At 25◦C and atmospheric pressure the volume change of mixing of binary liquid mixtures of species 1 and 2 is given by)2545(2121xxxxVwhere Vis in cm3-mol-1. At these conditions, the molar volumes of pure liquid 1 and 2 are V1= 110 and V2= 90 cm3-mol-1. Determine the partial molar volumes 1VE and2VEin a mixture containing 40 mole percent of

Answer:

(a) See below

(b) 103.935 °F; 102.235 °F

Explanation:

The equation relating the temperature to time is

[tex]T = T_{0} + \Delta T\left (1 - e^{-t/\tau} \right )[/tex]

1. Calculate the thermometer readings after  0.5 min and 1 min

(a) After 0.5 min

[tex]\begin{array}{rcl}T & = & T_{0} + \Delta T\left (1 - e^{-t/\tau} \right )\\ & = & 100 + 10\left (1 - e^{-0.5/1} \right )\\ & = & 100 + 10\left (1 - e^{-0.5} \right )\\ & = & 100 + 10 (1 - 0.6065)\\ & = & 100 + 10(0.3935)\\ & = & 100 + 3.935\\ & = & 103.935\,^{\circ}F\\\end{array}[/tex]

(b) After 1 min

[tex]\begin{array}{rcl}T & = & T_{0} + \Delta T\left (1 - e^{-t/\tau} \right )\\ & = & 100 + 10\left (1 - e^{-1/1} \right )\\ & = & 100 + 10\left (1 - e^{-1} \right )\\ & = & 100 + 10 (1 - 0.3679)\\ & = & 100 + 10(0.6321)\\ & = & 100 + 6.321\\ & = & 106.321\,^{\circ}F\\\end{array}[/tex]

2. Calculate the thermometer reading after 2.0 min

T₀ =106.321 °F

ΔT = 100 - 106.321 °F = -6.321 °F

  t = t - 1, because the cooling starts 1 min late

[tex]\begin{array}{rcl}T & = & T_{0} + \Delta T\left (1 - e^{-(t - 1)/\tau} \right )\\ & = & 106.321 - 6.321\left (1 - e^{-(2 - 1)/1} \right )\\ & = & 106.321 - 6.321\left (1 - e^{-1} \right )\\ & = & 106.321 - 6.321 (1 - 0.3679)\\ & = & 106.321 - 6.321 (0.6321)\\ & = & 106.321 - 3.996\\ & = & 102.325\,^{\circ}F\\\end{array}[/tex]

3. Plot the temperature readings as a function of time.

The graphs are shown below.

To plot the variation of the thermometer reading with time, we need to consider the first-order dynamics of the thermometer. The time constant of the thermometer is 1 min, and it takes approximately 5 time constants to reach steady state. To calculate the thermometer reading at specific times, we can use the formula for first-order dynamics.

To plot the variation of the thermometer reading with time, we need to consider the first-order dynamics of the thermometer. The time constant of the thermometer is 1 min, so it takes approximately 5 time constants (or 5 minutes) to reach steady state. After that, when the thermometer is suddenly placed in a bath at 110oF, it will start to change its reading towards the new temperature following the first-order dynamics. Then, at t = 1 min, it is returned to the bath at 100oF and again it starts to change its reading towards the new temperature.

(b) To calculate the thermometer reading at t = 0.5 min, we can use the formula for first-order dynamics: R = R0 + (R1 - R0)(1 - e^(-t/tau)), where R0 is the initial reading, R1 is the final reading, t is the time, and tau is the time constant. Plugging in the values, we get R = 100 + (110 - 100)(1 - e^(-0.5/1)) ≈ 105.3oF. To calculate the thermometer reading at t = 2.0 min, we use the same formula with t = 2.0 and the new R0 would be the reading at t = 1 min, which we can calculate similarly.

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The potential of the silver-silver chloride electrode in a saturated KCl solution can be predicted as 0.195V, whatever the potential of the calomel electrode, this is calculated by considering the potential difference between the calomel electrode in standard condition and in a saturated KCl solution.

