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castortr0y [4]
3 years ago
9

if 334.6 g of phosphoric acid is reacted with excess potassium hydroxide. the final mass K3PO4 produced is found to be 248g. wha

t is the percent yield?
Chemistry
1 answer:
Orlov [11]3 years ago
3 0

Answer:

                     %age Yield  =  34.21 %

Explanation:

                   The balance chemical equation for the decomposition of KClO₃ is as follow;

                            3 KOH + H₃PO₄ → K₃PO₄ + 3 H₂O

Step 1: Calculate moles of H₃PO₄ as;

Moles = Mass / M/Mass

Moles = 334.6 g / 97.99 g/mol

Moles = 3.414 moles

Step 2: Find moles of K₃PO₄ as;

According to equation,

                 1 moles of H₃PO₄ produces  =  1 moles of K₃PO₄

So,

              3.414 moles of H₃PO₄ will produce  =  X moles of K₃PO₄

Solving for X,

                      X = 1 mol × 3.414 mol / 1 mol

                      X = 3.414 mol of K₃PO₄

Step 3: Calculate Theoretical yield of K₃PO₄ as,

Mass = Moles × M.Mass

Mass = 3.414 mol × 212.26 g/mol

Mass = 724.79 g of K₃PO₄

Also,

%age Yield  =  Actual Yield / Theoretical Yield × 100

%age Yield  =  248 g / 724.79 × 100

%age Yield  =  34.21 %

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Firlakuza [10]

It's a physical change because the composition of the salt hasn't change. Only the change in substance/form has occured. To tell if something has undergone physical change is:

• Melting

• Boiling

• Freezing

• Condensing

• Breaking

• Bending

• Dissolving

• Molecules can change motion and proximity

To tell if something changed chemically:

• Molecules rearrange with other molecules to make new substance

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4 0
3 years ago
3.
sineoko [7]

Answer:

3. V = 0.2673 L

4. V = 2.4314 L

5. V = 0.262 L

6. V = 2.224 L

Explanation:

3. assuming ideal gas:

  • PV = RTn

∴ R = 0.082 atm.L/K.mol

∴ V1 = 225 L

∴ T1 = 175 K

∴ P1 = 150 KPa = 1.48038 atm

⇒ n = RT/PV

⇒ n = ((0.082 atm.L/K.mol)(175 K))/((1.48038 atm)(225 L))

⇒ n = 0.043 mol

∴ T2 = 112 K

∴ P2 = P1 = 150 KPa = 1.48038 atm

⇒ V2 = RT2n/P2

⇒ V2 = ((0.082 atm.L/K.mol)(112 K)(0.043 mol))/(1.48038 atm)

⇒ V2 = 0.2673 L

4. gas is heated at a constant pressure

∴ T1 = 180 K

∴ P = 1 atm

∴ V1 = 44.8 L

⇒ n = RT/PV

⇒ n = ((0.082 atm.L/K.mol)(180 K))/((1 atm)(44.8 L))

⇒ n = 0.3295 mol

∴ T2 = 90 K

⇒ V2 = RT2n/P

⇒ V2 = ((0.082 atm.L/K.mol)(90 K)(0.3295 mol))/(1 atm)

⇒ V2 = 2.4314 L

5.  V1 = 200 L

∴ P1 = 50 KPa = 0.4935 atm

∴ T1 = 271 K

⇒ n = RT/PV

⇒ n = ((0.082 atm.L/K.mol)(271 K))/((0.4935 atm)(200 L))

⇒ n = 0.2251 mol

∴ P2 = 100 Kpa = 0.9869 atm

∴ T2 = 14 K

⇒ V2 = RT2n/P2

⇒ V2 = ((0.082 atm.L/K.mol)(14 K)(0.2251 mol))/(0.9869 atm)

⇒ V2 = 0.262 L

6.a)  ∴ V1 = 24.6 L

∴ P1 = 10 atm

∴ T1 = 25°C = 298 K

⇒ n = RT/PV

⇒ n = ((0.082 atm.L/K.mol)(298 K))/((10 atm)(24.6 L))

⇒ n = 0.0993 mol

∴ T2 = 273 K

∴ P2 = 101.3 KPa = 0.9997 atm

⇒ V2 = RT2n/P2

⇒ V2 = ((0.082 atm.L/K.mol)(273 K)(0.0993 mol))/(0.9997 atm)

⇒ V2 = 2.224 L

3 0
3 years ago
Th e molar absorption coeffi cient of a substance dissolved in water is known to be 855 dm3 mol−1 cm−1 at 270 nm. To determine t
Olegator [25]

Answer : The percentage reduction in intensity is 79.80 %

Explanation :

Using Beer-Lambert's law :

A=\epsilon \times C\times l

A=\log \frac{I_o}{I}

\log \frac{I_o}{I}=\epsilon \times C\times l

where,

A = absorbance of solution

C = concentration of solution = 3.25mmol.dm^{3-}=3.25\times 10^{-3}mol.dm^{-3}

l = path length = 2.5 mm = 0.25 cm

I_o = incident light

I = transmitted light

\epsilon = molar absorptivity coefficient = 855dm^3mol^{-1}cm^{-1}

Now put all the given values in the above formula, we get:

\log \frac{I_o}{I}=(855dm^3mol^{-1}cm^{-1})\times (3.25\times 10^{-3}mol.dm^{-3})\times (0.25cm)

\log \frac{I_o}{I}=0.6947

\frac{I_o}{I}=10^{0.6947}=4.951

If we consider I_o = 100

then, I=\frac{100}{4.951}=20.198

Here 'I' intensity of transmitted light = 20.198

Thus, the intensity of absorbed light I_A = 100 - 20.198 = 79.80

Now we have to calculate the percentage reduction in intensity.

\% \text{reduction in intensity}=\frac{I_A}{I_o}\times 100

\% \text{reduction in intensity}=\frac{79.80}{100}\times 100=79.80\%

Therefore, the percentage reduction in intensity is 79.80 %

3 0
3 years ago
What is the number of moles in 526 L O2 at STP?
Katena32 [7]
526 L O2 x 1 mol O2 / 22.4 L = 23.5 mol O2
3 0
3 years ago
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A group of students were asked to design a device that converts light energy into heat. They built a small solar oven using a ti
horrorfan [7]

Answer:

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Explanation:

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