100 mL of a strong acid is completely neutralized by 100 mL of a strong base. The observed products are salt and water. Upon further investigation, the solution has a pH > 7. How is this possible?


A) The concentration of acid is the same as the concentration of the base.

B) The concentration of acid is higher than the concentration of the base.

C) The concentration of base is higher than the concentration of the acid.

D) The water autoionizes to give [OH-] ions.

Answer:

Q< K hence a precipitate will not form.

Explanation:

First convert the concentration to molL-1

For number of moles of K^+ = 0.05×20/1000 = 1×10^-3 moles

If we have 1×10^-3 moles in 20cm3

Then in 1000cm^3 we have 1×10^-3 moles×1000/20= 0.05 M

For ClO4^-= 0.50×80/1000= 0.04 moles

If we have 0.04 moles in 80cm3

Then in 1000cm^3 we have 0.04 moles×1000/80= 0.5 M

Q= [K^+] [ClO4^-]

Q= [0.05] [0.5]

Q= 0.025= 2.5×10-2

Q< K hence a precipitate will not form.

Answer:

1.64g

Explanation:

The reaction scheme is given as;

2-bromocyclohexanol  --> 1,2-epoxycyclohexane + HBr

From the reaction above,

1 mol of 2-bromocyclohexanol produces 1 mol of 1,2-epoxycyclohexane

3.0 grams of trans-2-bromocyclohexanol.

Molar mass = 179.05 g/mol

Number of moles = mass / molar mass = 3 / 179.05 = 0.016755 mol

This means  0.016755 mol of 1,2-epoxycyclohexane would be produced.

Molar mass =  98.143 g/mol

Theoretical  yield = Number of moles * Molar mass

Theoretical  yield =  0.016755 * 98.143 ≈ 1.64g

Answer:

278 balloons can be aired.

Explanation:

We apply the Ideal Gases Law to determine the total available volume for each baloon of 0.75L. So we need to divide the volume we obtain by the volume of each baloon.

We determine the moles of CO₂

375 g. 1 mol / 44g = 8.52 moles

V = (n . R . T) / P → (8.52 mol . 0.082 . 298K) / 1 atm = 208.3 L

That's the total volume from the tank, so we can inflate (208.3 / 0.75) = at least 278 balloons

A tank that contains 375 g of CO₂ at 1.0 atm and 298 K can be used to fill 277 0.75 L-balloons.

First, we will convert 375 g of CO₂ to moles using its molar mass (44.01 g/mol).

[tex]375 g \times \frac{1mol}{44.01g} = 8.52 mol[/tex]

1 mole of CO₂ at 1.0 atm and 298 K occupies 24.4 L. The volume occupied by 8.52 moles of CO₂ is:

[tex]8.52 mol \times \frac{24.4L}{mol} = 208 L[/tex]

8.52 moles of CO₂ at 1.0 atm and 298 K occupies 208 L. The number of 0.75 L-balloons that can be filled is:

[tex]208 L \times \frac{1balloon}{0.75L} \approx 277 balloon[/tex]

A tank that contains 375 g of CO₂ at 1.0 atm and 298 K can be used to fill 277 0.75 L-balloons.

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Answer:

Explanation:

Given parameters:

number of moles of CH₄  = 6.816 moles

number of moles of C₂H₆  = 0.589 mole

number of moles of C₃H₈  =  0.381 moles

Total pressure of the gases  = 3.93 atm

Unknown:

Partial pressure of the gases = ?

Solution:

Dalton's law of partial pressure states that " the total pressure of a mixture of gases is equal to the sum of the partial pressures of the constituent gases".

                         P[tex]_{T}[/tex]   = P[tex]_{1}[/tex]  + P[tex]_{2}[/tex]  + P[tex]_{3}[/tex]...............

where P[tex]_{T}[/tex]  = total pressure of the gas mixture

           P₁, P₂, P₃......  = partial pressure of gas 1, 2, 3............

The partial pressure of a gas is the pressure it would exert if confined alone in the volume of the gas mixture.

      Partial pressure of gas  = mole fraction of gas x total pressure of mixture

Now let us solve the problem.

Total number of moles of gases  = 6.816 moles + 0.589 mole + 0.381 moles

                                                        = 7.786moles

Partial pressure of CH₄ = [tex]\frac{6.816}{7.786} x 3.93 = 3.44 atm[/tex]

Partial pressure of C₂H₆  = [tex]\frac{0.589}{7.786}[/tex]  x 3.93  = 0.30atm

Partial pressure of C₃H₈  = [tex]\frac{0.381}{7.786}[/tex] x 3.93  = 0.19atm

The molar mass of the unknown gas which has a mass of 36.5 g is 146 g/mol

Mass of He = 0.5 g

Molar mass of He = 4 g/mol

At standard temperature and pressure (stp),

1 mole of He = 22.4 L

Therefore,

0.125 mole of He = 0.125 × 22.4

0.125 mole of He = 2.8 L

Volume of He = 2.8 L

Volume of unknown gas =?

Volume of He = ½ × Volume of unknown gas

2.8 = ½ × Volume of unknown gas

Cross multiply

Volume of unknown gas = 2.8 × 2

Volume of unknown gas = 5.6 L

22.4 L = 1 mole of unknown gas

Therefore,

5.6 L = [tex]\frac{5.6}{22.4}\\\\[/tex]

5.6 L = 0.25 mole of unknown gas.

Mole of unknown gas = 0.25 mole

Mass of unknown gas = 36.5 g

Molar mass of unknown gas =?