Firstly, the difference in electrode potential between calomel electrode in standard conditions and in a saturated KCl solution is (E ° = 0.268 V) - (E (saturated KCl) = 0.241 V) = 0.027 V. This difference indicates the change in potential when the electrode is in a saturated KCl solution compared to the standard conditions. Given that the E ° for the silver-silver chloride electrode is 0.222 V in standard conditions, the potential of the Ag/AgCl electrode in a saturated KCl solution can be predicted as E (Ag/AgCl, saturated KCl) = (E ° Ag/AgCl) - (ΔE Calomel) = 0.222 V - 0.027 V = 0.195 V. This implies that the potential of the silver-silver chloride electrode in a saturated KCl solution should be 0.195 V.

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Answer:

109.5° (d)

Explanation:

Use the table 10.1 to determine the

1. Electron geometry

2. Molecular geometry

3. Bond angles

-Four electron groups give a tetrahedral electron geometry

- two bonding groups and two lone pair give a bent molecular geometry

- the idealized bond angles for tetrahedral geometry are 109.5°.

Answer:

SrCO₃ will precipitate in the reaction

Explanation:

Na₂CO₃  +  SrCl₂   →  2 NaCl  +  SrCO₃ ↓

CO₃⁻² reacts with all the elements of group 2 of periodic table, to make an insoluble solid.

NaCl is a ionic salt, which is extremely soluble.

To solve this, you have to analyse all the reactions of cations and anions, with the qualitative inorganic analysis.

Answer:

(E)-3-methylpenta-1,3-diene

Explanation:

Bromine water in carbon tetrachloride is used to test for the presence of alkenes. Bromine was added to this unsaturated compound with double bonds, as the reddish color of the solution disappeared. This means bromine was added to the double bonds present in our structure.

Since three products were formed upon ozonolysis, this means we have two double bonds present in our structure, a diene. Besides, a total of 6 carbons are seen in the given structures, so molecular formula of the compound is [tex]C_6H_{10}[/tex].

Ozonolysis occurs when a double bond is broken and oxygen atoms are added to each end of the double bond, that's why we have a total of 3 products.

Our starting material is (E)-3-methylpenta-1,3-diene. You may wish to refer to the file uploaded below which represents the ozonolysis step of breaking the two double bonds and forming the 3 products shown.

Final answer:

The unknown compound undergoing the given reactions is likely to have the structures (1'R,2'S,3'R)-1-[(2',3'-O-isopropylidene)-4'-cyclopenten-1'-yl]-4-iodoimidazole and 06-(Benzotriazol-1-yl)-2-bromo-9-[2-deoxy-3,5-di-O-(t-butyldimethylsilyl)-B-D-ribofuranosyllpurine.

Explanation:

The unknown compound with empirical formula C3H5 reacted with Br2/CCl4 to undergo an addition reaction. This is indicated by the change in color of the bromine solution from orangish/red to clear. Further treatment with O3 followed by work-up with dimethylsulfide resulted in the identification of specific products. Based on this information, the most likely structures for the unknown compound are: (1'R,2'S,3'R)-1-[(2',3'-O-isopropylidene)-4'-cyclopenten-1'-yl]-4-iodoimidazole and 06-(Benzotriazol-1-yl)-2-bromo-9-[2-deoxy-3,5-di-O-(t-butyldimethylsilyl)-B-D-ribofuranosyllpurine.

Answer:

1) There were 7.65 grams of heptane burned

2) There reacted 38.94 grams of HCl

Explanation:

1) The combustion of heptane

Step 1: Data given

Mass of CO2 = 23.5 grams

Molar mass of CO2 = 44.01 g/mol

Step 2: The balanced equation:

C7H16  + 11O2 ⟶ 7CO2 + 8H2O

Step 3: Calculate moles of CO2

Moles CO2 = mass CO2 / molar mass CO2

Moles CO2 = 23.5 grams / 44.01 g/mol

Moles CO2 = 0.534 moles

Step 4: Calculate moles heptane

For 1 mole of Heptane , we need 11 moles of O2 to produce 7 moles of CO2 and 8 moles of H2O

