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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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A student titrating a sample of a powdered aspirin tablet to find out how much acetylsalicylic acid is in the sample finds that
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Answer:

305 mg

Explanation:

The equation of the reaction is;

C9H8O4 + NaOH = C9H7O4Na + H2O

Amount of NaOH reacted = 15.5/1000 L * 0.109 M = 1.69 * 10^-3 moles

Since the reaction has a 1:1 mole ration, 1.69 * 10^-3 moles of acetylsalicylic acid reacted.

Assuming all the acetylsalicylic acid reacted, then mass of acetylsalicylic acid present = 1.69 * 10^-3 moles * 180.158 g/mol = 0.305 g or 305 mg

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3 years ago
Explain how phosphorus can form a bond with 5 different chlorine atoms. In your answer, explain what
pav-90 [236]

Answer:

Phosphorus can have expanded octet, because it can shift it's lone pair electrons (3s orbital electrons) to empty 3d obital during excited state and thus can form 5 bonds.

Explanation:

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5 0
3 years ago
odium carbonate (Na2CO3Na2CO3) is used to neutralize the sulfuric acid spill. How many kilograms of sodium carbonate must be add
Arte-miy333 [17]

Answer : The mass of sodium carbonate added to neutralize must be, 6.54\times 10^3kg

Explanation :

First we have to calculate the moles of H_2SO_4.

\text{Moles of }H_2SO_4=\frac{\text{Mass of }H_2SO_4}{\text{Molar mass of }H_2SO_4}

Given:

Molar mass of H_2SO_4 = 98 g/mole

Mass of H_2SO_4 = 6.05\times 10^3kg=6.05\times 10^6g

Conversion used : (1 kg = 1000 g)

Now put all the given values in the above expression, we get:

\text{Moles of }H_2SO_4=\frac{6.05\times 10^6g}{98g/mol}=6.17\times 10^4mol

The moles of H_2SO_4 is, 6.17\times 10^4mol

Now we have to calculate the moles of Na_2CO_3

The balanced neutralization reaction is:

Na_2CO_3+H_2SO_4\rightarrow Na_2SO_4+H_2CO_3

From the balanced chemical reaction we conclude that,

As, 1 mole of H_2SO_4 neutralizes 1 mole of Na_2CO_3

So, 6.17\times 10^4mol of H_2SO_4 neutralizes

Now we have to calculate the mass of Na_2CO_3

\text{ Mass of }Na_2CO_3=\text{ Moles of }Na_2CO_3\times \text{ Molar mass of }Na_2CO_3

Molar mass of Na_2CO_3 = 106 g/mole

\text{ Mass of }Na_2CO_3=(6.17\times 10^4mol)\times (106g/mole)=6.54\times 10^6g=6.54\times 10^3kg

Thus, the mass of sodium carbonate added to neutralize must be, 6.54\times 10^3kg

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