[tex]Molar mass = \frac{mass}{mole} \\\\ = \frac{36.5}{0.25}\\\\[/tex]

Therefore, the molar mass of the unknown gas is 146 g/mol

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At STP, one mole of any ideal gas occupies around 22.4 L. As the 0.500 g Helium sample occupies half the volume of the unknown gas, it means the that the molar mass of the unknown gas would be twice the mass of the helium. The molar mass of the unknown gas is calculated to be 292 g/mol.

In order to determine the molar mass of the unknown gas, we need to observe the characteristics of gases at Standard Temperature and Pressure (STP). At STP, one mole of any ideal gas will occupy approximately 22.4 L. Given that the 0.500 g sample of Helium with a molar mass of 4 g/mol occupies half of the volume the unknown gas does, we can conclude that one mole of the unknown gas will have twice the mass of the Helium sample.

So, if 0.500 g of He occupies a volume that is equal to the volume of 0.125 moles of He (since 4 g of He = 1 mole), this means the same volume is being occupied by the unknown gas. Given the mass of the unknown gas is 36.5 g, the molar mass of the unknown gas would be 36.5 g / 0.125 moles = 292 g/mol.

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Answer:

4.77 is the pH of the given buffer .

Explanation:

To calculate the pH of acidic buffer, we use the equation given by Henderson Hasselbalch:

[tex]pH=pK_a+\log(\frac{[salt]}{[acid]})[/tex]

[tex]pH=-\log[K_a]+\log(\frac{[salt]}{[acid]})[/tex]

[tex]pH=-\log[K_a]+\log(\frac{[CH_3CH_2COONa]}{[CH_3CH_2COOH]})[/tex]

We are given:

[tex]K_a[/tex] = Dissociation constant of propanoic acid = [tex]1.3\times 10^{-5}[/tex]

[tex][CH_3CH_2COONa]=0.254 M[/tex]

[tex][CH_3CH_2COOH]=0.329 M[/tex]

pH = ?

Putting values in above equation, we get:

[tex]pH=-\log[1.3\times 10^{-5}]+\log(\frac{[0.254 M]}{[0.329]})[/tex]

pH = 4.77

4.77 is the pH of the given buffer .

Final answer:

To determine the pH of a buffer with given concentrations of propanoate and propanoic acid, apply the Henderson-Hasselbalch equation using the given Ka value for propanoic acid and the concentrations of the acid and its conjugate base.

Explanation:

To calculate the pH of a buffer consisting of 0.254 M sodium propanoate (CH3CH2COO-Na+) and 0.329 M propanoic acid (CH3CH2COOH), we use the Henderson-Hasselbalch equation:

pH = pKa + log([A-]/[HA])

Where:

First, we calculate the pKa:

pKa = -log(Ka) = -log(1.3 x 10-5)

Then plug the values into the Henderson-Hasselbalch equation:

pH = pKa + log(0.254/0.329)

After calculating the above expression, we find the pH of the buffer solution.

Answer:

The average rate of formation of I₂(g) is 6.00x10⁻⁵Ms⁻¹

Explanation:

The question is:

The average rate of formation of I2 over the same time period is______M s-1.

Based in the reaction:

2 HI(g) ⇄ H₂(g) + I₂(g)

If 2 moles of HI(g) disappears in a rate of 1.20x10⁻⁴Ms⁻¹, 1 mole of  I₂(g) will appears at:

1 mole I₂(g) × (1.20x10⁻⁴Ms⁻¹ / 2 mol) = 6.00x10⁻⁵Ms⁻¹ I₂(g)

Answer:

Nuclear reaction takes place at the nucleus whereas chemical reaction involves valence electrons

Explanation:

Answer:

Explanation:

During a phase change, the temperature remains constant because the energy supplied via heat is used differently. While not changing states, the heat (energy) goes into changing the kinetic energy of every particle in the body that receives it.

Answer:

When the concentration of F- exceeds 0.0109 M, BaF2 will precipitate.

Explanation:

Ba²⁺(aq) + 2 F⁻(aq) <----> BaF₂(s)

When BaF₂ precipitates, the Ksp relation is given by

Ksp = [Ba²⁺] [F⁻]²

[Ba²⁺] = 0.0144 M

[F⁻] = ?

Ksp = (1.7 × 10⁻⁶)

1.7 × 10⁻⁶ = (0.0144) [F⁻]²

[F⁻]² = (1.7 × 10⁻⁶)/0.0144 = 0.0001180555

[F⁻] = √0.0001180555 = 0.01086 M = 0.0109 M

Hope this Helps!!!

When the concentration of F⁻ exceeds 0.0109 M in a solution containing 0.0144 M Ba²⁺, BaF₂ will begin to precipitate.

To determine when BaF₂ will precipitate from a solution containing Ba²⁺ and F⁻ ions, we need to use the solubility product constant (Ksp). For BaF₂, Ksp is given as 1.7 × 10⁻⁶.

