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3 years ago

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Which sample of gas at STP has the same number of molecules as 6 liters of Cl2(g) at STP?

1 answer:

lara [203]3 years ago

3 0

Answer:

The correct answer is option 2 (6 liters of N2)

Explanation:

The complete question

Which sample of gas at STP has the same number of molecules as 6 liters of Cl2(g) at STP?

(1) 3 liters of O2(g)

(2) 6 liters of N2(g)

(3) 3 moles of O2(g)

(4) 6 moles of N2(g)

Step 1: Data given

Volume = 6 L

STP = 1 atm and 273 K

Step 2: Calculate moles of Cl2

p*V = n*R*T

n = (p*V)/ (R*T)

⇒with n = the number of moles Cl2 = TO BE DETERMINED

⇒with V = the volume of Cl2 = 6.0 L

⇒with p = the pressure of Cl2 = 1 atm

⇒with R = the gas constant = 0.08206 L*atm/mol*K

⇒ with T = the temperature = 273 K

n = (1.0 * 6.0 ) (0.08206 * 273)

n = 0.2678 moles Cl2

This means we need 0.2678 moles of a gas at STP

Option 3 has 3 moles of O2 ⇒ Not the same number of molecules

Option 4 has 6 moles of N2 ⇒ Not the same number of molecules

For 3 liters of O2 we'll have:

⇒with n = the number of moles O2 = TO BE DETERMINED

⇒with V = the volume of O2 = 3.0 L

⇒with p = the pressure of O2 = 1 atm

⇒with R = the gas constant = 0.08206 L*atm/mol*K

⇒ with T = the temperature = 273 K

n = (1.0 * 3.0 ) (0.08206 * 273)

n = 0,1339 moles ⇒ Not the same number of molecules

For 6 liters of N2 we'll have

⇒with n = the number of moles N2 = TO BE DETERMINED

⇒with V = the volume of N2 = 6.0 L

⇒with p = the pressure of N2 = 1 atm

⇒with R = the gas constant = 0.08206 L*atm/mol*K

⇒ with T = the temperature = 273 K

n = (1.0 * 6.0 ) (0.08206 * 273)

n = 0.2678 moles N2 ⇒ The same number of molecules

The correct answer is option 2 (6 liters of N2)

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Explanation:

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The buffer of the problem is HNO₂/NO₂⁻ <em>where nitrous acid is the weak acid and NO₂⁻ is its conjugate base.</em>

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NO₂⁻ + HCl → HNO₂ + Cl⁻

Ionic equation is:

NO₂⁻ + H⁺ + Cl⁻ → HNO₂ + Cl⁻

In the net ionic equation, you avoid the ions that don't react, that is:

<h3>NO₂⁻ + H⁺ → HNO₂</h3>

b. In the same way, the weak acid will react with the strong acid producing water and the conjugate base, thus:

HNO₂ + NaOH → NO₂⁻ + H₂O + Na⁺

The ionic equation is:

HNO₂ + Na⁺ + OH⁻ → NO₂⁻ + H₂O + Na⁺

And the net ionic equation is:

<h3>HNO₂ + OH⁻ → NO₂⁻ + H₂O</h3>

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Answer:

Explanation:

Given that:

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Initial concentration $C_o$ = 2.00 ms/l

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$time (t) = \dfrac{200}{0.3}$

$time (t) = 666.66 \ sec$

$time (t) = \dfrac{666.66 }{3600} hrs$

$time (t) = 0.185 hrs$

$\text{Using First Order Reaction:}$

$\dfrac{dc}{dt}=kc$

where;

$t = \dfrac{1}{k} \Big( \dfrac{C_o}{C_e}-1 \Big)$

$0.185 = \dfrac{1}{5.09} \Big ( \dfrac{200}{C_e}- 1 \Big)$

$0.942 = \Big ( \dfrac{200}{C_e}- 1 \Big)$

$1+ 0.942 = \Big ( \dfrac{200}{C_e} \Big)$

$\dfrac{200}{C_e} = 1.942$

$C_e = \dfrac{200}{1.942}$

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Thus; the concentration of species in the reactant = 102.98 mg/l

b). If the plug flow reactor has the same efficiency as CSTR, Then:

$t _{PFR} = \dfrac{1}{k} \Big [ In ( \dfrac{C_o}{C_e}) \Big ]$

$\dfrac{V_{PFR}}{Q_{PFR}} = \dfrac{1}{k} \Big [ In ( \dfrac{C_o}{C_e}) \Big ]$

$\dfrac{V_{PFR}}{Q_{PFR}} = \dfrac{1}{5.09} \Big [ In ( \dfrac{200}{102.96}) \Big ]$

$\dfrac{V_{PFR}}{Q_{PFR}} =0.196 \Big [ In ( 1.942) \Big ]$

$\dfrac{V_{PFR}}{Q_{PFR}} =0.196(0.663)$

$\dfrac{V_{PFR}}{0.3 hrs} =0.196(0.663)$

$\dfrac{V_{PFR}}{0.3*3600 sec} =0.196(0.663)$

$V_{PFR} =0.196(0.663)*0.3*3600$

$V_{PFR} = 140.34 \ m^3$

The volume of the PFR is ≅ 140 m³

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bagirrra123 [75]

For the compound B the following statement is correct-

B. It is an ether because it is unable to form a hydrogen bond, so it is less soluble in water.

The solubility of alcohol in water depends upon the capability of formation of hydrogen bond in the solute. Now in alcohol the -OH group is polar in nature which enhance the possibility of hydrogen bond formation and it is more soluble in water.

On the other hand although there presence a -O- functional group in ether. It is less soluble in water due to non polarity of the functional group.

From the given data it is seen that compound A is more soluble in water than compound B. Thus it may be predicted that compound A is alcohol and B is ether.

Henceforth, for the compound B the following statement is correct-

B. It is an ether because it is unable to form a hydrogen bond, so it is less soluble in water.

The reason of incorrect options:

A. compound B cannot be an alcohol as it is less soluble in water.

C. In ether the functional group is -O-, thus electronegative atom (O) is present.

D. As both the compound (alcohol and ether) has equal molecular mass thus the organic chain will be same in alcohol and the hydrogen bond interaction will be more prominent than the dispersion force between the -OH group.

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