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wlad13 [49]
4 years ago
15

Into which molecule are all the carbon atoms in glucose ultimately incorporated during cellular respiration?

Chemistry
1 answer:
pav-90 [236]4 years ago
3 0

Answer:

The correct option is: carbon dioxide (CO₂)

Explanation:

Cellular respiration refers to the process of generation of energy in the form of adenosine triphosphate (ATP) from various nutrients such as carbohydrates, fats, and proteins.

Aerobic cellular respiration involves the oxidation of nutrients such as glucose (C₆H₁₂O₆), by molecular oxygen (O₂) to give carbon dioxide (CO₂), water (H₂O) and heat energy.

Reaction involved: (glucose) C₆H₁₂O₆ (s) + 6 O₂ (g) → 6 CO₂ (g) + 6 H₂O (l) + heat energy

<u>Therefore, in a cellular respiration, all the carbon atoms of the glucose molecules, form carbon dioxide.</u>

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what is the molar mass of a gaseous flouride of sulfur containing 70.4% F and having a density of approximately 4.5g/L at 20 deg
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Answer:

The molar mass is 180.2 g/mol

Explanation:

<u>Step 1:</u> Data given

% of F = 70.4 %

Density = 4.5 g/L

Temperature = 20 °C

Pressure = 1 atm

<u>Step 2:</u> Calculate the number of moles

PV = nRT

 ⇒ with P = the pressure = 1.00 atm

⇒ with V = the volume = Assume this is 1L

⇒ with n = the number of moles = TO BE DETERMINED

⇒ R = the gas constant = 0.08206 L*atm/K*mol

⇒ T = the temperature : 20°C = 293 Kelvin

1 atm*1L= n(0.08206 L-atm/mol-K)*(293 K)

n = 0.04159 moles

<u>Step 3</u>: Calculate molar mass

Molar mass = Mass / moles

4.5 grams / 0.04159 moles = 108.2 g/mol

<u>Step 4:</u> Calculate moles of F

Moles = Mass / molar mass

Moles F = 70.4 g / 19 g/mol

Moles F =  3.70 moles

Moles S = 29.6g / 32.07 g/mol

Moles S = 0.923 moles S

<u>Step 5:</u> Divide by the smallest amount of moles

F = 3.70 / 0.923 = 4

S = 0.923 / 0.923 = 1

The empirical formula is SF4

The molar mass of SF4 = 32.07 + 4*19 = 108.07 g/mol

This means the empirical formula is the same as the molecular formula SF4

The molar mass is 180.2 g/mol

4 0
3 years ago
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Answer:

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Explanation:

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3 years ago
Estimate the Calorie content of 65 g of candy from the following measurements. A 15-g sample of the candy is placed in a small a
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Answer:

The calorie content of  65g of candy is 326.78 cal

Explanation:

Step 1: Data given

Mass of the candy = 15.00 grams

Mass of the container = 0.325 kg

Mass of water = 1.75kg

0.624 kg at an initial temperature of 15.0°C.

The specific heat of aluminium = 0.22 Cal/kg°C

The specific heat of water = 1 cal/kg°C

Step 2: Calculate calorie content for a 15 gram sample

ΔQ = Σm*c*ΔT

 ⇒ m = mass in grams

⇒ with c= the specific heat in Cal/kg°C

⇒ with ΔT = T2 -T1 = the change in temperatures in °C

ΔQ = m(bomb) * C(aluminium) * ΔT + m(cup) * C(aluminium) * ΔT + m(H2O) * c(H20) * ΔT

ΔQ = (m(bomb) + m(cup)) * c(aluminium)  + m(H2O)*c(H20) ) * ΔT

⇒ with mass of the bomb calorimeter = 0.325 kg

⇒ with mass of the cup = 0.624 kg

⇒ with c(aluminium) = the specific heat of aluminium = 0.22 Cal/kg°C

⇒ with mass of water = 1.75 kg

⇒ with c(water) = the heat capacity of water = 1 Cal/kg°C

⇒ with ΔT = the change in temperature = T2 - T1 = 53.5 - 15.0 = 38.5 °C

ΔQ = 0.325*0.22*38.5 + 0.624*0.22*38.5 + 1.75*1*38.5

ΔQ = ((0.325 + 0.624)*0.22 + 1.75*1)*38.5

ΔQ = 75.41 cal

Step 3: Calculate the calorie content for a 65 gram sample

For a 65g sample the calorie content will be more or less 4x higher than a 15 gram sample:

ΔQ = 75.41  * (65/15) = 326.78 cal

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How do we know stars are element factories?
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over time, elements got more dense as they collided and cooled, leading to other gases, metals...

then they explode, which lead to all kinds of further complex elements.

The stuff that makes up the universe a lot of it is from exploding star stuff, and the Earth was built up upon these remains and the atoms that make up organisms even could originally be from stars.

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