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2 years ago

What could be a reason for the difference between the theoretical and experimental heats of reaction

Chemistry

1 answer:

Fiesta28 [93]2 years ago

8 0

Answer:

Theoretical value is the value a scientist expects from an equation, assuming perfect or near-perfect conditions. Experimental value, on the other hand, is what is actually measured from an experiment. Rarely (in fact never) are these numbers the same.

Take the area of a sheet of paper. I know that the area of a rectangle can be found by multiplying the lengths of both sides together. I can assume an 8.5x11 sheet, so I calculate exactly 93.5 square inches. This is my theoretical value. When I actually do the measurements on my paper, it turns out my paper has been slightly cut on one end, or I’m measuring with a shoddy ruler. Therefore, I might measure only 92.8 square inches. This is my experimental value. See thats it’s close, but not exact.

Expanding on this concept, quantum mechanics is so widely accepted in the scientific community because many theoretical values calculated by mathmaticians concurred with experimental values to many, many decimal places. These would be constants such as Plank’s Constant, energy levels of harmonic potentials, and energy levels of the hydrogen atom.

Explanation:

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describes a type of reaction in which negative ions in two compounds switch places , forming two new compounds

Nadusha1986 [10]

Answer:

A double displacement reaction or double replacement reaction.

Explanation:

5 0

3 years ago

Read 2 more answers

The combustion of 0.570 g of benzoic acid (ΔHcomb = 3,228 kJ/mol; MW = 122.12 g/mol) in a bomb calorimeter increased the tempera

torisob [31]

Answer:

The temperature change from the combustion of the glucose is 6.097°C.

Explanation:

Benzoic acid;

Enthaply of combustion of benzoic acid = 3,228 kJ/mol

Mass of benzoic acid = 0.570 g

Moles of benzoic acid = $\frac{0.570 g}{122.12 g/mol}=0.004667 mol$

Energy released by 0.004667 moles of benzoic acid on combustion:

$Q=3,228 kJ/mol \times 0.004667 mol=15.0668 kJ=15,066.8 J$

Heat capacity of the calorimeter = C

Change in temperature of the calorimeter = ΔT = 2.053°C

$Q=C\times \Delta T$

$15,066.8 J=C\times 2.053^oC$

$C=7,338.92 J/^oC$

Glucose:

Enthaply of combustion of glucose= 2,780 kJ/mol.

Mass of glucose=2.900 g

Moles of glucose = $\frac{2.900 g}{180.16 g/mol}=0.016097 mol$

Energy released by the 0.016097 moles of calorimeter combustion:

$Q'=2,780 kJ/mol \times 0.016097 mol=44.7491 kJ=44,749.1 J$

Heat capacity of the calorimeter = C (calculated above)

Change in temperature of the calorimeter on combustion of glucose = ΔT'

$Q'=C\times \Delta T'$

$44,749.1 J=7,338.92 J/^oC\times \Delta T'$

$\Delta T'=6.097^oC$

The temperature change from the combustion of the glucose is 6.097°C.

6 0

3 years ago

Identify a cation. An atom that has gained a proton. An atom that has lost an electron. An atom that has gained an electron. An

iren2701 [21]

the answer would be B an atom that has lost an electron

8 0

2 years ago

Calcium chloride and silver nitrate react to form a precipitate of silver chloride in a solution of calcium nitrate. This is an

rewona [7]

B a Displacement reaction.

8 0

3 years ago

An aqueous solution is listed as being 33.8% solute by mass with a density of 1.15 g/mL, the molar mass of the solute is 145.6 g

vodomira [7]

Answer:

A) 2.69 M

B) 0.059

Explanation:

A) We have:

33.8% solute by mass= 33.8 g solute/100 g solution

molarity = mol solute/ 1 L solution

molarity= $\frac{33.8 g solute}{100 g solution}$ x $\frac{1.15 g solution}{1 ml}$ x $\frac{1 mol solute}{145.6 g solute}$ x $\frac{1000 ml}{1 L}$

molarity= 2.69 mol solute/L solution = 2.69 M

B) We know that there are 33.8 g of solute in 100 g of solution.

As the total solution is compounded by solute+solvent (in this case, solvent is water), the mass of water is the difference between the mass of the total solution and the mass of solute:

mass of water= 100 g - 33.8 g = 66.2 g

Now, we calculate the number of mol of both solute and water:

mol solute= 33.8 g solute x $\frac{1 mol solute}{145.6 g}$ = 0.232 mol

mol H20= 66.2 g H₂O x $\frac{1 mol H2O}{18 g}$

Finally, the mol fraction of solute (Xsolute) is calculated as follows:

Xsolute= $\frac{mol solute}{total mol}= \frac{mol solute}{mol solute + mol H2O}=\frac{0.232 mol}{0.232 mol + 3.677 mol}$

Xsolute= 0.059

4 0

3 years ago