Answer:
1.56 M
Explanation:
This is a dilution process and so a dilution formula is suitably used as follows C1V1 = C2V2 where
C1 = concentration of the stock solution
V1 = volume of the stock solution
C2 = concentration of the resulting (dilute) solution and
V2 = the volume of the resulting (dilute) solution
C1V1 = C2V2 (Making C2 subject of the formula)
C2 = C1V1/V2
Given: C1 = 5.736 M; V1 = 3 Ml; V2 = (3+8) 11 Ml
C2 = 5.736 x 3 / 11
C2 = 1.56 M
The concentration of the resulting solution will be "1.56 M".
Dilution process:The procedure of decreasing the concentration of such a particular solute through its solution has been known as dilution.
This same chemist could essentially add extra solvent to the mixture.
According to the question,
Stock solution's concentration, C₁ = 5.736 M
Stock solution's volume, V₁ = 3 mL
Resulting solution's volume, V₂ = 3 mL + 8 mL
= 11 mL
By using the dilution equation, we get
→ C₂ = [tex]\frac{C_1 V_1}{V_2}[/tex]
By substituting the above values,
= [tex]\frac{5.736\times 3}{11}[/tex]
= [tex]\frac{17.208}{11}[/tex]
= [tex]1.56[/tex] M
Thus the above answer is right.
Find out more information about concentration here:
https://brainly.com/question/16343005
A large plate is fabricated from a steel alloy that has a plane strain fracture toughness of 89 MPa (81.00 ksi). If the plate is exposed to a tensile stress of 336 MPa (48730 psi) during use, determine the minimum length of a surface crack that will lead to fracture. Assume a value of 0.92 for Y.
Explanation:
The given data is as follows.
[tex]K_{k}[/tex] = 89 MPa, [tex]\sigma[/tex] = 336 MPa
Y = 0.92
Now, we will calculate the length of critical interior flaw as follows.
[tex]a_{c} = \frac{1}{\pi}(\frac{K_{k}}{\sigma Y})^{2}[/tex]
= [tex]\frac{1}{\pi}(\frac{89}{336 \times 0.92})^{2}[/tex]
= [tex]\frac{656.38}{3.14}[/tex]
= 209.04 mm
Thus, we can conclude that minimum length of a surface crack that will lead to fracture is 209.04 mm.
Be sure to answer all parts. Complete the correct name for the following compounds. (a) Na3[Fe(CN)6] hexacyanoferrate (b) [Cr(en)2Cl2]I (ethylenediamine)chromium() (c) [Co(en)3]I3 (ethylenediamine)cobalt()
Answer:
1. Na3[Fe(CN)6]
Oxidation number of iron is +3
Sodium hexacyanoferrate (III)
2. [Cr (en)2Cl2]+
Oxidation number of chromium is +3
Dichlorobis (ethylenediamine)chromium (III) ion
3. [Co (en)3]Cl3
Oxidation number of cobalt is +3
Tris (ethylenediamine)cobalt (III) chloride
When performing qualitative tests in glass test tubes, such as the iodoform test or dinitrophenylhydrazine test, why should you avoid rinsing the glassware with acetone prior this experiment?
Answer:
It will lead to false positive (or negative results) for this classification tests.
Explanation:
In this tests, the functional group that is really being tested for is the Carbonyl group. In the iodoform, the presence of a Carbonyl group gives a reaction and in the dinitrophenylhydrazine test, aldehydes give a reaction and ketones do not.
So, rinsing the glassware with acetone is introducing ketone before the qualitative test has even began, thereby leading to false results for each of the two tests mentioned in the question.
Final answer:
Rinsing glassware with acetone before conducting qualitative tests such as the iodoform test or dinitrophenylhydrazine test should be avoided to prevent false positive results due to acetone's interference or reaction with the test compounds.