For 0.534 moles of CO2 we have 0.534/7 = 0.0763 moles of  heptane

Step 5: Calculate mass of heptane

Mass of heptane = moles heptane * molar mass heptane

Mass heptane = 0.0763 moles * 100.21 g/mol

Mass heptane = 7.65 grams

2) The reaction of limestone with hydrochloric acid

Step 1: Data given

Mass of CO2 = 23.5 grams

Step 2: The balanced equation:

CaCO3 + 2HCl ⟶ CaCl2 + CO2 + H2O

Step 3: Calculate moles of CO2

Moles CO2 = mass CO2 / molar mass CO2

Moles CO2 = 23.5 grams / 44.01 g/mol

Moles CO2 = 0.534 moles

Step 4: Calculate moles of HCl

For 1 mol of CaCO3 we need 2 moles of HCl to produce 1 mol of CaCl2, 1 mol of CO2 and 1 mol of H2O

For 0.534 moles of CO2 we have 2*0.534 = 1.068 moles of HCl

Step 5: Calculate mass of HCl

Mass HCl = moles HCl * molar mass HCl

Mass HCl = 1.068 moles * 36.46 g/mol

Mass HCl = 38.94 grams

To produce 23.5 g of CO₂, approximately 7.63 g of heptane were burned, and 38.94 g of HCl reacted. By using the molar masses and stoichiometric relationships from the balanced reactions, we can find the required masses. Moles of substances involved were calculated using the molar masses and balanced equations.

Determining Grams of Heptane and HCl Reacted

First, let's find out how many grams of heptane (C₇H₁₆) were burned to produce 23.5 g of CO₂ in the given balanced chemical reaction:

Combustion of Heptane:

Balanced equation: C₇H₁₆ + 11 O₂ ⟶ 7 CO₂ + 8 H₂O

Next, let's determine how many grams of HCl reacted to produce 23.5 g of CO₂ in the reaction between limestone and hydrochloric acid:

Reaction of Limestone with Hydrochloric Acid:

Balanced equation: CaCO₃ + 2 HCl ⟶ CaCl₂ + CO₂ + H₂O

Answer: 386.0 g/mol

Explanation:

As the relative lowering of vapor pressure is directly proportional to the amount of dissolved solute.

The formula for relative lowering of vapor pressure will be,

[tex]\frac{p^o-p_s}{p^o}=i\times x_2[/tex]

where,

[tex]\frac{p^o-p_s}{p^o}[/tex]= relative lowering in vapor pressure

i = Van'T Hoff factor = 1 (for non electrolytes)

[tex]x_2[/tex] = mole fraction of solute  =[tex]\frac{\text {moles of solute}}{\text {total moles}}[/tex]

Given : 21.47 g of compound X is present in 233.8 g of diethyl ether

moles of solute (X) = [tex]\frac{\text{Given mass}}{\text {Molar mass}}=\frac{21.47g}{Mg/mol}[/tex]

moles of solvent (diethyl ether) = [tex]\frac{\text{Given mass}}{\text {Molar mass}}=\frac{233.8g}{74g/mol}=3.160moles[/tex]

[tex]x_2[/tex] = mole fraction of solute =[tex]\frac{\frac{21.47g}{Mg/mol}}{\frac{21.47g}{Mg/mol}+3.160}[/tex]

[tex]\frac{463.57-455.55}{463.57}=1\times \frac{\frac{21.47g}{Mg/mol}}{\frac{21.47g}{Mg/mol}+3.160}[/tex]

[tex]0.017301=1\times \frac{\frac{21.47g}{Mg/mol}}{\frac{21.47g}{Mg/mol}+3.160}[/tex]

[tex]M=386.0g/mol[/tex]

The molecular weight of this compound is 386.0 g/mol

Electrolytes are substances -minerals- that conduct electricity once they are dissolved in water. Some examples include Na⁺ (Sodium), Ca⁺² (Calcium), PO₄⁻³ (Phosphate) and Mg⁺² (Magnesium).

They are important because of the rols they play in our organism: The electrical charges they carry are what allow muscles to contract and nerve impulses to be transmitted. Without proper muscle contractions or nerve transmissions we would die.

