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seropon [69]
4 years ago
9

A sample of salt (sodium chloride) is placed on the tip of a platinum wire. When the sample is heated on blue flame the flame tu

rned yellow. Explain how this can be done to electrons in sodium atoms
Chemistry
1 answer:
gulaghasi [49]4 years ago
8 0
So platinum is a transition metal. In general transition metals are reducers, which means they can give the electrons they have, to the sodium atoms. Also in chemistry we look at sub orbitals rather that shells(2,8,8). So due to the energy from heat, the d orbital split as electrons move to a higher energy level. Some of the electrons are given to the sodium ions and therefore the flame changes colour to yellow. 
The excitation of the electrons is caused by them getting energy and so moving up an energy level. This energy is released and the electron returns to it's original state. The energy released, however, does not release in the same direction, but in different/various directions. Therefore the colour of the light changes as some energy is released in the surrounding.
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Answer:

sickness.

Explanation:

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6 0
4 years ago
If I initially have 4.0 L of a gas at a pressure of 1.1 atm and a temperature of 298 K, what will the volume be if I increase th
DochEvi [55]

Answer:

1.34L

Explanation:

1 torr = 0.00132atm

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Using combined gas law

P1V1/T1 = P2V2/T2

(1.1×4.0)/298 = (3.41 × V2)/310

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4 0
3 years ago
What volume of a 0.2089 M KI solution contains enough KI to react exactly with the Cu(NO3)2 in 40.88 mL of a 0.3842 M solution o
tatyana61 [14]

Explanation:

As molarity is the number of moles placed in a liter of solution. Therefore, no. of mole = Molarity × volume of solution in liter

Hence, moles of Cu(NO_{3})_{2} will be calculated as follows.

No. of mole of Cu(NO_{3})_{2} = 0.3842 M \times 0.04388 L = 0.0168 mole

According to the given reaction, 2 mole of Cu(NO_{3})_{2} react with 4 mole of KI.

Therefore, for 0.0168 mole amount of Cu(NO_{3})_{2} required will be as follows.

Cu(NO_{3})_{2} = 0.0168 \times \frac{4}{2}

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Hence, volume of KI required will be calculated as follows.

               Volume = \frac{\text{no. of moles}}{Molarity}

    Volume of KI = \frac{0.0337}{0.2089}

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                            = 161.4 ml            (as 1 L = 1000 mL)

Thus, we can conclude that 161.4 ml  volume of a 0.2089 M KI is required for the given situation.

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4 years ago
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Answer:

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5 0
3 years ago
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6 0
3 years ago
Read 2 more answers
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