Explanation:
When performing qualitative tests in glass test tubes, such as the iodoform test or dinitrophenylhydrazine test, rinsing the glassware with acetone prior to the experiment should be avoided because it can lead to false positive results. These qualitative tests rely on specific chemical reactions to occur, and if residual acetone remained in the tube, it might react or interfere with the test reagents or the compound being tested, giving an inaccurate result. For instance, acetone itself can produce a yellow precipitate with dinitrophenylhydrazine, mimicking a positive result for the presence of certain carbonyl groups. Therefore, cleaning the glassware properly without using acetone, or ensuring that any acetone used is completely evaporated and the glassware is dried in a water-free environment, is crucial to the accuracy of these tests.
To quickly dry glassware without using acetone, rinsing with distilled water and allowing the glassware to dry overnight or using warm air or nitrogen gas for drying is recommended. Additionally, glassware cleanliness is paramount in chemical testing and experiments to avoid contamination that could affect the results.
Manganese commonly occurs in nature as a mineral. The extraction of manganese from the carbonite mineral rhodochrosite, involves a two-step process. In the first step, manganese (II) carbonate and oxygen react to form manganese (IV) oxide and carbon dioxide: 2MnCO3(s)+O2(g)→2MnO2(s)+2CO2(g) In the second step, manganese (IV) oxide and aluminum react to form manganese and aluminum oxide: 3MnO2(s)+4Al(s)→3Mn(s)+2Al2O3(s) Write the net chemical equation for the production of manganese from manganese (II) carbonate, oxygen and aluminum. Be sure your equation is balanced.
Answer: The net chemical equation for the formation of manganese from manganese (II) carbonate, oxygen and aluminum is written above.
Explanation:
The given chemical equation follows:
Equation 1: [tex]2MnCO_3(s)+O_2(g)\rightarrow 2MnO_2(s)+2CO_2(g)[/tex] ( × 3)
Equation 2: [tex]3MnO_2(s)+4Al(s)\rightarrow 3Mn(s)+2Al_2O_3(s)[/tex] ( × 2)
As, the net chemical equation does not include manganese (IV) oxide. So, to cancel out from the net equation, we need to multiply equation 1 by (3) and equation 2 by (2)
Now, the net chemical equation becomes:
[tex]6MnCO_3(s)+3O_2(g)+8Al(s)\rightarrow 6Mn(s)+4Al_2O_3(s)+6CO_2(g)[/tex]
Hence, the net chemical equation for the formation of manganese from manganese (II) carbonate, oxygen and aluminum is written above.
A chlorine atom is adsorbed on a small patch of surface (see sketch at right). This patch is known to contain possible adsorption sites. The atom has enough energy to move from site to site, so it could be on any one of them. Suppose a atom also becomes adsorbed onto the surface. Calculate the change in entropy. Round your answer to significant digits, and be sure it has the correct unit symbol.
The given question is incomplete. The complete question is as follows.
A chlorine atom is adsorbed on a small patch of surface (see sketch at right). This patch is known to contain 81 possible adsorption sites. The atom has enough energy to move from site to site, so it could be on any one of them. Suppose a Br atom also becomes adsorbed onto the surface. Calculate the change in entropy. Round your answer to significant digits, and be sure it has the correct unit symbol.
Explanation:
The change in entropy will be calculated by using the following formula.
[tex]\Delta S = k_{B} ln (\frac{W}{W_{o}})[/tex]
Initially, it is given tha Cl atom could be adsorbed on any of the 81 sites. When Br is added then there will be 80 possible sites when the Br can be adsorbed. This means that total possible sites are as follows.
[tex]81 \times 80[/tex]
= 6480
This shows that there are 6480 microstates which are accessible to the system.
So, change in entropy will be calculated using the Boltzmann constant as follows.
[tex]\Delta S = 1.381 \times 10^{-23} J/K \times ln (\frac{6480}{81})[/tex]
= [tex]1.381 \times 10^{-23} J/K \times 4.382[/tex]
= [tex]6.05 \times 10^{-23} J/K[/tex]
or, = [tex]6.1 \times 10^{-23} J/K[/tex] (approx)
Thus, we can conclude that the change in entropy is [tex]6.1 \times 10^{-23} J/K[/tex].