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ElenaW [278]
3 years ago
14

How to prove that a given colorless liquid is an acid​

Chemistry
2 answers:
tresset_1 [31]3 years ago
7 0

Answer:

1. Using Litmus papers. Litmus paper will turn blue if solution is acidic or it will turn blue if it is basic. ... If pH is below 7 or reddish then the given solution is acidic and if pH is higher than 7 or bluish then the solution is basic.

Explanation:

Pachacha [2.7K]3 years ago
5 0

Answer:

Explanation:

by using litmus papaer it will turn into red if solutions is acidic

and into blue if it's basic

by using Magnesium ribbon. Magnesium do not react with bases so hydrogen gas will produce when Magnesium is dipped in an acidic solution

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For salt formation procedure crude lidocaine should be dissolved in diethyl ether and 2 ml of 2.2 M sulfuric acid should be adde
mixas84 [53]

Answer:

20 equivalents of sulfuric acid are needed to per equivalent of lidocaine.

Explanation:

To do this calculation we should take in account the key information given:

<em>2 mL (0.02 L) of sulfuric acid (H2SO4) 2.2 M per 1 g of lidocaine (molecular weight=234.34 g/mol)</em>

So we can calculate from here the number of moles of each compound needed for the procedure.

Starting with the sulfuric acid, we know that the molar concentration is defined as [concentration] =n(moles number)/V(volume).

So n=[concentration]*V, and therefore :

n(H2SO4)=2.2 M * 0.02 L=0.044 mol

While the moles of lidocaine can be calculated as n(moles number)=m(mass amount)/mW(molecular weight), so:

n(lidocaine)=m(lidocaine)/mW(lidocaine)= 1 g /234,34 g/mol~0,0043 mol.

Now we can calculate the equivalents.

Take in account that for sulfuric acid, de number of equivalents per mole is 2 (because of the double dissociation of the acid), while the number of equivalents per mole of lidocaine is 1.

Then we can conclude that:

0,0043 equivalents of lidocaine need 0,088 equivalents of sulfuric acid

1 equivalent of lidocaine need (0,088/0.0043)=20,46~20 equivalents of sulfuric acid

Finally, we can conclude that for the described procedure, 20 equivalents of sulfuric acid per equivalent of lidocaine are needed.

4 0
3 years ago
Describe the chemical reaction based on the chemical equation below. Also, explain whether the equation is balanced.
Allushta [10]

Answer:

Chemical equation: NH₃(g) + O₂(g) → NO(g) + H₂O(l).

Oxidation reaction: N⁻³ → N⁺² + 5e⁻ /×4.

Reduction reaction: O₂⁰ + 4e⁻ → 2O⁻² /×5.

Oxidation reaction: 4N⁻³ → 4N⁺² + 20e⁻.

Reduction reaction: 5O₂⁰ + 20e⁻ → 10O⁻² /×5.

Balanced chemical equation: 4NH₃(g) + 5O₂(g) → 4NO(g) + 6H₂O(l).

Ammonia (NH₃) reacts with oxygen (O₂) and form nitrogen(II) oxide (NO) and water (H₂O). Catalyst is this reaction is platinum (Pt).

This is combustion reaction.

5 0
4 years ago
Read 2 more answers
Explica brevemente cuidado del aparato digestivo ​
kirill115 [55]

Answer:

For better digestive health, follow these simple tips:

Eat a high-fiber diet. ...

Be sure you're getting both soluble and insoluble fiber. ...

Minimize your intake of foods high in fat. ...

Select lean meats. ...

Add probiotics to your diet. ...

Follow a regular eating schedule. ...

Drink plenty of water.

Explanation:

6 0
3 years ago
Which of the following statements is not true regarding physical properties and changes?
ratelena [41]
I think it is B, but i am not sure
8 0
4 years ago
Read 2 more answers
What mass (in g) of potassium chlorate is required to supply the proper amount of oxygen needed to burn 117.3 g of methane
Brilliant_brown [7]

The mass (in g) of potassium chlorate required to supply the proper amount of oxygen needed to burn 117.3 g of methane is 1196.82 g

<h3>Combustion of methane</h3><h3 />

Methane burns in oxygen to produce carbon (iv) oxide and water according to the equation of the reaction below:

CH₄ + 2O₂  ----> CO₂ + 2H₂O

1 mole of methane requires 2 moles of oxygen for complete combustion

1 mole of methane has a mass of 16 g

moles of methane in 117.3 g = 117.3/16 = 7.33 moles of methane

7.33 moles of methane will require 2 * 7.33 moles of oxygen

7.33 moles of methane will require 14.66 moles of oxygen

<h3>Decomposition of potassium chlorate </h3>

The decomposition of potassium chlorate produces oxygen

The equation of the reaction is given below:

  • 2KClO3 → 2KCl + 3O2.

2 moles of potassium chlorate produces 3 moles of oxygen

14.66 moles of oxygen will be produced by 14.66 * 2/3  moles of potassium chlorate

14.66 moles of oxygen will be produced by 9.77 moles of potassium chlorate

1 mole of  potassium chlorate has a mass of 122.5

9.77 moles of potassium chlorate has a mass of 1196.82 g

Therefore, the mass (in g) of potassium chlorate required to supply the proper amount of oxygen needed to burn 117.3 g of methane is 1196.82 g

Learn more about mass and molar mass at: brainly.com/question/15476873

7 0
3 years ago
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