All categories

2 years ago

The human body needs at least 1.03 x 10-^2 mol O2 every minute. If all of this oxygen is used for the

Chemistry

1 answer:

Neko [114]2 years ago

3 0

<h3>Answer:</h3>

0.3093 g of glucose are consumed each minute by the body.

<h3>Explanation:</h3>

During cellular respiration glucose is broken down in presence of oxygen to yield energy, water and carbon dioxide.
The equation for the reaction taking place during cellular respiration is;

C₆H₁₂O₆ + 6O₂ → 6H₂O + 6CO₂

We are required to calculate the amount of glucose in grams;

<h3>Step 1: Calculate the moles of glucose broken down</h3>

From the equation, the mole ratio of glucose to Oxygen is 1 : 6

Moles of Oxygen in a minute is 1.03 × 10^-2 moles

Therefore, moles of glucose will be;

= (1.03 × 10^-2)÷6

= 1.717 × 10^-3 moles

<h3>Step 2: Mass of glucose </h3>

Mass is given by multiplying the number of moles with molar mass

mass = moles × molar mass

Molar mass glucose is 180.156 g/mol

Therefore;

Mass = 1.717 × 10^-3 × 180.156 g/mol

= 0.3093 g

Hence, 0.3093 g of glucose are consumed each minute by the body.

You might be interested in

In May 2016, William Trubridge broke the world record in free diving (diving underwater without the use of supplemental oxygen)

Maru [420]

Answer:

The volume that this same amount of air will occupy in his lungs when he reaches a depth of 124 m is - 0.27 L.

Explanation:

Using Boyle's law

${P_1}\times {V_1}={P_2}\times {V_2}$

Given ,

V₁ = 3.6 L

V₂ = ?

P₁ = 1.0 atm

P₂ = 13.3 atm (From correct source)

Using above equation as:

${P_1}\times {V_1}={P_2}\times {V_2}$

${1.0\ atm}\times {3.6\ L}={13.3\ atm}\times {V_2}$

${V_2}=\frac{{1.0}\times {3.6}}{13.3}\ L$

${V_2}=0.27\ L$

The volume that this same amount of air will occupy in his lungs when he reaches a depth of 124 m is - 0.27 L.

7 0

2 years ago

The chlorination of methane occurs in a number of steps that results in the formation of chloromethane and hydrogen chloride. Th

kenny6666 [7]

Answer:

Total pressure = 0,806 atm

Partial pressure of CH₄: 0,037 atm

Partial pressure of Cl₂: 0,396 atm

Partial pressure of CH₃Cl: 0,125 atm

Partial pressure of HCl: 0,125 atm

Partial pressure of Cl⁻: 0,125 atm

Explanation:

For the reaction:

2CH₄(g)+3Cl₂(g)⟶2CH₃Cl(g)+2HCl(g)+2Cl⁻(g)

295 mL≡ 0,295L of methane at STP are:

n = PV/RT

P = 1 atm; V = 0,295L; R = 0,082atmL/molK; T = 273K.

moles of methane: 0,0132 moles

For 725 mL of chlorine ≡ 0,725L

moles of chlorine at STP are: ≡ 0,0324 moles

For a complete reaction of 0,0132 moles of CH₄:

0,0132 mol CH₄× $\frac{3molCl_{2}}{2 molCH_{4}}$ = <em>0,0198 moles</em>

The reaction reaches 77%, moles of Cl₂ that react are: 0,0198×77% = 0,0153 mol

As you have 0,0324 moles of Cl₂, moles that will not react are:

0,0324 - 0,0153 = <em>0,0171 mol Cl₂</em>

As the reaction reaches 77% completion, moles of CH₄ that react are:

0,0132×77% =<em> 0,0102 moles of CH₄ And the moles that don't react are </em><em>0,00300 mol</em>

Thus, moles of each compound are:

0,0102 moles of CH₄× $\frac{3Cl_{2}}{2 molCH_{4}}$ = <em>0,0153 mol + 0,0171 mol = 0,0324 mol Cl₂</em>

0,0102 moles of CH₄× $\frac{2CH_{3}Cl}{2 molCH_{4}}$ = <em>0,0102 mol CH₄</em>

0,0102 moles of CH₄× $\frac{2HCl}{2 molCH_{4}}$ = <em>0,0102 mol HCl</em>

0,0102 moles of CH₄× $\frac{2Cl^{-}}{2 molCH_{4}}$ = <em>0,0102 mol Cl⁻</em>

Total pressure using:

P = nRT/V

Where: n = 0,0102mol×3+0,0324mol + 0,0030mol = 0,0660mol; R = 0,082 atmL/molK; T = 298K; V = 2L

Total pressure = 0,806 atm

Partial pressure of CH₄: 0,037 atm

Partial pressure of Cl₂: 0,396 atm

Partial pressure of CH₃Cl: 0,125 atm

Partial pressure of HCl: 0,125 atm

Partial pressure of Cl⁻: 0,125 atm

<em>-To obtain partial pressure you change the moles for each compound-</em>

<em />

I hope it helps!

5 0

2 years ago

Large sharks eat many other marine life animals. The sharks and the animals they eat are all part of which level of organization

Sladkaya [172]

Answer:

the answer is ecosystem i believe.

8 0

2 years ago

Read 2 more answers

6K + B2O3 → 3K2O + 2B

atroni [7]

Answer:

104.84 moles

Explanation:

Given data:

Moles of Boron produced = ?

Mass of B₂O₃ = 3650 g

Solution:

Chemical equation:

6K + B₂O₃ → 3K₂O + 2B

Number of moles of B₂O₃:

Number of moles = mass/ molar mass

Number of moles = 3650 g/ 69.63 g/mol

Number of moles = 52.42 mol

Now we will compare the moles of B₂O₃ with B from balance chemical equation:

B₂O₃ : B

1 : 2

52.42 : 2×52.42 = 104.84

Thus from 3650 g of B₂O₃ 104.84 moles of boron will produced.

6 0

2 years ago

The concentration of sodium chloride in an aqueous solution that is 2.23 M and that has a density of 1.01 g/mL is __________% by

amm1812

Answer:

The concentration of sodium chloride in an aqueous solution that is 2.23 M and that has a density of 1.01 g/mL is 12.90% by mass

Explanation:

2.23 M aqueous solution of NaCl means there are 2.23 moles of NaCl in 1000 mL of solution.

We know that density is equal to ratio of mass to volume.

Here density of solution is 1.01 g/mL.

So mass of 1000 mL solution = ( $1.01\times 1000$ ) g = 1010 g

molar mass of NaCl = 58.44 g/mol

So mass of 2.23 moles of NaCl = ( $2.23\times 58.44$ ) g = 130.3 g

% by mass is ratio of mass of solute to mass of solution and then multiplied by 100.

Here solute is NaCl.

So % by mass of 2.23 M aqueous solution of NaCl = $\frac{130.0}{1010}\times 100$ % = 12.90%

3 0

2 years ago