We do not need to drink Gatorade to get electrolytes. It's true that sport drinks contain electrolytes, but those electrolytes can also come from different sources in our everyday diet, such as salt or fruits and vegetables.

Answer:

False, False, False

Explanation:

Indicate whether each statement is true or false.

When the rate constant should be determined so we cannot calculate the enthalpy change.

The Exothermic reactions should not be faster.

here the relationship that lies between the rate constant and the activation energy that should be provided by the Arrhenius equation should be based on the Ea.

In the case when the reaction should be considered as the exothermic so it represent the sign of the enthalpy.

Also, the activation energy does not based upon the temperature.

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Answer: The daughter nuclide formed by the beta decay of given isotope is [tex]_{39}^{89}\textrm{Y}[/tex]

Explanation:

Beta decay is defined as the process in which beta particle is emitted. In this process, a neutron gets converted to a proton and an electron.

The released beta particle is also known as electron.

[tex]_Z^A\textrm{X}\rightarrow _{Z+1}^A\textrm{Y}+_{-1}^0\beta[/tex]

We are given:

Parent isotope = [tex]_{38}^{89}\textrm{Sr}[/tex]

The chemical equation for the beta decay process of [tex]_{38}^{89}\textrm{Sr}[/tex] follows:

[tex]_{38}^{89}\textrm{Sr}\rightarrow _{39}^{89}\textrm{Y}+_{-1}^0\beta[/tex]

Hence, the daughter nuclide formed by the beta decay of given isotope is [tex]_{39}^{89}\textrm{Y}[/tex]

Final answer:

The daughter nuclide from the beta decay of Strontium-89 is Yttrium-89, and the nuclear equation for the decay is 8938Sr → 8939Y + β-.

Explanation:

The question is asking to identify the daughter nuclide from the beta decay of Strontium-89 (8938Sr). In beta decay, a neutron in the nucleus is converted into a proton and a beta particle (an electron) is emitted. The atomic number increases by 1, but the mass number remains the same. Therefore, the daughter nuclide of 8938Sr undergoing beta decay would be 8939Y (Yttrium-89).

The nuclear equation for this decay process would be written as:

8938Sr → 8939Y + β-

Answer:

H-O-H  polar

O-C-O nonpolar

H-C-N   polar

Explanation:

Looking up the electronegativities of the atoms involved in this question, we have:

Atom   Electronegativity

H               2.2

C               2.55

N               3.04

O               3.44

All of the atoms differ in electronegativity resulting in individual dipole moments in H-O, O-C, H-C and C-N bonds. To find if the molecules will be polar we need to consider the structure of the compound to see if there is a resultant dipole moment.

In H-O-H, we have 2 lone pairs of electrons around the central oxygen atom which push the angle H-O-H  of the ideal tetrahedral structure to be smaller than 109.5 º  resulting in an overall dipole moment making it polar.

In O-C-O, we have two dipole moments that exactly cancel each other in the linear molecule  since the central carbon atom does not have lone pairs of electrons since it  has 2 double bonds. Therefore the molecule is nonpolar.

In H-C-N, again we have have a central carbon atom without lone pairs of electrons and the shape of the molecule is linear. But, now we have that the dipole moment in C-N is stronger than the H-C dipole because of the difference in electronegativity of nitrogen compared to hydrogen. The molecule has an overall dipole moment and it is polar.

Answer:

H-O-H is polar

O-C-O is non polar

H-C-N is polar

Explanation:

The electronegativity value of H is 2.1 and the Electronegavity value for O is 3.5 which makes the electronegativity difference of 1.4. O has 2 lone pair of electrons which makes the shape non- linear and polar.

The bond between C and O is a double bond. The Electronegativity of O is 3.5 while the Electronegativity value for C is 2.5 which gives the electronegativity difference of 1.0. There are no lone pairs on carbon and the dipole or partial charges on carbon and oxygen cancels out making O-C-O non-polar.

The Electronegativity value of H is 2.1, electronegativity value for C is 2.5 and the electronegativity value for N is 3.0 which makes the molecule to develop partial positive charge on H and partial negative charge on the C-N making the compound polar.

