The vapour pressure of water at 20 C is 2.34 kPa. Given that the heat of vaporisation is 2537.4 kJ/kg, use the Clausius-Clapeyron equation to give the vapour pressure at 40 C.

Answer:

See explanation

Explanation:

Fusion is the process of fusing two isotopes of Hydrogen namely Tritium  and Deuterium to produce Helium. To achieve this, tremendous heat required (about a million degrees Celsius). The same for the pressure. To achieve this, the hydrogen is made into a plasma through rarefaction so it becomes susceptible to magnetic fields. Magnetic confinement fusion is an approach to generating thermonuclear fusion power that uses magnetic fields to confine the hot fusion fuel in the form of a plasma. Electro-Magnets surround the chamber/reactor and are pulsed adiabatically (as in a bicycle pump) and the gas becomes extremely hot that may melt the surroundings.

As the ions in the plasma are charged (the plasma is so hot all the negatively-charged electrons are stripped off the atoms, leaving them with a positive charge) they respond to magnetic fields. Extra fields help shape the plasma and hold it stable.

It will take the microwave approximately 4.2 seconds to emit a total of 4650 J of energy, calculated by dividing the total energy by the power it emits per second, using two significant figures.

To calculate how long it will take for a microwave emitting 1,100 J of energy every second to emit a total of 4650 J, we can use the formula:

Time (s) = Total Energy (J) / Power (J/s).

Substituting the given values:

Time (s) = 4650 J / 1,100 J/s = 4.22727272727... seconds

Since the microwave emits energy at a rate of 1,100 J every second, we want to express the answer with the same number of significant figures as the least precise measurement, which is two significant figures. Therefore, the time should be rounded to 4.2 seconds.

To convert cm³ or mm³ to m³, one must divide by 1,000,000 or 1,000,000,000 respectively. Thus, 2.7 cm³ is equal to 2.7 x 10^-6 m³ and 2.0 mm³ is equal to 2.0 x 10^-9 m³.

To convert cubic centimeters (cm³) and cubic millimeters (mm³) to cubic meters (m³), you need to know the unit conversions. One square meter is equal to 1,000,000 cubic centimeters and 1,000,000,000 cubic millimeters.

This means you can convert 2.7 cm³ to meters by dividing by 1,000,000, yielding an answer of 0.0000027 m³ (to two significant figures, or 2.7 x 10^-6 m³).

Similarly, 2.0 mm³ can be converted to meters by dividing by 1,000,000,000, resulting in an answer of 0.000000002 m³ or 2.0 x 10^-9 m³.

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The conversion from cm3 and mm3 to m3 is done by multiplying the original number by 1e-6 for cm3 and 1e-9 for mm3. The results for the examples given are 2.7 cm3 equals 2.7e-6 m3, and 2.0 mm3 equals 2.0e-9 m3.

The task is to convert measurements from one unit (cubic centimeters or cubic millimeters) to another (cubic meters). It's important to understand the relevant conversion factors:

1 cm3 = 1e-6 m3

1 mm3 = 1e-9 m3

Applying these to your examples we get:

2.7 cm3 = 2.7 * 1e-6 m3 = 2.7e-6 m3

2.0 mm3 = 2.0 * 1e-9 m3 = 2.0e-9 m3

So these are the conversions using two significant figures.

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Answer:

Fischer projection is the method used for representing a three-dimensional organic molecule as a two dimensional molecule.  

This method can be used for determining the D- and L- configuration of the organic molecules.

In the Fisher projection of carbohydrate molecule, if the hydroxyl group attached to the last stereocenter of the molecule is placed on the right-hand side, then the carbohydrate is said to have D-configuration.

Whereas, if the hydroxyl group attached to the last stereocenter of the molecule is placed on the left-hand side, then the carbohydrate is said to have L-configuration.

Final answer:

To determine if a carbohydrate is D or L in a Fischer projection, look at the hydroxyl (-OH) group on the penultimate carbon; if it's to the right, it's a D-sugar, and if it's to the left, it's an L-sugar. This classification does not directly relate to the sugar's optical activity but its stereochemistry relative to glyceraldehyde.

