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Mumz [18]
3 years ago
12

Suppose you are driving in a convertible with the top down at constant speed. You point a rifle straight upward and fire it. In

the absence of air resistance, would the bullet land
(a) behind you,
(b) ahead of you, or
(c) in the barrel of the rifle?
Physics
1 answer:
lyudmila [28]3 years ago
4 0

Answer:

(c) in the barrel of the rifle

Explanation:

When we are moving in a convertible with constant speed so as per inertia all the objects are moving with same speed as that the speed of convertible.

So here we can say that bullet when shot from the rifle then its speed is same as that of convertible in horizontal direction

So here we can say that if air friction is absent then bullet will move same horizontal distance as the distance moved by you in the convertible

so after it it will reach again at the same position then the displacement of you and bullet is same in horizontal direction so we can say that correct answer will be

(c) in the barrel of the rifle

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2 years ago
Phosphorous reacts with chlorine gas to produce phosphorous pentachloride. Calculate the mass of product produced when 25.0 g of
MariettaO [177]

Answer : The mass of product produced if the reaction occurred with a 70.5 percent yield will be, 20.67 grams.

Explanation : Given,

Mass of P = 25 g

Mass of Cl_2 = 25 g

Molar mass of P = 30.97 g/mole

Molar mass of Cl_2 = 71 g/mole

Molar mass of PCl_5 = 208.24 g/mole

First we have to calculate the moles of P and Cl_2.

\text{Moles of }P=\frac{\text{Mass of }P}{\text{Molar mass of }P}=\frac{25g}{30.97g/mole}=0.807moles

\text{Moles of }Cl_2=\frac{\text{Mass of }Cl_2}{\text{Molar mass of }Cl_2}=\frac{25g}{71g/mole}=0.352moles

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

2P+5Cl_2\rightarrow 2PCl_5

From the balanced reaction we conclude that

As, 5 moles of Cl_2 react with 2 moles of P

So, 0.352 moles of Cl_2 react with \frac{2}{5}\times 0.352=0.1408 moles of P

That means, in the given balanced reaction, Cl_2 is a limiting reagent and it limits the formation of products and P is an excess reagent because the given moles are more than the required moles.

Now we have to calculate the moles of PCl_5.

As, 5 moles of Cl_2 react with 2 moles of PCl_5

So, 0.352 moles of Cl_2 react with \frac{2}{5}\times 0.352=0.1408 moles of PCl_5

Now we have to calculate the mass of PCl_5.

\text{Mass of }PCl_5=\text{Moles of }PCl_5\times \text{Molar mass of }PCl_5

\text{Mass of }PCl_5=(0.1408mole)\times (208.24g/mole)=29.32g

Now we have to calculate the mass of product produced (actual yield).

\%\text{ yield of }PCl_5=\frac{\text{Actual yield of }PCl_5}{\text{Theoretical yield of }PCl_5}\times 100

70.5=\frac{\text{Actual yield of }PCl_5}{29.32g}\times 100

\text{Actual yield of }PCl_5=20.67g

Therefore, the mass of product produced if the reaction occurred with a 70.5 percent yield will be, 20.67 grams.

3 0
3 years ago
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