain temperature, 0.900 mol SO 3 is placed in a 2.00 L container. 2 SO 3 ( g ) − ⇀ ↽ − 2 SO 2 ( g ) + O 2 ( g ) At equil
puteri [66]

Answer: The value of K_c is 0.0057

Explanation:

Initial moles of  SO_3 = 0.900 mole

Volume of container = 2.00 L

Initial concentration of SO_3=\frac{moles}{volume}=\frac{0.900moles}{2.00L}=0.450M  

equilibrium concentration of O_2=\frac{moles}{volume}=\frac{0.110mole}{2.00L}=0.055M [/tex]

The given balanced equilibrium reaction is,

                            2SO_3(g)\rightleftharpoons 2SO_2(g)+O_2(g)

Initial conc.              0.450 M               0        0

At eqm. conc.    (0.450 -2x) M         (2x) M   (x) M

The expression for equilibrium constant for this reaction will be,

K_c=\frac{[O_2][SO_2]^2}{[SO_3]^2}

K_c=\frac{x\times (2x)^2}{0.450-2x)^2}

we are given : x = 0.055

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

K_c=\frac{0.055\times (2\times 0.055)^2}{0.450-2\times 0.055)^2}

K_c=0.0057

Thus the value of the equilibrium constant is 0.0057

3 0
4 years ago
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