Explanation:

How to Determine if a Carbohydrate is D or L Using a Fischer Projection

When examining a Fischer projection of a monosaccharide, you can determine whether it is a D-sugar or an L-sugar by looking at the orientation of the hydroxyl (-OH) group on the penultimate carbon (second-last carbon) in the chain. The rule is straightforward: if the -OH group on this carbon is to the right side of the Fischer projection, the sugar is designated as a D-sugar. Conversely, if the -OH group is to the left side, the sugar is an L-sugar.

This method of classification is based on the relative configuration to glyceraldehyde, where D-glyceraldehyde has the -OH on the right at the chiral center farthest from the carbonyl group, hence all D-sugars follow this pattern. L-sugars are the mirror images (enantiomers) of the D-sugars, with their -OH groups flipped to the opposite side.

It's important to note that the D/L configuration does not directly correlate with the optical activity of the sugar (++/--) but rather describes its stereochemistry relative to glyceraldehyde. The D/L nomenclature is fundamental in distinguishing the stereochemistry of sugars and their derivatives.

Answer:

The volume of a 0.200 M KCl solution containing 5.00 10-2 mol of solute is 0,25 L

Explanation:

Molarity (M) means: moles of solute which are contained in 1 L of solution.  

In this case we have 0,2 moles which are in 1 L, so, as we have 5x10*-2 moles we have to apply a rule of three to find out the volume.

0,2 moles ........... 1 L

5x10*-2 moles ........... x

x= (5x10*-2 moles . 1 L) / 0,2 moles = 0.25L

(we can also say 250 mL)

Answer: The final volume of the oxygen gas is 4.04 L

Explanation:

To calculate the final temperature of the system, we use the equation given by Charles' Law. This law states that volume of the gas is directly proportional to the temperature of the gas at constant pressure and number of moles.

Mathematically,

[tex]\frac{V_1}{T_1}=\frac{V_2}{T_2}[/tex]

where,

[tex]V_1\text{ and }T_1[/tex] are the initial volume and temperature of the gas.

[tex]V_2\text{ and }T_2[/tex] are the final volume and temperature of the gas.

We are given:

[tex]V_1=3.24L\\T_1=29^oC=(29+273)K=302K\\V_2=?\\T_2=104^oC=(104+273)K=377K[/tex]

Putting values in above equation, we get:

[tex]\frac{3.24L}{302K}=\frac{V_2}{377K}\\\\V_2=4.04L[/tex]

Hence, the final volume of the oxygen gas is 4.04 L

Answer:

Re=8561.79

Friction factor is 0.014

Flow is turbulent flow.

Explanation:

Given that

Diameter ,d=1 in

d=0.0254 m     (1 in =0.0254 m )

Volume flow rate,Q = 2 gpm

We know that

[tex]1\ gpm=7.5\times 10^{-5}\ m^3/s[/tex]

[tex]2\ gpm=2\times 7.5\times 10^{-5}\ m^3/s[/tex]

[tex]2\ gpm=15\times 10^{-5}\ m^3/s[/tex]

We know that

Q= A x V

[tex]A=\dfrac{\pi}{4}\times 0.0254^2\ m^2[/tex]

[tex]A=0.00050\ m^2[/tex]

So

[tex]V=\dfrac{15\times 10^{-5}}{0.00050}[/tex] m/s

V=0.3 m/s

So Reynolds number(Re)

[tex]Re=\dfrac{\rho VD}{\mu }[/tex]

Properties of water at 25 C

[tex]\mu=8.9\times 10^{-4}\ Pa.s[/tex]

[tex]Re=\dfrac{1000\times 0.0254\times 0.3}{8.9\times 10^{-4}}[/tex]

Re=8561.79

Re>4000 ,It means that flow is turbulent flow.

Friction factor

If we assume that pipe is smooth

[tex]f=\dfrac{0.136}{Re^{0.25}}[/tex]

[tex]f=\dfrac{0.136}{8561.79^{0.25}}[/tex]

f=0.014.

Friction factor is 0.014

To identify whether the flow is laminar or turbulent and to determine the friction factor, you should first find the Reynolds number using the given parameters, such as the flow rate of 2.0 gpm and then refer to a Moody chart or similar source.