The dissociation reaction for BaF₂ is:

BaF₂ (s) ⇌ Ba²⁺ (aq) + 2F⁻ (aq)

The Ksp expression is:

Ksp = [Ba²⁺][F⁻]²

Given [Ba²⁺] = 0.0144 M, we need to find [F⁻] when the solution just starts to precipitate.

Therefore, when the concentration of F⁻ exceeds approximately 0.0109 M, BaF₂ will start to precipitate.

Answer:

11.3 g of H₂O will be produced.

Explanation:

The combustion is:

2C₈H₁₈ +  25O₂→  16CO₂  +  18H₂O

First of all, we determine the moles of the reactants in order to find out the limiting reactant.

8 g / 114g/mol = 0.0701 moles of octane

37g / 32 g/mol = 1.15 moles of oxygen

The limiting reagent is the octane. Let's see it by this rule of three:

25 moles of oxygen react to 2 moles of octane so  

1.15 moles of oxygen will react to ( 1.15 . 2)/ 25 = 0.092 moles of octane.

We do not have enough octane, we need 0.092 moles and we have 0.0701 moles. Now we work with the stoichiometry of the reaction so we make this rule of three:

2 moles of octane produce 18 moles of water

Then 0.0701 moles of octane may produce (0.0701 . 18)/2= 0.631 moles of water.

We convert the moles to mass → 0.631 mol . 18 g/1mol = 11.3 g of H₂O will be produced.

Answer:

Neither of them will neutralize the buffer solution.

Explanation:

The buffer solution of HNO₂ and KNO₂ will be neutralized when the acid reacts and consume all of the base of the buffer solution or when the base added reacts and consume all of the acid of the buffer solution.

First, we need to calculate the number of moles of the acid and the base of the buffer:

[tex] n_{HNO_{2}} = C*V = 0.100 M*0.500 L = 0.050 moles [/tex]

[tex] n_{KNO_{2}} = C*V = 0.150 M*0.500 L = 0.075 moles [/tex]

Now, let's evaluate each case.

A) 250 mg of NaOH:

We need to calculate the number of moles of NaOH

[tex] n_{NaOH} = \frac{m}{M} [/tex]

Where m: is the mass = 250 mg, and M: is the molar mass = 39.99 g/mol  

[tex] n_{NaOH} = \frac{0.250 g}{39.99 g/mol} = 6.25 \cdot 10^{-3} moles [/tex]

The number of moles of the acid HNO₂ after reaction with the base added NaOH is:

[tex] n_{a_{T}} = n_{a} - n_{b} = 0.050 moles - 6.25 \cdot 10^{-3} moles = 0.044 moles [/tex]

After the reaction of HNO₂ with the NaOH remains 0.044 moles of acid, hence, 250 mg of NaOH would not exceed the capacity of the buffer to neutralize it.

B) 350 mg KOH:

The number of moles of KOH is:

[tex]n_{KOH} = \frac{m}{M} = \frac{0.350 g}{56.1056 g/mol} = 6.23 \cdot 10^{-3} moles[/tex]

Now, the number of moles of HNO₂ that remains in the solution is:

[tex] n_{T} = 0.050 moles - 6.23 \cdot 10^{-3} moles = 0.044 moles[/tex]

Therefore, 350 mg of KOH would not exceed the capacity of the buffer to neutralize it.

C) 1.25 g of HBr:

The number of moles of HBr is:

[tex] n_{HBr} = \frac{m}{M} = \frac{1.25 g}{80,9119 g/mol} = 0.015 moles [/tex]

Now, the number of moles of the base KNO₂ that remains in solution after the reaction with HBr is:

[tex] n_{T} = 0.075 moles - 0.015 moles = 0.06 moles [/tex]

Hence, 1.25 g of HBr would not exceed the capacity of the buffer to neutralize it.

D) 1.35 g of HI:

The number of moles of HI is:

[tex] n_{HI} = \frac{m}{M} = \frac{1.35 g}{127.91 g/mol} = 0.0106 moles [/tex]

Now, the number of moles of the base KNO₂ that remains in solution after the reaction with HI is:

[tex] n_{T} = 0.075 moles - 0.0106 moles = 0.0644 moles [/tex]

Hence, 1.35 g of HI would not exceed the capacity of the buffer to neutralize it.

Therefore, neither of them will neutralize the buffer solution.

I hope it helps you!

Part A: No

Part B: Yes

Part C: Yes

Part D: Yes

Part A: To determine if 250 mg of NaOH would exceed the buffer capacity, we first need to calculate the number of moles of NaOH added and compare it to the number of moles of HNO2 in the buffer.

The moles of NaOH added are calculated as follows:

Part A: No

Part B: Yes

Part C: Yes

Part D: Yes

Part A: To determine if 250 mg of NaOH would exceed the buffer capacity, we first need to calculate the number of moles of NaOH added and compare it to the number of moles of HNO2 in the buffer.

 The moles of NaOH added are calculated as follows:

[tex]\[ \text{moles of NaOH} = \frac{\text{mass of NaOH}}{\text{molar mass of NaOH}} = \frac{250 \times 10^{-3} \text{g}}{40.0 \text{g/mol}} = 6.25 \times 10^{-3} \text{mol} \][/tex]

The volume of the buffer solution is 500.0 mL, which is equivalent to 0.500 L. The moles of HNO2 in the buffer are calculated by: [tex]\[ \text{moles of HNO2} = \text{concentration of HNO2} \times \text{volume of buffer} = 0.100 \text{M} \times 0.500 \text{L} = 5.00 \times 10^{-2} \text{mol} \][/tex]

 Since the moles of NaOH added (6.25 × 10^-3 mol) are less than the moles of HNO2 in the buffer (5.00 × 10^-2 mol), the buffer can neutralize the added NaOH. Therefore, the answer is No.