Answer:

Decomposition of aluminium oxide forms  aluminium atoms and  oxygen atoms.

Explanation:

Decomposition reaction:

When a single compound break down into two or more simpler products.

For example "AB" reactant undergoes decomposition to form "A" and "B" products.

The chemical reaction is as follows.

[tex]AB\rightarrow A+B[/tex]

The given compound is aluminium oxide.

The decomposition reaction of aluminium oxide is a follows.

[tex]Al_{2}O_{3}\rightarrow Al+O_{2}[/tex]

The balanced equation is as follows.

[tex]2Al_{2}O_{3}\rightarrow 4Al+3O_{2}[/tex]

Therefore, Decomposition of aluminium oxide forms aluminium atoms and  oxygen atoms.

Answer:

a) The electron mobility for this metal is 0.00455 m²V/s

b) There are 4.12 * 10^28 free electrons per cubic meter

Explanation:

Step 1: Data given

electrical resistivity of 3.3 * 10^−8(Ω⋅m)

A current of 25 A is passed through a specimen of this metal 15 mm thick.

A magnetic field of 0.95 tesla.

A Hall voltage of −2.4×10−7V is measured

Step 2: Calculate the conductivity

σ= 1/ ρ

 ⇒ with ρ = electrical resistivity of 3.3*10^−8Ω*m

σ = 1/3.3*10^−8

σ = 30303030.3 = 3*10^7/ Ω*m

step 3: Calculate the electron mobility

µe = Vdσ / IxBz

⇒ with µe = the electron mobility in m²V/s

⇒ V = voltage of  2.4*10^−7V

⇒ d = 0.015 meter

⇒ σ = the conductivity = 3*10^7/ Ω*m

⇒ Ix = the current = 25 A

⇒ Bz =  magnetic field = 0.95 tesla

µe = (2.4 * 10^-7 * 0.015 * 3*10^7) / (25*0.95)

µe = 0.00455 m²V¨/s

The electron mobility is 0.00455 m²V¨/s

Step 4: Calculate the number of free electrons per cubic meter

n = σ/e*µe

⇒ e= the charge of an electron = 1.6 * 10^-19 Coulombs

⇒ σ = The conductivity = 3*10^7/ Ω*m

⇒ µe = The electron mobility = 0.00455 m²V¨/s

n = 3*10^7/ Ω*m /(1.6 * 10^-19 * 0.00455)

n = 4.12 * 10^28 / m³

There are 4.12 * 10^28 free electrons per cubic meter

The electron mobility of the metal is 3.18 × 10^-3 m^2/Vs. The number of free electrons per cubic meter is 1.52 × 10^29 electrons/m^3.

To calculate the electron mobility and the number of free electrons per cubic meter for the given metal, we can use the equations involving Hall effect and resistivity. The formula for electron mobility is given by μ = -VH / (B * I * t), where VH is the Hall voltage, B is the magnetic field, I is the current, and t is the thickness. Substituting the values, we get the electron mobility as 3.18 × 10^-3 m^2/Vs.

The formula for the number of free electrons per cubic meter is given by n = 1 / (e * R), where e is the charge of an electron (1.6 × 10^-19 C) and R is the resistivity. Substituting the values, we get the number of free electrons per cubic meter as 1.52 × 10^29 electrons/m^3.

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The applied pressure at the end of this process is P = -2.859atm

Explanation:

Write down the given values

there are concentration of two solutions

S1 = 0.053 M

S2 = 0.170 M

Temperature (T) = 23°C

that is written as 298 K

it can be solved by multiplying ans subtracting the values

P=MR1-MR2

0.053 M * 0.082 * 298 - 0.170 M * 0.082 * 298

= 1.295 - 4.15412

P = -2.859atm

Final answer:

Reverse osmosis is a technique that involves applying pressure to force water molecules from a high concentration solution to a low concentration solution through a semipermeable membrane. This method is commonly used to convert saltwater into freshwater. The concentration of the solutions will determine the amount of pressure required.

