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igomit [66]
3 years ago
14

Why is blood classified as class evidence and not individual evidence?

Chemistry
1 answer:
Evgen [1.6K]3 years ago
3 0

Answer:

Explanation:

Blood type is considered to be class evidence. Although it may not specifically identify the suspect, explain how it still could be useful in helping to investigate a crime. used to rule out a certain suspect. Explain how the product rule can be used to determine whether two blood samples come from the same source.

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A substance is basic or alkaline if it's pH is greater than 7.0.

8 0
2 years ago
What is the order in which electrons start filling the orbitals?
liberstina [14]

The answer is actually A, 1s 2s 2p 3s 3p 4s 3d 4p.

7 0
3 years ago
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Which statement correctly uses a description of time?
Eva8 [605]

Answer:

A

Explanation:

it is because it began a new millennium

3 0
3 years ago
If have a volume of 18 L of a gas at a temperature of 272 K and a pressure of 90 atm, what will be the pressure of the gas if ra
Solnce55 [7]

Answer:

P₂ ≅ 100 atm (1 sig. fig. based on the given value of P₁ = 90 atm)

Explanation:

Given:

P₁ = 90 atm                    P₂ = ?

V₁ = 18 Liters(L)              L₂ = 12 Liters(L)      

=> decrease volume => increase pressure

=> volume ratio that will increase 90 atm is (18L/12L)                                                                  

T₁ = 272 Kelvin(K)          T₂ = 274 Kelvin(K)

=>  increase temperature => increase pressure

=> temperature ratio that will increase 90 atm is (274K/272K)

n₁ = moles = constant    n₂ = n₁ = constant

P₂ = 90 atm x (18L/12L) x (274K/272K) = 135.9926471 atm (calculator)

By rule of sig. figs., the final answer should be rounded to an accuracy equal to the 'measured' data value having the least number of sig. figs. This means P₂ ≅ 100 atm based on the given value of P₁ = 90 atm.

3 0
2 years ago
Menthol is a flavoring agent extracted from peppermint oil. it contains c, h, and o. in one combustion analysis, 10.00 mg of the
marishachu [46]
Combustion reaction for menthol is as follows;
CxHyOz + O₂ ---> xCO₂ +  H₂O
Mass of CO₂ formed - 28.16 mg
Therefore number of moles formed - 28.16/ 44 g/mol = 0.64 mmol
Mass of water formed  - 11.53 mg
number of water moles formed - 11.53 mg/18 g/mol = 0.64 mmol
From CO₂,
1 mol of CO₂ - 1 mol of C and 2 mol of O
therefore number of C moles - 0.64 mmol
                                 O moles - 1.28 mmol
from H₂O
1 mol of H₂O - 2 mol of H and 1 mol of O
number of H moles - 1.28 mmol
                 O moles - 0.64 mmol

Mass of menthol initially - 10 mg
in reactions, the masses of products are equal to the masses of reactants. The excess mass to the products formed is due to O₂ in air
Original mass of menthol - 10 mg
mass of water and CO₂ - 11.53 mg + 28.16 mg = 39.69
Difference in mass - 39.69 - 10 = 29.69 mg
This difference comes from O moles in air - 29.69 mg/ 16 g/mol = 1.8556 mmol
then O moles coming from menthol - (1.28 + 0.64) - 1.8556 = 0.064 mmol

In menthol 
C moles - 0.64 mmol
H moles - 1.28 mmol
O moles - 0.064 mmol
ratios of C:H:O
C          H         O
0.64    1.28      0.064
x1000  x1000    x1000 to get whole numbers
640       1280      64
10          20          1
Simplest ratio of C:H:O is 10:20:1
therefore empirical formula of menthol is C₁₀H₂₀O


5 0
3 years ago
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