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3 years ago

11

What happens when the kinetic energy of molecules increases so much that electrons are released by the atoms, creating a swirlin

g gas of positive ions and negative electrons?

1 answer:

maxonik [38]3 years ago

3 0

When that happens, you get a plasma — the fourth state of matter.

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CuCl2 + 2NaNO3 mc023-1.jpg Cu(NO3)2 + 2NaCl What is the percent yield of NaCl if 31.0 g of CuCl2 reacts with excess NaNO3 to pro

MatroZZZ [7]

The stoichiometric ratio of CuCl2 to NaCl is 1 is to 2. The stoichiometric ratio of 31.0 g CuCl2 is 26.95 grams of NaCl by converting the amount of CuCl2 to mole and multiplying by 0.5 and molar mass of NaCl.This amount is equal to 78.65% yield.

3 0

3 years ago

Read 2 more answers

When performing labratory experiments, which of the following items is ALWAYS necessary to use?

densk [106]

All of the the answers are are correct but a fume hood is more for if you are dealing with chemicals that can produce fumes the are deadly to people

3 0

3 years ago

List and explain the four most important factors that control the relative reactivity of a carbonyl-containing functional group

ioda

Resonance, leaving group, carbonyl carbon delta+, and steric effect is the most crucial variables that affect the relative reactivity of a functional group containing a carbonyl in an addition or substitution process.

Discussion:

1. Carbonyl Carbon Delta+: The carbonyl group becomes more electrophilic and accelerates nucleophilic assault when the carbonyl carbon delta+ is bigger.

2. Resonance: When the carbonyl is transformed into the tetrahedral adduct, it may be lost. Loss of resonance increases the energy of the transition state for this nucleophilic assault because resonance has the function of stabilizing. Therefore, a carbonyl functional group's resistance to nucleophilic attack increases as resonance in the group increases in importance.

3. Leaving group: Tetrahedral adduct fragmentation is encouraged by a better LG.

4. Steric effects: The nucleophilic attack on carbonyl carbon is delayed when sterically impeded.

Learn more about carbonyl here:

brainly.com/question/21440134

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8 0

1 year ago

What volume (in mL) of a 0.200 MHNO3 solution is required to completely react with 27.6 mL of a 0.100 MNa2CO3 solution according

ladessa [460]

Answer:

There is 27.6 mL of a 0.200 M HNO3 solution required

Explanation:

<u>Step 1: </u>The balanced equation is:

Na2CO3(aq)+2HNO3(aq)→2NaNO3(aq)+CO2(g)+H2O(l)

This means for 1 mole Na2CO3 consumed, there is consumed 2 mole of HNO3 and there is produced 2 moles of NaNO3, 1 mole of CO2 and 1 mole of H2O

<u>Step 2: </u>Calculating moles of Na2CO3

moles of Na2CO3 =volume of Na2CO3 * Molarity of Na2CO3

moles of Na2CO3 = 27.6 *10^-3 * 0.1 M = 0.00276 moles

<u>Step 3: </u>Calculating moles of HNO3

In the balanced equation, we can see that for 1 mole of Na2CO3 consumed, there are consumed 2 moles of HNO3.

So for 0.00276 moles consumed of Na2CO3, there are consumed 0.00552 moles of HNO3.

This means 0.00276 moles of the base Na2CO3 would react with 0.00552 moles of the acid HNO3

<u>Step 4: </u>Calculating the volume of HNO3

volume of HNO3 = moles of HNO3 / Molarity of HNO3

volume of HNO3 = 0.00552 moles / 0.200 M = 0.0276 L

0.0276 L = 27.6 ml

There is 27.6 mL of a 0.200 M HNO3 solution required

4 0

3 years ago

n acid with a pKa of 8.0 is present in a solution with a pH of 6.0. What is the ratio of the protonated to the deprotonated form

Nady [450]

Answer : The ratio of the protonated to the deprotonated form of the acid is, 100

Explanation : Given,

$pK_a=8.0$

pH = 6.0

To calculate the ratio of the protonated to the deprotonated form of the acid we are using Henderson Hesselbach equation :

$pH=pK_a+\log \frac{[Salt]}{[Acid]}$

$pH=pK_a+\log \frac{[Deprotonated]}{[Protonated]}$

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

$6.0=8.0+\log \frac{[Deprotonated]}{[Protonated]}$

$\frac{[Deprotonated]}{[Protonated]}=0.01$

As per question, the ratio of the protonated to the deprotonated form of the acid will be:

$\frac{[Protonated]}{[Deprotonated]}=100$

Therefore, the ratio of the protonated to the deprotonated form of the acid is, 100

5 0

3 years ago