Explanation:

When a compartment containing a dilute solution is connected to another compartment containing a concentrated solution by a semipermeable membrane, water molecules move from the dilute solution to concentrated solution. This phenomenon is called osmosis. By applying pressure in the higher concentration solution, water molecules migrate from a high concentration solution to a low concentration solution through a semipermeable membrane. This method is called reverse osmosis water filter system. In this technique, the membrane must be able to tolerate the high pressure, and prevent solute molecules to pass through. This technology certainly works, and it has been used to convert salt (ocean or sea) water into fresh water. With this technique, the water with higher concentration is discharged. Thus, this technology is costly in regions where the water cost is high.

Answer:

Number of particles = 2.0 g*(6.0 x 10^23 particles/mol) / 20.18 g/mol

Option C is correct

Explanation:

Step 1: Data give

Mass of Ne = 2.0 grams

Molar mass of neon = 20.18 g/mol

Number of Avogadro = 6.0 *10^23 /mol

Step 2: Calculate number of moles of neon

Moles Ne = Mass of ne / Molar mass of ne

Moles Ne = 2.0 / 20.18 g/mol

Moles Ne = 0.099 moles

Step 3: Calculate nulber of particles

Number of particles = 6.022*10^23 / mol * 0.099 moles = 5.96 *10^22

Number of particles = 6.022*10^23 * (2.0g/ 20.18g/mol)

Number of particles = 2.0 g*(6.0 x 10^23 particles/mol) / 20.18 g/mol

Option C is correct

Answer:

Initial we have 0.183 + 3*0.183 = 0.732 moles of gas (those will all be consumed).

In the products we have 0.366 moles of gas

Explanation:

Step 1: Data given

Number of moles of N2 = 0.183 mol

Number of moles of H2 = 0.549 mol

Step 2: The balanced equation:

N2 + 3H2 → 2NH3

Step 3: Calculate the limiting reactant

For 1 mol of N2 we need 3 moles of H2 to produce 2 moles of NH3

There is no limiting reactant. But will be completely consumed

Step 4: Calculate moles of NH3

For 1 mol of N2 we have 2 moles of NH3 produced

For 0.183 moles of N2 we have 2*0.183 = 0.366 moles of NH3 produced

Initial we have 0.183 + 3*0.183 = 0.732 moles of gas (those will all be consumed).

In the products we have 0.366 moles of gas

Answer:

Initially, there are 0.732 mol of gas particles. After the reaction is complete, there are 0.366 mol of gas particles. Therefore, the ratio of product to reactant prticles is 5.0

Explanation:

Answer:

A. yes

594°C

Explanation:

There is some info missing. I think this is the original question.

For many purposes we can treat ammonia (NH₃) as an ideal gas at temperatures above its boiling point of -33 °C. Suppose the pressure on a 6.0 m³ sample of ammonia gas at 16.0°C is tripled. Is it possible to change the temperature of the ammonia at the same time such that the volume of the gas doesn't change? If you answered yes, calculate the new temperature of the gas. Round your answer to the nearest °C.

Given data

V₁ = V₂ = 6.0 m³

T₁ = 16.0°C + 273.15 = 289.2 K

T₂ = ?

P₂ =  3 P₁

Assuming constant volume and ideal gas behavior, we can find the new temperature using the Gay-Lussac's law.

[tex]\frac{P_{1}}{T_{1}} =\frac{P_{2}}{T_{2}} \\T_{2}=\frac{P_{2}.T_{1}}{P_{1}} =\frac{3P_{1}.T_{1}}{P_{1}}=3T_{1}=3 \times 289.2K = 867.6K[/tex]

°C = 867.6K - 273.15 = 594°C

Yes, it is possible to change the temperature of the ammonia gas while keeping the volume constant. The new temperature is approximately -88.6 °C.

According to Gay-Lussac's law, for a given amount of gas at constant volume, the pressure and temperature are directly proportional. Therefore, if the pressure is tripled, the temperature of the ammonia gas can be changed in the same proportion to keep the volume constant.

We can use the formula P1/T1 = P2/T2 to calculate the new temperature (T2) when the pressure (P) is tripled. Given that the temperature (T1) is the boiling point of ammonia, we can plug in the values and solve for T2.

Using this formula, we can find that the new temperature is approximately -88.6 °C.

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