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rjkz [21]
3 years ago
15

Individual M has longer limbs and muscles than Individual N. Both individuals lift the same amount of weight from the ground to

their shoulders an equal number of times. If all repetitions were completed in synchrony, which statement about the amount of work completed by the individuals is true? At the end of the workout:
A. Individual M will have done less work because the object was moved a greater distance
B. both individuals will have done the same amount of work because force was applied for the same duration of time.
C. Individual M will have done more work because the muscle contracted faster to achieve synchrony with Individual N
D. Individual N would have done less work because the object was moved a shorter distance
Chemistry
1 answer:
Elena L [17]3 years ago
5 0

Answer:

D. Individual N would have done less work because the object was moved a shorter distance

Explanation:

When a body is moved a given distance (x) by a specific force (F), the work done on the body is equivalent to F*x. This shows that work is proportional to the applied force and the distance traveled by the body. Therefore, the higher the distance the higher the work done and the lower the distance the lower the work done. The work done by Individual N is lower due to the shorter limbs that correspond to shorter distance.

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Yuri [45]

Answer:

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5 0
2 years ago
Liquid acetone is ______________.
MaRussiya [10]
A is the answer maybe
5 0
3 years ago
Light
Ratling [72]

Answer:

Light energy acts as a catalyst that speeds up the rate of photosynthesis

Explanation:

4 0
3 years ago
Butane (C4H10) has a heat of vaporization of 22.44 kJ/mol and a normal boiling point of -0.4 ∘C. A 250 mL sealed flask contains
erastova [34]

Given that:

  • The heat of vaporization = 22.44 kJ/mol = 22440 J/mol
  • normal boiling point which is the initial temperature = 0.4° C = (273 + (-0.4))K  = 272.6 K
  • volume  = 250 mL = 0.250 L
  • Mass of butane = 0.8 g
  • the final temperature = -22° C = (273 + (-22)) K = 251 K

The first step is to determine the vapor pressure at the final temperature of 251K by using the Clausius-Clapeyron equation. This is following by using the ideal gas equation to determine the numbers of moles of butane gas. After that, the mass of butane present in the liquid is determined by using the relation for the number of moles.

Using Clausius-Clapeyron Equation:

\mathbf{In (\dfrac{P_2}{P_1} )= -\dfrac{\Delta H_{vap}}{R}(\dfrac{1}{T_2} - \dfrac{1}{T_1})}

where;

P1 and P2 correspond to the temperature at T1 and T2.

∴

replacing the values into the given equation, we have;

\mathbf{In \dfrac{P_2}{1\  atm} = -\dfrac{22440 \ J/mol}{8.314 \ J/mol.K}(\dfrac{1}{251 \ K} - \dfrac{1}{272.6 \ K})}

\mathbf{In \dfrac{P_2}{1\  atm} =-(0.852053785)}

\mathbf{P_2=0.427 \ atm}

As such, at -22° C; the vapor pressure = 0.427 atm

Now, using the ideal gas equation:

PV = nRT

where:

  • P = Pressure
  • V = volume
  • n = number of moles of butane
  • R = universal gas constant
  • T = temperature

∴

Making (n) the subject of the formula:

\mathbf{n = \dfrac{PV}{RT}}

\mathbf{n = \dfrac{0.427 atm \times 0.250 L}{(0.08206 \ L.atm/k.mol) \times 251}}

\mathbf{n =0.00518 mol}

We all know that the standard molecular weight of butane = 58.12 g/mol

∴

Using the relation for the number of moles which is:

\mathbf{number \  of \  moles = \dfrac{mass}{molar mass}}

mass = 0.00518 mole × 58.12 g/mol

mass = 0.301 g

∴

The mass of butane in the flask = 0.301 g

But the mass of the butane present as a liquid in the flask is

= 0.8 g - 0.301 g

= 0.499 g

In conclusion, the mass of the butane present as a liquid in the flask is 0.499 g

Learn more about vapourization here:

brainly.com/question/17039550?referrer=searchResults

7 0
3 years ago
Chlorine atoms react with methane, forming HCl and CH3. The rate constant for the reaction was determined to be 3.600×107 at 278
Ratling [72]

<u>Answer:</u> The activation energy for the reaction is 40.143 kJ/mol

<u>Explanation:</u>

To calculate activation energy of the reaction, we use Arrhenius equation for two different temperatures, which is:

\ln(\frac{K_{317K}}{K_{278K}})=\frac{E_a}{R}[\frac{1}{T_1}-\frac{1}{T_2}]

where,

K_{317K} = equilibrium constant at 317 K = 3.050\times 10^{8}M^{-1}s^{-1}

K_{278K} = equilibrium constant at 278 K = 3.600\times 10^{7}M^{-1}s^{-1}

E_a = Activation energy = ?

R = Gas constant = 8.314 J/mol K

T_1 = initial temperature = 278 K

T_2 = final temperature = 317 K

Putting values in above equation, we get:

\ln(\frac{3.050\times 10^8}{3.600\times 10^{7}})=\frac{E_a}{8.314J/mol.K}[\frac{1}{278}-\frac{1}{317}]\\\\E_a=40143.3J/mol=40.143kJ/mol

Hence, the activation energy for the reaction is 40.143 kJ/mol

8 0
3 years ago
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