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3 years ago

In which state of matter are the particles least able to move?

Chemistry

1 answer:

amid [387]3 years ago

3 0

The solid state where particles are more compact

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Anhydrous CoCl2 is blue; CoCl2•6H2O is red. When CoCl2•6H2O is heated and water is clearly being driven off, it becomes uniforml

Sedaia [141]

Answer: On losing 6 moles of water, cobalt chloride forms unstable violet-coloured ions, before generating its stable blue-coloured anhydrous form.

Explanation:

The hydrated cobalt chloride loses its 6 water of crystallization, then dissociates into ions: cobalt ions and chlorine ions that appear violet, and quickly combined to form the stable anhydrous Cobalt chloride with blue colour.

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3 years ago

What is an ionic compound

Gwar [14]

Ionic compound is formed by negative and positive ions, these ions form ionic bond

6 0

3 years ago

Identify the type of reaction in the chemical reaction below:

Morgarella [4.7K]

Complete question is;

Identify the type of reaction in the chemical reaction below:

2P205 ➡️ 4P + 502

single replacement

synthesis

decomposition

combustion

double replacement

Answer:

Decomposition

Explanation:

We. An see in the question that the compound 2P205 is broken down into simpler substances which are phosphorus (P) and oxygen (O).

Now, this is a decomposition reaction because a decomposition reaction is one in which a compound is broken down into simpler substances

6 0

2 years ago

How to do them 11-15 most difficult ones for me

BigorU [14]

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3 years ago

A 10 mm Brinell hardness indenter is used for some hardness testing measurements of a steel alloy. a) Compute the HB of this mat

Lady_Fox [76]

Explanation:

(a) The given data is as follows.

Load applied (P) = 1000 kg

Indentation produced (d) = 2.50 mm

BHI diameter (D) = 10 mm

Expression for Brinell Hardness is as follows.

HB = $\frac{2P}{\pi D [D - \sqrt{(D^{2} - d^{2})}]}$

Now, putting the given values into the above formula as follows.

HB = $\frac{2P}{\pi D [D - \sqrt{(D^{2} - d^{2})}]}$ = $\frac{2 \times 1000 kg}{3.14 \times 10 mm [D - \sqrt{((10 mm)^{2} - (2.50)^{2})}]}$

= $\frac{2000}{9.98}$

= 200

Therefore, the Brinell HArdness is 200.

(b) The given data is as follows.

Brinell Hardness = 300

Load (P) = 500 kg

BHI diameter (D) = 10 mm

Indentation produced (d) = ?

d = $\sqrt{(D^{2} - [D - \frac{2P}{HB} \pi D]^{2})}$

= $\sqrt{(10 mm)^{2} - [10 mm - \frac{2 \times 500 kg}{300 \times 3.14 \times 10 mm}]^{2}}$

= 4.46 mm

Hence, the diameter of an indentation to yield a hardness of 300 HB when a 500-kg load is used is 4.46 mm.

5 0

2 years ago