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A rock climber hangs freely from a nylon rope that is 15 m long and has a diameter of 8.3 mm. If the rope stretches 5.1 cm, what

irinina [24]

Answer:

Mass of the climber = 69.38 kg

Explanation:

Change in length

$\Delta L=\frac{PL}{AE}$

Load, P = m x 9.81 = 9.81m

Young's modulus, Y = 0.37 x 10¹⁰ N/m²

Area

$A=\frac{\pi (8.3\times 10^{-3})^2}{4}=5.41\times 10^{-5}m^2$

Length, L = 15 m

ΔL = 5.1 cm = 0.051 m

Substituting

$0.051=\frac{9.81m\times 15}{5.41\times 10^{-5}\times 0.37\times 10^{10}}\\\\m=69.38kg$

Mass of the climber = 69.38 kg

6 0

3 years ago

A rope, under a tension of 221 N and fixed at both ends, oscillates in a second-harmonic standing wave pattern. The displacement

mamaluj [8]

Answer:

sup qwertyasdfghjk

Explanation:

3 0

3 years ago

Who’s of the following is classified as a salt: ammonia or sodium chloride?

Temka [501]

Sodium Chloride could be classified as a salt.

8 0

3 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

Speed (S) of a moving object is calculated from

bonufazy [111]

Answer:distance divided by time

Explanation:

7 0

3 years ago

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