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3 years ago

Given the atomic weights of carbon, 12.01; hydrogen, 1.01; and oxygen, 16.0, what is the molar mass of glucose?

2 answers:

8 0

You just multiply the atomic weight by how many mol of each element are in a mol of the molecule.

<span>(6mol carbon * 12.01g/mol carbon) + (12 mol Hydrogen * 1.01g/mol) + (6 mol oxygen * 16.0g/mol oxygen) = 180.18 g / mol glucose.
I hope my answer helped you.</span>

Elan Coil [88]3 years ago

3 0

The molar mass of any molecule is the sum of its constituent elements atomic masses. In glucose, there are six carbon atoms, twelve hydrogen atoms, and six oxygen atoms. If we show this mathematically:
6(12.01 g/mol) + 12(1.01 g/mol) + 6(16.00 g/mol) = 180.18 grams/mole is the molar mass of glucose
The answer is C) 180.18 grams

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Assume that the maximum number of ATPs is produced (38). At pH 7, and in the presence of excess Mg2 , how much of the energy in

Lubov Fominskaja [6]

In one mole of glucose 38 ATP energy is stored this accounts for only 40 per-cent of the total energy in glucose.

Explanation:

In standard conditions, during the cellular respiration 1 mole of Glucose in the presence of oxygen produces 36 or 38 ATPs. This accounts for only 40% of the total energy as the remaining 60 per-cent of the energy is dissipated as heat.

I mole of glucose enters the glycolysis step of aerobic cellular respiration which after oxidative phosphorylation and Electron transport chain would give 38 ATP molecules.

It can be said that only 38.3% of energy is put in ATP molecules.

8 0

3 years ago

Nitrogen and hydrogen combine to form ammonia in the Haber process. Calculate (in kJ) the standard enthalpy change ∆H° for the r

Kazeer [188]

Answer:

∆H° rxn = - 93 kJ

Explanation:

Recall that a change in standard in enthalpy, ∆H°, can be calculated from the inventory of the energies, H, of the bonds broken minus bonds formed (H according to Hess Law.

We need to find in an appropiate reference table the bond energies for all the species in the reactions and then compute the result.

N₂ (g) + 3H₂ (g) ⇒ 2NH₃ (g)

1 N≡N = 1(945 kJ/mol) 3 H-H = 3 (432 kJ/mol) 6 N-H = 6 ( 389 kJ/mol)

∆H° rxn = ∑ H bonds broken - ∑ H bonds formed

∆H° rxn = [ 1(945 kJ) + 3 (432 kJ) ] - [ 6 (389 k J]

∆H° rxn = 2,241 kJ -2334 kJ = -93 kJ

be careful when reading values from the reference table since you will find listed N-N bond energy (single bond), but we have instead a triple bond, N≡N, we have to use this one .

8 0

3 years ago

During combustion, methane yields carbon dioxide and water. The unbalanced equation for this reaction is:CH4(g)+O2(g) → CO2(g)+

almond37 [142]

Answer:

CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l)

Mole ratios for the balanced equation be:

1:2, 2:1, 1:1, 1:2, 2:2

Explanation:

CH4(g)+O2(g) → CO2(g)+ H2O(l)

To balance a chemical equation, you must have the same mole of each element in both sides of the reaction (reactant side and product side)

CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l)

2 C

8 H

8 O

In both side.

It takes 1 mole of methane to react with 2 mole of oxygen in order to produce 1 mol of dioxide and 2 mole of water.

Mole ratios for the balanced equation be:

1:2, 2:1, 1:1, 1:2, 2:2 - We should compare each compound

1 mol methane → 2 mole of oxygen

2 mole of oxygen → 1 mol of methane

2 mole of oxygen → 1 mol of dioxide

1 mole of dioxide → 1 mol of Methane

2 mole of water → 2 mole of oxygen (the same as opposite)

1 mol methane → 2 mole of water

2 mol of water → 1 mol of methane

4 0

3 years ago

A solution is prepared by mixing 2.17 g of an unknown non-electrolyte with 225.0 g of chloroform. The freezing point of the resu

Deffense [45]

Answer:

The molar mass of the unknown non-electrolyte is 64.3 g/mol

Explanation:

Step 1: Data given

Mass of an unknown non-electrolyte = 2.17 grams

Mass of chloroform = 225.0 grams

The freezing point of the resulting solution is –64.2 °C

The freezing point of pure chloroform is – 63.5°C

kf = 4.68°C/m

Step 2: Calculate molality

ΔT = i*kf*m

⇒ ΔT = The freezing point depression = T (pure solvent) − T(solution) = -63.5°C + 64.2 °C = 0.7 °C

⇒i = the van't Hoff factor = non-electrolyte = 1

⇒ kf = the freezing point depression constant = 4.68 °C/m

⇒ m = molality = moles unknown non-electrolyte / mass chloroform

0.7 °C = 1 * 4.68 °C/m * m

m = 0.150 molal

Step 3: Calculate moles unknown non-electrolyte

molality = moles unknown non-electrolyte / mass chloroform

Moles unknown non-electrolyte = 0.150 molal * 0.225 kg

Moles unknown non-electrolyte = 0.03375 moles

Step 4: Calculate molecular mass unknown non-electrolyte

Molar mass = mass / moles

Molar mass = 2.17 grams / 0.03375 moles

Molar mass = 64.3 g/mol

The molar mass of the unknown non-electrolyte is 64.3 g/mol

6 0

3 years ago

When an object is lifted 6 meters off the ground, it gains a certain amount of potential energy. If the same object is lifted 18

AlexFokin [52]

Answer:

The answer to your question is the letter C. three times as much

Explanation:

Data

First step = 6 m

Second step = 18 m

Potential energy is the energy stored that depends on its position.

Formula

Pe = mgh

m = mass; g = gravity; h = height

Potential energy of the first step

Pe1 = 6mg

Potential energy of the second step

Pe2 = 18mg

-Divide the Pe2 by the Pe1

Pe2/Pe1 = 18mg/6mg

= 3

7 0

2 years ago