Find the spring constant for a spring that stretches 1.4cm when a 638 g weight is attached to it. How much work is done as the spring extends?

Answers

Answer 1

Answer:

Spring constant: [tex]\rm 447\; N \cdot m^{-1}[/tex].

Work done: [tex]\rm 0.0438\; J[/tex].

Explanation:

Convert all values to SI units.

Length change: [tex]\rm 1.4 \; cm = 0.014\; m[/tex];Mass of the weight: [tex]\rm 638\; g = 0.638\; kg[/tex].

Assume that the spring-mass system is vertical and is placed on the surface of the earth. The gravitational acceleration constant will be equal to [tex]\rm 9.81\; N\cdot kg^{-1}[/tex].

Gravitational pull on the weight:

[tex]W = m\cdot g = \rm 0.638\; kg \times 9.81\; N\cdot kg^{-1} = 6.25878\; N[/tex].

That's also the size of the force on the spring. [tex]F = \rm 6.25878\; N[/tex].

The spring constant is the size of the force required to deshape the spring (by stretching, in this case) by unit length.

[tex]\displaystyle k_{\text{spring}} = \frac{F}{\Delta x} = \rm \frac{6.25878\; N}{0.014\; m} = 447.056\; N\cdot m^{-1}[/tex].

Assume that there's no energy loss in this process. The work done on the spring is the same as the elastic potential energy that it gains:

[tex]\begin{aligned} &\text{EPE} \\=& \frac{1}{2}k\cdot x^{2} \\=&\rm \frac{1}{2} \times 447.056\; N\cdot m^{-1} \times (0.014\; m)^{2}\\ =& \rm 0.0438\; J\end{aligned}[/tex].


Related Questions

An organic chemist measures the temperature T of a solution in a reaction flask. Here is the result. T = 149.206 °C Convert T to Sl units. Round your answer to 3 decimal places.

Answers

If an organic chemist measures the temperature T of a solution in a reaction flask. The temperature in SI units is approximately 422.356 K.

What is the temperature in SI units?

To convert temperature from Celsius (°C) to SI units (Kelvin, K),  lets make use  of the following formula:

K = °C + 273.15

Given that T = 149.206 °C, we can calculate the temperature in Kelvin:

K = 149.206 + 273.15

= 422.356 K

Therefore the temperature in SI units is approximately 422.356 K.

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The temperature T in SI units is 422.356 Kelvin (K).

Temperature is a measure of the average kinetic energy of the particles in a substance. It represents how hot or cold an object or environment is.

To convert the temperature T from degrees Celsius (°C) to SI units, we need to use the Kelvin scale. The Kelvin scale is an absolute temperature scale where 0 Kelvin (0 K) represents absolute zero, the lowest possible temperature.

Calculation:

T = 149.206 °C

T (in Kelvin) = T (in °C) + 273.15

T (in Kelvin) = 149.206 °C + 273.15

T (in Kelvin) = 422.356 K.

Therefore, the temperature T in SI units is 422.356 Kelvin (K).

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Consider the gas reaction: H2 (g)2(g) 2HI (g) The equilibrium constant at 731 K is 50.3. Equal amounts of all three gases (0.100 M) are introduced in a container, calculate the concentration of each gas after the system reaches equilibrium. Express your results with the right number of significant figures.

Answers

Answer : The concentration of [tex]H_2,I_2[/tex] and [tex]HI[/tex] at equilibrium 0.033 M, 0.033 M and 0.234 M respectively.

Solution :  Given,

[tex]K_c=50.3[/tex]

Concentration = 0.100 M

The given equilibrium reaction is,

                              [tex]H_2(g)+I_2(g)\rightleftharpoons 2HI(g)[/tex]

Initially conc.       0.100     0.100       0.100

At equilibrium  (0.100-x)   (0.100-x)  (0.100+2x)

The expression of [tex]K_c[/tex] will be,

[tex]K=\frac{[HI]^2}{[H_2][I_2]}[/tex]

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

[tex]50.3=\frac{(0.100+2x)^2}{(0.100-x)\times (0.100-x)}[/tex]

By solving the term x, we get

[tex]x=0.067\text{ and }0.158[/tex]

From the values of 'x' we conclude that, x = 0.158 can not more than initial concentration. So, the value of 'x' which is equal to 0.158 is not consider.

