There are three sets of sketches below, showing the same pure molecular compound (water, molecular formula H_2 O) at three different temperatures. The sketches are drawn as if a sample of water were under a microscope so powerful that individual atoms could be seen. Only one sketch in each set is correct. Use the slider to choose the correct sketch in each set. You may need the following information: melting point of H_2 O: 0.0 degree C boiling point of H_2 O: 100.0 degree C

Answer:

27.8 minutes

Explanation:

The reaction follows a first order

Rate = k[A] = change in concentration/time

k = 9×10^-3s^-1

Let the original concentration of A be y

Concentration of A at time t = 6.25% × y = 0.0625y

Change in concentration = y - 0.0625y = 0.9375y

0.009 × 0.0625y = 0.9375y/t

t = 0.9375y/0.0005625y = 1666.7sec = 1666.7/60 = 27.8 minutes

It takes approximately 12.8 minutes for the concentration of the reactant [A] to drop to 6.25% of the original concentration.

To determine the time it takes for the concentration of a reactant [A] in a first-order reaction to drop to 6.25% of its original concentration, we can use the first-order kinetics equation:

ln([A]₀/[A]) = kt

where ln is the natural logarithm, [A]₀ is the initial concentration of A, [A] is the concentration of A at time t, k is the rate constant, and t is the time in seconds. To find t when [A] is 6.25% of [A]₀, we can set [A]/[A]₀= 0.0625 (since 6.25% is equivalent to 0.0625 in decimal form).

Using the given rate constant 9.00×10⁻³ s⁻¹, the equation becomes:

ln(1/0.0625) = (9.00×10⁻³) × t

Solving this gives t = 768 seconds. Since there are 60 seconds in a minute, this is equivalent to 12.8 minutes.

Therefore, it takes approximately 12.8 minutes for [A] to drop to 6.25% of its original concentration in a first-order reaction at 45 °C with a rate constant of 9.00×10⁻³ s⁻¹.

Answer: The partial pressure of gas A is 6.34 atm and that of gas B is 17.1 atm

Explanation:

To calculate the pressure of the gas, we use the equation given by ideal gas, which follows:

[tex]PV=nRT[/tex]       ......(1)

where,

P = pressure of the gas

V = Volume of the gas

T = Temperature of the gas

R = Gas constant = [tex]0.0821\text{ L. atm }mol^{-1}K^{-1}[/tex]

n = number of moles of gas

We are given:

[tex]V=7.07L\\T=30.4^oC=[30.4+273]K=303.4K\\n=1.80mol[/tex]

Putting values in equation 1, we get:

[tex]p_A\times 7.07L=1.80mol\times 0.0821\text{ L atm }mol^{-1}K^{-1}\times 303.4K\\\\p_{A}=\frac{1.80\times 0.0821\times 303.4}{7.07}=6.34atm[/tex]

We are given:

[tex]V=7.07L\\T=30.4^oC=[30.4+273]K=303.4K\\n=4.86mol[/tex]

Putting values in equation 1, we get:

[tex]p_B\times 7.07L=4.86mol\times 0.0821\text{ L atm }mol^{-1}K^{-1}\times 303.4K\\\\p_{B}=\frac{4.86\times 0.0821\times 303.4}{7.07}=17.1atm[/tex]

Hence, the partial pressure of gas A is 6.34 atm and that of gas B is 17.1 atm

Answer:

The partial pressure of gas A is 6.34 atm

The partial pressure of gas B is 17.12 atm

Explanation:

Step 1 :Data given

Volume of cylinder = 7.07 L

Number of moles gas A = 1.80 moles

Number of moles gas B = 4.86 moles

Temperature =30.4 ° C = 303.55 K

Step 2: Calculate pressure of gas A

p*V = n*R*T

p =(n*R*T)/V

⇒ with p = the partial pressure of gas A

⇒ with V = The volume of the cylinder = 7.07 L

⇒ with n = the number of moles gas A = 1.80 moles

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

⇒ with T = the temperature = 303.55 K

p = (1.80 *0.08206 *303.55)/7.07

p = 6.34 atm

Step 3: Calculate pressure of gas B

p*V = n*R*T

p =(n*R*T)/V

⇒ with p = the partial pressure of gasB

⇒ with V = The volume of the cylinder = 7.07 L

⇒ with n = the number of moles gas B = 4.86 moles

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

⇒ with T = the temperature = 303.55 K

p = (4.86 *0.08206 *303.55)/7.07

p = 17.12 atm

The partial pressure of gas A is 6.34 atm

The partial pressure of gas B is 17.12 atm

Answer:

We need 7.01 grams of KOH (option 5)

Explanation:

Step 1: Data given

Volume aqueous KOH solution = 250 mL = 0.250 L

Molarity = 0.500 M

Molar mass of KOH = 56.10 g/mol

Step 2: Calculate moles KOH

Moles KOH = molarity * volume

Moles KOH = 0.500 M * 0.250 L

Moles KOH = 0.125 moles

Step 3: Calculate mass of KOH

Mass KOH = moles KOH * molar mass KOH

Mass KOH = 0.125 moles * 56.10 g/mol

Mass KOH = 7.01 grams

We need 7.01 grams of KOH

Answer:

2.165 x 10^3 nm.

Explanation:

Using Rybergs equation,

1/lambda = R * (1/n1^2 - 1/n2^2)

Where,

R = rybergs constant = 109737.32 cm^-1

n1 = 7

n2 = 4

= 109737.32 * (1/7^2 - 1/4^2)

= 4619.05

Lambda = 2.165 x 10^-4 cm

Since 100 cm = 1m, 1 nm = 10^-9 m

= 2.165 x 10^-4 cm * 1 m/100 cm * 1 nm/10^-9 m

= 2.165 x 10^3 nm.

