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Free_Kalibri [48]
3 years ago
11

In this experiment, the metal cations in the solutions were initially in the ground state. When placed in the flame, the metals

then (absorbed, emitted) energy as (electricity, heat, EM radiation). When this occurred, electrons made transitions from (low, high) energy levels to (low, high) energy levels. The metals were then in the (ground, excited) state. The electrons in these metals then made transitions from (low, high) energy levels to (low, high) energy levels, resulting in the (absorption, emission) of energy as (electricity, heat, EM radiation).
Chemistry
2 answers:
Andrej [43]3 years ago
7 0
Initially, you should know that what the text describes is a test used to analyze and identify some metal ions in a compound. This is the principle behind it: first the ions are excited, by the heat of the flame, they absorb energy and the electrons are promoted to higher energy levels; after this, the electrons will fall back down to lower energy levels releasing energy as light. The flame color that is seen is related with the frequency of the light emitted.

Here is the text with the right words.

In this experiment, the metal cations in the solutions were initially in the ground state. When placed in the flame, the metals then (absorbed) energy as (heat). When this occurred, electrons made transitions from (low) energy levels to (high) energy levels. The metals were then in the (excited) state. The electrons in these metals then made transitions from (high) energy levels to (low) energy levels, resulting in the (emission) of energy as (EM radiation).

tangare [24]3 years ago
6 0
In this experiment, the metal cations in the solutions were initially in the ground state. When placed in the flame, the metals then (absorbed) energy as (heat). When this occurred, electrons made transitions from (low) energy levels to (high) energy levels. The metals were then in the (ground) state. The electrons in these metals then made transitions from (low) energy levels to (high) energy levels, resulting in the (emission) of energy as (heat).
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What part of the atom do the different 35 and 37 represent?
loris [4]
It would be atomic masses of the same atoms, and that atoms will be isotopes.
4 0
4 years ago
How many ml of 0. 10 m naoh should the student add to 20 ml of 0. 10 m hf or if she wished to prepare a buffer with a ph of 3. 5
Lemur [1.5K]

The required volume of 0.10M NaOH solution that student will add is 20 mL.

<h3>How do we calculate the volume?</h3>

Volume of any solution which is required to prepare buffer will be calculated by using the below equation as:

M₁V₁ = M₂V₂, where

  • M₁ & V₁ are the molarity and volume of NaOH solution.
  • M₂ & V₂ are the molarity and volume of HF solution.

On putting values, we get

V₁ = (0.1)(20) / (0.1) = 20mL

Hence required volume of NaOH solution is 20mL.

To know more about molarity & volume, visit the below link:

brainly.com/question/24305514

#SPJ4

3 0
2 years ago
Copper(II) sulfide, CuS, is used in the development of aniline black dye in textile printing. What is the maximum mass of CuS wh
Naya [18.7K]

Answer:

1.82 g   is the maximum mass of CuS.

Explanation:

Considering:

Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}

Or,

Moles =Molarity \times {Volume\ of\ the\ solution}

Given :

<u>For CuCl_2 : </u>

Molarity = 0.500 M

Volume = 38.0 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 38.0×10⁻³ L

Thus, moles of CuCl_2 :

Moles=0.500 \times {38.0\times 10^{-3}}\ moles

<u>Moles of CuCl_2  = 0.019 moles </u>

<u>For (NH_4)_2S : </u>

Molarity = 0.600 M

Volume = 42.0 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 42.0×10⁻³ L

Thus, moles of (NH_4)_2S :

Moles=0.600 \times {42.0\times 10^{-3}}\ moles

<u>Moles of (NH_4)_2S  = 0.0252 moles </u>

According to the given reaction:

CuCl_2_{(aq)}+(NH_4)_2S_{(aq)}\rightarrow CuS_{(s)}+2NH_4Cl_{(aq)}

1 mole of CuCl_2 reacts with 1 mole of (NH_4)_2S

So,  

0.019 mole of CuCl_2 reacts with 0.019 mole of (NH_4)_2S

Moles of (NH_4)_2S = 0.019 mole

Available moles of (NH_4)_2S = 0.0252 mole

<u>Limiting reagent is the one which is present in small amount. Thus, CuCl_2 is limiting reagent.</u>

The formation of the product is governed by the limiting reagent. So,

1 mole of CuCl_2 gives 1 mole of CuS

0.019 mole of CuCl_2 gives 0.019 mole of CuS

Moles of CuS formed = 0.019 moles

Molar mass of CuS = 95.611 g/mol

Mass of CuS = Moles × Molar mass = 0.019 × 95.611 g = 1.82 g

<u>1.82 g   is the maximum mass of CuS.</u>

5 0
4 years ago
Which of the following measurements is expressed to three significant figures?
matrenka [14]

The answer is C. 7.30 x 10^-7

8 0
3 years ago
How does the law of conservation of mass apply to this: C2H4+O2→ 2H2O+2CO2
Ahat [919]

Answer:

Add a coefficient of 3 in front of the 02 to balance the 6 O atoms in the products

Explanation:

Add a coefficient of 3 in front of the 02 to balance the 6 O atoms in the products

6 0
3 years ago
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