The Reynolds number (NR) and friction factor of the flow can be determined using the given parameters (Flow rate, pipe diameter, and water temperature). The Reynolds number is an indicator that can reveal whether flow is laminar or turbulent. For flow in a tube of uniform diameter, the Reynolds number is defined as an equation related to the properties of the fluid and the characteristics of the flow.

In this case, if the Reynolds number (NR) is below 2000, the flow is considered laminar. If NR is above 3000, the flow becomes turbulent. For values of NR between 2000 and 3000, it may be either or both, depending on factors such as the roughness of the pipe's internal surface and the flow velocity.

The friction factor is a measure of the total resistance created by the force on a fluid as it moves through a pipe. The calculation of this factor also considers the Reynolds number and pipe roughness.

To determine whether your given conditions (2.0 gpm water flow through a 1-inch Schedule 40 steel pipe at 25◦C) will result in laminar or turbulent flow and the corresponding friction factor, please consider calculating the Reynolds number first using the suitable formula and the given conditions and then consulting a Moody chart or a similar source to find the corresponding friction factor.

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Answer:

The balanced chemical equation is Mg + 2HCl ⇒ MgCl2 + H2

The molarity of the hydrochloric acid solution is HCl 0.04 M and the pH = 1.4.

The volume of hydrogen gas produced by the reaction of 0.510 g of Mg will be 0.482 L.

Explanation:

First, for the balanced equation you have to consider the oxidation state of the elements to find subscripts. Then you can find the correct coeficients. Mg= +2, Cl = -1.

Mg + HCL ⇒ MgCl2 + H2

For the molarity of the solution you have to notice tha if 0.510 grams of Mg reacts with 0,5 L of hydroclhoric acid, and from the previous equation 1 mol of Mg reacts with 2 mol HCl.

The atomic mass of Mg = 24.31 grs/mol

24.31 grs------------ 1 mol Mg

0.510 grs------------ x=0.02 mol Mg.

If 1 mol of Mg reacts with 2 mol HCl, then 0.02 mol of Mg will react with

0.04 mol HCl. So, the molarity of the solution is 0.04 M HCl.

Then to calculate the pH we use the formula pH = - log [H+]

⇒ pH = -log [0.04]⇒ pH=1.4.

Finally, from the balanced equation and the findings described, and considering that at 25°C and 1.00 atm 1 mol of gas has volume of 24.1 L.

1 mol H2----------- 24.1 L

0.02 mol H2----- x= 0.482L.

Answer:

C. fluid density & viscosity

Explanation:

In 1850, Darcy-Weisbach experimentally deduced an equation to calculate shear losses ("friction"), in a tube with permanent flow and constant diameter:

hf = (f x L x V^2) / (D x 2g)

where:

hf: shear losses

f:  shear loss factor (pipe roughness)

g: gravity acceleration

D: tube diameter

L: tube length  

V: fluid average speed in the tube

To calculate the loss factor “f” in the Poiseuille laminar region, he proposed in 1846 the following equation:

f = 64 / Re

Where:

Re: Reynolds number

The influence of the parameters on f is quantitatively different according to the characteristics of the current.

In any straight pipeline that transports a liquid at a certain temperature, there is a critical speed below which the regimen is laminar. This critical value that marks the transition between the two regimes, laminar and turbulent, corresponds to a Re = 2300, although in practice, between 2000 and 4000 the situation is quite inaccurate. Thus:

Re <2000: laminar regimen

2000 <Re <4000: critical or transition zone

Re> 4000: turbulent regime

Final answer:

The friction factor for fluid flow in a pipe does not depend upon the pipe length, pipe roughness, fluid density & viscosity, or mass flow rate of fluid.

Explanation:

The friction factor for fluid flow in a pipe does not depend upon the pipe length (A), pipe roughness (B), fluid density & viscosity (C), or mass flow rate of fluid (D).

This is because the friction factor, also known as the Darcy-Weisbach factor, is determined by the characteristics of the flow itself, such as the Reynolds number, which is a dimensionless quantity that relates the inertia of the fluid to the viscous forces acting on it.

The friction factor can be calculated using the Colebrook-White equation or obtained from Moody's diagram based on the relative roughness of the pipe and the Reynolds number.

