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Paul [167]
2 years ago
15

A student investigates the motion of a toy vehicle. The student graphs 40 seconds of data from the investigation.

Chemistry
1 answer:
Stolb23 [73]2 years ago
8 0

Although we do not have the graph needed to answer this question, we can confirm that if the toy were to be accelerating, one should see this on the graph as a curved line.

<h3>How can we know if the toy was accelerating?</h3>

When graphing motion in terms of distance over time, we can use the shape of the line in the graph to determine if the speed is constant or if there is an acceleration present. If the line is straight, the speed is constant, but if there is a curve, there will be acceleration.

Therefore, we can confirm that if the toy were to be accelerating, one should see this on the graph as a curved line.

To learn more about motion visit:
brainly.com/question/11049671?referrer=searchResults

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A 5.5 g sample of a substance contains only carbon and oxygen. Carbon makes up 35% of the mass of the substance. The rest is mad
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We have been given the condition that carbon makes up 35% of the mass of the substance and the rest is made up of oxygen. With this, it can be concluded that 65% of the substance is made up of oxygen. If we let x be the mass of oxygen in the substance, the operation that would best represent the scenario is,

<span>                                       x = (0.65)(5.5 g)</span>

<span>                                       <em> </em><span><em>x = 3.575 g</em></span></span>

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If 0.580 moles of a monoprotic weak acid (ka = 7.4 10-5 is titrated with naoh, what is the ph of the solution at the half-equiva
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Match the chemical with the statement that best describes the observed change in the rate of the reaction when the concentration
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Answer:

Explanation:

When the amount of H2O2 is doubled while KI is kept constant, the rate of reaction doubles.

When the amount of KI is doubled and the amount of H2O2 is halved, the rate stays nearly constant.

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H2O2 + I– -> IO– + H2O (Step 1)

H2O2 + IO– -> I– + H2O + O2 (Step 2)

It can be seen that the iodine ion (provided by the KI solution) is a product as well as a reactant.

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8 0
3 years ago
5. What concentration of acid must be added to change the pH of 1 mM phosphate buffer from 7.4 to 7.3 (pKas of the phosphate buf
mr_godi [17]

Explanation:

According to the Henderson-Hasselbalch equation, the relation between pH and pK_{a} is as follows.

               pH = pK_{a} + log \frac{base}{acid}

where,     pH = 7.4 and pK_{a} = 7.21

As here, we can use the pK_{a} nearest to the desired pH.

So,      7.4 = 7.21 + log \frac{base}{acid}

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1 mM phosphate buffer means [HPO_{4}] + [H_{2}PO_{4}] = 1 mM

Therefore, the two equations will be as follows.

           \frac{HPO_{4}}{H_{2}PO_{4}} = 1.55 ............. (1)

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Now, putting the value of [HPO_{4}] from equation (1) into equation (2) as follows.

             1.55[H_{2}PO_{4}] + [tex][H_{2}PO_{4}] = 1 mM

                        2.55 [H_{2}PO_{4}] = 1 mM

                             [H_{2}PO_{4}] = 0.392 mM

Putting the value of [H_{2}PO_{4}] in equation (1) we get the following.

                     0.392 mM + [HPO_{4}] = 1 mM

                          [HPO_{4}] = (1 - 0.392) mM

                              [HPO_{4}] = 0.608 mM

Thus, we can conclude that concentration of the acid must be 0.608 mM.

7 0
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