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lozanna [386]
3 years ago
10

During an autopsy and subsequent toxicology reports, it is discovered that a man died from arsenic poisoning. When investigating

the crime scene, a bottle of Gatorade was discovered next to the dead body. What type of lab technique will need to be performed first to see if there is any arsenic mixed in with the Gatorade?
Physics
1 answer:
Vaselesa [24]3 years ago
4 0

Spectrophotometry is a laboratory technique that can be performed first to check if there is presence of arsenic mixed in with Gatorade. In spectrophotometry, arsenic concentration will be measure in a specific solution by calculating the amount of light absorbed by inorganic arsenic.

Various analytical reagents are used such as molybdate reagent and variamine blue as a chromogenic reagent where a violet colored indicates presence of arsenic. 

Furthermore, new analytical reagents has been development and proposed like vanillin-2- amino nicotinic acid (VANA) and 2, 4-Dihydroxy benzophenone-2-amino thiophenol (BPBT). These are used for direct non-extractive spectrophotometric determination of arsenic. The reagent reacts with arsenic in acidic medium to form light greenish yellow colored.

 

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A 440-g cylinder of brass is heated to 97.0 degree Celsius and placed in a 2 points
nydimaria [60]

Answer:

Specific heat of brass is 0.40 J g⁻¹ °C⁻¹ .

Explanation:

Given :

Mass of brass, m₁ = 440 g

Temperature of brass, T₁ = 97° C

Mass of water, m₂ = 350 g

Temperature of water, T₂ = 23° C

Specific heat of water, C₂ = 4.18 J g⁻¹ °C⁻¹

Equilibrium temperature, T = 31° C

Let C₁ be the specific heat of brass.

Heat loss by brass = Heat gain by water

m₁ x C₁ x ( T₁ -T ) = m₂ x C₂ x ( T - T₁ )

Substitute the suitable values in above equation.

440 x C₁ x (97 - 31) = 350 x 4.18 x (31 - 23)

C₁ = \frac{11704}{29040}

C₁ = 0.40 J g⁻¹ °C⁻¹

4 0
3 years ago
A mass MM uniform solid cylinder of radius RR and a mass MM thin uniform spherical shell of radius RR roll without slipping. If
vampirchik [111]

Answer:

vcyl / vsph = 1.05

Explanation:

  • The kinetic energy of a rolling object can be expressed as the sum of a translational kinetic energy plus a rotational kinetic energy.
  • The traslational part can be written as follows:

       K_{trans} = \frac{1}{2}* M* v_{cm} ^{2}  (1)

  • The rotational part can be expressed as follows:

       K_{rot} = \frac{1}{2}* I* \omega ^{2}  (2)

  • where I = moment of Inertia regarding the axis of rotation.
  • ω = angular speed of the rotating object.
  • If the object has a radius R, and it rolls without slipping, there is a fixed relationship between the linear and angular speed, as follows:

       v = \omega * R (3)

  • For a solid cylinder, I = M*R²/2 (4)
  • Replacing (3) and (4)  in (2), we get:

       K_{rot} = \frac{1}{2}* \frac{1}{2} M*R^{2} * \frac{v_{cmc} ^{2}}{R^{2}} = \frac{1}{4}* M* v_{cmc}^{2}  (5)

  • Adding (5) and (1), we get the total kinetic energy for the solid cylinder, as follows:

       K_{cyl} = \frac{1}{2}* M* v_{cmc} ^{2}  +\frac{1}{4}* M* v_{cmc}^{2}  =  \frac{3}{4}* M* v_{cmc} ^{2} (6)

  • Repeating the same steps for the spherical shell:

        I_{sph} = \frac{2}{3} * M* R^{2} (7)  

       K_{rot} = \frac{1}{2}* \frac{2}{3} M*R^{2} * \frac{v_{cms} ^{2}}{R^{2}} = \frac{1}{3}* M* v_{cms}^{2}  (8)

      K_{sph} = \frac{1}{2}* M* v_{cms} ^{2}  +\frac{1}{3}* M* v_{cms}^{2}  =  \frac{5}{6}* M* v_{cms} ^{2} (9)

  • Since we know that both masses are equal each other, we can simplify (6) and (9), cancelling both masses out.
  • And since we also know that both objects have the same kinetic energy, this means that (6) are (9) are equal each other.
  • Rearranging, and taking square roots on both sides, we get:

       \frac{v_{cmc}}{v_{cms}} =\sqrt{\frac{10}{9} } = 1.05 (10)

  • This means that the solid cylinder is 5% faster than the spherical shell, which is due to the larger moment of inertia for the shell.
3 0
3 years ago
NEED HELP ASAP!!!!
podryga [215]
The answer is B
I rhink
6 0
3 years ago
At the top of a pole vault, an athlete actually can do work pushing on the pole before releasing it. Suppose the pushing force t
Salsk061 [2.6K]

Answer:

The work done on the athlete is approximately 2.09 J

Explanation:

From the definition of the work done by a variable force:

\displaystyle{\int_{x_i}^{x_j}F(x)dx}

and substituting with the function of our problem:

\displaystyle{\int_{0}^{0.19}(140x-190x^2)dx\approx2.09\mathrm{J}}

5 0
4 years ago
A rightward force of 12.0 N is applied to a 2.0-kg object to accelerate it across a horizontal
ad-work [718]

Answer:

below

Explanation:

Net accelerating force becomes  12-8 = 4 N

F = ma

4 = 2 * a

a = 2 m/s^2

8 0
1 year ago
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