Answer:

0.09 g of precipitate can be formed.

Explanation:

The chemical equation can be written as follows:

AgNO₃ + KCl → AgCl(↓) + NO₃⁻ + K⁺

From the equation, we know that 1 mol AgNO₃ reacts with 1 mol KCl to produce 1 mol AgCl.

The problem gives us data to calculate the initial number of moles of silver nitrate and potassium chloride:

n° of moles of silver nitrate = concentration * volume

n° of moles of silver nitrate = 0.10 mol/l * 0.006 l = 6 x 10⁻⁴ mol AgNO₃

n° of moles of KCl = 0.15 mol/l * 0.005 l = 7.5 x 10⁻⁴ mol KCl

Since 1 mol AgNO₃ reacts with 1 mol KCl, 6 x 10⁻⁴ mol AgNO₃ will react with 6 x 10⁻⁴ mol KCl and produce 6 x 10⁻⁴ mol AgCl.

1.5 x 10⁻⁴ mol KCl is in excess.

The molar mass of AgCl is 143.32 g/mol, then, 6 x 10⁻⁴ mol AgCl will have a mass of (6 x 10⁻⁴ mol AgCl * 143.32 g / 1 mol) 0.09 g.

Answer:

a) The two compounds you will expect in these solution are 1-aminobutane and its conjugate acid.

b) The greatest total amount is of 1-aminobutane.

c) The least total amount is of the conjugate acid.

Explanation:

The equilibrium in water of 1-aminobutane is:

CH₃(CH₂)₃NH₃⁺ ⇄ CH₃(CH₂)₃NH₂ + H⁺

a) The two compounds you will expect in these solution are 1-aminobutane and its conjugate acid.

b) The equlibrium constant is: K = 1,66x10⁻¹¹.

That means you will have 1-aminobutane:Conjugate acid in a ratio of 6x10¹⁰ : 1 .

The greatest total amount is of 1-aminobutane

c) Thus, The least total amount is of the conjugate acid.

I hope it helps!

Answer : The correct option is, (b) 20 kPa

Explanation :

The Raoult's law for liquid phase is:

[tex]p_A=x_A\times p^o_A[/tex]     .............(1)

where,

[tex]p_A[/tex] = partial vapor pressure of A

[tex]p^o_A[/tex] = vapor pressure of pure substance A

[tex]x_A[/tex] = mole fraction of A

The Raoult's law for vapor phase is:

[tex]p_A=y_A\times p_T[/tex]      .............(2)

where,

[tex]p_A[/tex] = partial vapor pressure of A

[tex]p_T[/tex] = total pressure of the mixture

[tex]y_A[/tex] = mole fraction of A

Now comparing equation 1 and 2, we get:

[tex]x_A\times p^o_A=y_A\times p_T[/tex]

[tex]p_T=\frac{x_A\times p^o_A}{y_A}[/tex]    ............(3)

First we have to calculate the total pressure of the mixture.

Given:

[tex]x_A=0.4[/tex] and [tex]x_B=1-x_A=1-0.4=0.6[/tex]

[tex]y_A=0.7[/tex] and [tex]y_B=1-y_A=1-0.7=0.3[/tex]

[tex]p^o_A=70kPa[/tex]

Now put all the given values in equation 3, we get:

[tex]p_T=\frac{0.4\times 70kPa}{0.7}=40kPa[/tex]

Now we have to calculate the vapor pressure of B.

Formula used :

[tex]x_B\times p^o_B=y_B\times p_T[/tex]

[tex]p^o_B=\frac{y_B\times p_T}{x_B}[/tex]

Now put all the given values in this formula, we get:

[tex]p^o_B=\frac{0.3\times 40kPa}{0.6}=20kPa[/tex]

Therefore, the vapor pressure of B is 20 kPa.

Answer:

Effectiveness and cold stream output temperature of the heat exchange Increases. So, Answer is b) Increases.

Explanation:

We have a heat exchanger, and it is required to compare the effectiveness and cold stream output if the length is increased.

