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2 years ago

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A gas mixture at 0°C and 1.0atm contains 0.010mol of H2, 0.015mol of O2, and 0.025mol of N2. Assuming ideal behavior, what is th

e partial pressure of hydrogen gas (H2) in the mixture?
A. About 0.010atm, because there is 0.010mol of H2 in the sample.

B. About 0.050atm, because there is 0.050mol of gases at 0°C and 1.0atm.

C. About 0.20atm, because H2 comprises 20% of the total number of moles of gas.

D. About 0.40atm, because the mole ratio of H2:O2:N2 is 0.4:0.6:1.

1 answer:

Olegator [25]2 years ago

7 0

Answer:

PH₂ = 0.2 atm

C) About 0.20atm, because H2 comprises 20% of the total number of moles of gas.

Explanation:

To determine the partial pressure of hydrogen gas (H2) in the mixture,

Partial pressure H₂ = Ptotal * xH₂

xH₂ = Mole fraction of H₂ = ∩H₂ / ( ∩H₂ + ∩O₂ + ∩N₂)

xH₂ = 0.01 / (0.01 + 0.015 + 0.025)

xH₂ = 0.01/0.05

xH₂ = 0.2

therefore

PH₂ = pT * xH₂

PH₂ = 1.0 atm * 0.2

PH₂ = 0.2 atm

so the correct option is C) About 0.20atm, because H2 comprises 20% of the total number of moles of gas.

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Identify the most and the least acidic compound in each of the following sets.

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Answer:

See explanation

Explanation:

Our answer options for this question are:

a. 2-chlorobutanoic acid:_______ 2-chlorobutanoic acid:_______ 3-chlorobutanoic acid:______.

b. 2,4-dinitrobenzoic acid:______ p-nitrobenzoic acid:______ p-bromobenzoic acid:_______.

c. p-cyanobenzoic acid:________ benzoic acid:_______ p-aminobenzoic acid:______

We have to check each set of molecules

<u>a. 2-chlorobutanoic acid,</u> <u>3-chlorobutanoic acid</u>

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In this case, the difference between these molecules is the position of "Cl". If the chlorine atom is closer to the acid group, we will have a higher inductive effect. So, the bond O-H would be weaker and we will have more acidity. So, the molecule with more acidity is <u>2-chlorobutanoic acid</u> and the less acidic would be <u>3-chlorobutanoic acid.</u>

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<u>b. 2,4-dinitrobenzoic acid,</u> <u>p-nitrobenzoic acid,</u> <u>p-bromobenzoic acid</u>

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In this case, we have several structural differences. In all the structure, we have deactivating groups ( $Br$ and $NO_2$ ). If we have a deactivating group the acidity will increase. In the case of "Br", we have a weak deactivating, so, this will be the less acidic one (<u>p-bromobenzoic acid)</u>

in <u>2,4-dinitrobenzoic acid</u> we have two deactivating groups, therefore, this would be the most acid compound.

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2 years ago

Check all the boxes that describe the electron sea model. It is the simplest metal bonding model. It is the most complicated met

Mrac [35]

Answer:

Another view could be from this definition: (Electron Sea Model) : The metallic bond consists of a series of metals atoms that have all donated their valence electrons to an electron cloud that permeates the structure. This electron cloud is frequently referred to as an electron sea. It might help to visualize the electron sea model as if it were a box of marbles that are surrounded by water. The marbles represent the metal atoms and the water represents the electron sea.

Explanation:

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2 years ago

Read 2 more answers

A student mixed together aqueous solutions of Y and Z. A white precipitate(solid)formed. which could not be Y and Z

maks197457 [2]

<span>(A)hydrochloric acid + silver nitrate
HCl(aq) + AgNO3(aq) -----> AgCl(s) +HNO3(aq)

</span><span>(B)hydrochloric acid + sodium hydroxide
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</span><span>(C)calcium chloride + silver nitrate
CaCl2(aq) + AgNO3(aq) ----> </span>AgCl(s) +Ca(NO3)2(aq)

<span>(D)sodium chloride + silver nitrate
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AgCl is a white precipitate.
In (B) no precipitate was formed, so answer is B.

4 0

2 years ago

A sample consisting of 1.0 mol of perfect gas molecules with CV = 20.8 J K−1 is initially at 4.25 atm and 300 K. It undergoes re

Marat540 [252]

Answer : The value of final volume, temperature and the work done is, 8.47 L, 258 K and -873.6 J

Explanation :

First we have to calculate the value of $\gamma$ .

$\gamma=\frac{C_p}{C_v}$

As, $C_p=R+C_v$

So, $\gamma=\frac{R+C_v}{C_v}$

Given :

$C_v=20.8J/K\\\\R=8.314J/K$

$\gamma=\frac{8.314+20.8}{20.8}=1.4$

Now we have to calculate the initial volume of gas.

Formula used :

$P_1V_1=nRT_1$

where,

$P_1$ = initial pressure of gas = 4.25 atm

$V_1$ = initial volume of gas = ?

$T_1$ = initial temperature of gas = 300 K

n = number of moles of gas = 1.0 mol

R = gas constant = 0.0821 L.atm/mol.K

$(4.25atm)\times V_1=(1.0mol)\times (0.0821L.atm/mol.K)\times (300K)$

$V_1=5.80L$

Now we have to calculate the final volume of gas by using reversible adiabatic expansion.

$P_1V_1^{\gamma}=P_2V_2^{\gamma}$

where,

$P_1$ = initial pressure of gas = 4.25 atm

$P_2$ = final pressure of gas = 2.50 atm

$V_1$ = initial volume of gas = 5.80 L

$V_2$ = final volume of gas = ?

$\gamma$ = 1.4

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

$(4.25atm)\times (5.80L)^{1.4}=(2.50atm)\times V_2^{1.4}$

$V_2=8.47L$

Now we have to calculate the final temperature of gas.

Formula used :

$P_2V_2=nRT_2$

where,

$P_2$ = final pressure of gas = 2.50 atm

$V_2$ = final volume of gas = 8.47 L

$T_2$ = final temperature of gas = ?

n = number of moles of gas = 1.0 mol

R = gas constant = 0.0821 L.atm/mol.K

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

$(2.50atm)\times (8.47L)=(1.0mol)\times (0.0821L.atm/mol.K)\times T_2$

$T_2=257.9K\approx 258K$

Now we have to calculate the work done.

$w=nC_v(T_2-T_1)$

where,

w = work done = ?

n = number of moles of gas =1.0 mol

$T_1$ = initial temperature of gas = 300 K

$T_2$ = final temperature of gas = 258 K

$C_v=20.8J/K$

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

$w=(1.0mol)\times (20.8J/K)\times (258-300)K$

$w=-873.6J$

Therefore, the value of final volume, temperature and the work done is, 8.47 L, 258 K and -873.6 J

8 0

2 years ago

What is the molarity of a HNO3 solution prepared by adding 290.7 mL of water to 350.0 mL of 12.3 M HNO3?

Lisa [10]

Answer:

6.72M of HNO3

Explanation:

In the problem you are diluting the original HNO3 solution by the addition of some water. The final volume is:

290.7mL + 350.0mL = 640.7mL

And you are diluting the solution:

640.7mL / 350.0mL = 1.8306 times

As the original concentration was 12.3M, the final concentration will be:

12.3M / 1.8306 =

<h3>6.72M of HNO3</h3>

5 0

2 years ago