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maw [93]
2 years ago
3

Which grouping of the three phases of bromine is listed in order from left to right for increasing distance between bromine mole

cules?
(1) gas, liquid, solid (3) solid, gas, liquid(2) liquid, solid, gas (4) solid, liquid, gas
Chemistry
2 answers:
ololo11 [35]2 years ago
4 0
4) solid, liquid and gas
MissTica2 years ago
3 0

Answer : The correct option is 4.

Explanation :

As we know that the molecules are tightly packed in solid, loosely packed in liquid and very loosely packed in gas state. when the inter-molecular spacing between the molecules is very less. So, it is considered as solid.

When the inter-molecular spacing between the molecules is moderate. So, it is considered as liquid.

When the inter-molecular spacing between the molecules is very large. So, it is considered as gas.

Therefore, the increasing order of three phase of bromine on the basis of molecule distance is,

Br(solid) < Br(liquid) < Br(gas)

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Fluorine gas is placed in contact with calcium metal at high temperatures to produce calcium fluoride powder. What is the formul
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Answer:

Upper F subscript 2 (g) plus upper C a (s) right arrow with delta above upper C a upper F subscript 2 (s).

Explanation:

This is a chemical reaction problem.

In expressing any chemical reaction, we need to understand that there are reactants and products.

  • The reactants are the species on the left hand side that are combining.
  • The products are the species on the right hand side that are formed.
  • Every chemical reaction is obeys the law of conservation of matter i.e equal number of matter on both sides.

Using the statement of this problem, we can deduce that;

 Reactants are Fluorine gas and Calcium metal

  Product is Calcium Fluoride

Note: A metal is a solid(s) and powder is a solid(s). A gas is denoted as (g). They depict the state of the species reacting.

                    F₂_{(g)}     +    Ca_{(s)}               →           CaF₂_{(s)}

We can see that equal number of atoms are on both sides of the expression.

7 0
2 years ago
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What should be done if particles of precipitate appear in the filtrate?
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<span>The instructor should be questioned to see if the filtrate is able to be recycled. This precipitate can contaminate the filtrate, rendering it useless for repeated experiments. If it is able to be recycled, a second pass through the filter might be required to remove the precipitate.</span>
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What is the kinetic energy acquired by the electron in hydrogen atom, if it absorbs a light radiation of energy 1.08x101 J. (A)
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Explanation:

The given data is as follows.

            Energy of radiation absorbed by the electron in hydrogen atom = 1.08 \times 10^{-17} J

As energy is absorbed as a photon. Hence, frequency will be calculated will be as follows.

                                    E = h \nu

               1.08 \times 10^{-17} J = 6.626 \times 10^{-34} Js \times \nu

               \nu = 0.163 \times 10^{17} s^{-1}

or,                \nu = 1.63 \times 10^{16} s^{-1}    

It is known that,        \nu = \frac{c}{\lambda}

                1.63 \times 10^{16} s^{-1} = \frac{3 \times 10^{8} m/s}{\lambda}                  

                   \lambda = 1.84 \times 10^{-8} m

And, according to De-Broglie equation \lambda = \frac{h}{p}

as,        p = m \times \nu

So,          \lambda = \frac{h}{m \times \nu}

            m \times \nu = \frac{6.626 \times 10^{-34} Js}{1.84 \times 10^{-8} m}          

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Now, on squaring both the sides we get the following.

           (m \times \nu)^{2} = (3.6 \times 10^{-26} J/m)^{2}    

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where,   m = mass of electron

So,           m \times \nu^{2} = \frac{12.96 \times 10^{-52}}{m}

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Since,  K.E = \frac{1}{2}m \nu^{2}

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                 = 0.71 \times 10^{-21} J

Thus, we can conclude that kinetic energy acquired by the electron in hydrogen atom is 7.1 \times 10^{-22} J.

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