Heat exchangers are engineering devices used to transfer energy. Thermal energy is transferred from Fluid 1 - Hot fluid (HF) to a Fluid 2 - Cold Fluid (CF). Both fluids 1 and 2 can flow with different values of mass flow rate and different specific heat. When the streams go inside the heat exchanger Temperature of Fluid 1 (HF) will decrease, at the same time Temperature of the Fluid 2 (CF) will increase.

In this case, we need to analyze the behavior taking into account different lengths of heat exchangers. If the length of the heat exchanger increases, it means the transfer area will increases. Heat transfer will increase if the transfer area increases. In this sense, the increasing length is the same than increase heat transfer.

If the heat transfer increases, it means Fluid 1 (HF) will reduce its temperature, and at the same time Fluid 2 (CF) will increase its temperature.

Finally, Answer is b) Effectiveness and cold stream output temperature increases when the length of the heat exchanger is increased.

Answer:

Ka = [tex]4.04 \times 10^{-11}[/tex]

Explanation:

Initial concentration of weak acid = [tex]4.5 \times 10^{-4}\ M[/tex]

pH = 6.87

[tex]pH = -log[H^+][/tex]

[tex][H^+]=10^{-pH}[/tex]

[tex][H^+]=10^{-6.87}=1.35 \times 10^{-7}\ M[/tex]

HA dissociated as:

[tex]HA \leftrightharpoons H^+ + A^{-}[/tex]

(0.00045 - x)    x     x

[HA] at equilibrium = (0.00045 - x) M

x = [tex]1.35 \times 10^{-7}\ M[/tex]

[tex]Ka = \frac{[H^+][A^{-}]}{[HA]}[/tex]

[tex]Ka = \frac{(1.35 \times 10^{-7})^2}{0.00045 - 0.000000135}[/tex]

0.000000135 <<< 0.00045

[tex]Therefore, Ka = \frac{(1.35 \times 10^{-7})^2}{0.00045 } = 4.04 \times 10^{-11}[/tex]

Answer:

0.8 mL.

Explanation:

You need to know that the equivalence in mcg to mg is 1000 mcg are 1 mg; so 500 mcg are 0.5 mg. Now you know that there are 0.5 mg per 2 mL, so if you divide all by 2, you will know that you have 0.25 mg per mL. Now you applied a rule of three: ((0.2 mg)(1 ml))/(.25 mg) = 0.8 mL. So the nurse need to administer 0.8 mL to provide 0.2 mg of dose.

Answer : The  value of [tex]\Delta G^o[/tex] for the reaction is -959.1 kJ

Explanation :

The given balanced chemical reaction is,

[tex]2H_2S(g)+3O_2(g)\rightarrow 2H_2O(g)+2SO_2(g)[/tex]

First we have to calculate the enthalpy of reaction [tex](\Delta H^o)[/tex].

[tex]\Delta H^o=H_f_{product}-H_f_{reactant}[/tex]

[tex]\Delta H^o=[n_{H_2O}\times \Delta H_f^0_{(H_2O)}+n_{SO_2}\times \Delta H_f^0_{(SO_2)}]-[n_{H_2S}\times \Delta H_f^0_{(H_2S)}+n_{O_2}\times \Delta H_f^0_{(O_2)}][/tex]

where,

[tex]\Delta H^o[/tex] = enthalpy of reaction = ?

n = number of moles

[tex]\Delta H_f^0[/tex] = standard enthalpy of formation

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

[tex]\Delta H^o=[2mole\times (-242kJ/mol)+2mole\times (-296.8kJ/mol)}]-[2mole\times (-21kJ/mol)+3mole\times (0kJ/mol)][/tex]

[tex]\Delta H^o=-1035.6kJ=-1035600J[/tex]

conversion used : (1 kJ = 1000 J)

Now we have to calculate the entropy of reaction [tex](\Delta S^o)[/tex].

[tex]\Delta S^o=S_f_{product}-S_f_{reactant}[/tex]

[tex]\Delta S^o=[n_{H_2O}\times \Delta S_f^0_{(H_2O)}+n_{SO_2}\times \Delta S_f^0_{(SO_2)}]-[n_{H_2S}\times \Delta S_f^0_{(H_2S)}+n_{O_2}\times \Delta S_f^0_{(O_2)}][/tex]

where,

[tex]\Delta S^o[/tex] = entropy of reaction = ?

n = number of moles

[tex]\Delta S_f^0[/tex] = standard entropy of formation

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

[tex]\Delta S^o=[2mole\times (189J/K.mol)+2mole\times (248J/K.mol)}]-[2mole\times (206J/K.mol)+3mole\times (205J/K.mol)][/tex]

[tex]\Delta S^o=-153J/K[/tex]

Now we have to calculate the Gibbs free energy of reaction [tex](\Delta G^o)[/tex].

