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butalik [34]
3 years ago
12

The neutralization of H3PO4 with KOH is exothermic. 55.0 mL of 0.213 M H3PO4 is mixed with 55.0 mL of 0.640 M KOH initially at 2

2.43 °C. Predict the final temperature of the solution if its density is 1.13 g/mL and its specific heat is 3.78 J/(g·°C) Assume that the total volume is the sum of the individual volumes.
RXN: H3 PO4(aq) + 3KOH(aq) 3 H2O(l) + K3PO4(aq) + 173.2 kj
Chemistry
1 answer:
Lilit [14]3 years ago
7 0

Answer:

26.75ºC

Explanation:

By the reaction, we can notice that the reaction releases 173.2 kJ (173200 J) per mol of H₃PO₄. This heat can also be calculated by:

Heat released per mol of H₃PO₄ = msolution*csolution*ΔT/nH₃PO₄

Where msolution is the mass of the solution, csolution is the specific heat, ΔT is the variation at temperature (final - initial) and nH₃PO₄ is the number of moles of H₃PO₄.

msolution = Volume*density

msolution = (55 + 55)*1.13

msolution = 124.3 g

nH₃PO₄ = Volume * concentration

nH₃PO₄ = 0.055*0.213 = 0.011715 mol

173200 = 124.3*3.78*(T - 22.43)/0.011715

469.9674(T - 22.43) = 2029.038

T - 22.43 = 4.32

T = 26.75ºC

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Solid potassium chlorate decomposes upon heating to form
olga55 [171]

Answer:

32.6%

Explanation:

Equation of reaction

2KClO₃ (s) → 2KCl (s) + 3O₂ (g)

Molar mass of 2KClO₃ = 245.2 g/mol ( 122.6 × 2)

Molar volume of Oxygen at s.t.p = 22.4L / mol

since the gas was collected over water,

total pressure = pressure of water vapor + pressure of  oxygen gas

0.976 = 0.04184211 atm + pressure of oxygen gas at 30°C

pressure of oxygen = 0.976 - 0.04184211 = 0.9341579 atm = P1

P2 = 1 atm, V1 = 789ml, V2 = unknown, T1 = 303K, T2 = 273k at s.t.p

Using ideal gas equation

\frac{P1V1}{T1} = \frac{P2V2}{T2}

V2 = \frac{P1V1T2}{T1P2}

V2 = 664.1052 ml

245.2   yielded 67.2 molar volume of oxygen

0.66411 will yield = \frac{245.2 * 0.66411}{67.2}  = 2.4232 g

percentage of potassium chlorate in the original mixture = \frac{2.4232 * 100}{7.44} = 32.6%

3 0
3 years ago
If you mix 50mL of 0.1 M TRIS acid with 60 mL of0.2 M<br> TRIS base, what will be the resulting pH?
Katyanochek1 [597]

<u>Answer:</u> The pH of resulting solution is 8.7

<u>Explanation:</u>

To calculate the number of moles for given molarity, we use the equation:

\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}

  • <u>For TRIS acid:</u>

Molarity of TRIS acid solution = 0.1 M

Volume of solution = 50 mL

Putting values in above equation, we get:

0.1M=\frac{\text{Moles of TRIS acid}\times 1000}{50mL}\\\\\text{Moles of TRIS acid}=0.005mol

  • <u>For TRIS base:</u>

Molarity of TRIS base solution = 0.2 M

Volume of solution = 60 mL

Putting values in above equation, we get:

0.2M=\frac{\text{Moles of TRIS base}\times 1000}{60mL}\\\\\text{Moles of TRIS base}=0.012mol

Volume of solution = 50 + 60 = 110 mL = 0.11 L    (Conversion factor:  1 L = 1000 mL)

  • To calculate the pH of acidic buffer, we use the equation given by Henderson Hasselbalch:

pH=pK_a+\log(\frac{[salt]}{[acid]})

pH=pK_a+\log(\frac{[\text{TRIS base}]}{[\text{TRIS acid}]})

We are given:

pK_a = negative logarithm of acid dissociation constant of TRIS acid = 8.3

[\text{TRIS acid}]=\frac{0.005}{0.11}

[\text{TRIS base}]=\frac{0.012}{0.11}

pH = ?

Putting values in above equation, we get:

pH=8.3+\log(\frac{0.012/0.11}{0.005/0.11})\\\\pH=8.7

Hence, the pH of resulting solution is 8.7

6 0
4 years ago
How does the process of waste removal in a unicellular organism compare to the waste removal in a multicellular organism
Sonja [21]
The digestive system of multicellular is made up of several organs that work together to break down food so that it may be used in the body. Unicellular organisms have organelles, similar to organs, which help to digest food that will be used by the cell
6 0
3 years ago
Can anyone help me on this ques pls, much appreaciated
Sloan [31]

Answer:

Option D is Correct answer

4 0
3 years ago
Read 2 more answers
How many grams of iki would it take to obtain a 100 ml solution of 0.300 m iki? how many grams of iki would it take to create a
GalinKa [24]

0.300 M IKI represents the concentration which is in molarity of a potassium iodide solution. This means that for every liter of solution there are 0.300 moles of potassium iodide. Knowing that molarity is a ratio of solute to solution.

By using a conversion factor:

100 ml x (1L / 1000 mL) x (0.300 mol Kl / 1 L) x (166.0g / 1 mol Kl) = 4.98 g

Therefore, in the first conversion by simply converting the unit of volume to liter, Molarity is in L where the volume is in liters. The next step is converted in moles from volume by using molarity as a conversion factor which is similar to how density can be used to convert between volume and mass. After converting to moles it is simply used as molar mass of Kl which is obtained from periodic table to convert from mole to grams.

In order to get the grams of IKI to create a 100 mL solution of 0.600 M IKI, use the same formula as above:

100 ml x (1L / 1000 mL) x (0.600 mol Kl / 1 L) x (166.0g / 1 mol Kl) = 9.96 g

3 0
3 years ago
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