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
rewona [7]
3 years ago
8

1. Some athletes have as little as 3.0% body fat. If such a person has a body mass of 65 kg, how many pounds of body fat does th

at person have? In liposuction, a doctor removes fat deposits from a person’s body. If body fat has a density of 0.94 g/mL and 3.0 L of fat is removed, how many pounds of fat were removed from the patient?
Chemistry
2 answers:
noname [10]3 years ago
4 0

Answer:

4.299009 pounds of fat was present in the person's body.

6.2170284 pounds of fat were removed from the patient

Explanation:

Percentage of fat in the body = 3.05 of mass

Let the amount of fat be x

Mass of the athlete = 65 kg

3.0\%=\frac{x}{65 kg}\times 100

x = 1.95 kg = 4.299009 pounds (1 kg = 2.20462 pounds)

4.299009 pounds of fat was present in the person's body.

Density of fat = 0.94 g/L

Volume of fat to be removed = V = 3.0 L = 3000 mL[/tex]

Let the mass of fat to be removed be y

Density=\frac{Mass}{Volume}

0.94 g/mL=\frac{y}{3000 mL}

y = 2,820 g = 2.820 kg=6.2170284 pounds

6.2170284 pounds of fat were removed from the patient

77julia77 [94]3 years ago
3 0
1. 3.0% ----> 3.0 kg fat= 100 kg body weigh
also remember that 1 kg= 2.20 lbs

65 kg  \frac{3.0 kg}{100 kg} x  \frac{2.20 lb}{1 kg} = 4.29 lbs

2. 0.94 g/mL----> 0.94 grams= 1 mL
1 Liters= 1000 mL
1kg= 1000 grams

3 Liters  \frac{1000 mL}{1 L} x   \frac{0.94 grams}{1 mL} x  \frac{1 kg}{1000 g} x   \frac{2.20 lbs}{1 kg}  = 6.20 lbs
You might be interested in
An element has an average atomic mass of 1.008 amu. It consists of two isotopes , one having a mass of 1.007 amu, and one having
dimulka [17.4K]

Answer:

The most abundant isotope is 1.007 amu.

Explanation:

Given data:

Average atomic mass = 1.008 amu

Mass of first isotope = 1.007 amu

Mass of 2nd isotope = 2.014 amu

Most abundant isotope = ?

Solution:

First of all we will set the fraction for both isotopes

X for the isotopes having mass  2.014 amu

1-x for isotopes having mass 1.007 amu

The average atomic mass is 1.008 amu

we will use the following equation,

2.014x + 1.007  (1-x) = 1.008  

2.014x + 1.007  - 1.007 x = 1.008  

2.014x - 1.007x  =  1.008  -  1.007

1.007 x = 0.001

x= 0.001/ 1.007

x= 0.0009

0.0009 × 100 = 0.09 %

0.09 % is abundance of isotope having mass  2.014 amu because we solve the fraction x.

now we will calculate the abundance of second isotope.

(1-x)

1-0.0009 = 0.9991

0.9991 × 100= 99.91%

6 0
3 years ago
the quantity of antimony in an ore can be determined by an oxidation-reduction titration with an oxidizing agent. The ore is dis
Basile [38]

Answer:

BrO₃⁻(aq) + 3Sb³⁺(aq) + 6H⁺(aq) → Br⁻(aq) + 3Sb⁵⁺(aq) + 3H₂O(l)

Explanation:

At a redox equation, one substance is being oxidized (losing electrons), and the other is being reduced (gaining electrons). In the given reaction:

BrO₃⁻(aq) + Sb³⁺(aq) → Br⁻(aq) + Sb⁵⁺(aq)

When it's at an acidic solution, it must be ions H⁺ on the reactant, which will form water with the oxygen, so the complete reaction is:

BrO₃⁻(aq) + Sb³⁺(aq) + H⁺(aq) → Br⁻(aq) + Sb⁵⁺(aq) + H₂O(l)

As we can see, the antimony is being oxidized (go from +3 to +5), and the Bromo is being reduced. The oxidation number of brome in the reactant, knowing that the oxidation number of O is -2, is:

x + 3*(-2) = -1

x = +5

So, it's going from +5 to -1, and the half-reactions are:

BrO₃⁻(aq) + 6e⁻ → Br⁻(aq)

Sb³⁺(aq) → Sb⁵⁺(aq) + 2e⁻

The number of electrons must be the same, so the second equation must be multiplied by 3:

3Sb³⁺(aq) → 3Sb⁵⁺(aq) + 6e⁻

Thus, the equation will be:

BrO₃⁻(aq) + 3Sb³⁺(aq) + H⁺(aq) → Br⁻(aq) + 3Sb⁵⁺(aq) + H₂O(l)

Now, we verify the amount of the elements, which must be equal on both sides. So, we multiply H₂O by 3, and H⁺ by 6, and the balanced reaction will be:

BrO₃⁻(aq) + 3Sb³⁺(aq) + 6H⁺(aq) → Br⁻(aq) + 3Sb⁵⁺(aq) + 3H₂O(l)

6 0
3 years ago
If 33.6 grams of KCl are dissolved in 192 grams of water, what is the concentration of the solution in percent by mass?21.2% KCl
Lelu [443]
The answer here would be 14.9 %KCl and here is how I can explain why:
33.5g / 225.6g x 100% = 14.9% 
<span>you are looking for % of KCl in the solution, you have to add the mass of the KCl and water to get the total mass of the solution
Hope this helps a lot
</span>

7 0
3 years ago
The number of atoms in 2.50 mol Cr
Slav-nsk [51]

Answer:

1.51 x 10²⁴atoms

Explanation:

Given parameters:

Number of moles of Cr = 2.50mol

Solution;

what is a mole;

  a mole of any substance is the amount of substance that contains the avogadro's number of particles which is 6.02 x 10²³ particles

The particles here can be atoms, protons, neutrons, electrons e.t.c

                   1 mole contains 6.02 x 10²³ atoms

                  2.5 moles will contain  2.5 x 6.02 x 10²³   = 1.51 x 10²⁴atoms

8 0
3 years ago
What is the density of carbon dioxide at STP?
Marina86 [1]

0.00196 g

The density of carbon dioxide at STP is 0.00196 g/mL. Density has the equation: ρ is density. m is mass.

4 0
3 years ago
Other questions:
  • Which of these experimental designs could lead to bias?
    12·2 answers
  • What will happen if a peeled banana is put on a hotplate?
    14·1 answer
  • Dead leaves and plants on the ground form what??
    7·1 answer
  • HELP ASAP. NEED THIS DONE
    8·1 answer
  • What is the chemical formula for borax
    13·2 answers
  • A sample of 17.0 M concentrated H2SO4 stock solution with a volume of 25.0 cm3 was diluted to final concentration of 5.0 M H2SO4
    13·1 answer
  • Calculate the number of formula units in 1.87 mol of NH4Cl
    12·1 answer
  • 5) Determine the mass if lithium hydroxide that is produced when 12.87 g of lithium
    9·1 answer
  • se tienen 10 litros de nitrogeno gaseoso bajo una presion de 560 mm Hg. Determina el volumen del nitrogeno a una presion de 760
    13·1 answer
  • What is the temperature change of an 36-gram sample of water that had absorbed 695 Joules of heat? C = 4.18 J/g K
    6·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!