Part B: Similarly, for 350 mg of KOH, we calculate the moles of KOH:

[tex]\[ \text{moles of KOH} = \frac{350 \times 10^{-3} \text{g}}{56.1 \text{g/mol}} = 6.24 \times 10^{-3} \text{mol} \][/tex]

 Since the moles of KOH are approximately equal to the moles of NaOH added in Part A and are less than the moles of HNO2 in the buffer, the buffer can neutralize the added KOH. Therefore, the answer is No.

 Part C: For 1.25 g of HBr, we calculate the moles of HBr:

[tex]\[ \text{moles of HBr} = \frac{1.25 \times 10^{-3} \text{g}}{80.9 \text{g/mol}} = 1.55 \times 10^{-2} \text{mol} \][/tex]

 Since the moles of HBr added (1.55 × 10^-2 mol) are greater than the moles of HNO2 in the buffer (5.00 × 10^-2 mol), the buffer cannot neutralize the added HBr. Therefore, the answer is Yes.

 Part D: For 1.35 g of HI, we calculate the moles of HI:

[tex]\[ \text{moles of HI} = \frac{1.35 \times 10^{-3} \text{g}}{127.9 \text{g/mol}} = 1.06 \times 10^{-2} \text{mol} \][/tex]

Since the moles of HI added (1.06 × 10^-2 mol) are greater than the moles of HNO2 in the buffer (5.00 × 10^-2 mol), the buffer cannot neutralize the added HI. Therefore, the answer is Yes.

Correction: In Part B, the moles of KOH calculated are slightly less than the moles of HNO2 in the buffer, so the buffer can neutralize the added KOH. The initial answer provided was incorrect; it should be No, not Yes. The corrected answer is No for Part B."

The volume of the buffer solution is 500.0 mL, which is equivalent to 0.500 L. The moles of HNO2 in the buffer are calculated by:

[tex]\[ \text{moles of HNO2} = \text{concentration of HNO2} \times \text{volume of buffer} = 0.100 \text{M} \times 0.500 \text{L} = 5.00 \times 10^{-2} \text{mol} \][/tex]

Since the moles of NaOH added (6.25 × 10^-3 mol) are less than the moles of HNO2 in the buffer (5.00 × 10^-2 mol), the buffer can neutralize the added NaOH. Therefore, the answer is No.

 Part B: Similarly, for 350 mg of KOH, we calculate the moles of KOH:

[tex]\[ \text{moles of KOH} = \frac{350 \times 10^{-3} \text{g}}{56.1 \text{g/mol}} = 6.24 \times 10^{-3} \text{mol} \][/tex]

 Since the moles of KOH are approximately equal to the moles of NaOH added in Part A and are less than the moles of HNO2 in the buffer, the buffer can neutralize the added KOH. Therefore, the answer is No.

Part C: For 1.25 g of HBr, we calculate the moles of HBr:

[tex]\[ \text{moles of HBr} = \frac{1.25 \times 10^{-3} \text{g}}{80.9 \text{g/mol}} = 1.55 \times 10^{-2} \text{mol} \][/tex]

 Since the moles of HBr added (1.55 × 10^-2 mol) are greater than the moles of HNO2 in the buffer (5.00 × 10^-2 mol), the buffer cannot neutralize the added HBr. Therefore, the answer is Yes.

Part D: For 1.35 g of HI, we calculate the moles of HI:

[tex]\[ \text{moles of HI} = \frac{1.35 \times 10^{-3} \text{g}}{127.9 \text{g/mol}} = 1.06 \times 10^{-2} \text{mol} \][/tex]

 Since the moles of HI added (1.06 × 10^-2 mol) are greater than the moles of HNO2 in the buffer (5.00 × 10^-2 mol), the buffer cannot neutralize the added HI. Therefore, the answer is Yes.

 Correction: In Part B, the moles of KOH calculated are slightly less than the moles of HNO2 in the buffer, so the buffer can neutralize the added KOH. The initial answer provided was incorrect; it should be No, not Yes. The corrected answer is No for Part B."

Answer:

4 Sb (+3) +  2 Br (2) O (3) ----> 4 SbBr + 3 O(-2)

In the equation, place the valences in parentheses and the stoichiometric balance numbers so that the reaction is balanced, place them in bold.

Explanation:

In this chemical equation it seems to me that the valences of BrO are wrong.

That is why I will complete the equation, demonstrating the chemical reaction, and also use the appropriate valences.

Assuming that Sb has a valence of +3 and that the Br that appears in the form of oxide with oxygen (valence -2) has a valence of +3 (because -3 does not exist as valence or oxidation state of bromine).

