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Annette [7]
3 years ago
11

When standardizing an iodine solution, 30.37 ml of the solution was required to completely oxidize 10.00 ml of a 5 mg/ml solutio

n of ascorbic acid. what is the molarity of the titrant??
Chemistry
1 answer:
Svetach [21]3 years ago
4 0
First, we need to determine the reaction that happens when standardizing. The reactions would be:

 KIO3(aq) + 6 H+ (aq) + 5 I- (aq) ------> 3 I2(aq) + 3 H2O(l) + K+ (aq)
C6H8O6(aq) + I2(aq) ------> C6H6O6(aq) + 2 I- (aq) + 2 H+ (aq)

5g/L C6H8O6 (0.01 L) ( 1 mol / 176.12 g) ( 1 mol I2 / 1 mol C6H8O6 ) ( 1 mol KIO3 / 3 mol I2 ) = 0.00009 mol KIO3

Molarity = 0.00009 mol KIO3 / 0.03037 L = 0.003 M

Hope this answers the question. Have a nice day.
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_Al(NO3)3 +_(NH4)3 P04_AlPO4+_ NH, NO3<br>balancing equations​
astra-53 [7]

Answer:

Al(NO₃)₃ + (NH₄)₃PO₄ —> AlPO₄ + 3NH₄NO₃

The coefficients are: 1, 1, 1, 3

Explanation:

__Al(NO₃)₃ + __(NH₄)₃PO₄ —> __AlPO₄ + __NH₄NO₃

The above equation can be balance as illustrated below:

Al(NO₃)₃ + (NH₄)₃PO₄ —> AlPO₄ + NH₄NO₃

There are 12 atoms of H on the left side and 4 atoms on the right side. It can be balance by writing 3 before NH₄NO₃ as shown below:

Al(NO₃)₃ + (NH₄)₃PO₄ —> AlPO₄ + 3NH₄NO₃

Now the equation is balanced.

The coefficients are: 1, 1, 1, 3

4 0
3 years ago
A dry gas at a temperature of 67.5 C and a pressure of 882 torr occupies a volume of 242.2 mL. What will be the volume of the ga
masya89 [10]

Answer:

265 mL is the new volume for the gas

Explanation:

We decompose the Ideal Gases Law in order to find the answer of this question: P . V = n . R . T

We can propose the formula for the 2 situations, where  n remains constant.

R refers to 0.082 L.atm/mol.K which is physic constant.

We convert the temperature to Absolute value:

67.5°C + 273 = 340.5 K

80°C + 273 = 353 K

We convert the volume to L → 242.2 mL . 1 L/1000 mL = 0.2422 L

We convert the pressure values to atm:

882 Torr . 1 atm/ 760 Torr = 1.16 atm

840 Torr . 1atm / 760 Torr = 1.10 atm

P₁. V₁ / T₁ = P₂ . V₂ / T₂     → Let's replace data:

1.16 atm . 0.2422L / 340.5K = 1.10 atm . V₂ / 353 K

(1.16 atm . 0.2422L / 340.5K) . 353K = 1.10 atm . V₂

V₂ = 0.291 L.atm / 1.10 atm → 0.2647 L ≅ 265 mL

4 0
3 years ago
Which value gives the number of particles in 1 mol of a substance?
Alexxx [7]
The number of particles (molecules or atoms) is: 6.022 x 10²³ particles (atoms or molecules).

1 mol of H₂O has 6.022 x 10²³ molecules.
1 mol of Al has 6.022 x 10²³ atoms. 
7 0
3 years ago
Consider the reaction of solid aluminum iodide and potassium metal to form solid potassium iodide and aluminum metal.The balance
ratelena [41]

Answer:

674.26 g of AlI₃

Explanation:

We'll begin by calculating the theoretical yield of aluminum (Al). This can be obtained as follow:

Percentage yield of Al = 67.8%

Actual yield of Al = 30.25 g

Theoretical yield of Al =?

Percentage yield = Actual yield /Theoretical yield × 100/

67.8% = 30.25 / Theoretical yield

67.8 / 100 = 30.25 / Theoretical yield

0.678 = 30.25 / Theoretical yield

Cross multiply

0.678 × Theoretical yield = 30.25

Divide both side by 0.678

Theoretical yield = 30.25 / 0.678

Theoretical yield of Al = 44.62 g

Next, we shall determine the mass of AlI₃ that reacted and the mass of Al produced from the balanced equation. This can be obtained as follow:

AlI₃(s) + 3K(s) → 3KI(s) + Al(s)

Molar mass of AlI₃ = 27 + (3×127)

= 27 + 381 = 408 g/mol

Mass of AlI₃ from the balanced equation = 1 × 408 = 408 g

Molar mass of Al = 27 g/mol

Mass of Al from the balanced equation = 1 × 27 = 27 g

Summary:

From the balanced equation above,

408 g of AlI₃ reacted to produce 27 g of Al.

Finally, we shall determine the mass of

AlI₃ required to produce 44.62 g of Al. This can be obtained as follow:

From the balanced equation above,

408 g of AlI₃ reacted to produce 27 g of Al.

Therefore, Xg of AlI₃ will react to produce 44.62 g of Al i.e

Xg of AlI₃ = (408 × 44.62)/27

Xg of AlI₃ = 674.26 g

Thus, 674.26 g of AlI₃ is needed for the reaction.

8 0
3 years ago
Label the parts of the atom.<br><br> Which atomic particle determines the identity of the atom?
zavuch27 [327]
A: Electron
B: Neutron
C: Proton
3 0
3 years ago
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