A 2.00 L container of gas has a pressure of 1.00 atm at 300 K. The temperature of the gas is halved to 150K, and the measured pressure of the same 2.00 Liter sample is 0.420 atm. Which of the following is the best explanation for these observations? A. Pressure is proportional to temperature for a fixed volume of gas. B. The molecules of the gas occupy a significant portion of the volume. C. The molecules of the gas have negligible volume of their own. D. The molecules have significant attractive forces at 150 K. E. The gas is closer to an ideal gas at 150 K.

Answers

Answer 1

Answer:

The best explanation for the observation is that, the Pressure is proportional to temperature for a fixed volume of gas. (Option A)

From the question given above, the following data were obtained:

Initial volume (V₁) = 2 L

Initial pressure (P₁) = 1 atm

Initial temperature (T₁) = 300 K

Final temperature (T₂) = 150 K

Final volume (V₂) = 2 L = constant

Final pressure (P₂) = 0.420 atm

From the above, we can see that the volume is constant.

Applying the combine gas equation, we can conclude as follow:

P₁V₁ / T₁ = P₂V₂ / T₂

V₁ = V₂

P₁ / T₁ = P₂ / T₂

P/T = constant

P = constant × T

Thus, we can conclude that the pressure is proportional to the temperature at constant volume. This simply implies that the pressure will increase if the temperature increase and it will also decrease if the temperature decreases.

The correct answer to the question is Option A.

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Answer 2

Answer:

Answer:

Explanation:

PV=nRT

PV/nT

V/T -> (1)/(300)=(x)/(150)

x=.420

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The gas-phase reaction follows an elementary rate law and is to be carried out first in a PFR and then in a separate experiment in a CSTR. When pure A is fed to a 10 dm 3 PFR at 300 K and a volumetric flow rate of 5 dm 3 /s, the conversion is 80%. When a mixture of 50% A and 50% inert (I) is fed to a 10 dm 3 CSTR at 320 K and a volumetric flow rate of 5 dm 3 /s, the conversion is also 80%. What is the activation energy in cal/mol

Answers

Answer:

The activation energy is $=8.1\,kcal\,mol^(-1)$

Explanation:

The gas phase reaction is as follows.

$A \rightarrow B+C$

The rate law of the reaction is as follows.

$-r_(A)=kC_(A)$

The reaction is carried out first in the plug flow reactor with feed as pure reactant.

From the given,

Volume "V" = $10dm^(3)$

Temperature "T" = 300 K

Volumetric flow rate of the reaction $v_(o)=5dm^(3)s$

Conversion of the reaction "X" = 0.8

The rate constant of the reaction can be calculate by the following formua.

$V= (v_(0))/(k)[(1+\epsilon )ln((1)/(1-X)-\epsilon X)]$

Rearrange the formula is as follows.

$k= (v_(0))/(V)[(1+\epsilon )ln((1)/(1-X)-\epsilon X)]............(1)$

The feed has Pure A, mole fraction of A in feed $y_{A_(o)}$ is 1.

$\epsilon =y_{A_(o)}\delta$

$\delta$ = change in total number of moles per mole of A reacte.

$=1(2-1)=1$

Substitute the all given values in equation (1)

$k=(5m^(3)/s)/(10dm^(3))[(1+1)ln (1)/(1-0.8)-1 * 0.8] = 1.2s^(-1)$

Therefore, the rate constant in case of the plug flow reacor at 300K is $1.2s^(-1)$

The rate constant in case of the CSTR can be calculated by using the formula.

$(V)/(v_(0))= (X(1+\epsilon X))/(k(1-X)).............(2)$

The feed has 50% A and 50% inerts.

Hence, the mole fraction of A in feed $y_{A_(o)}$ is 0.5

$\epsilon =y_{A_(o)}\delta$

$\delta$ = change in total number of moles per mole of A reacted.