Answer:

Percentage dissociated = 0.41%

Explanation:

The chemical equation for the reaction is:

[tex]C_3H_6ClCO_2H_{(aq)} \to C_3H_6ClCO_2^-_{(aq)}+ H^+_{(aq)}[/tex]

The ICE table is then shown as:

                               [tex]C_3H_6ClCO_2H_{(aq)} \ \ \ \ \to \ \ \ \ C_3H_6ClCO_2^-_{(aq)} \ \ + \ \ \ \ H^+_{(aq)}[/tex]

Initial   (M)                     1.8                                       0                               0

Change  (M)                   - x                                     + x                           + x

Equilibrium   (M)            (1.8 -x)                                  x                              x

[tex]K_a = \frac{[C_3H_6ClCO^-_2][H^+]}{[C_3H_6ClCO_2H]}[/tex]

where ;

[tex]K_a = 3.02*10^{-5}[/tex]

[tex]3.02*10^{-5} = \frac{(x)(x)}{(1.8-x)}[/tex]

Since the value for [tex]K_a[/tex] is infinitesimally small; then 1.8 - x ≅ 1.8

Then;

[tex]3.02*10^{-5} *(1.8) = {(x)(x)}[/tex]

[tex]5.436*10^{-5}= {(x^2)[/tex]

[tex]x = \sqrt{5.436*10^{-5}}[/tex]

[tex]x = 0.0073729 \\ \\ x = 7.3729*10^{-3} \ M[/tex]

Dissociated form of  4-chlorobutanoic acid = [tex]C_3H_6ClCO_2^- = x= 7.3729*10^{-3} \ M[/tex]

Percentage dissociated = [tex]\frac{C_3H_6ClCO^-_2}{C_3H_6ClCO_2H} *100[/tex]

Percentage dissociated = [tex]\frac{7.3729*10^{-3}}{1.8 }*100[/tex]

Percentage dissociated = 0.4096

Percentage dissociated = 0.41%     (to two significant digits)

Answer:

0.00091%

Explanation:

The fraction of 4-chlorobutanoic acid that would dissociatr in aqueos solution is a function of Ionization percentage. It is obtained by

Ka = [dissociated acid] / [original acid] x 100%

The equation of the reaction is

C₂HClCOOH  + H20 (aq)  = C₂HClCOO⁻  + H₃O⁺     pka =4.69

But pKa = - log Ka

   4.69  = - Log Ka

10⁻⁴⁶⁹   =    Ka

Taking the antilog of the equation we get the ionization constant

                    Ka   = 0.0000204 M

                           =2.04 x 10⁻⁵M

At the beginning of the reaction we have the following concentrations

1.8M C₂HClCOOH  :   0M (zero molar) C₂HClCOO⁻  :  0M (zero molar) H₃O⁺

At equilibrium, we have,

(1.8M -x)                                         xM C₂HClCOO⁻   and    xM H₃0⁺

Therefore,

Ka  =  [C2HClCOO-] [H30+] / [C2HClCOOH],

Inputing the value of Ka

                  Ka .[C2HClCOOH] = [CHCLCOO-] [H3O+]

                0.0000204 (1.8 - X) =  (x).(x)

                                            x² =  (0.00003672 - 0.0000204 X)

                                                 =  (3.672 x10-5 - 2.04 x10⁻⁵X)

                    x² +2.04x10⁻⁵ x     =  3.672 x 10⁻⁵

x² + 2.04 x 10⁻⁵x - 3.672 x 10⁻⁵ = 0

                                           x      = 0.00001632

                                                   = 1.632 x 10⁻⁵

Inputing back into equation 1

1.8 - x                  = [H3O+]

1.8 - 0.00001632 = 1.7999837

It therefore implies that only 0.00001632M of 4-chloroutanoic acid dissciated at equilibrium, we can now calculate the percentage dissociation by

Percentage dissociation = 0.00001632 / 1.8M x 100%

                                         = (1.632 x 10⁻⁵/1.8 ) x 100%

                                         = 0.00090667%

                                          = 0.00091%

Answer:

0.1 mol O2

Explanation:

1 mol of any gas at STP = 22.4 L

2.24L/22.4L = 0.1

1mol * 0.1 = 0.1 mol

you have 0.1 mol of O2

The number of moles of [tex]O_2[/tex]  is required.

The number of moles of [tex]O_2[/tex] is 0.2 moles.

V = Volume = 2.24 L

[tex]\rho[/tex] = Density of [tex]O_2[/tex] at STP = 1.429 g/L

M = Molar mass of [tex]O_2[/tex] = 16 g/mol

Density is given by

[tex]\rho=\dfrac{m}{V}\\\Rightarrow m=\rho V\\\Rightarrow m=1.429\times 2.24\\\Rightarrow m=3.20096\ \text{g}[/tex]

Molar mass is given by

[tex]M=\dfrac{m}{n}\\\Rightarrow n=\dfrac{m}{M}\\\Rightarrow n=\dfrac{3.20096}{16}\\\Rightarrow n=0.2\ \text{moles}[/tex]

The number of moles of [tex]O_2[/tex] is 0.2 moles.

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Answer: PLease make me Brainliest

"They can bring anything from tropical warm and humid days to arctic cold depending on the type of air mass. Fronts form the boundaries of air masses with differing properties. The most severe weather usually occurs when dry-cold continental polar air clashes with warm-humid maritime tropical air."

Final answer:

When a continental polar air mass meets a continental tropical air mass, the interaction can create a weather front that often leads to precipitation and can spawn extratropical cyclones, resulting in variable weather conditions including rain, snow, and thunderstorms.

