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PIT_PIT [208]
3 years ago
5

What type of scientific inquiry is friction ridge analysis

Chemistry
2 answers:
Ilia_Sergeevich [38]3 years ago
8 0

Answer:

Towards this goal, this project aims to develop a statistical measure of the uncertainty of the decisions made on the friction ridge evidence (i.e., evidential value of fingerprint comparison), which ultimately can be referred to as a scientific basis of the identification decisions made in friction ridge analysis.

Explanation:

MariettaO [177]3 years ago
8 0

Answer:

statistics friction ridge analysis summary. this project aims to develop a statistical measure of the uncertainty of the decision made on the friction ridge evidence, which ultimately can be referred to as a scientific basis of the identification decision made in friction ridge analysis.

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How many grams of copper (II) hydroxide can be prepared from 2.4 grams of copper (II) nitrate (Cu(NO3)2 ) and excess sodium hydr
andreyandreev [35.5K]

Answer:

percentage yield = 67%

Explanation:

Mass of Cu(NO₃)₂  = 15.25 g

Mass of NaOH   = 12.75 g

Percentage yield = ?

Solution:

Cu(NO₃)₂ + 2NaOH   →  Cu(OH)₂ + 2NaNO₃

Moles of Cu(NO₃)₂:

Number of moles = mass/ molar mass

Number of moles = 15.25 g /187.56 g/mol

Number of moles = 0.08 mol

Moles of NaOH :

Number of moles = mass/ molar mass

Number of moles = 12.75 g / 40 g/mol

Number of moles = 0.32 mol

Now we will compare the moles of Cu(OH)₂ with NaOH and Cu(NO₃)₂.      NaOH             :      Cu(OH)₂

                               2                   :          1

                               0.32              :           1/2×0.32 = 0.16 mol

                            Cu(NO₃)₂         :           Cu(OH)₂

                                  1                  :               1

                             0.08                :              0.08

The number of moles produced by  Cu(NO₃)₂  are less so it will limiting reactant.

Mass of Cu(OH)₂:

Mass = number of moles × molar mass

Mass = 0.08 mol × 97.6 g/mol

Mass = 7.808 g

Theoretical yield = 7.808 g

Percent yield:

percentage yield = Actual yield/ theoretical yield ×  100

percentage yield = 5.23 g/  7.808 g ×  100

percentage yield = 0.67 ×  100

percentage yield = 67%

5 0
2 years ago
2.(04.01 LC)
Mila [183]

Answer:

2KClO3 —> 2KCl + 3O2

The coefficients are 2, 2, 3

Explanation:

From the question given above, we obtained the following equation:

KClO3 —> 2KCl + 3O2

The above equation can be balance as follow:

There are 2 atoms of K on the right side and 1 atom on the left side. It can be balance by putting 2 in front of KClO3 as shown below:

2KClO3 —> 2KCl + 3O2

Now, the equation is balanced.

Thus, the coefficients are 2, 2, 3

8 0
2 years ago
Which of these is a chemical property?<br><br> boiling point<br> odor<br> ability to rust<br> color
NISA [10]

Answer:

ability to rust

Explanation:

i'm like 90% sure thats correct

6 0
3 years ago
Question 1 of 10
kati45 [8]
I believe the answer is C.
3 0
2 years ago
Solution C is a 1.00 L buffer solution that is 1.420 M in acetic acid and 0.67 M in sodium acetate. Acetic acid has a pKa of 4.7
Maslowich

Answer:

The correct answer is 0.10.

Explanation:

Based on the given question, in a buffer solution of 1 liter, the molarity of acetic acid is 1.420 M and the molarity of sodium acetate is 0.67. The pKa value of acetic acid given is 4.74, now the pH of buffer is,  

pH of buffer = pKa + log ([CH3COONa]/[CH3COOH])

= 4.74 + log (0.67/1.420)

= 4.74 + (-0.326)

= 4.41  

Now 0.10100 mol of HCl is added, the HCl reacts with sodium acetate to give,  

CH3COONa + HCl = CH3COOH + NaCl

Now the concentration of CH3COONa becomes = 0.67-0.101 = 0.57 M, and the new concentration of CH3COOH becomes = 1.420 + 0.101 = 1.52 M

Now the new pH will be,  

= pKa + log (0.57/1.52)

= 4.74 + (-0.426)

= 4.31

The pH change is 4.41-4.31 = 0.10  

3 0
3 years ago
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