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2 years ago

You burn 270 grams of glucose during respiration. How

Chemistry

1 answer:

BlackZzzverrR [31]2 years ago

8 0

Answer:

396 g OF CO2 WILL BE PRODUCED BY 270 g OF GLUCOSE IN A RESPIRATION PROCESS.

Explanation:

To calculate the gram of CO2 produced by burning 270 g of gucose, we first write out the equation for the reaction and equate the two variables involved in the question;

C6H12O6 + 6O2 -------> 6CO2 + 6H2O

1 mole of C6H12O6 reacts to form 6 moles of CO2

Then, calculate the molar mass of the two variables;

Molar mass of glucose = ( 12 *6 + 1* 12 + 16* 6) g/mol = 180 g/mol

Molar mass of CO2 = (12 + 16 *2) g/mol = 44 g/mol

Next is to calculate the mass of glucose and CO2 involved in the reaction by multiplying the molar mass by the number of moles

1* 180 g of glucose yields 6 * 44 g of CO2

180 g of glucose = 264 g of CO2

If 270 g of glucose were to be used, how many grams of CO2 will be produced;

so therefore,

180 g of glucose = 264 g of CO2

270 g of glucose = x grams of CO2

x = 264 * 270 / 180

x = 71 280 / 180

x = 396 g of CO2.

In other words, 396 g of CO2 will be produced by respiration from 270 g of glucose.

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A solution was prepared by dissolving 0.800 g of sulfur S8, in 100.0 g of acetic acid, HC2H3O2. Calculate the freezing point and

Romashka [77]

Answer: The freezing point of solution is 16.5°C and the boiling point of solution is 118.2°C

Explanation:

To calculate the molality of solution, we use the equation:

$Molality=\frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ in grams}}$

Where,

$m_{solute}$ = Given mass of solute $(S_8)$ = 0.800 g

$M_{solute}$ = Molar mass of solute $(S-8)$ = 256.52 g/mol

$W_{solvent}$ = Mass of solvent (acetic acid) = 100.0 g

Putting values in above equation, we get:

$\text{Molality of solution}=\frac{0.800\times 1000}{256.52\times 100.0}\\\\\text{Molality of solution}=0.0312m$

Calculation for freezing point of solution:

Depression in freezing point is defined as the difference in the freezing point of water and freezing point of solution.

$\Delta T_f=\text{freezing point of acetic acid}-\text{Freezing point of solution}$

To calculate the depression in freezing point, we use the equation:

$\Delta T_f=iK_fm$

or,

$\text{Freezing point of acetic acid}-\text{Freezing point of solution}=iK_fm$

where,

Freezing point of acetic acid = 16.6°C

i = Vant hoff factor = 1 (for non-electrolyte)

$K_f$ = molal freezing point depression constant = 3.59°C/m

m = molality of solution = 0.0312 m

Putting values in above equation, we get:

$16.6^oC-\text{freezing point of solution}=1\times 3.59^oC/m\times 0.0312m\\\\\text{Freezing point of solution}=16.5^oC$

Hence, the freezing point of solution is 16.5°C

Calculation for boiling point of solution:

Elevation in boiling point is defined as the difference in the boiling point of solution and freezing point of pure solution.

The equation used to calculate elevation in boiling point follows:

$\Delta T_b=\text{Boiling point of solution}-\text{Boiling point of acetic acid}$

To calculate the elevation in boiling point, we use the equation:

$\Delta T_b=iK_bm$

or,

$\text{Boiling point of solution}-\text{Boiling point of acetic acid}=iK_fm$

where,

Boiling point of acetic acid = 118.1°C

i = Vant hoff factor = 1 (for non-electrolyte)

$K_f$ = molal boiling point elevation constant = 3.08°C/m

m = molality of solution = 0.0312 m

Putting values in above equation, we get:

$\text{Boiling point of solution}-118.1^oC=1\times 3.08^oC/m\times 0.0312m\\\\\text{Boiling point of solution}=118.2^oC$

Hence, the boiling point of solution is 118.2°C

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lisabon 2012 [21]

Explanation:

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Bromine, a liquid at room temperature, has a boiling point of 58°C and a melting point of -7.2°C. Bromine can be classified as a

1. compound.

2. impure substance.

3. mixture.

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pure substance.

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otez555 [7]

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1 year ago

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