Explanation:

If a continental polar air mass (cP) clashed with a continental tropical air mass (cT), a significant weather event would likely occur. The cP air mass is characterized as cold and dry, and is a stable air mass that can bring cooler temperatures to affected regions. On the other hand, a cT air mass is hot and dry. When these two air masses meet, the differing temperatures and densities of the air can cause the formation of a weather front, often leading to precipitation and potentially stormy conditions.

The collision often happens along what's called the polar front, where a cyclonic shear is created due to the opposing streams of air. This interaction can lead to the formation of extratropical cyclones, which are large low-pressure systems that can cause a wide array of weather conditions, including rain, snow, thunderstorms, and in some cases, severe weather outbreaks.

Given that cP air is denser than cT air, the warmer, less dense cT air would rise above the cP air mass. This rising motion can lead to cooling and condensation of water vapor, resulting in cloud formation and precipitation. This is a key mechanism through which weather disturbances are generated in regions where such air masses clash, particularly in the midlatitudes, which are notorious for their variable weather patterns.

Answer:

11.88

Explanation:

The capacity of container to be able to hold any amount of moles is dependent on size of container.

Mole is defined as the unit of amount of substance . It is the quantity measure of amount of substance of how many elementary particles are present in a given substance.

It is defined as exactly 6.022×10²³ elementary entities. The elementary entity can be a molecule, atom ion depending on the type of substance. Amount of elementary entities in a mole is called as Avogadro's number.

It is widely used in chemistry as a suitable way for expressing amounts of reactants and products.For the practical purposes, mass of one mole of compound in grams is approximately equal to mass of one molecule of compound measured in Daltons. Molar mass has units of gram per mole . In case of molecules, where molar mass  in grams present in one mole  of atoms is its atomic mass.

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Answer:

Explanation:

Hi there,

To get started, recall the Heat-specific heat capacity equation of a substance:

Q=mCΔT

Our final temperature is 47.8 °C, since this is the way it is worded in the response (from temperature X to temperature Y)

Quite simply, we can go ahead and plug in mass, specific heat capacity, and change in temperature as all the units match up!

[tex]Q= (525 g)(0.21 \ cal\ / g*C^{o} )(47.8C^{o}-13.0 C^{o})=3836.7 \ cal[/tex]

Study well and persevere.

thanks,

Final answer:

3823.8 calories of heat are required to raise the temperature of 525g of Aluminum from 13.0°C to 47.8°C, calculated using the formula Q = mcΔT with a specific heat of 0.21 cal/g°C.

Explanation:

To find how many calories of heat are required to raise the temperature of 525g of Aluminum from 13.0°C to 47.8°C, we use the formula for heat transfer Q = mcΔT, where 'm' is the mass of the substance, 'c' is the specific heat of the substance, and ΔT is the change in temperature.

First, we calculate the change in temperature (ΔT): ΔT = final temperature - initial temperature = 47.8°C - 13.0°C = 34.8°C. Using the provided values: m (mass) = 525g c (specific heat of Aluminum) = 0.21 cal/g°C ΔT (change in temperature) = 34.8°C

We plug these values into the equation to find the heat (Q): Q = m·c·ΔT Q = (525g)(0.21 cal/g°C)(34.8°C) Q = 3823.8 calories.

Therefore, 3823.8 calories of heat are required to raise the temperature of 525g of Aluminum from 13.0°C to 47.8°C.

Answer:

pH after the addition of 10 ml NaOH = 4.81

pH after the addition of 20.1 ml NaOH = 8.76

pH after the addition of 25 ml NaOH = 8.78

Explanation:

(1)

Moles of butanoic acid initially present = 0.1 x 20 = 2 m moles  = 2 x 10⁻³ moles,

Moles of NaOH added = 10 x 0.1 = 1 x 10⁻³ moles

                          CH₃CH₂CH₂COOH + NaOH ⇄ CH₃CH₂CH₂COONa + H₂O

Initial conc.            2 x 10⁻³                 1 x 10⁻³           0            

Equilibrium             1 x 10⁻³                   0                  1 x 10⁻³

Final volume = 20 + 10 = 30 ml = 0.03 lit

So final concentration of Acid = [tex]\frac{0.001}{0.03} = 0.03mol/lit[/tex]

Final concentration of conjugate base [CH₃CH₂CH₂COONa][tex]=\frac{0.001}{0.03} = 0.03 mol/lit[/tex]

Since a buffer solution is formed which contains the weak butanoic acid and conjugate base of that acid .

Using Henderson Hasselbalch equation to find the pH

[tex]pH=pK_{a}+log\frac{[conjugate base]}{[acid]} \\\\=-log(1.54X10^{-5} )+log\frac{0.03}{0.03} \\\\=4.81[/tex]

During titration of butanoic acid with NaOH, we can calculate the pH at various points using the Henderson-Hasselbalch equation for buffer scenarios. After 10.00mL of NaOH, the pH will be 4.74. After 20.10 mL, the pH will be 8.27, and after 25.00 mL, the pH will be 12.30.

This involves calculating the pH at various stages during a titration procedure. Here the titration involves a weak acid, butanoic acid, with a strong base, NaOH. We can simplify the reaction as follows: CH₃CH₂CH₂COOH + OH- --> CH₃CH₂CH₂COO- + H₂O.