Thus, the concentration of [tex]H_2[/tex] and [tex]I_2[/tex] at equilibrium = (0.100-x) = 0.100 - 0.067 = 0.033 M

The concentration of [tex]HI[/tex] at equilibrium = (0.100+2x) = 0.100 + 2(0.067) = 0.234 M

The metal Cadmium tends to form Cd2+ ions. The following observations are made: i) When a strip of zinc metal is placed in Cd2+(aq), cadmium metal is deposited onto the strip, ii) When a strip cadmium metal is placed into Ni2+(aq), nickel metal is deposited onto the strip; A) Write net-ionic equations to explain each of these observations. B) What can you conclude about the position of cadmium in the activity series relative to zinc and nickel?

Answers

A)

The first reaction can be represented by the following equation:

i) Zn(s) + Cd²⁺(aq) → Zn²⁺(aq) + Cd(s)

The second reaction can be represented by the following equation:

ii) Cd(s) + Ni²⁺(aq) → Cd²⁺(aq) + Ni(s)

B) The activity series is a tool used to predict displacement reactions. More reactive metals (above in the series) can displace less reactive ones (below in the series) from the salts they are part of in solutions.

In i) we can see zinc displacing cadmium. Therefore, cadmium is below zinc in the activity series.

In  ii) we can see cadmium displacing nickel. Thus, cadmium is above nickel in the activity series.

Final answer:

The student's question involves writing net-ionic equations for redox reactions with cadmium and determining its position in the activity series. Zinc can replace cadmium ions, and cadmium can replace nickel ions, indicating that cadmium is less reactive than zinc but more reactive than nickel.

Explanation:

The student is asking about redox reactions involving the metal cadmium, zinc, and nickel, which can be explained using net-ionic equations and the activity series of metals.

A) Net-ionic equations:

B) Conclusion about cadmium's position:

Based on the observations, cadmium is less reactive than zinc but more reactive than nickel within the activity series. This is deduced because zinc can replace cadmium ions and cadmium can replace nickel ions in solution, indicating a middle position for cadmium in the series between zinc and nickel.

a Draw a Lewis structure for CH_4 Explicitly draw all H atoms. Include all valence lone pairs in your answer. Include all nonzero formal charges. C opy P aste H ** H н. H H

Answers

Answer :  The Lewis-dot structure of [tex]CH_4[/tex] is shown below.

Explanation :

Lewis-dot structure : It shows the bonding between the atoms of a molecule and it also shows the unpaired electrons present in the molecule.

In the Lewis-dot structure the valance electrons are shown by 'dot'.

The given molecule is, [tex]CH_4[/tex]

As we know that carbon has '4' valence electrons and hydrogen has '1' valence electron.

Therefore, the total number of valence electrons in [tex]CH_4[/tex] = 4 + 4(1) = 8

According to Lewis-dot structure, there are 8 number of bonding electrons and 0 number of non-bonding electrons.

Now we have to determine the formal charge for each atom.

Formula for formal charge :

[tex]\text{Formal charge}=\text{Valence electrons}-\text{Non-bonding electrons}-\frac{\text{Bonding electrons}}{2}[/tex]

[tex]\text{Formal charge on C}=4-0-\frac{8}{2}=0[/tex]

[tex]\text{Formal charge on }H_1=1-0-\frac{2}{2}=0[/tex]

[tex]\text{Formal charge on }H_2=1-0-\frac{2}{2}=0[/tex]

[tex]\text{Formal charge on }H_3=1-0-\frac{2}{2}=0[/tex]

[tex]\text{Formal charge on }H_4=1-0-\frac{2}{2}=0[/tex]

Hence, the Lewis-dot structure of [tex]CH_4[/tex] is shown below.

The factor 0.01 corresponds to which prefix? A) milli B) deci C) deka D) centi

Answers

The answer is: D) centi

The factor 0.01 corresponds to the prefix D. 'centi'.

In the metric system, prefixes are used to denote different multiples or fractions of a base unit. Here's a breakdown of the prefixes relevant to this question:

milli-: This prefix represents the factor $10⁻³ $ or 0.001.

deci-: This prefix represents the factor $10⁻¹ $ or 0.1.

deka-: This prefix represents the factor $10¹ or 10.

centi-: This prefix represents the factor $10⁻² $ or 0.01.

