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4vir4ik [10]
3 years ago
10

A piece of ebonite and acetate are both rubbed with wool. What will occur when the two pieces (ebonite and acetate) are brought

together
Chemistry
1 answer:
lesya [120]3 years ago
5 0

Answer: The options are not given, here are the options from another website.

A. They will attract each other

b. They will repel each other

C. Nothing will happen to them

d. They will get heated up

The answer is they will attract each other

Explanation:

This is because the charge from the rubbing of ebonite and acetate to a wool is called triboeletric effect.

When ebonite is rubbed with a wool it will produce negative charge, the electrons around it will be negative but when acetate is rubbed with it, it will produce positive charge. This two will attract each other because unlike charges(negative and positive) attract each other.

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Two elements are studied. One has atomic number X and one has atomic number X+2. It is known that element X is a halogen. To wha
vladimir2022 [97]

Answer:

Group 1 (or IA)

Explanation:

If element X is a halogen, then it belongs to the group 17 (or VIIA, under a different notation).

For each extra unit of atomic number, the group number increases by 1. That means that the X+1 element would belong to the group 18 (or VIIIA). <em>The X+2 element would thus belong in the group 1 </em>(or IA) one period higher (higher as in numeric value, not as in position in the periodic table).

7 0
3 years ago
Calculate the acid dissociation constant Ka of a 0.2 M solution of weak acid that is 0.1% ionized is ________.
mars1129 [50]

Answer: acid dissociation constant Ka= 2.00×10^-7

Explanation:

For the reaction

HA + H20. ----> H3O+ A-

Initially: C. 0. 0

After : C-Cx. Cx. Cx

Ka= [H3O+][A-]/[HA]

Ka= Cx × Cx/C-Cx

Ka= C²X²/C(1-x)

Ka= Cx²/1-x

Where x is degree of dissociation = 0.1% = 0.001 and c is the concentration =0.2

Ka= 0.2(0.001²)/(1-0.001)

Ka= 2.00×10^-7

Therefore the dissociation constant is

2.00×10^-7

7 0
3 years ago
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Name the two possible products in the precipitation reaction of copper (II) chloride and sodium phosphate. Use the charges on th
satela [25.4K]

Answer:

General equation for a double-displacement reaction:  

AB + CD --> AC + BD

• sodium chloride – NaCl copper sulfate – CuSO₄  

NaCl + CuSO₄ --> Na₂SO₄ + CuCl₂

The products formed are sodium sulfate and copper (II) chloride.

Copper (II) chloride forms a blue colored solution.

• sodium hydroxide – NaOH copper sulfate – CuSO₄  

NaOH + CuSO₄ --> Na₂SO₄ + Cu(OH)₂

The products formed are sodium sulfate and copper (II) hydroxide.

Copper (II) hydroxide forms a blue colored solution.

• sodium phosphate – Na₂HPO₂ copper sulfate – CuSO₄  

Na₂HPO₄ + CuSO₄ --> Na₂SO₄ + CuHPO₄

The products formed are sodium sulfate and copper (II) hydrogen phosphate.

Copper (II) hydrogen phosphate forms a blue colored solution.

• sodium chloride – NaCl silver nitrate – AgNO₃  

NaCl + AgNO₃--> AgCl + NaNO₃

The products formed are silver chloride and sodium nitrate.

Silver chloride forms a white precipitate.

• sodium hydroxide – NaOH silver nitrate – AgNO₃  

NaOH + AgNO₃   --> NaNO₃ + AgOH

The products formed are silver hydroxide and sodium nitrate.

Silver hydroxide forms a white precipitate.

• sodium phosphate – Na₂HPO₄ silver nitrate – AgNO₃

Na₂HPO₄ + AgNO₃  --> NaNO₃ +  Ag₂HPO₄

The products formed are sodium nitrate and silver hydrogen phosphate.

Silver hydrogen phosphate forms a colorless solution.

Explanation:

5 0
3 years ago
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Answer:

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Explanation:

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