Answer : All possible values of 'ml' for the following orbitals are:

(a) At l = 3,  [tex]m_l=+3,+2,+1,0,-1,-2,-3[/tex]

(b) l = 0, 1

At l = 0,  [tex]m_l=0[/tex]

At l = 1,  [tex]m_l=+1,0,-1[/tex]

(c) At l = 1,  [tex]m_l=+1,0,-1[/tex]

Explanation:

There are 4 quantum numbers :

Principle Quantum Numbers : It describes the size of the orbital. It is represented by n. n = 1,2,3,4....

Azimuthal Quantum Number : It describes the shape of the orbital. It is represented as 'l'. The value of l ranges from 0 to (n-1). For l = 0,1,2,3... the orbitals are s, p, d, f...

Magnetic Quantum Number : It describes the orientation of the orbitals. It is represented as m_l. The value of this quantum number ranges from [tex](-l\text{ to }+l)[/tex]. When l = 2, the value of [tex]m_l[/tex] will be -2, -1, 0, +1, +2.

Spin Quantum number : It describes the direction of electron spin. This is represented as [tex]m_s[/tex]The value of this is [tex]+\frac{1}{2}[/tex] for upward spin and [tex]-\frac{1}{2}[/tex] for downward spin.

(a) l = 3 then the value of 'ml' is,

At l = 3,  [tex]m_l=+3,+2,+1,0,-1,-2,-3[/tex]

(b) n = 2 then the value of 'ml' is,

l = 0, 1

At l = 0,  [tex]m_l=0[/tex]

At l = 1,  [tex]m_l=+1,0,-1[/tex]

(c) n = 6 and l = 1 then the value of 'ml' is,

n = 6

l = 0, 1, 2, 3, 4, 5

At l = 1,  [tex]m_l=+1,0,-1[/tex]

The  orbital refers to a region in space where there is a high probability of finding an electron.

The term orbital refers to a region in space where there is a high probability of finding an electron. Within each energy level, there are orbitals.

Let us consider each of the levels shown;

(a) l = 3

The ml values for this orbital are; -3, -2, -1, 0, 1, 2, 3

(b) n = 2

The ml values for this orbital are; -1, 0, 1

(c) n = 6, l = 1

The ml values for this orbital are; -1, 0, 1

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Answer : The balanced chemical reaction will be:

[tex]H_2O(l)\rightleftharpoons H^+(aq)+OH^-(aq)[/tex]

Or,

[tex]H_2O(l)+H_2O(l)\rightleftharpoons H_3O^+(aq)+OH^-(aq)[/tex]

Explanation :

Autoionization of water  : The autoionization of water means that the reaction water with water means self ionization.

In the autoionization of water, one water molecule loses an hydrogen ion and another one gains it.

The balanced chemical reaction will be:

[tex]H_2O(l)\rightleftharpoons H^+(aq)+OH^-(aq)[/tex]

Or,

[tex]H_2O(l)+H_2O(l)\rightleftharpoons H_3O^+(aq)+OH^-(aq)[/tex]

The autoionization of water is a chemical reaction where two water molecules react to form hydronium and hydroxide ions. This is represented as H₂O(l) + H₂O(l) = H3O+(aq) + OH¯(aq). The ion-product constant for water (Kw) is a measure of this process.

The autoionization of water is a process in which water molecules, acting as both an acid and a base, react with each other to form ions. This can be represented by the chemical reaction: H₂O(l) + H₂O(l) = H3O+(aq) + OH¯(aq). In this equation, the (l) denotes the liquid state of water and (aq) designates aqueous, or water-dissolved, ions.

The ion-product constant for water, Kw, is equal to the product of the concentrations of the hydronium and hydroxide ions: Kw = [H3O+][OH¯]. At 25 °C, the value of Kw is approximately 1.0 × 10-14, indicating very slight ionization. This value increases with temperature, indicating an endothermic reaction.

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Answer:A, C, D, E, F

Explanation:

A. True: Oxidizing agents are electron acceptors. They accept electrons and the get reduced. This means their oxidation number reduces

B. False: Reducing agents do not accept H+ ions. Reducing agents remove oxygen from another substance or give hydrogen to it.

C. True: oxidizing agents oxidizes other molecules but they accept electrons and get reduced themselves. If a molecule accepts electrons it has been reduced.

D True: Redox reactions MAY and may not involve the transfer of hydrogen ions depending on the reactants (H+). But redox in terms of acid and base means the donating and receiving of protons(H+)

E. True: A molecule that has gained H atoms is said to be reduced. Oxidizing agents are always the proton acceptor.

F. True: Oxidizing agents May and may not accept H+. In terms of acid and base oxidizing agents accept protons(H+)

Regarding redox reactions, statements C, D, and E are generally true. Statement B and F can be true in specific scenarios but not universally, while statement A is incorrect because oxidizing agents actually donate electrons.

In redox reactions, oxidation and reduction occur simultaneously. We look at the given statements:

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A first-order decomposition reaction has a rate constant of 0.00440 yr−1. Hence the correct answer is 473.418years. Rounded to the correct number of significant figures, the time it takes for the reactant to reach 12.5% of its original value is approximately 470 years.

A first-order reaction follows the exponential decay equation:

[tex]\[ [A][/tex] = [tex][A]_0 \times e^{-kt}[/tex]

Where:

[tex]\([A]\)[/tex] is the concentration of reactant at time [tex]\(t\)[/tex]

[tex]\([A]_0\)[/tex] is the initial concentration of the reactant.

[tex]\(k\)[/tex] is the rate constant.

[tex]\(t\)[/tex] is time.