As we know that,

[tex]\Delta G^o=\Delta H^o-T\Delta S^o[/tex]

At room temperature, the temperature is 500 K.

[tex]\Delta G^o=(-1035600J)-(500K\times -153J/K)[/tex]

[tex]\Delta G^o=-959100J=-959.1kJ[/tex]

Therefore, the value of [tex]\Delta G^o[/tex] for the reaction is -959.1 kJ

Answer:

Molality = 1.13 m

Explanation:

Molality is defined as the moles of the solute present in 1 kilogram of the solvent.

Given that:

Mass of [tex]CH_3OH[/tex] = 26.5 g

Molar mass of [tex]CH_3OH[/tex] = 32.04 g/mol

The formula for the calculation of moles is shown below:

[tex]moles = \frac{Mass\ taken}{Molar\ mass}[/tex]

Thus,

[tex]Moles= \frac{26.5\ g}{32.04\ g/mol}[/tex]

[tex]Moles\ of\ CH_3OH= 0.8271\ moles[/tex]

Mass of water = 735 g = 0.735 kg ( 1 g = 0.001 kg )

So, molality is:

[tex]m=\frac {0.8271\ moles}{0.735\ kg}[/tex]

Molality = 1.13 m

Explanation:

Rules for counting significant figures:

1) 101 cm - 99.1 cm = 1.9 cm ≈ 2.0 cm

1.9 cm ≈ 2.0 cm

There are two significant figures that is 2 and 0.

2) 0.5 atm + 9.8 atm = 10.3 atm ≈ 10.0 atm

There are three significant figures that is 3 , 0 and 0.

Answer:

1. Percentage by weight = 0.5023 = 50.23 %

2. molar fraction =0.153

Explanation:

We know that

Molar mass of HClO4 = 100.46 g/mol

So the mass of 5 Moles= 5 x 100.46

       Mass (m)= 5 x 100.46 = 502.3 g

Lets assume that aqueous solution of HClO4  and the density of solution is equal to density of water.

Given that concentration HClO4 is 5 M it means that it have 5 moles of HClO4 in 1000 ml.

We know that

Mass = density x volume

Mass of 1000 ml  solution = 1 x 1000 =1000     ( density = 1 gm/ml)

            m'=1000 g

1.

Percentage by weight = 502.3 /1000

Percentage by weight = 0.5023 = 50.23 %

2.

We know that

molar mass of water = 18 g/mol

mass of water in 1000 ml = 1000 - 502.3 g=497.9 g

So moles of water = 497.7 /18 mole

moles of water = 27.65 moles

So molar fraction = 5/(5+27.65)

molar fraction =0.153

The atomic mass of Cr53 can be calculated by establishing an equation that takes into account the atomic masses and abundances of all isotopes of Chromium. Solve the equation for 'x' (representing the atomic mass of Cr53) to find your answer.

The atomic mass of a given isotope of an element is determined by the weighted average of the masses of its isotopes, each multiplied by the abundance of that isotope. Since the atomic masses and abundances of Cr50, Cr52, and Cr54 are already provided, we only need to account for Cr53's mass. Using Chromium's atomic mass (51.9961 u), we can establish an equation to solve for Cr53's atomic mass.

Here's our equation:

51.9961 u = (49.9460 u * Cr50's abundance) + (51.9405 u * 0.8379) + (x * Cr53's abundance) + (53.9389 u * 0.0237)

First calculate the contribution of Cr50, Cr52, and Cr54 to the atomic weight, then subtract this from the total atomic weight (51.9961 u). Divide this value by the Cr53's abundance to get the atomic weight of Cr53.

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The atomic mass of the Cr53 isotope is approximately 23.83 u, calculated based on its natural abundance and atomic mass in the given chromium isotope mixture.

