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2 years ago

14

Which is the strongest acid? Ch3ch2ch2chfch2co2h ch3ch2ch2cbr2ch2co2h ch3ch2ch2cf2ch2co2h ch3chbrch2ch2ch2co2h ch2clch2ch2ch2ch2

co2h?

1 answer:

liq [111]2 years ago

8 0

The strength of an acid increases if the stability of conjugate base increases

The stability of a conjugate base increases with the presence of electron with drawing group (electronegative group)

Thus more the electronegativity of an atom attached to a carboxylic acid higher the strength of acid

In these examples CH3CH2CH2CF2CH2COOH contains to electronegative flourine atoms which stabilizes the conjugate base hence this will be the strongest acid among the given acids

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When nitrogen dioxide (NO2) from car exhaust combines with water in the air, it forms nitric acid (HNO3), which causes acid rain

yKpoI14uk [10]

<u>Answer:</u>

<u>For a:</u> The number of molecules of nitrogen dioxide is $4.52\times 10^{23}$

<u>For b:</u> The mass of nitric acid formed is 54.81 grams

<u>For c:</u> The mass of nitric acid formed is 206 grams

<u>Explanation:</u>

The given chemical reaction follows:

$3NO_2(g)+H_2O(l)\rightarrow 2HNO_3(aq.)+NO(g)$

<u>For a:</u>

By Stoichiometry of the reaction:

1 mole of water reacts with 3 moles of nitrogen dioxide

So, 0.250 moles of water will react with $\frac{3}{1}\times 0.250=0.75mol$ of nitrogen dioxide

According to mole concept:

1 mole of a compound contains $6.022\times 10^{23}$ number of molecules.

So, 0.75 moles of nitrogen dioxide will contain $0.75\times 6.022\times 10^{23}=4.52\times 10^{23}$ number of molecules

Hence, the number of molecules of nitrogen dioxide is $4.52\times 10^{23}$

<u>For b:</u>

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

Given mass of nitrogen dioxide = 60.0 g

Molar mass of nitrogen dioxide = 46 g/mol

Putting values in equation 1, we get:

$\text{Moles of nitrogen dioxide}=\frac{60.0g}{46g/mol}=1.304mol$

By Stoichiometry of the reaction:

3 moles of nitrogen dioxide produces 2 mole of nitric acid

So, 1.304 moles of nitrogen dioxide will produce = $\frac{2}{3}\times 1.304=0.870$ moles of nitric acid

Now, calculating the mass of nitric acid from equation 1, we get:

Molar mass of nitric acid = 63 g/mol

Moles of nitric acid = 0.870 moles

Putting values in equation 1, we get:

$0.870mol=\frac{\text{Mass of nitric acid}}{63g/mol}\\\\\text{Mass of nitric acid}=(0.870mol\times 63g/mol)=54.81g$

Hence, the mass of nitric acid formed is 54.81 grams

<u>For c:</u>

<u>For nitrogen dioxide:</u>

Given mass of nitrogen dioxide = 225 g

Molar mass of nitrogen dioxide = 46 g/mol

Putting values in equation 1, we get:

$\text{Moles of nitrogen dioxide}=\frac{225g}{46g/mol}=4.90mol$

<u>For water:</u>

Given mass of water = 55.2 g

Molar mass of water = 18 g/mol

Putting values in equation 1, we get:

$\text{Moles of water}=\frac{55.2g}{18g/mol}=3.06mol$

By Stoichiometry of the reaction:

3 moles of nitrogen dioxide reacts with 1 mole of water

So, 4.90 moles of nitrogen dioxide will react with = $\frac{1}{3}\times 4.90=1.63mol$ of water

As, given amount of water is more than the required amount. So, it is considered as an excess reagent.

Thus, nitrogen dioxide is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

3 mole of nitrogen dioxide produces 2 moles of nitric acid

So, 4.90 moles of nitrogen dioxide will produce $\frac{2}{3}\times 4.90=3.27mol$ of nitric acid

Now, calculating the mass of nitric acid from equation 1, we get:

Molar mass of nitric acid = 63 g/mol

Moles of nitric acid = 3.27 moles

Putting values in equation 1, we get:

$3.27mol=\frac{\text{Mass of nitric acid}}{63g/mol}\\\\\text{Mass of nitric acid}=(3.27mol\times 63g/mol)=206g$

Hence, the mass of nitric acid formed is 206 grams

7 0

2 years ago

4.85 x 10^-4 tons to cups

fiasKO [112]

Answer:

2.08753919851 cups

0.464942810185139 cups

2.08753919851 cups

Explanation:

i looked it up

7 0

2 years ago

5. A 25 kg object travels a distance of 90 m in 30 s. Find the momentum

xz_007 [3.2K]

Answer:

★ v = d/t

★ v = 90/30

★ v =3

<u>Acc</u><u> </u><u>to</u><u> </u><u>question</u>

★ momentum = mass * velocity

★ m = 25×3

★ m = 75kgm/s

Hope it help

6 0

2 years ago

Read 2 more answers

What is the intermolecular force of NH3

Dominik [7]

ANSWER:
London dispersion and hydrogen bonds.

EXPLANATION :
Every molecule experiences London dispersion as an intermolecular force.since the ammonia ion has hydrogen atoms bonded to nitrogen,a very electronegative atom,the molecule is also polar since the nitrogen atom more strongly pulls on the electrons from the hydrogen atoms than the hydrogens themselves do.

This effect is similar to that water,where the oxygen pulls the electrons of the hydrogen atoms with a greater magnitude,resulting in the oxygen having a partial negative charge and the hydrogens having a partial positive charge relative to each other.

This polarity shows that the molecule has dipole-dipole intermolecular forces but since the polarity is from a result of highly electronegative atoms (such as nitrogen,oxygen,fluorine) and hydrogen atoms actually bonded to them,the polarity is categorized in it's own intermolecular force called a hydrogen bond.

I HOPE IT HELPS:)

3 0

3 years ago

A student heats a sample of hydrate once, and the mass of the sample and the evaporating dish is 16.428 g. After a second heatin

jolli1 [7]

Answer:

12.371 g

Explanation:

Given :

$m_{evaporating\ dish}=1.135\ g$

$m_{evaporating\ dish}+m_{Hydrate\ sample}=25.637\ g$

$m_{evaporating\ dish}+m_{First\ heated\ sample}=16.428\ g$

$m_{evaporating\ dish}+m_{Second\ heated\ sample}=13.266\ g$

Mass of salt hydrate:

$m_{evaporating\ dish}=1.135\ g$

$m_{evaporating\ dish}+m_{Hydrate\ sample}=25.637\ g$

$m_{Hydrate\ sample}=25.637-m_{evaporating\ dish}\ g=25.637-1.135\ g=24.502\ g$

Mass of salt anhydrous:

$m_{evaporating\ dish}=1.135\ g$

$m_{evaporating\ dish}+m_{Second\ heated\ sample}=13.266\ g$

$m_{Second\ heated\ sample}=m_{salt\ anhydrous}=13.266-m_{evaporating\ dish}\ g=13.266-1.135\ g=12.131\ g$

Mass of water:

$m_{water}=m_{Hydrate\ sample}-m_{salt\ anhydrous}=24.502-12.131\ g=12.371\ g$

$m_{water}=12.371\ g$

4 0

3 years ago