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3 years ago

Tommy was walking at a rate of 4

1 answer:

8 0

A positive acceleration indicates that the object sped up. This means that if you compare the first speed to the second, the second speed should be higher.

A negative acceleration indicates that the object has slowed down. This means that if you compare the first speed to the second, the second speed should be lower.

If an acceleration is 0, it means that it neither slowed down nor sped up.

Now let us analyze your problem by listing down the speed and the time:

At noon: 4 mi/hr
12:30 : 6 mi/hr
2:30 : 2 mi/hr

From noon to 12:30, you will notice that there is an increase in speed. This means that Tommy had a positive acceleration. (Rules out D.)

From 12:30 to 2:30, there is a decrease in speed. This would indicate that Tommy had a negative acceleration. (Rules out C.)

No speed was the same, so acceleration was never 0. (Rules out A.)

From the assumptions above, we can now deduce that the answer is B.

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How would you expect this object to behave if it starts out not moving? (the object is a circle, and the arrows show forces acti

Oduvanchick [21]

B. It would have a net force of 200N to the left. Meaning the the forces are unbalanced and the net force would cause it to accelerate (speed up) in the direction of the net force.

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3 years ago

Read 2 more answers

(b) Which solution has a higher percent ionization of the acid, a 0.10M solution of HC2H3O2(aq) or a 0.010M solution of HC2H3O2(

boyakko [2]

Answer: The percent ionization is higher for 0.010 M solution of $HC_2H_3O_2$

Explanation:

$HC_2H_3O_2\rightarrow H^+C_2H_3O_2^-$

cM 0M 0M

$c-c\alpha$ $c\alpha$ $c\alpha$

So dissociation constant will be:

$K_a=\frac{(c\alpha)^{2}}{c-c\alpha}$

a) Given c= 0.10 M and $\alpha$ = ?

$K_a=1.8\times 10^{-5}$

Putting in the values we get:

$1.8\times 10^{-5}=\frac{(0.10\times \alpha)^2}{(0.10-0.10\times \alpha)}$

$(\alpha)=0.013$

$\%(\alpha)=0.013\times 100=1.3\%$

b) Given c= 0.010 M and $\alpha$ = ?

$K_a=1.8\times 10^{-5}$

Putting in the values we get:

$1.8\times 10^{-5}=\frac{(0.010\times \alpha)^2}{(0.010-0.010\times \alpha)}$

$(\alpha)=0.041$

$\%(\alpha)=0.041\times 100=4.1\%$

The percent ionization is higher for 0.010 M solution of $HC_2H_3O_2$

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3 years ago

For the following reaction, 3.70 grams of oxygen gas are mixed with excess carbon monoxide . The reaction yields 8.25 grams of c

Elden [556K]

Answer:

81% Yield

Explanation:

2CO + O₂ => 2CO₂

Excess 3.70g O₂ => 8.25g CO₂ (actural yield)

(3.70g O₂)/(32g O₂/mol O₂)

= 0.1156 mol O₂ => 2(0.1156) mol CO₂

= 10.175g (Theoretical Yield)

%Yield = (Actual Yield / Theoretical Yield)100%

= (8.25g/10.175g)100% = 81% Yield

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How much does one mole of carbon weigh (round to two decimal places)? Make sure your answer has units.

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The number of atoms present in carbon, specifically Carbon-12, is the number of atoms present in 1 gram of the substance. One moles of the substance weighs approximately 12.01 g/mole. The unit suggests that every mole of carbon weighs 12.01 grams.

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