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Rufina [12.5K]
2 years ago
5

A soccer ball is traveling at a velocity of 50 m/s. The kinetic energy of the ball is 500 J. What is the mass of the soccer ball

? (Formula: )
0.1 kg
0.2 kg
0.4 kg
0.5 kg
Chemistry
2 answers:
Arlecino [84]2 years ago
8 0

Answer : The mass of the soccer ball is, 0.4 Kg

Explanation : Given,

Kinetic energy  of the ball = 500 J

Velocity of the ball = 50 m/s

Formula used :

K.E=\frac{1}{2}\times m\times v^2

where,

K.E = kinetic energy  of the ball

m = mass of the soccer ball

v = velocity of the ball

Now put all the given values in this formula, we get  mass of the soccer ball.

500J=\frac{1}{2}\times m\times (50m/s)^2

m=0.4Kg

Therefore, the mass of the soccer ball is, 0.4 Kg

Lilit [14]2 years ago
7 0

Hey there Fam

the only thing dat brought me here is Soccer ball XD

anyways hahahaha the answer is

C)0.4

thank you

YEEEEET

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Consider the following hypothetical reaction: 2 P + Q → 2 R + S The following mechanism is proposed for this reaction: P + P Q
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Answer: the answer is option (D). k[P]²[Q]

Explanation:

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2P + Q → 2R + S

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P + P ⇄ T     (fast)

Q + T → R + U    (slow)

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we would be applying the Rate law  to determine the mechanism.

The mechanism above is a three step process where the slowest step seen is the rate determining step. From this, we can see that this slow step involves an intermediate T as reactant and is expressed in terms of a starting substance P.

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3 0
2 years ago
1) Aluminum sulphate can be made by the following reaction: 2AlCl3(aq) + 3H2SO4(aq) Al2(SO4)3(aq) + 6 HCl(aq) It is quite solubl
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Answer:

88.9%

Explanation:

Step 1:

The balanced equation for the reaction. This is given below:

2AlCl3(aq) + 3H2SO4(aq) —> Al2(SO4)3(aq) + 6HCl(aq)

Step 2:

Determination of the masses of AlCl3 and H2SO4 that reacted and the mass of Al2(SO4)3 produced from the balanced equation.

Molar mass of AlCl3 = 27 + (35.5x3) = 133.5g/mol

Mass of AlCl3 from the balanced equation = 2 x 133.5 = 267g

Molar mass of H2SO4 = (2x1) + 32 + (16x4) = 98g/mol

Mass of H2SO4 from the balanced equation = 3 x 98 = 294g

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= 54 + 3[32 + 64]

= 54 + 3[96] = 342g/mol

Mass of Al2(SO4)3 from the balanced equation = 1 x 342 = 342g

Summary:

From the balanced equation above,

267g of AlCl3 reacted with 294g of H2SO4 to produce 342g of Al2(SO4)3.

Step 3:

Determination of the limiting reactant. This is illustrated below:

From the balanced equation above,

267g of AlCl3 reacted with 294g of H2SO4.

Therefore, 25g of AlCl3 will react with = (25 x 294)/267 = 27.53g of H2SO4.

From the calculations made above, we see that only 27.53g out 30g of H2SO4 given were needed to react completely with 25g of AlCl3.

Therefore, AlCl3 is the limiting reactant and H2SO4 is the excess.

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Determination of the theoretical yield of Al2(SO4)3.

In this case we shall be using the limiting reactant because it will produce the maximum yield of Al2(SO4)3 since all of it is used up in the reaction.

The limiting reactant is AlCl3 and the theoretical yield of Al2(SO4)3 can be obtained as follow:

From the balanced equation above,

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Therefore, the theoretical yield of Al2(SO4)3 is 32.02g

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Determination of the percentage yield of Al2(SO4)3.

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Explanation:

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