To determine the atomic mass of the Cr53 isotope, we can use the information provided about the natural abundances and atomic masses of the chromium isotopes.

Let's denote the natural abundance of Cr50 as x. Since the ratio of Cr50 to Cr53 is 1:0.4579, the natural abundance of Cr53 would be 0.4579x. The total natural abundance of Cr50 and Cr53 is 1, so we have the equation:

x + 0.4579x = 1

Solving for x, we find that x is approximately 0.6852.

Now, we can calculate the total contribution of Cr53 to the atomic mass:

Atomic mass of Cr53 = Natural abundance of Cr53 * Atomic mass of Cr53

Atomic mass of Cr53 = 0.4579 * 51.9961

Calculating this gives us the atomic mass of the Cr53 isotope.

Atomic mass of Cr53 = 23.8256 u

In summary, the atomic mass of the Cr53 isotope is approximately 23.83 u.

The mass of the insoluble solid is determined by first calculating the volume of the liquid solvent, then using that information to find the volume of the solid, and finally multiplying the volume of the solid by its density.

To determine the mass of the solid, we first calculate the volume of the liquid solvent. The volume of a substance can be calculated using the formula volume = mass/density. Thus, the volume of the liquid solvent is 26.6 g / 0.865 g/mL = approximately 30.8 mL.

Now, knowing that the total volume of the solid and liquid together is 91.0 mL, we can find out the volume of the solid as follows: 91.0 mL (total volume) - 30.8 mL (volume of the liquid) = 60.2 mL.

We can then use the given density of the solid (1.75 g/mL) to calculate its mass. According to the formula mass = volume x density, the mass of the solid is therefore 60.2 mL x 1.75 g/mL = 105.35g.

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Answer: The volume of bone for given sample is [tex]30.49cm^3[/tex]

Explanation:

To calculate volume of a substance, we use the equation:

[tex]\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}[/tex]

Density of bone = [tex]820mg/cm^3=0.82g/cm^3[/tex]      (Conversion factor:  1 g = 1000 mg)

Mass of bone = 25 g

Putting values in above equation, we get:

[tex]0.82g/cm^3=\frac{25g}{\text{Volume of bone}}\\\\\text{Volume of bone}=30.49cm^3[/tex]

Hence, the volume of bone for given sample is [tex]30.49cm^3[/tex]

Answer:

19.3 L

Explanation:

V= n × 22.4

where V is volume and n is moles

First, to find the moles of CO2, divide 38.0 by the molecular weight of CO2 which is 44.01

n= m/ MM

n= 38/ 44.01

n= 0.86344012724

V= 0.86344012724 × 22.4

V= 19.3410588502 L

V= 19.3 L

Explanation:

Nitrogen triiodide (NI₃)

Valence electrons of nitrogen = 5

Valence electrons of iodine = 7

The total number of the valence electrons  = 5 + 3(7) = 26

The Lewis structure is drawn in such a way that the octet of each atom and duet for the hydrogen in the molecule is complete. So,  

The Lewis structure is shown in image below.

Nitrogen posses one lone pair and thus the geometry is pyramidal.

The Lewis structure of NI3 is drawn by counting the valence electrons, creating single bonds between nitrogen and iodine, completing octets for iodine atoms, and placing remaining electrons on the nitrogen to complete its octet. This results in a structure where nitrogen is single-bonded to three iodine atoms, each surrounded by three lone pairs, and nitrogen has one lone pair.

To draw the Lewis structure for NI3 (nitrogen triiodide), follow these steps:

Count the total number of valence electrons. Nitrogen has 5 valence electrons, and each iodine has 7 valence electrons, totaling (5 + 3*7) = 26 valence electrons.

Draw a single bond between the nitrogen atom and each iodine atom. This will use up 6 of the valence electrons (2 for each bond).

Complete the octets for the iodine atoms by adding six more electrons to each iodine in the form of electron pairs, using up 18 of the remaining valence electrons.

Place any remaining electrons (2) on the nitrogen atom to complete its octet.

Examine the structure. Each iodine has 8 electrons, and nitrogen has 8 electrons, making the structure complete.

