Ask question

Ask question

All categories

3 years ago

11

Manufacturers of headache remedies routinely claim that their own brands are more potent pain relievers than the competing brand

s. Their way of making the comparison is to compare the number of molecules in the standard dosage. Tylenol uses 325 mg of acetaminophen (C8H9NO2) as the standard dose, while Advil uses 2.00 x 102 mg of ibuprofen (C13H18O2). Find the number of molecules of pain reliever in the standard doses of
(a) Tylenol and

(b) Advil.

1 answer:

Licemer1 [7]3 years ago

4 0

a) 12.95*10^20 molecules

b) 5.84*10^20 molecules

Explanation:

a)Tylenol

Mass of Tylenol = 325 mg

(m)=325*10^-3 g

Molecular weight of Tylenol(M) = 151.16256 g/mol

the number of moles present

n = m/M

= 325*10^-3 g /151.1656 g/mol

= 2.15 *10^-3 mol

the number of molecules present

N= nNa

=(2.15*10^ -3 mol)(6.023*10^23 molecules/mol)

N=12.95*10 ^20 molecules

b)Advil

Mass of Advil = 200 mg

m=200*10^-3 g

Molecular weight of Advil(M) =206.28082 g/mol

the number of moles

n=m/M

=200*10^ -3 g/ 206.28082 g/mol

=0.96955*10^ -3 mol

the number of molecules resent

N=nNa

=(0.96955*10^ -3)(6.023*10^ 23 molecules/ mol)

N=5.84*10^20 molecules

You might be interested in

A fence 8 ft high (w) runs parallel to a tall building and is 24 ft (d) from it. Find the length (L) of the shortest ladder t

AveGali [126]

Answer:

25.3 ft

Explanation:

The illustration of the problem is shown the attached image.

The length of the ladder can be calculated using the Pythagoras theorem:

$Hypotenuse^2 = opposite^2 + adjacent^2$

The hypotenuse is the length of the ladder.

$Hypotenuse = \sqrt{opposite^2 + adjacent^2}$

$L^2 = BC^2 + (24 + AE)^2$ ........1

Triangle ABC is similar to triangle AEF, hence:

$\frac{BC}{8} = \frac{AE + 24}{AE}$

BC = $\frac{8(AE + 24)}{AE}$ .................2

Substitute 2 into 1

$L^2$ = $(\frac{8(AE + 24)}{AE})^2$ + $(24 + AE)^2$

Let AE = x

$L^2$ = $(\frac{8x + 192}{x})^2 + (24 + x)^2$

= $(8 + \frac{192}{x})^2 + (24 + x)^2$

Minimize L with respect to x.

2 $\frac{dL}{dx}$ = $2(8 + \frac{192}{x})(-\frac{192}{x^2}) + 2(24 + x)$

=

5 0

3 years ago

spaceship of mass m travels from the Earth to the Moon along a line that passes through the center of the Earth and the center o

satela [25.4K]

Answer:

the correct result is r = 3.71 10⁸ m

Explanation:

For this exercise we will use the law of universal gravitation

F = $- \frac{m_{1} m_{2} }{r^2}$

We call the masses of the Earth M, the masses of the moon m and the masses of the rocket m ', let's set a reference system in the center of the Earth, the distance from the Earth to the moon is d = 3.84 108 m

rocket force -Earth

F₁ = - \frac{m' M }{r^2}

rocket force - Moon

F₂ = - \frac{m' m }{(d-r)^2}

in the problem ask for what point the force has the relation

2 F₁ = F₂

let's substitute

2 $2 \frac{M}{r^2} = \frac{m}{(d-r)^2}$

(d-r) ² = $\frac{m}{2M}$ r²

d² - 2rd + r² = \frac{m}{2M} r²

r² (1 -\frac{m}{2M}) - 2rd + d² = 0

Let's solve this quadratic equation to find the distance r, let's call

a = 1 - \frac{m}{2M}

a = 1 - $\frac{7.36 10^{22} }{2 \ 5398 10^{24}}$ = 1 - 6.15 10⁻³

a = 0.99385

a r² - 2d r + d² = 0

r = $\frac {2d \frac{+}{-} \sqrt{4d^2 - 4 a d^2}} {2a}$

r = [2d ± 2d $\sqrt{1-a}$ ] / 2a

r = $\frac{d}{a}$ (1 ± √ (1.65 10⁻³)) = $\frac{d}{a}$ (1 ± 0.04)

r₁ = \frac{d}{a} 1.04

r₂ = \frac{d}{a} 0.96

let's calculate

r₁ = $\frac{3.84 10^8}{0.99385}$ 1.04

r₁ = 401.8 10⁸ m

r₂ = \frac{3.84 10^8}{0.99385} 0.96

r₂ = 3.71 10⁸ m

therefore the correct result is r = 3.71 10⁸ m

3 0

2 years ago

N experiment is performed in deep space with two uniform spheres, one with mass 27.0 and the other with mass 107.0 . They have e

Reptile [31]

Answer:

Explanation:

Apply the law of conservation of energy

$KE_i+PE_i=KE_f+PE_f$

$Gm_1m_2[\frac{1}{r_f} -\frac{1}{r_1} ]=\frac{1}{2} (m_1v_1^2+m_2v_2^2)$

from the law of conservation of the linear momentum

$m_1v_1=m_2v_2$

Therefore,

$Gm_1m_2[\frac{1}{r_f} -\frac{1}{r_1} ]=\frac{1}{2} (m_1v_1^2+m_2v_2^2)$

$=\frac{1}{2} [m_1v_1^2+m_2[\frac{m_1v_1}{m_2} ]^2]\\\\=\frac{1}{2} [m_1v_1^2+\frac{m_1^2v_1^2}{m_2} ]\\\\=\frac{m_1v_1^2}{2} [\frac{m_1+m_2}{m_2} ]$

$v_1^2=[\frac{2Gm_2^2}{m_1+m_2} ][\frac{1}{r_f} -\frac{1}{r_1} ]$

Substitute the values in the above result

$v_1^2=[\frac{2Gm_2^2}{m_1+m_2} ][\frac{1}{r_f} -\frac{1}{r_1} ]$

$=[\frac{2(6.67\times 10^-^1^1)(107)^2}{27+107} ][\frac{1}{26} -\frac{1}{41}] \\\\=1.6038\times 10^-^1^0\\\\v_1=\sqrt{1.6038\times 106-^1^0} \\\\=1.2664 \times 10^-^5m/s$

B) the speed of the sphere with mass 107.0 kg is

$v_2=\frac{m_1v_1}{m_2}$

$=[\frac{27}{107} ](1.2664 \times 10^-^5)\\\\=3.195\times 10^-^6m/s$

C) the magnitude of the relative velocity with which one sphere is

$v_r=v_1+v_2\\\\=1.2664\times 10^-^5+3.195\times10^-^6\\\\=15.859\times10^-^6m/s$

D) the distance of the centre is proportional to the acceleration

$\frac{x_1}{x_2} =\frac{a_1}{a_2} \\\\=\frac{m_2}{m_1} \\\\=3.962$

Thus,

$x_1=3.962x_2$

and

$x_2=0.252x_1$

When the sphere make contact with eachother

Therefore,

$x_1+x_2+2r=41\\x_1+0,252x_1+2r=41\\1.252x_1+2r=41\\x_1=32.747-1.597r$

And

$x_1+x_2+2r=41\\3.962x_2+x_2+2r+41\\4.962x_2+2r=41\\x_2=8.262-0.403r$

The point of contact of the sphere is

$32.747-1.597r=8.262-0.403r\\\\r=\frac{24.485}{1.194} \\\\=20.506m$

3 0

3 years ago

A field measuring 12 meters by 16 meters is to have a brick paver walkway installed all around it, increasing the total area to

kolezko [41]

Answer:

1.5 m

Explanation:

Length. L = 12 m

Width, W = 16 m

Area, A = 12 x 16 = 192 m^2

Let the width of pavement be d.

The new length, L' = 12 + 2d

the new width, W' = 16 + 2d

New Area, A' = L' x W' = (12 + 2d)(16 + 2d) = 192 + 56 d + 4d^2

Difference in area = A' - A

285 = 192 + 56 d + 4d^2 - 192

93 = 56 d + 4d^2

4d^2 + 56 d - 93 = 0

$d = \frac{-56\pm \sqrt{56^{2}+4\times 4\times 93}}{8}$

$d=\frac{-56\pm 87.72}{8}$ \

d = 1.5 m

Thus, the width of the pavement is 1.5 m.

6 0

3 years ago

Two forces, F? 1 and F? 2, act at a point, as shown in the picture. (Figure 1) F? 1 has a magnitude of 9.20 N and is directed at

Stolb23 [73]

Answer:

a. Fx = -8.089 N b. Fy = 3.525 N c. 8.824 N d. 336.45°

Explanation:

Since F₁ = 9.2 N and acts at 57° above the negative axis in the second quadrant, its x-component is -F₁cos57° and its y- component is F₁sin57°

Since F₁ = 5.2 N and acts at 53.7° below the negative axis in the third quadrant, its x-component is -F₂cos53.7° and its y- component is -F₂sin53.7°

Part A

What is the x component Fx of the resultant force?

The x component of the resultant force Fx = -F₁cos57° + -F₂cos53.7° = -9.2cos57° + (-5.2cos53.7°) = (-5.011 - 3.078) N = -8.089 N

Part B

What is the y component Fy of the resultant force?

The y component Fy of the resultant force = F₁sin57° + -(F₂sin53.7°) = 9.2sin57° - 5.2sin53.7° = (7.716 - 4.191) N = 3.525 N

Part C

What is the magnitude F of the resultant force?

The magnitude F of the resultant force = √(Fx² + Fy²)

F = √(-8.089² N + 3.525² N) = √65.432 + 12.426 = √77.858 = 8.824 N

Part D

What is the angle ? that the resultant force forms with the negative x axis?

The angle the resultant force makes with the negative x axis is given by

θ = tan⁻¹(Fy/Fx) = tan⁻¹(3.525/-8.089) = tan⁻¹-0.4358 = -23.55°.

To measure it from the negative x axis, we add 360. So, our angle = 360 -23.55 = 336.45°

7 0

3 years ago