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bazaltina [42]
3 years ago
12

Eat at Joe's! Did you know that this neon sign is experiencing a change of state? When a neon sign lights up, like the one you s

ee here, something is happening to the neon atoms inside the tube. How is this colorful sign related to kinetic energy and phase change? A) The electrical current in the tube generates the neon gas that glows in the dark. B) Electrical energy excites the neon atoms. They spin faster in the tube producing electromagnetic waves that we see as colors. C) Neon atoms accelerate once electrical energy is turned on to the tube. The faster the atoms move in the tube, the brighter the glow we see. D) Electrical energy accelerates the electrons in the neon gas. The gas ionizes and becomes plasma, containing both positive and negative ions.
Chemistry
2 answers:
vlabodo [156]3 years ago
5 0
D) <span> Electrical energy accelerates the electrons in the neon gas. The gas ionizes and becomes plasma, containing both positive and negative ions.</span>
kykrilka [37]3 years ago
3 0
The answer is D. E<span>lectrical energy accelerates the electrons in the neon gas. The gas ionizes and becomes plasma, containing both positive and negative ions. </span>
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2.00 L of 0.800 M NaNO3 must be prepared from a solution known to be 2.50 M in concentration.How many mL are required? Plus don'
Morgarella [4.7K]

Answer:

1.36 × 10³ mL of water.

Explanation:

We can utilize the dilution equation. Recall that:

\displaystyle M_1V_1= M_2V_2

Where <em>M</em> represents molarity and <em>V</em> represents volume.

Let the initial concentration and unknown volume be <em>M</em>₁ and <em>V</em>₁, respectively. Let the final concentration and required volume be <em>M</em>₂ and <em>V</em>₂, respectively. Solve for <em>V</em>₁:

\displaystyle \begin{aligned} (2.50\text{ M})V_1 &= (0.800\text{ M})(2.00\text{ L}) \\ \\ V_1 & = 0.640\text{ L} \end{aligned}

Therefore, we can begin with 0.640 L of the 2.50 M solution and add enough distilled water to dilute the solution to 2.00 L. The required amount of water is thus:
\displaystyle 2.00\text{ L} - 0.640\text{ L} = 1.36\text{ L}

Convert this value to mL:
\displaystyle 1.36\text{ L} \cdot \frac{1000\text{ mL}}{1\text{ L}} = 1.36\times 10^3\text{ mL}

Therefore, about 1.36 × 10³ mL of water need to be added to the 2.50 M solution.

8 0
2 years ago
Match the following with their correct molecular weight. 2-butanone Propyl acetate 4-methyl-2-pentanone Butyl acetate Methanol E
Hatshy [7]

Answer:

2-butanone = 72.11 g/mol (option F)

Propyl acetate  = 102.13 g/mol (option C)

4-methyl-2-pentanone = 100.16 g/mol (option D)

Butyl acetate = 116.16 g/mol (option B)

Methanol = 32.04 g/mol (option E)

Ethanol  = 46.07 g/mol (option A)

Explanation:

Step 1: Data given

Molar mass of C = 12.01 g/mol

Molar mass of H = 1.01 g/mol

Molar mass of O = 16.00 g/mol

Step 2:

2-butanone = C4H8O

⇒ 4*12.01 + 8*1.01 + 16.00 = 72.11 g/mol (option F)

Propyl acetate = C5H10O2

⇒ 5*12.01 + 10*1.01 + 2*16.00 = 102.13 g/mol (option C)

4-methyl-2-pentanone = C6H12O

⇒ 6*12.01 + 12*1.01 + 16.00 = 100.16 g/mol (option D)

Butyl acetate = C6H12O2

⇒ 6*12.01 + 12*1.01 + 2*16.00 = 116.16 g/mol (option B)

Methanol = CH3OH = CH4O

⇒ 12.01 + 4*1.01 + 16.00 = 32.04 g/mol (option E)

Ethanol = C2H5OH = C2H6O

⇒ 2*12.01 + 6*1.01 + 16.00 = 46.07 g/mol (option A)

4 0
3 years ago
1. Write the chemical equation when excess chloride ion is added to aqueous solution of cobalt chloride. Note observed color of
Aneli [31]

The element cobalt can form compounds in two different oxidation states, +2 and +3. 


The +2 state is more common.

The ion Co2+ (aq) is pink.

Other compounds of cobalt(II), which include both anhydrous Co2+ and complex ions, are commonly blue.

If an aqueous solution contains both cobalt(II) and chloride ions, the blue ion CoCl42- forms, in equilibrium with the pink Co2+ (aq) ion.

<span>CoCl42- (aq) <===========> Co2+ (aq) + 4Cl1-(aq)</span>

5 0
3 years ago
A vial is filled with room temperature water, and blue cold water has been placed at the bottom. A warm water bag sits next to t
MrRa [10]

Explanation:

In the context, a vial which is used in store medical samples is filled with water at room temperature. And the vial is kept on a cold water. Also a water bag containing warm water is kept near the vial.

The cold water kept at the bottom of the vial is having lower kinetic energy while warm water will have higher kinetic energy than the others. Since the water in the vial is at room temperature and it is in touch with the cold blue water, the water in the vial will loose or give its temperature to the cold blue water through conduction as well as convection process since temperature always flows from a hot body towards the cold body.

On the other hand, the warm water placed next tot he vial will give its temperature to the atmosphere.  

5 0
3 years ago
If the value of kc at 25oc is 3.7108, and the equilibrium concentrations for n2 and h2 are 0.000105 m and 0.0000542 m, respectiv
frez [133]

Equation of decomposition of ammonia:

N2+3H2->2NH3

Euilibrium constant:

Kc=(NH3)^2/((N2)((H2)^3))

As concentration of N2=0.000105, H2=0.0000542

so equation will become:

3.7=(NH3)^2/(0.000105)*(0.0000542)^3

NH3=√(3.7*0.000105*(0.0000542)^3)

NH3=7.8×10⁻⁹

So concentration of ammonia will be 7.8×10⁻⁹.

5 0
3 years ago
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