It is given that the rate constant[tex]\(k\)[/tex] is 0.00440 y[tex]r^(^-^1^)[/tex] and we want to find out how long it takes for the reactant concentration to reach 12.5% of its original value, which means [tex]\([A][/tex] = [tex]0.125 \times [A]_0\).[/tex]

The equation to solve for time[tex]\(t\)[/tex]:

[tex]\[ t[/tex] = [tex]-\frac{1}{k} \ln\left(\frac{[A]}{[A]_0}\right)[/tex]

Substitute the given values:

[tex]\[ t = -\frac{1}{0.00440 \, \text{yr}^{-1}} \ln\left(\frac{0.125 \times [A]_0}{[A]_0}\right)[/tex]

Simplifying:

[tex]\[ t = -\frac{1}{0.00440 \, \text{yr}^{-1}} \ln(0.125) \][/tex]

Now calculate the value:

[tex]\[ t = -\frac{1}{0.00440 \, \text{yr}^{-1}} \times (-2.07944) \][/tex]

[tex]\[ t \approx 473.418 \, \text{years} \][/tex]

Rounded to the correct number of significant figures, the time it takes for the reactant to reach 12.5% of its original value is approximately 470 years.

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A first-order decomposition reaction is when the rate of the reaction is proportional to the concentration of the reactant. To find the time it takes for the reactant to reach 12.5% of its original value, we can use the equation ln([reactant] / [initial]) = -kt, where t represents time and k is the rate constant. Using the given rate constant, we find that it would take approximately 32.48 years for the reactant to reach 12.5% of its original value.

A first-order decomposition reaction is one in which the rate of the reaction is proportional to the concentration of the reactant. The rate equation for a first-order reaction is given by: rate = k[reactant], where k is the rate constant and [reactant] is the concentration of the reactant.

In this case, the rate constant is given as 0.00440 yr-1. To find the time it takes for the reactant to reach 12.5% of its original value, we can use the equation ln([reactant] / [initial]) = -kt, where [initial] is the initial concentration. Rearranging the equation, we have t = ln([reactant] / [initial]) / -k. Plugging in the percentage values, we get t = ln(0.125) / -0.00440 yr-1.

Calculating this value, we find that t ≈ 32.48 years.

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The amount of ice added was approximately 38.5 grams.

To calculate this, we can use the principle of conservation of energy. The heat lost by the water as it cools down to the final temperature (15 °C) is equal to the heat gained by the ice as it melts and then warms up to the final temperature.

First, we calculate the heat lost by the water:

[tex]\[ Q_{\text{water}} = m_{\text{water}} \times c_{\text{water}} \times \Delta T \][/tex]

Where:

[tex]\( m_{\text{water}} = 64.3 \, \text{g} \)[/tex] (mass of water)

[tex]\( c_{\text{water}} = 4.184 \, \text{J/g°C} \)[/tex] (specific heat of water)

[tex]\( \Delta T = 55°C - 15°C = 40°C \)[/tex] (change in temperature)

[tex]\[ Q_{\text{water}} = 64.3 \, \text{g} \times 4.184 \, \text{J/g°C} \times 40°C = 10707.712 \, \text{J} \][/tex]

Next, we calculate the heat gained by the ice:

[tex]\[ Q_{\text{ice}} = m_{\text{ice}} \times L_f + m_{\text{ice}} \times c_{\text{water}} \times \Delta T \][/tex]

Where:

[tex]\( L_f = 334 \, \text{J/g} \)[/tex] (heat of fusion of water)

[tex]\( m_{\text{ice}} \)[/tex] is the mass of ice we want to find

[tex]\( c_{\text{water}} = 4.184 \, \text{J/g°C} \)[/tex]  (specific heat of water)

[tex]\( \Delta T = 15°C \)[/tex] (change in temperature, from 0 °C to 15 °C)

Let's set up the equation using [tex]\( m_{\text{ice}} \):[/tex]

[tex]\[ 10707.712 \, \text{J} = m_{\text{ice}} \times 334 \, \text{J/g} + m_{\text{ice}} \times 4.184 \, \text{J/g°C} \times 15°C \][/tex]

Now, we solve for [tex]\( m_{\text{ice}} \):[/tex]

[tex]\[ 10707.712 \, \text{J} = m_{\text{ice}} \times (334 \, \text{J/g} + 62.76 \, \text{J/g}) \]\[ 10707.712 \, \text{J} = m_{\text{ice}} \times 396.76 \, \text{J/g} \]\[ m_{\text{ice}} = \frac{10707.712 \, \text{J}}{396.76 \, \text{J/g}} \approx 27.0 \, \text{g} \][/tex]

However, this is the amount of ice needed to cool the water to 0 °C. To find the total amount of ice needed to cool the water to 15 °C, we add the ice that will melt at 0 °C to the ice that will further cool down to 15 °C:

[tex]\[ m_{\text{total ice}} = m_{\text{ice}} \text{ at 0 °C} + m_{\text{ice}} \text{ cooling from 0 °C to 15 °C} \]\[ m_{\text{total ice}} = 27.0 \, \text{g} + (27.0 \, \text{g} \times 15/334) \approx 38.5 \, \text{g} \][/tex]

So, approximately 38.5 grams of ice were added to the water.

Complete Question:

A quantity of ice at 0 °C was added to 64.3 g of water in a glass at 55 °C. The final temperature of the system was 15 °C. How much ice was added?

A)The melting point of water is 0 °C.

B)The heat of fusion of water is 334 J g–1 .

C)The specific heat of liquid water is 4.184 J g–1 °C –1

Answer:

c. an element.

Explanation:

An element -

It refers to the substance , which has same type of atoms , with exactly same number of protons , is referred to as an element .

In term of chemical species , elements are the smallest one , and can not be bifurcated down to any further small substance by the means of any chemical reaction .

Hence , from the given information of the question ,

The correct term is an element  .