$=0.5(2-1)=0.5$

Substitute the all values in formula (2)

$(10dm^(3))/(5dm^(3))=(0.8(1+0.5(0.8)))/(k(1-0.8))=2.8s^(-1)$

Therefore, the rate constant in case of CSTR comes out to be $2.8s^(-1)$

The activation energy of the reaction can be calculated by using formula

$k(T_(2))=k(T_(1))exp[(E)/(R)((1)/(T_(1))-(1)/(T_(2)))]$

In the above reaction rate constant at the two different temperatures.

Rearrange the above formula is as follows.

$E= R *((T_(1)T_(2))/(T_(1)-T_(2)))ln(k(T_(2)))/(k(T_(1)))$

Substitute the all values.

$=1.987cal/molK((300K *320K)/(320K *300K))ln (2.8)/(1.2)=8.081 *10^(3)cal\,mol^(-1)$

$=8.1\,kcal\,mol^(-1)$

Therefore, the activation energy is $=8.1\,kcal\,mol^(-1)$

Which is not the name of a family on the periodic tablea) Halogens
b) Noble Gases
c) Alkali Earth Metals
d) Actinides

Answers

Answer:

I think it's D

Explanation:

Actinides is the correct answer
your noble gases are in the 8th column
Your Halogens are your 7th column
and you alkali earth metals are the 2nd column

What is the change in electrons for nitrogen in the following reaction?S + NO3 - -> SO2 + NO

A. Gain 2 electrons
B. Gain 3 electrons
C. Lose 2 electrons
D. Lose 3 electrons

Answers

Nitrogen changes from +5 in $NO_3$ - to +2 in NO. This means nitrogen has gained 3 electrons. Option B

To determine the change in electrons for nitrogen in the given reaction, we need to compare the oxidation state of nitrogen in the reactant ( $NO_3$ -) and the product (NO).

In the reactant, -, nitrogen is in the +5 oxidation state. This is because oxygen has an oxidation state of -2, and there are three oxygen atoms in $NO_3$ -. Therefore, nitrogen must have an oxidation state of +5 to balance the overall charge of $NO_3$ -.

In the product, NO, nitrogen is in the +2 oxidation state. This is because oxygen has an oxidation state of -2, and there is only one oxygen atom in NO. Therefore, nitrogen must have an oxidation state of +2 to balance the overall charge of NO.

By comparing the oxidation states of nitrogen in the reactant and the product, we can determine the change in electrons. The change in oxidation state corresponds to the change in the number of electrons gained or lost by the nitrogen atom.

In this case, nitrogen changes from +5 in $NO_3$ - to +2 in NO. This means nitrogen has gained 3 electrons.

Therefore, the correct answer is B) Gain 3 electrons.

The nitrogen atom undergoes a reduction because it gains electrons, reducing its oxidation state from +5 to +2 in the reaction.

Option B

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Calculate the solubilities of the following compounds in a 0.02 M solution of barium nitrate using molar concentrations, first ignoring ionic strength and activities. a. silver iodate
b. barium sulfate
c. Repeat the above calculations using ionic strength and activities.

Answers

Answer:

a. 1.7 × 10⁻⁴ mol·L⁻¹; b. 5.5 × 10⁻⁹ mol·L⁻¹

c. 2.3 × 10⁻⁴ mol·L⁻¹; 5.5 × 10⁻⁸ mol·L⁻¹

Explanation:

a. Silver iodate

Let s = the molar solubility.

AgIO₃(s) ⇌ Ag⁺(aq) + IO₃⁻(aq); Ksp = 3.0 × 10⁻⁸

E/mol·L⁻¹: s s

$K_(sp) =\text{[Ag$^(+)$][IO$_(3)$$^(-)$]} = s* s = s^(2) = 3.0* 10^(-8)\ns = \sqrt{3.0* 10^(-8)} \text{ mol/L} = 1.7 * 10^(-4) \text{ mol/L}$

