CHEMISTRY COLLEGE

Glucose, C 6 H 12 O 6 , is used as an energy source by the human body. The overall reaction in the body is described by the equation C 6 H 12 O 6 ( aq ) + 6 O 2 ( g ) ⟶ 6 CO 2 ( g ) + 6 H 2 O ( l ) Calculate the number of grams of oxygen required to convert 58.0 g of glucose to CO 2 and H 2 O . mass of O 2 : 61.76 g Calculate the number of grams of CO 2 produced.

Answers

Answer 1

Answer:

Answer:

$m_(O_2)=61.87gO_2$

$m_(CO_2)=85.07gCO_2$

Explanation:

Hello,

Considering the given reaction's stoichiometry, grams of oxygen result:

$m_(O_2)=58.0gC_6H_(12)O_6*(1molC_6H_(12)O_6)/(180gC_6H_(12)O_6)*(6molO_2)/(1molC_6H_(12)O_6)*(32gO_2)/(1molO_2)\nm_(O_2)=61.87gO_2$

Moreover, the mass of produced carbon dioxide turns out:

$m_(CO_2)=58.0gC_6H_(12)O_6*(1molC_6H_(12)O_6)/(180gC_6H_(12)O_6)*(6molCO_2)/(1molC_6H_(12)O_6)*(44gCO_2)/(1molCO_2)\nm_(O_2)=85.07gCO_2$

Best regards.

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Calculate the entropy change for the gas. (2b)Calculate the entropy change when 14 g of nitrogen expand from a volume of 10 L to a volume of 30 L at the same temperature. Assuming ideal behaviour for the nitrogen gas

Answers

Explanation:

because T is const so

deltaS=Q/T=nRLn(V2/V1)

=(14/28)x8.314xLn(30/10)=4,567 J/k

How many unpaid electrons are in 1s2 2s2 2p6 3s2 3p6

Answers

Answer:

As we can see, the first shell has its 2 electrons

the second shell has its 8 electrons

we know that the third shell also needs 8 electrons and we can see that the third shell also has 8 electrons.

Therefore, all the shells of this atom are filled and this atom is stable

So we can say that there are no unpaired electrons

Kindly mark Brainliest, thanks

Answer:

No unpaired electrons

Explanation:

This is because it is diamagnetic. Hope that helps:)

In addition to displacing halide ions, the acetylide ion also adds to carbonyl groups. 2-Methyl-3-butyn-2-ol (MBI) is an acetylenic alcohol used in the manufacture of products for the agrochemical and specialty chemical industry. It can be synthesized by the addition of acetylene to acetone to form the alkoxide ion and, as a second step, protonation of the alkoxide ion to produce the alcohol. Complete the mechanism for 2-methyl-3-butyn-2-ol production by drawing in the products of each step and the missing curved arrows. Sodium amide deprotonates the terminal alkyne to form sodium ethynide. Draw all missing reactants and/or products in the appropriate boxes by placing atoms on the canvas and connecting them with bonds. Add charges where needed. Electron flow arrows should start on an atom or a bond and should end on an atom, bond, or location where a new bond should be create