Answer:

C. an element.

Explanation:

Answer:

4.4 × 10⁴ g/mol

Explanation:

The osmotic pressure (π) is a colligative property that can be calculated using the following expression.

π = M × R × T

where,

M: molarity

R: ideal gas constant

T: absolute temperature (27°C + 273.15 = 300 K)

Let's use it to find the molarity of the protein.

M = π / R × T

M = 0.053 atm / (0.082 atm.L/mol.K) × 300 K

M = 2.2 × 10⁻³ M

The molarity of the protein is:

M = mass of the protein / molar mass of the protein × liters of solution

molar mass of the protein = mass of the protein / M  × liters of solution

molar mass of the protein = 22 g / 2.2 × 10⁻³ mol/L  × 0.226 L

molar mass of the protein = 4.4 × 10⁴ g/mol

The molar mass of the protein is 4.4 * 10⁴ g/mol

It  is a colligative property that can be calculated using the following expression.

π = M × R × T

where,

[tex]M = \frac{\pi}{R * T} \\\\M = \frac{0.053 atm}{(0.082 atm.L/mol.K) * 300 K}\\\\ M = 2.2 * 10^{-3} M[/tex]

M = mass of the protein / Molar mass of the protein * liters of solution

Molar mass of the protein = mass of the protein / M  * liters of solution

Molar mass of the protein = [tex]\frac{22 g}{2.2 * 10^{-3} mol/L * 0.226 L}[/tex]

Molar mass of the protein = [tex]4.4 * 10^4 g/mol[/tex]

Thus, the molar mass of the protein is 4.4 * 10⁴ g/mol.

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Answer:

a) Ba (Z = 56), [Xe].6s² - 2e ⇒ Ba²⁺, [Xe].6s⁰

b) O (Z = 8), [He].2s².2p⁴ + 2e ⇒ O²⁻, [He].2s².2p⁶

c) Pb (Z = 82), [Xe].4f¹⁴.5d¹⁰.6s².6p² - 2e ⇒ Pb²⁺, [Xe].4f¹⁴.5d¹⁰.6s².6p⁰

Explanation:

The condensed electron configurations of given elements are below

a) Ba (Z = 56), [Xe].6s²

b) O (Z = 8), [He].2s².2p⁴

c) Pb (Z = 82), [Xe].4f¹⁴.5d¹⁰.6s².6p²

Since atoms tend to donate/receive more electrons to achieve the saturated or half-saturated orbital. So in our case it happens as below

a) Ba (Z = 56), [Xe].6s² - 2e ⇒ Ba²⁺, [Xe].6s⁰

b) O (Z = 8), [He].2s².2p⁴ + 2e ⇒ O²⁻, [He].2s².2p⁶

c) Pb (Z = 82), [Xe].4f¹⁴.5d¹⁰.6s².6p² - 2e ⇒ Pb²⁺, [Xe].4f¹⁴.5d¹⁰.6s².6p⁰

Answer:

4.94°C, the temperature for freezing the solution

Explanation:

Freezing point depression to solve this.

Formula = T° freezing pure solvent - T° freezing solution = Kf . m

With the data given, let's determine m (molality)

Molality → mol/kg (moles of solute in 1kg of solvent)

We need to convert the 344 g to kg → 344 g . 1kg/1000 g = 0.344 kg

Let's determine the moles of solute (naphtalene)

5 g / 128 g/mol = 0.039 mol

Molality → 0.039 mol / 0.344 kg → 0.113

Let's go back to the formula:

5.5°C - T° freezing of solution = 4.90°C /m. 0.113 m

T° freezing of solution = - ( 4.90°C /m. 0.113 m - 5.5°C)

T° freezing of solution =  4.94 °C

The freezing point of the solution is 4.94 °C.

The term freezing point refers to the point in which a liquid is converted to a solid.

We know that;

ΔT = K m i

K = freezing constant

m = molality of the solution

i = Van't Hoff factor

ΔT = 4.90°C/m × (5.00 g /128 g/mol)/0.344 Kg × 1

ΔT =0.56°C

Freezing point of solution = 5.5°C - 0.56°C = 4.94 °C

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Answer:

a and b are correct

Explanation:

This because both are aqueous solutions,therefore, identity of solvent is same that is water.

And because both solutions are non electrolyte they would not ionize in solution, and for the same concentration, the freezing point of both solution would also be  same. Since depression in freezing point is a colligative property this means that it depends on number of solute particles not nature of particles .

Hence answer is that  their freezing points and  Identity of the solvent shall remain the same.

Answer:

1.82 cm

Explanation:

Utilize the equation [tex]\frac{P_{1}V_{1} }{T_{1} } = \frac{P_{2}V_{2} }{T_{2} }[/tex] to calculate the change in volume and size of an air bubble.

P1 = pressure at 50m = [tex]P_{A}[/tex] + ρ*g*h  (where [tex]P_{A}[/tex] = atmospheric pressure, ρ = density of water, g = acceleration due to gravity, h =  height/depth)

P1 = 1.01 x 10⁵ Pa + (ρ x g x h)

    = 1.01 x 10⁵ Pa + (1000 kg/m³ x  9.8 m/s² x 50 m )

    =  1.01 x 10⁵ Pa + 4.9 x 10⁵ Pa

    = 5.91 x 10⁵ Pa

V1 = [tex]\frac{4}{3}\pi r_{1} ^{3}[/tex]     [tex]r_{1}[/tex] = 10 cm =  1 x 10⁻² m

T1 = 10 °C = 10 + 273 = 283 K

P2 = [tex]P_{A}[/tex] = 1.01 x 10⁵ Pa because at the surface, pressure is equal to atmospheric pressure

V2 = [tex]\frac{4}{3}\pi r_{2} ^{3}[/tex]  [tex]r_{2}[/tex] = ??