(a) After 10.00 mL of NaOH is added, the system isn't at equivalence. Here, the reaction hasn't fully completed and a buffer solution is present. Using the Henderson-Hasselbalch equation, we can find the pH: pH = pKa + log([base]/[acid]). After calculating, we can find pH = 4.74.

(b) After 20.10 mL NaOH is added, the system reaches past equivalence. The pH can be determined by finding pOH using the remaining OH- concentration and then subtracting from 14. After the calculation, the pH = 8.27.

(c) For 25.00 mL of NaOH, the system is beyond equivalence and extra OH- ions increase pH. The pH calculation is like previous step and the result will be pH = 12.30.

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Answer:

London dispersion forces

Explanation:

The London dispersion force is the weakest kind of intermolecular force. The London dispersion force is a temporary attractive force that occurs when the electrons in two adjacent atoms occupy positions that make the atoms form temporary dipoles. This force is sometimes called an induced dipole-induced dipole attraction.

These London dispersion forces are mostly seen in the halogens (e.g., Cl2 and I2), the noble gases (e.g., Xe and Ar), and in many non-polar molecules, such as carbon dioxide and propane. London dispersion forces are part of the van der Waals forces, and are very weak intermolecular attractions.

Answer:

1.38×10^25 molecules

Explanation:

Applying n= (no. of molcules)/NA

23 = N/6.02×10^23

= 1.38×10^25 molecules

The [H+] in the solution is 6.94752 * 10^−7 M.

To calculate the [H+] in the given solution, we need to first calculate the concentration of HX and then use the equilibrium constant (Ka) to find the concentration of [H+]. The equation for the dissociation of HX is: HX ⇌ H+ + X-. Since the dissociated amount is small compared to 0.803 M, we can assume that the concentration of HX is approximately 0.803 M. Now, using the equation for Ka and the concentration of HX, we can find the concentration of [H+]:

Ka = [H+][X-] / [HX]

[H+] = Ka * [HX] = (8.64 * 10^−7) * 0.803 = 6.94752 * 10^−7 M

Therefore, the [H+] in the solution is 6.94752 * 10^−7 M.

Answer:

B) 655, 485, 433, 409 nm

Explanation:

655, 485, 433, 409 nm

The greater the energy change involved in a transition, the shorter the wavelength.

Transitions to n = 2 from n = 3, 4, 5 and 6 give lines of wavelength 655, 485, 433 and 409 nm. Other answer choices represent the Lyman. Brackett, and Pfund series.

The Electronic configuration, of an element is 1s² 2s² 2p¹. The number of valance electrons present in the element is 3.

Electronic configuration refers to the arrangement of electrons within the orbitals of an atom or ion. It describes how the electrons are distributed among the various energy levels, sublevels, and orbitals within an atom.

The electronic configuration is often represented using a shorthand notation known as the Aufbau principle or the electron configuration notation. In this notation, the principal energy levels (denoted by the quantum number n) are represented by numbers (1, 2, 3, etc.), and the sublevels (s, p, d, f) are represented by letters. The superscript numbers indicate the number of electrons occupying each sublevel.

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Final answer:

The element with the electron configuration 1s2 2s2 2p1 has a total of 3 valence electrons, from the occupied 2s and 2p subshells in the second energy level.

Explanation:

The Electronic configuration, of an element is 1s² 2s² 2p¹. The number of valance electrons present in the element is 3.

Electronic configuration refers to the arrangement of electrons within the orbitals of an atom or ion. It describes how the electrons are distributed among the various energy levels, sublevels, and orbitals within an atom.

The electronic configuration is often represented using a shorthand notation known as the Aufbau principle or the electron configuration notation. In this notation, the principal energy levels (denoted by the quantum number n) are represented by numbers (1, 2, 3, etc.), and the sublevels (s, p, d, f) are represented by letters. The superscript numbers indicate the number of electrons occupying each sublevel.

Answer:

absorbed

Explanation:

Answer:

90/38 Sr

Explanation:

Gamma decay involves rearrangements in an atomic nucleus usually following a nuclear change. When a nuclear change occurs, the daughter nuclei is in excited state, a high energy state. The nucleons quickly rearrange themselves and rapidly lower the energy of the daughter nucleus. The corresponding amount of excited state energy is emitted as the daughter nuclei return to ground state as short range, high energy electromagnetic gamma rays.

Hence, the number of nucleons in the nucleus remain the same before and after a gamma decay.

The reaction will consume 3 moles of iron(III) oxide and will produce 3 moles of aluminum oxide and 12 moles of iron, following the mole ratio from the balanced chemical equation.

Starting from the balanced chemical equation, we can see that the ratio of "Fe2O3" (iron(III) oxide) to "Al" (aluminum) is 1:2. This ratio allows us to calculate the amount of moles that will react or produce. We can apply these ratios in the following way:

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Answer:

vhgbvbhdf

Explanation:

(a) It is given that the gas is ideal. Formula for change in entropy of the gas according to the first law of thermodynamics is as follows.

       [tex]\Delta U = dQ - dW[/tex]

For isothermal process, [tex]\Delta U = 0[/tex] at constant temperature.

So,   [tex]\Delta U = dQ - dW[/tex]

      [tex]0 = dQ - dW[/tex]

or,       dQ = dW = 730 J

Now, according to the second law of thermodynamics the entropy change is as follows.

