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Kipish [7]
3 years ago
7

A 150-lb person burns about 2700 Calories to run a marathon. How much energy is burned in kJ?

Chemistry
2 answers:
iris [78.8K]3 years ago
7 0

Answer:

150-lb person burned 11.30 kilo Joules of energy to ran a marathon.

Explanation:

Number of calories burned by 150-lb person = 2700 calories

1 cal = 0.001 kcal

2700 cal=2700\times 0.001 kcal=2.700 kcal

1 kilo calorie = 4.184 kilo Joules

Energy associated with 2.700 kcal:

2.700\times 4.184 kJ=11.30 kJ

150-lb person burned 11.30 kilo Joules of energy to ran a marathon.

Alex17521 [72]3 years ago
5 0
The question above is simply asking us to convert the value from calories into kilojoules. Therefore, we need some factor to multiply to be able to convert it to such units. For calories to kJ, the factor would be 4.184.

2700 cal ( 4.184 kJ/cal) = 11296.8 kJ
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A pan containing 40 grams of water was allowed to cool from a temperature of 91.0C. If the amount of heat repressed is 1,300 jou
Vinvika [58]

Answer:

83°C

Explanation:

The following were obtained from the question:

M = 40g

C = 4.2J/g°C

T1 = 91°C

T2 =?

Q = 1300J

Q = MCΔT

ΔT = Q/CM

ΔT = 1300/(4.2x40)

ΔT = 8°C

But ΔT = T1 — T2 (since the reaction involves cooling)

ΔT = T1 — T2

8 = 91 — T2

Collect like terms

8 — 91 = —T2

— 83 = —T2

Multiply through by —1

T2 = 83°C

The final temperature is 83°C

3 0
4 years ago
Learn how changes in binding free energy affect binding and the ratio of unbound and bound molecules.
Delvig [45]

C)[D]/[ED] = 5.20

D)[D]/[ED] = 5.20

E)[D']_T = 1.495* 10 ^-7 M

F)[D'] / [ED']  = 0.0579

Explanation:

E = 250 nM =2.5* 10 ^-7 mol/L , T=298.15 K

Dissociation constant of K_D = 1.3 μM (1.3 *10 ^-6 M)

E + D ⇄ ED → K_a = [ED] / [D][E]   (association constant)

ED ⇄ E + D → K_D = [E][D] / [ED]  (dissociation constant)

C)

[E] =2.5*10^-7 mol/L

K_D = 1.3* 10^-6 M

K_D = [E][D] / [ED] → [D]/[ED] = K_D / [E]

= [D]/[ED] = 1.3* 10 ^-6 / 2.5 *10^-7

= 13/25 * 10

=130/25 = 5.20

[D]/[ED] = 5.20

D)

ΔG =RTln Kd

ΔG_2 for E and D = 1.987 * 298.15 * ln 1.3*10^-6

ΔG_2 592.454 * [ln 1.3 +ln 10^-6]

ΔG_1 = 592.424 [0.2623 - 13.8155]

ΔG_2 = -592.424 * 13.553

ΔG_1 = -8184.633 cal/ mol

ΔG_1 = -8184.633  * 4.18 J/mol = -34244.508 J?mol

ΔG_1 = -34.245 KL/mol

so, ΔG_2 = ΔG_1 - 10.5 KJ/mol

ΔG_2 = -34.245 - 10.5

ΔG_2 = -44.745KJ / mol

ΔG_2 =RT ln K_D

-44.745 *10^3

=8.314 *298.15 lnK_D

lnK_D' = - 44745 / 2478.81 g

ln K_D' = -18.051

K_D' = -18.051

K_D' = e^-18.051

[D]/[ED] = 5.20

E)

[E] = 2.5* 10 ^-7 mol/ L = a

K_D' = [E][D] / [ED']                                  E +D' → ED'

K_D' = a/2(x-(a/2) / (a/2)

KD' = x - a/2

=2.447 *10^-8 = (2.5/2) * 10^-7

x=2.447 * 10^-8 + 1.25 * 10^-7

x = 2.447 *10^-8 + 1.25 * 10 ^-7

x= 10^-7 [1.25 + 0.2447]

x = 1.4947 * 10^-7

[D']_T = 1.495* 10 ^-7 M

F)

K_D' = [E][D'] / [ED']

[D'] / [ED'] = KD' / [E]

[D'] / [ED'] = 1.447 *10^-8 / 2.5* 10^-7

[D'] / [ED'] = 0.5788 * 10^-1

[D'] / [ED']  = 0.0579

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3 years ago
What kind of covalent bond forms between two fluorine atoms?
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A single covalent bond 
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Scientists found a fern fossil in some rock layers of the Grand Canyon. They estimated that a sample of one of the fossils origi
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This question is dealing with the half-life of carbon-14 which can be used to determine the age of a substance according to the following equation:

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No = # of carbon-14 atoms initially = 1000 atoms
t1/2 = half-life = 5730 years for carbon-14

We can now input all of the information into the formula to find the age of the fossil:

t = [ln (250/1000)/-ln2] x 5730 years

t = 11460 years

The fossil should be found to be roughly 11,460 years old.
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The energy from radiation can be used to cause the rupture of chemical bonds. A minimum energy of 242 kJ/mol is required to brea
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The question ask to determine the longest wavelength of radiation that possesses the energy to break the bond and the minimum energy of 242kj/mol is required to break the chlorine-chlorine bond so the energy would be that the minimum wave length is 495nm and the high wavelength to it is the color green end of its wave
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4 years ago
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