b. Barium sulfate

BaSO₄(s) ⇌ Ba²⁺(aq) + SO₄²⁻(aq); Ksp = 1.1 × 10⁻¹⁰

I/mol·L⁻¹: 0.02 0

C/mol·L⁻¹: +s +s

E/mol·L⁻¹: 0.02 + s s

$K_(sp) =\text{[Ba$^(2+)$][SO$_(4)$$^(2-)$]} = (0.02 + s) * s \approx 0.02s = 1.1* 10^(-10)\ns = (1.1* 10^(-10))/(0.02) \text{ mol/L} = 5.5 * 10^(-9) \text{ mol/L}$

c. Using ionic strength and activities

(i) Calculate the ionic strength of 0.02 mol·L⁻¹ Ba(NO₃)₂

The formula for ionic strength is

$\mu = (1)/(2) \sum_(i) {c_(i)z_(i)^(2)}\n\n\mu = (1)/(2) (\text{[Ba$^(2+)$]}\cdot (2+)^(2) + \text{[NO$_(3)$$^(-)$]}*(-1)^(2)) = (1)/(2) (\text{0.02}* 4 + \text{0.04}*1)= (1)/(2) (0.08 + 0.04)\n\n= (1)/(2) *0.12 = 0.06$

(ii) Silver iodate

a. Calculate the activity coefficients of the ions

$\log \gamma = -0.51z^(2)√(I) = -0.051(1)^(2)√(0.06) = -0.51* 0.24 = -0.12\n\gamma = 10^(-0.12) = 0.75$

b. Calculate the solubility

AgIO₃(s) ⇌ Ag⁺(aq) + IO₃⁻(aq)

$K_(sp) =\text{[Ag$^(+)$]$\gamma_(Ag^(+))$[IO$_(3)$$^(-)$]$\gamma_{IO_(3)^(-)}$} = s*0.75* s * 0.75 =0.56s^(2)= 3.0 * 10^(-8)\ns^(2) = (3.0 * 10^(-8))/(0.56) = 5.3 * 10^(-8)\n\ns =2.3 * 10^(-4)\text{ mol/L}$

(iii) Barium sulfate

a. Calculate the activity coefficients of the ions

$\log \gamma = -0.51z^(2)√(I) = -0.051(2)^(2)√(0.06) = -0.51*16* 0.24 = -0.50\n\gamma = 10^(-0.50) = 0.32$

b. Calculate the solubility

BaSO₄(s) ⇌ Ba²⁺(aq) + SO₄²⁻(aq

$K_(sp) =\text{[Ba$^(2+)$]$\gamma_( Ba^(2+))$[SO$_(4)$$^(2-)$]$\gamma_{ SO_(4)^(2-)}$} = (0.02 + s) * 0.32* s* 0.32 \approx 0.02*0.10s\n2.0* 10^(-3)s = 1.1 * 10^(-10)\ns = (1.1* 10^(-10))/(2.0 * 10^(-3)) \text{ mol/L} = 5.5 * 10^(-8) \text{ mol/L}$

Draw the structure of a compound with the molecular formula CgH1002 that exhibits the following spectral data. (a) IR: 3005 cm-1, 1676 cm-1, 1603 cm-1
(b) H NMR: 2.6 ppm (singlet, I = 3H), 3.9 ppm (singlet, I = 3H), 6.9 ppm (doublet, I = 2H), 7.9 ppm (doublet, I = 2H)
(c) 13C NMR: 26.2, 55.4, 113.7, 130.3, 130.5, 163.5, 196.6 ppm ?

Answers

Answer:

The answer you are looking for is A

Describe how you would make 250 ml of a 3 M solution of sodium acetate (NaOAc = 82.03 g/mol). First figure out how much sodium acetate you would need, then describe how you would make the solution if you were given a bottle of solid sodium acetate, a volumetric flask, and DI water.

Answers

Answer:

You need to do the following conversion to pass from 3M in 250 mL to g of sodium acetate

$3 M (mol/L)*(1L/1000 mL)*(250 mL)*(82.03 g/1 mol)=61.52 g$

Explanation:

First, you need to dissolve 61.52 g of solid sodium acetate (MW 82.03 g/mol) in 200 ml of DI water. Then, using a volumetric flask add water to bring the total volume of the solution to 250 mL.

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