T2 = 20 °C = 20 + 273 = 293 K

[tex]\frac{P_{1}V_{1} }{T_{1} } = \frac{P_{2}V_{2} }{T_{2} }[/tex]

V₂ = P₁V₁T₂

         P₂T₁

[tex]\frac{4}{3}\pi r_{2} ^{3}[/tex] = P₁ x [tex]\frac{4}{3}\pi r_{1} ^{3}[/tex]  x T₂

                P₂T₁

cancel out common terms

[tex]r_{2}[/tex]³ = 5.91 x 10⁵ Pa x (1 x 10⁻² m)³ x  293 k

              1.01 x 10⁵ Pa x 283 k

 [tex]r_{2}[/tex]³ = 757.9 x 10⁻⁹

   [tex]r_{2}[/tex] = 9.1 x 10⁻³ m

   [tex]r_{2}[/tex] = 0.91 cm  

Therefore, bubbles diameter = 2r = 1.82 cm

Answer:

1.82 cm

Explanation:

The pressure done by a column of a liquid is called the hydrostatic pressure (Ph) and it can be calculated by:

Ph = Patm + ρgh

Where Patm is the atmospheric pressure under the column (101325 Pa), ρ is the density of the liquid (1000 kg/m³ for water), g is the gravity acceleration (9.8 m/s²), and h is the depth (50 m), so:

Ph = 101325 + 1000*9.8*50

Ph = 591325 Pa

Because the bubble is in equilibrium with the surroundings, its pressure is the same as the surroundings. Supposing a perfect sferic bubble, its volume is:

V = (4/3)*π*r³

Where r is the radius, which is half of the diameter, so r = 0.5 cm.

V = (4/3)*π*(0.5)³

V = 0.52 cm³

According to the ideal gas law, the multiplication of the pressure (P) by the volume (V) divided by the temperature (T) of a gas is constant, so if 1 is the state where the bubble is 50 m depth, and 2 the state at the surface:

P1*V1/T1 = P2*V2/T2

P1 = Ph = 591325 Pa

V1 = 0.52 cm³

T1 = 10°C + 273 = 283 K

P2 = 101325 Pa (atmosferic pressure)

T2 = 20°C + 273 = 293 K

591325*0.52/283 = 101325*V2/293

101325V2 = 318,354.3357

V2 = 3.14 cm³

V2 = (4/3)*π*r³

(4/3)*π*r³ = 3.14

r³ = 0.75

r = ∛0.75

r = 0.91 cm

The diameter is then 2*r = 1.82 cm.

Answer:

Explanation:

Science is sustainable and reliable, as it is thoroughly tested and updated, impartial individuals will be observing scientific proof, and science only alters when new research justifies shift.The main concerns of many, if not most are scientific and engineering activities stability and transition. Efforts are made to ensure whatever theory is totally valid after a thorough investigation.

Answer: methanol and acetone are not suitable solvents for extracting organic compounds because they are miscible with water virtually in all proportions

Explanation:

Methane and acetone both are polar solvents, which means they are soluble in water. Hence, not suitable for the extraction from aqueous solutions.

Methanol is the simplest alcohol which is volatile, colorless, and inflammable.

Acetone is an organic compound whose chemical formula is CH3CH3CO.

Acetone and methanol both contain polar ends and are miscible in water.

Polar solvents are not suitable for the extraction of any compound.

Thus, Methanol and acetone are not suitable for extracting organic compounds from aqueous solutions.

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Answer:

The mass of the element is 28.085 amu

Explanation:

Step 1: Data given

Masses of isotopes:

27.977 amu ⇒

28.976 amu ⇒ 4.68%

29.973 amu ⇒  3.09%

Step 2: Calculate the abundance of the other isotope

100% - 4.68% - 3.09 % = 92.23 %

Step 3: calculate the mass othe element

0.9223 * 27.977 + 0.0468 * 28.976 + 0.0309*29.973 = total mass of the element

Total mass of the element = 28.085 amuj

The mass of the element is 28.085 amu

The weighted average atomic mass of the given element is calculated using the masses and relative abundances of its isotopes. The mass is found to be 28.085 amu. The correct answer is Option C.

The average atomic mass of an element is calculated by multiplying each isotope's mass by its relative abundance (as a decimal), and then adding up these products. We can use the given information: the element has three stable isotopes with masses of 27.977 amu, 28.976 amu, and 29.973 amu. The abundance of the first isotope can be determined by subtracting the abundances of the other two isotopes from 100% as only these three isotopes are stable. We have: (27.977 * x) + (28.976 * 0.0468) + (29.973 * 0.0309) = average atomic mass of the element. Solving this equation x(assuming it to be in percentage form)= 1- 0.0468 - 0.0309 = 0.9223 (or 92.23% in percentage form). Substituting the value of x we get: (27.977 * 0.9223) + (28.976 * 0.0468) + (29.973 * 0.0309) = 28.085 amu. Hence, the correct answer is (C) 28.085

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Answer: The volume of NaOH required to reach the endpoint is 100 mL

Explanation:

To calculate the volume of NaOH, we use the equation given by neutralization reaction:

[tex]n_1M_1V_1=n_2M_2V_2[/tex]

where,

[tex]n_1,M_1\text{ and }V_1[/tex] are the n-factor, molarity and volume of acid which is [tex]CH_3COOH[/tex]

[tex]n_2,M_2\text{ and }V_2[/tex] are the n-factor, molarity and volume of base which is NaOH.