How many gram-moles of formic acid, CH2O2, are in 225 g of the compound? Numbe g-mol CH,0, How many pound-moles of formic acid are in 225 g of the compound? Number Ib-moles CH,O,

Answers

Explanation:

(a)     Mass given = 225 g

        Molecular mass of [tex]CH_{2}O_{2}[/tex] = [tex]12 + 2 + 16 \times 2[/tex] = 12 + 2 + 32 = 46 g/mol

As, it is known that number of moles is equal to the mass divided by molar mass.

Mathematically,        No. of moles = [tex]\frac{mass}{\text{molecular mass}}[/tex]

Hence, calculate the number of moles of formic acid as follows.

                     No. of moles = [tex]\frac{225 g}{46 g/mol}[/tex]

                                           = 4.89 mol or g mol

Hence, in 225 g of the compound there are 4.89 mol or g mol.

(b)   It is known that 1 g = 0.0022 pound

So, pounds present in 4.89 g will be calculated as follows.

                       4.89 \times 0.0022 pound

                     = 0.0107 pound mole

Hence, in 225 g of the compound there are  0.0107 pound mole.

A mixture of methanol and methyl acetate contains 15.0 weight percent methanol. Determine the number of gmols of methanol in 110.0 kilograms of the mixture.

Answers

Answer:

There are 550.5 moles  of methanol in 110.0 kilograms of the mixture.

Explanation:

A solution 15% weight of methanol means there is 15g of methanol per 100g of the mixture or 0.1kg of the mixture. Also, the molar mass of methanol (CH3OH) is:

[tex]m_{C} + 4xm_{H} + m_{O} = 12.0g/mol + 4x1.0g/mol +  16.0g/mol = 32.0g/mol[/tex]

Thus, dividing 15g by molar mass

[tex]15.0g / 32.0\frac{g}{mol} = 0.5moles[/tex] we find there is 0.5 moles of methanol per 0.1Kg of the mixture. Calculating the number of mols of methanol in 110.0 kilograms of the mixture:

[tex]\frac{110.1Kgx0.5moles}{0.1Kg} = 550.5 moles[/tex]

Therefore, there is 550.5 moles  of methanol in 110.0 kilograms of the mixture.


A house has an area of 234 m2. What is its area in each unit?

A)km2

B)dm2

C)cm2

Answers

Answer:

a) [tex]0.000324 km^2[/tex]

b) [tex]32400 dm^2[/tex]

c) [tex]3.24x10^6 cm^2[/tex]

Explanation:

To do the different conversions we need to know the follow:

[tex]1 km->1000 m so 1 km^2-> 1000^2 m^2= 1x10^6 m^2[/tex]

[tex]324 m^2*(1 km^2/1x10^6 m^2) = 0.000324 km^2[/tex]

[tex]1m-> 10 dm so 1m^2->10^2 dm^2=100 dm^2[/tex]

[tex]324 m^2*(100 dm^2/1 m^2)=32400 dm^2[/tex]

[tex]1m-> 100 cm so 1m^2->100^2 cm^2=10000 cm^2[/tex]

[tex]324 m^2*(10000 cm^2/1 m^2)=3.24x10^6 cm^2[/tex]

Answer :

(A) [tex]234m^2=2.34\times 10^8km^2[/tex]

(B)  [tex]234m^2=2.34\times 10^4dm^2[/tex]

(C) [tex]234m^2=2.34\times 10^6cm^2[/tex]

Explanation :

The conversion used for area from [tex]m^2[/tex] to [tex]km^2[/tex] is:

[tex]1m^2=10^6km^2[/tex]

The conversion used for area from [tex]m^2[/tex] to [tex]dm^2[/tex] is:

[tex]1m^2=100dm^2[/tex]

The conversion used for area from [tex]m^2[/tex] to [tex]cm^2[/tex] is:

[tex]1m^2=10000cm^2[/tex]

Part (A):

As, [tex]1m^2=10^6km^2[/tex]

So, [tex]234m^2=\frac{234m^2}{1m^2}\times 10^6km^2=2.34\times 10^8km^2[/tex]

Part (B):

As, [tex]1m^2=100dm^2[/tex]

So, [tex]234m^2=\frac{234m^2}{1m^2}\times 100dm^2=2.34\times 10^4dm^2[/tex]

Part (C):

As, [tex]1m^2=10000cm^2[/tex]

So, [tex]234m^2=\frac{234m^2}{1m^2}\times 10000cm^2=2.34\times 10^6cm^2[/tex]

