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Volgvan
3 years ago
11

An aqueous solution is analyzed and found to contain potassium ions and sulfite ions. write the equation for the dissolution of

the salt that produced this solution. if the solution contains 0.700 mol of potassium ions, how many moles of sulfite ions are present? answer in units of ol.
Chemistry
1 answer:
enot [183]3 years ago
7 0
1) T<span>he dissolution of the salt potassium sulfite:
K</span>₂SO₃(aq) → 2K⁺(aq) + SO₃²⁻(aq).
Potassium has +1 charge because it lost one electron to accomplish stabile electron configuration of noble gas argon.
2) From dissolution reaction: n(K⁺) : n(SO₃²⁻) = 2 : 1.
n(K⁺) = 0.700 mol.
0.700 mol : n(SO₃²⁻) = 2 : 1.
n(SO₃²⁻) = 0.700 mol ÷ 2.
n(SO₃²⁻) = 0.350 mol; amount of sulfite anions.
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Answer:

89,4%

Explanation:

If you have a solutio of 2,5g of acetanilide in 50mL of water and you warm this solution to 100°C you will dissolve all acetanilide because the maximum solubility in 50mL will be:

5,5g / 100mL → 2,75g / 50mL.

Then, if you cold the water to 0°C the solubility in 50mL will be:

0,53g / 100mL → 0,265g / 50mL.

That means you will precipitate:

2,5g - 0,265g = <em>2,235g of acetanilide</em>

The theoretical percent recovery will be:

2,2365g / 2,5g ×100 = <em>89,4%</em>

<em></em>

I hope it helps!

7 0
3 years ago
The number of molecules in 1.00 liter of O2 gas at 56◦C and 821 torr is
enot [183]

Answer:

2. 2.41 x 10^22 molec

Explanation:

3 0
3 years ago
Which of the following explains the conservation of mass during cellular respiration?
lord [1]

Answer:

The total number of atoms when glucose and oxygen reacts stays the same when carbondioxide and water are produced.

Explanation:

Chemical reaction:

C₆H₁₂O₆  +  6O₂  →  6CO₂ + 6H₂O

We can see that the number of atoms of each element remain same on both side of reaction so law of conservation of mass is followed by this reaction. Six number of carbon atoms twelve number of hydrogen atoms and eighteen number of oxygen atoms are present on both side.

There are two types of respiration:

1. Aerobic respiration  

2. Anaerobic respiration

Aerobic respiration

It is the breakdown of glucose molecule in the presence of oxygen to yield large amount of energy. Water and carbon dioxide are also produced as a byproduct.

Glucose + oxygen → carbon dioxide + water + 38ATP

Anaerobic Respiration

It is the breakdown of glucose molecule in the absence of oxygen and produce small amount of energy. Alcohol or lactic acid and carbon dioxide are also produced as byproducts.  

Glucose→ lactic acid/alcohol + 2ATP + carbon dioxide

This process use respiratory electron transport chain as electron acceptor instead of oxygen. It is mostly occur in prokaryotes. Its main advantage is that it produce energy (ATP) very quickly as compared to aerobic respiration.  

8 0
4 years ago
What mass of NH3 in grams must be used to produce 1.81 tons of HNO3 by the Ostwald process, assuming an 80.0 percent yield in ea
Alex17521 [72]

The three reactions involved in the Ostwald process for the conversion of NH3 to HNO3 are:

4NH3(g) + 5O2(g) ==> 4NO(g) + 6H2O(l)   ------(1)

2NO(g) + O2(g) ==> 2NO2(g)  ------------------------(2)

3NO2(g) + H2O(l) ==> 2HNO3(aq) + NO(g) ---(3)

Mass of HNO3 produced = 1.81 tons

In grams, the mass of HNO3 = 1.81*2000*453.6 = 1642032 g

Molar mass of HNO3 = 63 g/mol

Thus, # moles of HNO3 produced = 1642032/63 = 26064 moles

Based on the stoichiometry of reaction (3):

Theoretical moles of NO2 = 1.5(Moles HNO2 ) = 1.5(26064) = 39096

Assuming 80% yield:

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Based on the stoichiometry in reaction (2):

Theoretical moles of NO = 31277 moles

Assuming 80% yield:

Actual moles of NO = 0.8*31277 = 25022 moles

Bases on stoichiometry in reaction(3):

Moles of NH3 = 25022 moles

Assuming 80% yield:

Actual moles of NH3 required = 0.8*25022 = 20018 moles

Mass of NH3 required = 20018 moles * 17 g/mole = 340306 g = 340.31 kg

7 0
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