A thermometer having first-order dynamics with a time constant of 1 min is placed in a temperature bath at 100oF. After the thermometer reaches steady state, it is suddenly placed in a bath at 110oF at t = 0 and left there for 1 min, after which it is immediately returned to the bath at 100oF.(a) Plot the variation of the thermometer reading with time (No hand-sketched plot please).(b) Calculate the thermometer reading at t = 0.5 min and at t = 2.0 min.

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

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:

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:

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:

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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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:

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.

learn more about temperature here: https://brainly.com/question/20510182

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