If you follow these steps, the resulting Lewis structure for NI3 will show a central nitrogen atom single-bonded to three iodine atoms, with each iodine atom surrounded by three lone pairs of electrons, and one lone pair of electrons on the nitrogen atom.

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Answer:

Vw = 2.80907 E-3 m³/Kg

Explanation:

specific volume is the inverse of density:

⇒ Vw = 1 / ρ

∴ water at:

⇒ P = 1400 KPa * ( 1000 Pa / KPa ) = 1400000 Pa

⇒ T = 200°C = 473 K

Water at these conditions is found as saturated steam, specific volume would be:

⇒ Vw = R*T / P

∴ R = 8.3144 Pa.m³/ Kg.K

⇒ Vw = (( 8.3144 ) * ( 473)) / 1400000

⇒ Vw = 2.80907 E-3 m³/Kg

Answer:

Number of contraction cycles that could theoretically be fueled by the complete combustion of one mole of glucose is around 17

Explanation:

Energy released during the complete combustion of 1 mole glucose = 2870 kJ

Energy required/muscle contraction cycle = 67 kJ/contraction

Energy conversion efficiency = 39%

Actual amount of energy converted to contraction per mole of glucose is:

[tex]=\frac{39}{100} *2870 kJ=1119.3 kJ[/tex]

Total contraction cycles fueled by the above energy is:

[tex]=\frac{1119.3\ kJ}{67\ kJ/contraction} =16.7\ i.e.\ around\ 17\ contractions[/tex]

Answer: Option (b) is the correct answer.

Explanation:

Sublimation is defined as the process in which a solid substance changes directly into gaseous phase without undergoing into liquid phase.

For example, when naphthalene balls are kept in a trunk or cupboard then after some days or months it changes into gas. Hence, they become consumed.

Also during this process energy is being absorbed by the solid substance because for breaking bonds energy is always absorbed by a substance. Only then it can change into liquid or vapor state.

So, a chemical process in which energy is being absorbed by a reaction is known as endothermic reaction. And, if heat energy is being released by the reaction then it is known as an exothermic reaction.

Therefore, we can conclude that when a solid sublimes, a solid converts to a gas. This phase change is endothermic because energy is absorbed when the intermolecular forces break.      

Explanation:

For an isothermal process equation will be as follows.

                W = nRT ln[tex]\frac{P_{1}}{P_{2}}[/tex]

It is given that mass is 10 kg/s or 10,000 g/s (as 1 kg = 1000 g). So, calculate number of moles of water as follows.

                    No. of moles = [tex]\frac{mass}{\text{molar mass}}[/tex]

                                           = [tex]\frac{10000 g/s}{18 g/mol}[/tex]

                                           = 555.55 mol/s

                                           = 556 mol/s (approx)

As T = [tex]50^{o}C[/tex] or (50 + 273.15) K = 323.15 K. Hence, putting the given values into the above formula as follows.

                  W = nRT ln[/tex]\frac{P_{1}}{P_{2}}[/tex]

                      = [tex]556 mol/s \times 8.314 J/ K mol K \times 323.15 K \times ln\frac{1}{10}[/tex]    

                     = [tex]556 mol/s \times 8.314 J/ K mol K \times 323.15 K \times -2.303[/tex]    

                     = -3440193.809 J/s

Negative sign shows work is done by the pump. Since, 1 J = 0.001 kJ. Therefore, converting the calculated value into kJ as follows.

                     [tex]3440193.809 J/s \times \frac{0.001 kJ}{1 J}[/tex]

                          = 3440.193 kJ/s

                          = 3451 kJ/s (approx)

Thus, we can conclude that the pump work is 3451 kJ/s.

Answer:

N₂ is the limiting reactant

Explanation:

The balanced reaction between N₂ gas and H₂ gas is:

N₂ + 3H₂ → 2NH₃

In order to determine the limiting reactant, we have to calculate the number of moles of each rectant, using their molecular weight:

Lastly, we multiply the number of moles of N₂ by 3, and the number of moles of H₂ by 1; due to the coefficients in the balanced reaction. Whichever number is lower, belongs to the limiting reactant.

N₂ => 10.7

H₂ => 50.0

Thus N₂ is the limiting reactant