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weeeeeb [17]
1 year ago
6

An enzyme works best at 98.6°f. the equation used to describe it requires the temperature to be in k. what is the correct tempe

rature? use (f×0.555) 255.37=k or (k−255.37)×1.8=f. 37 k 37 k 236 k 236 k 310 k 310 k 482 k
Physics
1 answer:
Marina86 [1]1 year ago
5 0

310.093k.

Given,

Best temperature for enzymes to work = 98.6°F

Unknown :

We need to convert the temperature from fahrenheit to kelvin.

Writing down the equation to be used in the problem's solution correctly:

K = (F x 0.55) + 255.37

F = (K - 255.37) x 1.8

The temperatures are expressed in kelvin (K) and fahrenheit (F), respectively.

There first equation would be significantly more quicker to use :

K = (98.6 x 0.555) + 255.37

= 54.723 + 255.37

= 310.093K

➤ The activity and rate of an enzyme reaction typically increase with temperature, while a decrease in temperature slows the enzymatic reaction. Each enzyme's activity peaks at its individual optimum temperature, and it decreases above and below that point.

Find more on enzymes at : brainly.com/question/1596855

#SPJ4

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A cylinder with a piston contains 0.200 mol of nitrogen at 1.50×105 Pa and 320 K . The nitrogen may be treated as an ideal gas.
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Answer:

Q = -105 J

Also we know that for cyclic process change in internal energy is always ZERO

Explanation:

First gas is compressed isobarically such that its volume is half of initial volume

So its temperature is also half

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Q = nC_p \Delta T

for diatomic gas we have

C_p = \frac{7}{2} R

so we will have

Q = 0.200(\frac{7}{2}R)(160 - 320)

Q = -930.7 J

Now in adiabatic process heat is not transferred

so in this process

Q = 0

so we have

T_1V_1^{1.4-1} = T_2V_2^{1.4-1}

(160)(\frac{V}{2})^{0.4} = T_2(V)^{0.4}

T_2 = 121.26 K

Now it is again reached to original pressure

so temperature will become initial temperature

so heat given in that part

Q_3 = nC_v\Delta T

here we know that

C_v = \frac{5}{2}R

Q_3 = (0.200)(\frac{5}{2}R)(320 - 121.26)

Q_3 = 825.76 J

So total heat given to the system is

Q = -930.7 + 0 + 825.76

Q = -105 J

Also we know that for cyclic process change in internal energy is always ZERO

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3 years ago
What is the time period of vibration?
Y_Kistochka [10]

Answer:

A time period is denoted by 'T' . It is the time to complete one cycle of vibration. As the frequency of a wave increases, the time period of the wave decreases. The unit for time period is 'seconds

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The basal metabolic rate is the rate at which energy is produced in the body when a person is at rest. A 75-kg (165-lb) person o
CaHeK987 [17]

Answer:

(a) Q = 142.67 W

(b) Basal Metabolic Rate = 178.33 W

Explanation:

(a)

We can find the heat radiated by the person by using Stefan-Boltzman's law:

Q = \sigma A (T^4 - T_{s}^4)\\

where,

Q = heat radiated per second = ?

σ = Stefan-Boltzman Constant = 5.6703 x 10⁻⁸ W/m².k⁴

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Ts = Temperature of room = 18° C +273 = 291 k

Therefore,

Q = (5.6703\ x\ 10^{-8}\ W/m^2.k^4)(2\ m^2)[(303\ k)^4-(291\ k)^4]<u></u>

<u>Q = 142.67 W</u>

<u></u>

(b)

Since the heat calculated in part (a) is 80 percent of basal metabolic rate. Therefore,

Q = (0.8)(Basal\ Metabolic\ Rate)\\Basal\ Metabolic\ Rate = \frac{Q}{0.8}\\\\Basal\ Metabolic\ Rate = \frac{142.67\ W}{0.8}

<u>Basal Metabolic Rate = 178.33 W</u>

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