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alina1380 [7]
3 years ago
13

List as many experimental techniques as you can that may be used to determine Ksp for a sparingly soluble salt. Select all that

apply.
a) Determine the concentration of ions in a saturated solution with a titrimetric procedure and use this value to calculate the Ksp.
b) Determine the concentration of ions in a saturated solution with an electrical conductivity measurement and use this value to calculate the Ksp.
c) Evaporating a given volume of a saturated solution of salt and finding the mass of the salt that dissolved to directly find the Ksp.
d) Determine the concentration of ions in a saturated solution with an EMG procedure and use this value to calculate the Ksp.
e) Determine the concentration of ions in a saturated solution with a colorimetric procedure and use this value to calculate the Ksp.
Chemistry
1 answer:
max2010maxim [7]3 years ago
5 0

Answer: a) and b)

Explanation:

We can both use titrimetric method and electrical conductivity meter meter to determine the concentration of the ions after which we can calculate the value of Ksp

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8 0
3 years ago
IV.2. The following problem considers the combustion of butane in a torch. Molecular weight of butane 58.0 g/mol. 2C4H10 + 1302
Anarel [89]

Answer:

(a) Oxygen

(b) 0.84 g

(c) 2.54 g

Explanation:

(a)

Explanation:

The formula for the calculation of moles is shown below:

moles = \frac{Mass\ taken}{Molar\ mass}

Given: For butane

Given mass = 1.00 g

Molar mass of butane = 58.0 g/mol

Moles of butane = 1.00 g / 58.0 g/mol = 0.0172 moles

Given: For O_2

Given mass = 3.00 g

Molar mass of O_2 = 32.0 g/mol

Moles of O_2 = 3.00 g / 32.0 g/mol = 0.09375 moles

According to the given reaction:

2C_4H_{10}+13O_2\rightarrow 8CO_2+10H_2O

2 moles of butane react with 13 moles of O_2

1 mole of butane react with 13/2 moles of O_2

0.0172 moles  of butane react with (13/2)*0.0172 moles of O_2

Moles of O_2 required = 0.1118 moles

Available moles of CuSO_4 = 0.09375 moles

<u>Limiting reagent is the one which is present in small amount. Thus, O_2 is limiting reagent. (0.09375 < 0.1118 )</u>

(b)

The formation of the product is governed by the limiting reagent. So,

13 moles of O_2 react with  2 moles of butane

1 mole of O_2 react with  2/13 moles of butane

0.09375 mole of O_2 react with  (2/13)*0.09375  moles of butane

Moles of butane used = 0.0144 moles

Molar mass of butane = 58.0 g/mol

<u>Mass of butane used = Moles × Molar mass = 0.0144 × 58.0 g = 0.84 g</u>

(c)

13 moles of O_2 on reaction forms 8 moles of carbon dioxide

1 mole of O_2 on reaction forms 8/13 moles of carbon dioxide

0.09375 mole of O_2 on reaction forms (8/13)*0.09375 moles of carbon dioxide

Moles of carbon dioxide obtained = 0.05769 moles

Molar mass of CO_2 = 44.0 g/mol

<u>Mass of CO_2 = Moles × Molar mass = 0.05769 × 44.0 g = 2.54 g</u>

6 0
3 years ago
Which of the following WOULD NOT affect the level at which a cargo ship floats
kow [346]
Answer: D. The depth of the water does not affect the level at which a cargo ship floats.
3 0
3 years ago
How many molecules are in 128g of H2O?
Lelechka [254]

Answer:

4.28x10^24 molecules

Explanation:

From Avogadro's hypothesis, 1mole of any substance contains 6.02x10^23 molecules. From the above, we understood that 1mole of H2O also contains 6.02x10^23 molecules.

1mole of H2O = (2x1) + 16 = 2 + 16 = 18g

Now, if 18g of H2O contains 6.02x10^23 molecules,

Then 128g of H2O will contain = (128x 6.02x10^23) /18 = 4.28x10^24 molecules

3 0
3 years ago
In a redox reaction what happens to a substance's charge when it is oxidized
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<span>oxidizing substance removes electrons from another substance, which are then added to itself, the oxidizing substance becomes “reduced” (more negative). And because it “accepts” electrons .</span>
5 0
3 years ago
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