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2 years ago

Imagine you are a detective examining a crime scene. You are trying to

Chemistry

2 answers:

Sidana [21]2 years ago

4 0

Answer:

C. pieces of hair found at the crime scene.

Explanation:

using the pieces of hair, you can get the DNA of the person. This will give u a better lead in solving the crime.

Hope it helps u ....

sveta [45]2 years ago

3 0

Answer:

C. Pieces of hair found at the crime scene.

Explanation:

Pieces of hair found at the crime scene is more useful to find the criminal who committed the crime, because it helps in indetifying criminal's DNA through the process of DNA fingerprinting. The hair is analyzed by the chemical makeup of the hair and extracting DNA from the hair follicle, pieces of hair must have hair follicle attached to find DNA of the criminal.

Nuclear DNA can be successfully extracted from the hair root. Cut hair samples are useless without hair follicle attached. Its DNA will be useful only few months for testing. The DNA last in hairs about 1,000 to 10,000 years.

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Cacodyl, which has an intolerable garlicky odor and is used in the manufacture of cacodylic acid, a cotton herbicide, has a mass

Papessa [141]

Answer:

The molecular formula of cacodyl is C₄H₁₂As₂.

Explanation:

<u>Let's assume we have 1 mol of cacodyl</u>, in that case we'd have 209.96 g of cacodyl and the<u> following masses of its components</u>:

209.96 g * 22.88/100 = 48.04 g C

209.96 g * 5.76/100 = 12.09 g H

209.96 g * 71.36/100 = 149.83 g As

Now we convert those masses into moles:

48.04 g C ÷ 12 g/mol = 4.00 mol C

12.09 g H ÷ 1 g/mol = 12.09 mol H

149.83 g As ÷ 74.92 g/mol = 2.00 mol As

Those amounts of moles represent the amount of each component in 1 mol of cacodyl, thus, the molecular formula of cacodyl is C₄H₁₂As₂.

3 0

1 year ago

A student uses visible spectrophotometry to determine the concentration of CoCl2(aq) in a sample solution. First the student pre

k0ka [10]

Answer:

$4\times 10^{-9} J$ is the approximate energy of one photon of this light.

Explanation:

Energy of the photon can be calculated by

$E=h\nu=\frac{h\times c}{\lambda}$ (Planck's equation)

where,

E = energy of photon

h = Planck's constant = $6.63\times 10^{-34}Js$

c = speed of light = $3\times 10^8m/s$

$\lambda$ = wavelength of light =

$\nu$ = frequency of the light

we have , $\lambda =510 nm=510\times 10^{-9}m$

Now put all the given values in the above formula, we get the energy of the photons.

$E=\frac{(6.63\times 10^{-34}Js)\times (3\times 10^8m/s)}{510\times 10^{-9}m}$

$E=3.9\times 10^{-19}J\approx 4\times 10^{-9} J$

$4\times 10^{-9} J$ is the approximate energy of one photon of this light.

5 0

2 years ago

3. What's the empirical formula of a molecule containing 18.7% lithium, 16.3% carbon, and 65.0% oxygen?

Hunter-Best [27]

Answer:

Empirical formula is Li₂CO₃.

Explanation:

Percentage of oxygen= 65.0%

Percentage of lithium = 18.7%

Percentage of carbon= 16.3%

Empirical formula = ?

Solution:

Number of gram atoms of C = 16.3/12 = 1.4

Number of gram atoms of Li = 18.7/6.94 = 2.7

Number of gram atoms of O = 65.0/ 16 = 4.1

Atomic ratio:

Li : C : O

2.7/1.4 : 1.4/1.4 : 4.1/1.4

2 : 1 : 3

Li : C : O = 2 : 1 : 3

Empirical formula is Li₂CO₃.

8 0

2 years ago

The decomposition of AB given here in this balanced equation 2AB (g)⟶ A2 (g) + B2 (g), has rate constants of 8.58 x 10-9 L/mol s

denis-greek [22]

Answer:

3.24 × 10^5 J/mol

Explanation:

The activation energy of this reaction can be calculated using the equation:

ln(k2/k1) = Ea/R x (1/T1 - 1/T2)

Where; Ea = the activation energy (J/mol)

R = the ideal gas constant = 8.3145 J/Kmol

T1 and T2 = absolute temperatures (K)

k1 and k2 = the reaction rate constants at respective temperature

First, we need to convert the temperatures in °C to K

T(K) = T(°C) + 273.15

T1 = 325°C + 273.15

T1 = 598.15K

T2 = 407°C + 273.15

T2 = 680.15K

Since, k1= 8.58 x 10-9 L/mol, k2= 2.16 x 10-5 L/mol, R= 8.3145 J/Kmol, we can now find Ea

ln(k2/k1) = Ea/R x (1/T1 - 1/T2)

ln(2.16 x 10-5/8.58 x 10-9) = Ea/8.3145 × (1/598.15 - 1/680.15)

ln(2517.4) = Ea/8.3145 × 2.01 × 10^-4

7.831 = Ea(2.417 × 10^-5)

Ea = 3.24 × 10^5 J/mol

8 0

1 year ago

How many moles of ions are in 285 ml of 0.0150 m mgcl2?

Solnce55 [7]

MgCl₂)= Mg²⁺ + 2Cl⁻
V(MgCl₂)=285cm³=0,285dm³
c(MgCl₂)=0,015 mol/dm³
n(MgCl₂)=c·V= 0,015 mol/dm³ · 0,285dm³ = 0,0042 mol
n(Mg²⁺)=n(MgCl₂)=0,0042 mol
n(Cl⁻)=2n(MgCl₂)=0,0084 mol

7 0

2 years ago

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