1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Elena L [17]
3 years ago
8

A sleeping 68 kg man has a metabolic power of 65 w . you may want to review ( pages 323 - 327) . part a how many calories does h

e burn during an 8.0 hour sleep
Chemistry
1 answer:
Kazeer [188]3 years ago
8 0

<u>Given:</u>

Mass = 68 kg

Power = 65 W

<u>To determine:</u>

Calories burned during an 8.0 hr sleep

<u>Explanation:</u>

Energy is expressed as:

Energy = Power * Time

Now, the unit of power = Watt (W)

1 W = 1 J/s i.e. 65 W = 65 J/s

Time = 8 hrs = 8 * 3600 s = 28800 s

Energy = 65 J/s * 28800 s = 1872000 J

Convert Joules (J) to calories (cal)

1 cal = 4.184 J

The calculated energy = 187200 J * 1 cal/4.184 = 447418 cal

Ans: The man burns nearly 447.4 kcal during an 8 hr sleep




You might be interested in
Which of the following laboratory procedures best illustrates the law of conservation of mass?
fomenos

The  laboratory  procedure that best  illustrate the law of conservation  is

heating 100 g of CaCo3  to produce  56 g of CaO  (answer C)

<u><em>explanation</em></u>

According to the law  of mass conservation ,  the mass of the  reactant  must   be equal  to the mass  of the product.

 According  to  option c Heating  100 g CaCO3  to produces  56 g CaO  (  40 +16=56)

The remaining mass  = 100-56  =  44  which  would the mass of CO2  [  12 + (16 x2)]= 44   since  CaCO3  decomposes to produce CaO  and  CO2


Therefore  the  mass  of reactant=   100g

                       mass  of product =  56 g +44 g =100

Therefore the laboratory    procedure for  decomposition  of CaCO<em>3</em>  illustrate the law of mass  conservation since the mass of reactant = mass of product.

   


4 0
2 years ago
Read 2 more answers
Why would it be difficult to sift the iron fillings from the sand
slamgirl [31]

Explanation:

Sand and iron fillings have the same size of particles.

It would be difficult to use a mesh to separate them. Also, sometimes , the iron filing clings well to the surface of the sand particles.

The best way to separate them is to use a bar magnet. The magnetic properties of the iron fillings will get it attracted to the bar magnet leaving the sand particles behind.

learn more:

Extraction brainly.com/question/5881840

#learnwithBrainly

5 0
3 years ago
PLS HELP
AnnyKZ [126]

Answer:

linear ,trigonal planar,tethradral geometry

4 0
2 years ago
What are the three major branches of natural science?
Shtirlitz [24]
Physical science , earth science,and life science are the branches of natural science
5 0
3 years ago
Read 2 more answers
B) At what pH is H2 at 10 atm at equilibrium with this solution and pure nickel?
scZoUnD [109]
19372)19292
Hahaha ss
Qowieuww
6 0
2 years ago
Other questions:
  • Convert mass to moles for both reactants.<br> 2.50g CuCl2<br> 0.50g Al
    8·1 answer
  • WHat are the 7 parts of the water cycle
    15·1 answer
  • Ammonia rapidly reacts with hydrogen chloride, making ammonium chloride. Calculate the number of grams of excess reactant when 6
    7·1 answer
  • Silicon is lustrous and is a conductor of heat and electricity. Which class of elements could this element belong to?
    12·1 answer
  • Draw structural formula of a cyclic secondary alcohol with the molecular formula c3h6o.
    5·1 answer
  • A chemist carefully measures the amount of heat needed to raise the temperature of a 278.0g sample of C5H12S from 0.8degree cels
    8·1 answer
  • HCN<br> H:C:N:<br> 3.<br> Is this Lewis Structure correct?
    11·1 answer
  • Who else just copy and pastes the questions from edge?
    9·1 answer
  • Un recipiente cerrado, de 4,25 L, con tapa móvil, contiene H2S(g) a 740 Torr y 50,0°C. Se introduce en ese recipiente N2(g) a te
    12·1 answer
  • Does carbon dissolve in water
    5·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!