We are given:

[tex]n_1=1\\M_1=0.250M\\V_1=50.0mL\\n_2=1\\M_2=0.125M\\V_2=?mL[/tex]

Putting values in above equation, we get:

[tex]1\times 0.250\times 50.00=1\times 0.125\times V_2\\\\V_2=\frac{1\times 0.250\times 50.0}{1\times 0.125}=100mL[/tex]

Hence, the volume of NaOH required to reach the endpoint is 100 mL

Answer:

13.301

Explanation:

To calculate the pH of the solution, we must obtain the pOH of the solution as illustrated below:

The dissociation equation is given below

Ba(OH)2 <==> Ba^2+ + 2OH^-

Since Ba(OH)2 dissociate to produce 2moles of OH^-, the concentration of OH^- = 2x0.1 = 0.2M

pOH = - Log[OH^-]

pOH = - Log 0.2

pOH = 0.699

But

pH + pOH = 14

pH = 14 — pOH

pH = 14 — 0.699

pH = 13.301

Answer:

The pH of this barium hydroxide solution is 13.30

Explanation:

Step 1: Data given

Concentration Ba(OH)2 = 0.10 M

Step 2: Calculate [OH-]

Ba(OH)2 ⇒ Ba^2+ + 2OH-

[OH-] = 2*0.10 M

[OH-] = 0.20 M

Step 3: Calculate pOH

pOH = -log[OH-]

pOH = -log(0.20)

pOH = 0.70

Step 4: Calculate pH

pH + pOH = 14

pH = 14 -pOH

pH = 14 - 0.70

pH = 13.30

The pH of this barium hydroxide solution is 13.30

Answer:

∆=2dn/n

Explanation:

Where∆ = lamda

the construction of the 10.20 cm into feet so which which Alina dispersion form 400-700 NM assuming the dielectric has a refractive index of 1.32.

construction of the interference which is described by

∆=2dn/n

the concept of interference which can be described by the calculation of the thickness of the wage at both end to do this report you to the wavelengths of absorption band and the thickness of dielectric constant.

That is

∆=2dn/n

d=∆R/2n

Where ∆ is equal wavelength d = thickness

n = interference order

R = refractive index of dielectric medium

range of linear dispersion is 400nm to 700nm

∆1 = 400nm

∆2 = 700nm

n = 1.32

d = ∆1R/2n

d = 400*10^-9m*1.32/2*1

d = 2.64*10^-7m

For ∆2

d = 700*10^-9m*1.32/2*1

d=4.62*10^-7m

Details of the construction is simply calculation of the thickness of the wedge at both ends

Final answer:

To build a 10.0-cm interference wedge with linear dispersion from 400 to 700 nm using a dielectric with refractive index 1.32 involves precision crafting of a wedge where the thickness incrementally increases, causing a linear change in path length and hence wavelength dispersion.

Explanation:

Constructing a 10.0-cm interference wedge with a linear dispersion from 400 to 700 nm involves creating a transparent object with a slight and uniform increase in thickness from one end to the other, using a dielectric material with a known refractive index, which in this case is 1.32.

The wedge must be carefully designed so that the path length difference between the top and bottom surfaces changes by exactly 300 nm (the range from 400 to 700 nm) over the 10 cm length. This incremental change in path length creates a linear dispersion of wavelengths when light is shone through the wedge.

At the thinnest point, light with a wavelength of 400 nm should experience constructive interference, whereas at the thickest point, light with a wavelength of 700 nm should interfere constructively. The process involves precision cutting and polishing to ensure consistent graduation of thickness across the wedge.

Answer:

% yield = 73.48 %

Explanation:

The fermentation reaction is:

C₆H₁₂O₆  →  2C₂H₅OH + 2CO₂          

The percent yield of C₂H₅OH is given by:

[tex] \% yield = \frac{m_{E}}{m_{T}} * 100 [/tex]

where [tex]m_{E}[/tex]: is the obtained mass of C₂H₅OH = 67.7g and [tex]m_{T}[/tex]: is the theoretical mass of C₂H₅OH.    

The theoretical mass of C₂H₅OH is calculated knowing that 1 mol of C₆H₁₂O₆ produces 2 moles of C₂H₅OH:  

[tex] m_{T} = mol * M [/tex]

where M: is the molar mass of C₂H₅OH =  46.068 g/mol

[tex] m_{T} = 2 moles * 46.068 g/mol = 92.136 g [/tex]                

Hence, the percent yield of C₂H₅OH is:

[tex] \% yield = \frac{67.7 g}{92.136 g}*100 = 73.48 \% [/tex]

I hope it helps you!                  

Taking into account definition of percent yield, the percent yield for the reaction is 73.58%.

In first place, the balanced reaction is:

C₆H₁₂O₆ → 2 C₂H₅OH + 2 CO₂

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

The molar mass of the compounds is:

Then, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:

The percent yield is the ratio of the actual return to the theoretical return expressed as a percentage.

The percent yield is calculated as the experimental yield divided by the theoretical yield multiplied by 100%:

[tex]percent yield=\frac{actual yield}{theoretical yield} x100[/tex]

where the theoretical yield is the amount of product acquired through the complete conversion of all reagents in the final product, that is, it is the maximum amount of product that could be formed from the given amounts of reagents.

In this case, you know:

Replacing in the definition of percent yield:

[tex]percent yield=\frac{67.7 g}{92 g} x100[/tex]

Solving:

percent yield= 73.58%

Finally, the percent yield for the reaction is 73.58%.

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The question happens to be in an incorrect order but the correct question can be seen below;

What concentration of [tex]SO^{2-}_3[/tex] is in equilibrium with [tex]Ag_2SO_{3(S)}[/tex] and [tex]7.10*10^{-3}M[/tex] [tex]Ag^+[/tex]? (The [tex]K_{sp}[/tex] of  

Answer:

[tex]2.96*10^{-10}M[/tex]

Explanation:

The concentration of [tex]SO^{2-}_3[/tex]  can be determined by using the solubility concept.