           [tex]\Delta S = \frac{dQ}{dT}[/tex]

                        = [tex]\frac{730 J}{293.15 K}[/tex]

                        = 2.490 J/K

Therefore, the entropy change of the gas is 2.490 J/K.

(b)  In the given process, at constant temperature the gas will be compressed slowly because then kinetic energy of the gas molecules will also be constant. The volume decreases so that the movement of molecules increases as a result, entropy of molecules will also increase.

This means that [tex]\Delta S = 2.490 J/K > 0[/tex]

An ideal gas is a hypothetical gas in which there are no intermolecular attractions between the molecules of a gas. The collision between the molecules is perfectly elastic.

The answers are as follows:

(a) The ideal gas is given, in which the change of entropy of the gas can be calculated by the first law of thermodynamics.

[tex]\Delta[/tex] U = dQ - dW

In an isothermal process, the \Delta U = 0 at constant temperature, such that:

[tex]\Delta[/tex] U = dq - dW

0 = dQ - dW

dQ = dW = 730 J (given)

Now, based on the second law of thermodynamics, the entropy change will be:

[tex]\Delta S = \dfrac {\text {dQ}}{\text {dW}}\\\\\Delta S = \dfrac {730 \text J}{293.15 \text K}[/tex]

[tex]\Delta[/tex]S = 2.490 J/K

Thus, the entropy change for the gas is 2.490 J/K.

(b) At constant temperature, the gas will be compressed slowly because the kinetic energy of the molecules will be constant. The decrease in the volume will be due to the increase in the movement of the molecules, which will cause an increase in the entropy. This means [tex]\Delta[/tex] S will be greater than  2.490 J/K.

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Answer: 80mL

Explanation:

Its on ck12

5*16=80

To prepare a 1.00 M HCl solution from a 16.0 M HCl stock solution, Maria needs to add 75 mL of water to the 5.00 mL of the stock solution.

To prepare a 1.00 M HCl solution from a 16.0 M HCl stock solution, Maria needs to dilute the stock. She should add water to the 5.00 mL of the stock solution. Let's calculate the volume of water she needs to add.

First, we can use the equation (stock concentration) × (stock volume) = (final concentration) × (final volume) to find the volume of the diluted solution. Plugging in the values, we have (16.0 M) × (5.00 mL) = (1.00 M) × (final volume). Solving for the final volume, we get:

final volume = (16.0 M × 5.00 mL) / 1.00 M

final volume = 80 mL

To find the volume of water Maria needs to add, we subtract the volume of the stock solution from the final volume:

volume of water = final volume - volume of stock solution = 80 mL - 5.00 mL = 75 mL

Therefore, Maria needs to add 75 mL of water to the 5.00 mL of 16.0 M HCl solution to prepare a 1.00 M HCl solution.

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This is an incomplete question, here is a complete question.

The reaction below has a Kp value of 3.3 × 10⁻⁵. What is the value of Kc for this reaction at 700 K?

[tex]2SO_3(g)\rightleftharpoons 2SO_2(g)+O_2(g)[/tex]

Answer : The value of [tex]K_p[/tex] is, [tex]5.79E^{-7}[/tex]

Explanation :

The relation between [tex]K_c[/tex] and [tex]K_p[/tex] is:

[tex]K_p=K_c\times (RT)^{\Delta n}[/tex]

where,

[tex]K_p[/tex] = equilibrium constant at constant pressure  = [tex]3.3\times 10^{-5}[/tex]

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

T = temperature = 700 K

[tex]\Delta n[/tex] = change in number of gaseous moles = Product moles - Reactant moles = (2+1) - (2) = 3 - 1 = 1 mol

[tex]K_c[/tex] = equilibrium constant

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

[tex]K_p=K_c\times (RT)^{\Delta n}[/tex]

[tex]3.3\times 10^{-5}=K_c(0.0821\times 700)^{1}[/tex]

[tex]K_c=5.7\times 10^{-7}=5.7E^{-7}[/tex]

Therefore, the value of [tex]K_p[/tex] is, [tex]5.7E^{-7}[/tex]

The value of the equilibrium constant of the reaction (Kc) is [tex]5.74 \times 10^{-7}[/tex]

The given parameters;

The change in the number of gaseous moles is calculated as follows;

[tex]\Delta n = product \ - reactant\\\\\Delta n = (2+1) - 2 = 1 \ mole[/tex]

The value of the equilibrium constant (Kc) is calculated as follows;

[tex]K_p = K_c \times (RT)^{\Delta n}\\\\3.3 \times 10^{-5} = K_c \times (0.0821 \times 700)^1\\\\3.3 \times 10^{-5} = 57.47K_c\\\\K_c = \frac{3.3 \times 10^{-5} }{57.47} \\\\K_c = 5.74 \times 10^{-7} \[/tex]

Thus, the value of the equilibrium constant of the reaction (Kc) is [tex]5.74 \times 10^{-7}[/tex]

"Your question is not complete, it seems to be missing the following information;"

The reaction below has a Kp value of 3.3 × 10⁻⁵. What is the value of Kc for this reaction at 700 K?

[tex]2SO_3(g) \ ---> \ 2SO_2(g) \ + \ O_2[/tex]

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