Answer:
Explanation:
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In this case, it is possible to comprehend these mass-particles problems by means of the concept of mole, molar mass and the Avogadro's number because one mole of any substance has 6.022x10²³ particles and have a mass equal to the molar mass.
In such a way, for C₆H₁₂O₆, whose molar mass is about 180.16 g/mol, the referred mass would be:
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Answer:
See explanation and image attached
Explanation:
My aim is to convert 1-bromobutane to butanal. The first step is to react the 1-bromobutane substrate with water. This reaction occurs by SN2 mechanism to yield 1-butanol. Hence reagent A is water.
1-butanol is now reacted with an oxidizing agent such as acidified K2Cr2O7 (reagent B) to yield butanal. Note that primary alkanols are oxidized to alkanals.
These sequence of reactions are shown in the image attached.
I think because its the only one to be liquid at normal temperatures.
The empirical formula :
C₁₀H₁₆N₄SO₇
<h3>Further explanation</h3>
Given
6.4 g sample
Required
The empirical formula
Solution
mass C :
= 12/44 x 8.37 g
= 2.28
mass H :
= 2/18 x 2.75 g
= 0.305
mass N = 1.06
mass S :
= 32/64 x 1.23
= 0.615
mass O = 6.4 - (2.28+0.305+1.06+0.615) = 2.14 g
Mol ratio :
= C : H : N : S : O
= 2.28/12 : 0.305/1 : 1.06/14 : 0.615/32 : 2.14/16
= 0.19 : 0.305 : 0.076 : 0.019 : 0.133 divided by 0.019
= 10 : 16 : 4 : 1 : 7
The empirical formula :
C₁₀H₁₆N₄SO₇
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