Given ionic solid is [tex]Ag_2SO_{3(S)}[/tex] ;

The Equilibrium Equation for the ionic compound will be:

[tex]Ag_2SO_{3(S)}[/tex] ⇄[tex]2Ag_{(aq)}[/tex] + [tex]SO^{2-}_3_{(aq)}[/tex]

Now, the solubility product  ([tex]K_{sp}[/tex]) of the ionic compound will be;

[tex]K_{sp}[/tex] [tex]= [Ag^+]^{2}[SO^{2-}_3][/tex]

Given that;

the concentration [tex]Ag^+[/tex] is [tex]7.10*10^{-3}M[/tex] ; &

solubility product of the given ionic solid is  [tex]1.5*10^{-14}[/tex]

∴

  [tex]K_{sp}[/tex] [tex]= [Ag^+]^{2}[SO^{2-}_3][/tex]

[tex]1.5*10^{-14}[/tex] [tex]= (7.10*10^{-3})^2[/tex]  [tex][SO^{2-}_3][/tex]

[tex][SO^{2-}_3][/tex] = [tex]\frac{1.5*10^{-14}}{ (7.10*10^{-3})^2}[/tex]

          = 2.97560008 × 10⁻¹⁰

          ≅ [tex]2.96*10^{-10}M[/tex]

Thus, the concentration of [tex][SO^{2-}_3][/tex] is [tex]2.96*10^{-10}M[/tex]

Answer:

The order of the reaction is 4.

The rate law of the reaction will be :

[tex]R=k[NO]3[H_2]^1[/tex]

Rate constant of the reaction: k =[tex] 6\times 10^5 M^{-3}s^{-1}[/tex]

Explanation:

[tex]2NO(g) +2H_2 (g)\rightarrow N_2(g) + 2H_2O(g)[/tex]

Let the stoichiometric coefficient of the NO and [tex]H_2[/tex] in rate law be x and y .

Rate of the reaction is given  by :

[tex]R=k[NO]^[H_2]^y[/tex]

1) When , [tex][NO]=0.0100 M, [H_2]=0.0100 M[/tex]

R =  0.00600 M/s

[tex]0.00600 M/s=k[0.0100 M]^x[0.0100 M]^y[/tex]..[1]

2) When , [tex][NO]=0.0200 M, [H_2]=0.0300 M[/tex]

R =  0.144 M/s

[tex]0.144 M/s=k[0.0200 M]^x[0.0300 M]^y[/tex]..[2]

3) When , [tex][NO]=0.0100 M, [H_2]=0.0200 M[/tex]

R =  0.0120 M/s

[tex]0.0120 M/s=k[0.0100 M]^x[0.0200 M]^y[/tex]..[3]

Dividing [1] and [3]

[tex]\frac{0.00600 M/s}{0.0120 M/s}=\frac{k[0.0100 M]^x\times [0.0100 M]^y}{k[0.0100 M]^x\times [0.0200 M]^y}[/tex]

y = 1

Dividing [1] and [2]

[tex]\frac{0.00600 M/s}{0.144 M/s}=\frac{k[0.0100 M]^x\times [0.0100 M]^1}{k[0.0200 M]^x\times [0.0300 M]^1}[/tex]

x = 3

The order of the reaction = x + y = 3 + 1 = 4

The rate law of the reaction will be :

[tex]R=k[NO]3[H_2]^1[/tex]

Rate constant of the reaction: k

When , [tex][NO]=0.0100 M, [H_2]=0.0100 M[/tex]

R =  0.00600 M/s

[tex]0.00600 M/s=k[0.0100 M]^3[0.0100 M]^1[/tex]..[1]

[tex]k=\frac{0.00600 M/s}{[0.0100 M]^3[0.0100 M]^1}=6\times 10^5 M^{-3}s^{-1}[/tex]

Final answer:

The reaction between nitrogen monoxide and hydrogen is second-order with respect to NO, first-order with respect to H2, and third-order overall. The rate law is rate = k[NO]²[H₂], where k is the rate constant specific to the reaction conditions.

Explanation:

The rate law expression for the reaction between hydrogen and nitrogen monoxide to form dinitrogen monoxide and water vapor can be determined using the given rate equation. By analyzing the equation rate = k[NO]²[H₂], we can deduce the orders of reaction with respect to each reactant and the overall order.

The rate law shows that the reaction rate is directly proportional to the square of the concentration of nitric oxide (NO) and linearly proportional to the concentration of hydrogen (H₂). Therefore, the order of reaction with respect to NO is 2 (second order), and with respect to H₂, it is 1 (first order). The overall order of the reaction is the sum of these individual orders, which is 2 + 1 = 3 (third order).

The rate constant (k) would be determined experimentally by measuring the reaction rates at known concentrations of reactants. It is specific to the reaction's conditions, such as temperature and pressure, and would be stated in units that correspond to a third-order reaction.

Answer:

Explanation:

This limit is a consequence of Heisenberg´s uncertainty principle:

Δp x Δx > =  h

This state that the product of the uncertainty in momentum ( or  velocity since p = mv ) times the uncertainty in position, Δx , must be greater or equal to Planck´s constant ( 6.626 x 10⁻³⁴ J·s ).

Later models refined this equation to:

Δp x Δx > =   h/4π

This is the consequence of duality wave matter of the electron and Schrodinger´s equation, in which we can talk of probabilities of finding an electron and not confined to specific distances from the nucleus as in the Bohr atom.

Now think of think of this relation in terms of the uncertainty it describes. If we know the position of the electron with great exactitude, the velocity of the particle will be very high since the mass of hte electron is very small.

This a principle in nature and has nothing to do with the precision of our instruments for particles at the subatomic level.

The reason we do not observe this effect  with everyday objects is that the obbects have masses so large compare to  subatomic particles that the term mΔv becomes large enough, allowing us to know the position and velocity of macroscopic objects with small uncertainties:

Δp  x  Δx > = h/4π,  Δp  very large  ( because the mass is very big ) then  Δx is very small

The same does not have with small masses of the subatomic levels.

Answer:

Molar concentration of Fe(NO3)3 . 9H2O = 0.12M

Explanation:

Fe(NO3).9H2O --> Fe(NO3)3 + 9H2O

By stoichiometry,

1 mole of Fe(NO3)3 will be absorb water to form 1 mole of Fe(NO3)3 . 9H2O

Therefore, calculating the mass concentration of Fe(NO3)3;

Molar mass of Fe(NO3)3 = 56 + 3*(14 + (16*3))

= 242 g/mol

Mass concentration of Fe(NO3)3 = molar mass * molar concentration

= 242 * 0.2

= 48.4 g/L

Molar mass of Fe(NO3)3 . 9H2O = 56 + 3*(14 + (16*3)) + 9* ((1*2) + 16)

= 242 + 162 g/mol

= 404g/mol

Concentration of Fe(NO3)3 . 9H2O = mass concentration/molar mass

= 48.4 /404

= 0.12 mol/l

Molar concentration of Fe(NO3)3 . 9H2O = 0.12M

Answer:

- 7.4 ºC

Explanation:

The change in melting  temperature is given by:

ΔTm = Kf m

where kf is the molal freezing  point depresion  constant, and m is the molality.

The molality of a solution is calculated as

m = mol solute/ kg solvent

Since we have the % composition it is easy to calculate the molality :

In 100 g of solution we have  24.9 ethylene glycol

mol glycol = 24.9 g / 62.07 g/ mol = 0.40 mol

molality = 0.40 mol / 0.1 kg = 4 m

Km for water is 1.86 ºC/m,

therefore,  

ΔTm = Kf m = 1.86 ºC/m x 4 = 7.4 ºC

Tm = -7.4 ºC for a solution

Answer:

[tex]\large \boxed{\text{55.8 u}}[/tex]

Explanation:

1. Calculate the volume of the unit cell

V = l³ = (2.866 × 10⁻⁸ cm)³ = 2.354 × 10⁻²³ cm³

2. Calculate the mass of a unit cell

[tex]\text{Mass} = 2.866 \times 10^{-23}\text{ cm}^{3} \times \dfrac{\text{7.87 g}}{\text{1 cm}^{3}} = 1.853 \times 10^{-22} \text{ g}[/tex]

3. Calculate the mass of one atom

A body-centred unit cell contains two atoms.

[tex]\text{Mass of 1 atom} = \dfrac{1.853 \times 10^{-22} \text{ g}}{\text{2 atoms}} \times \dfrac{\text{1 u}}{1.661 \times 10^{-24}\text{ g}} = \textbf{55.8 u}\\\\\text{The molar mass of Fe from the Periodic Table is $\large \boxed{\textbf{55.845 g/mol}}$}[/tex]

The ratio of acetate to acetic acid is approximately 1:1, while the ratio of ethylamine to ethylammonium ion is close to 0:1.

The ratio of acetate to acetic acid can be determined using the Henderson-Hasselbalch equation:

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

Since the pH is 7.4, which is close to the pKa of acetic acid (4.76), the ratio of acetate (A-) to acetic acid (HA) will be close to 1:1. The ratio of ethylamine to ethylammonium ion can be determined using the same equation. In this case, the pKa is the pKa of ethylamine (10.64). If the pH is 7.4, the ethylamine (C2H5NH2) will be mostly in its protonated form, ethylammonium ion (C2H5NH3+), giving a ratio close to 0:1.

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Answer:

a) The limiting reactant is O2.

b) 7.57 grams of P4O10 is produced

c) 7.53 grams P4O6 remains

Explanation:

Step 1: Data given

Mass of P4 = 7.55 grams

Mass of O2 = 7.55 grams

Molar mass of P4 = 123.90 g/mol

Molar mass of O2 = 32 g/mol

Step 2: The balanced equations:

P4 + 3O2-→P4O6

P4O6 + 2O2 → P4O10

Step 3: Calculate moles P4

Moles P4 = mass P4 / molar mass P4

Moles P4 = 7.55 grams / 123.90 g/mol

Moles P4 = 0.0609 moles

Step 4: Calculate moles O2

Moles O2 = 7.55 grams / 32.0 g/mol

Moles O2 = 0.236 moles

Step 5: Calculate the limiting reactant

For 1 mol P4 we need 3 moles O2 to produce 1 mol P4O6

P4 is the limiting reactant. It will completely be consumed. (0.0609 moles)

O2 is in excess. There will react 3*0.0609 = 0.1827 moles

There will remain 0.236 - 0.1827 = 0.0533 moles O2

Step 6: Calculate moles P4O6

For 1 mol P4 we need 3 moles O2 to produce 1 mol P4O6

For 0.0609 moles P4 we will have 0.0609 moles P4O6

Step 7: Calculate limting reactant

There remain 0.0533 moles O2 and there are 0.0609 moles P4O6 produced

For 1 mol P4O6 we need 2 moles O2 to produce 1 mol P4O6

The limiting reactant is O2. It will completely be reacted (0.0533 moles)

There will react 0.0533/2 = 0.02665 moles

There will remain 0.0609 - 0.02665 = 0.03425 moles P4O6

This is 0.03425 moles * 219.88 g/mol = 7.53 grams P4O6

Step 8: Calculate moles P4O10

For 1 mol P4O6 we need 2 moles O2 to produce 1 mol P4O6

For 0.0533 moles O2, we'll have 0.0533/2 = 0.02665 moles P4O10

Step 9: Calculate mass P4O10

Mass P4O10 = 0.02665 moles * 283.89 g/mol

Mass P4O10 = 7.57 grams