A woman informs her engineer husband that "hot water will freeze faster than cold water." He calls this statement nonsense but he does so in a polite sensitive manner. She answers by saying that she has actually timed the freezing process for ice trays in the home refrigerator and found that hot water does indeed freeze faster. As a friend, you are asked to settle the argument and preserve family harmony. Is there any logical explanation for the woman's observation

Answers

Answer:

Explanation:

It could be as a result of evaporation from the hot water causing it to have less mass than usual. Evaporation will allow the volume of the hot water to decrease and by so doing will cool faster than cold water which is at a higher mass. This is how I will explain my observations to them.

Final answer:

The concept of hot water freezing faster than cold water is known as the Mpemba effect and is influenced by various factors including evaporation and thermal energy transfer. Specific heat capacity plays a role in cooling processes, affecting the rate at which materials like metal are cooled when in contact with ice or water.

Explanation:

The observation that hot water freezes faster than cold water under certain conditions has been noted historically and is often referred to as the Mpemba effect. While it may seem counterintuitive, various experiments have shown that hot water can indeed freeze more quickly than cold water due to a number of factors, including evaporation (which reduces the mass of hot water), thermodynamic properties of water (such as specific heat capacity and the behavior of water molecules at different temperatures), and the environment in which the water is being frozen (like the surrounding air temperature and properties of the freezing surface).

Regarding the specific scenarios described, mixing hot and cold water or the implications of adding cold water to a hot liquid like tea or coffee, the result is affected by the thermal energy transfer between the substances. The outcome is influenced by the specific heat of the materials, the current temperatures, and the mass of the substances involved. For instance, pouring hot water into cold water would result in heat transfer from the hot water to both the cold water and its surroundings.

As for the question of cooling a hot piece of metal, the method of cooling (using ice versus cold water) will affect the rate at which the metal cools down, with the specific heat capacities playing a crucial role. Water generally has a higher specific heat capacity than ice, making it more efficient at absorbing heat.

Cyclopropane, C3H6, is used as a general anesthetic. If a sample of cyclopropane stored in a 2.36-L container at 10.0 atm and 25.0°C is transferred to a 7.79-L container at 5.56 atm, what is the resulting temperature? Enter your answer in the provided box

Answers

Explanation:

The given data is as follows.

     [tex]V_{1}[/tex] = 2.36 L,    [tex]T_{1}[/tex] = [tex]25^{o}C[/tex] = (25 + 273) K = 298 K,

      [tex]P_{1}[/tex] = 10.0 atm,   [tex]V_{2}[/tex] = 7.79 L,

       [tex]T_{2}[/tex] = ?,       [tex]P_{2}[/tex] = 5.56 atm  

And, according to ideal gas equation,  

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

Hence, putting the given values into the above formula to calculate the value of final temperature as follows.          

 [tex]\frac{10.0 atm \times 2.36 L}{298 K} = \frac{5.56 atm \times 7.79 L}{T_{2}}[/tex]

            [tex]T_{2}[/tex] = [tex]\frac{43.3124}{0.0792}[/tex] K

                     = 546.87 K

Thus, we can conclude that the final temperature is 546.87 K.

Final answer:

The resulting temperature when transferring cyclopropane from one container to another is approximately 12.4 °C.

Explanation:

To find the resulting temperature when transferring cyclopropane from one container to another, we can use the ideal gas law: PV = nRT. We are given the initial and final pressures and volumes, and we need to find the final temperature.

First, we can calculate the initial number of moles using the ideal gas law: n1 = (P1 * V1) / (R * T1). Then, we can use this number of moles to calculate the final temperature using the ideal gas law: T2 = (P2 * V2) / (n1 * R).

Substituting the given values, we have: T2 = (5.56 atm * 7.79 L) / ((10.0 atm * 2.36 L) / (0.0821 atm L/mol K)). Solving this equation, we find that the resulting temperature is approximately 12.4 °C.

Consider two concentric spheres whose radii differ by t = 1 nm, the "thickness" of the interface. At what radius (in nm) of sphere does the volume of a shell equal that of the interior sphere? Assume the shell thickness to be t = 1 nm.

Answers

Answer:

Radius of the interior sphere = 3.847 nm

Explanation:

The volume of the shell (Vs) is equal to the difference of the volume of the outer sphere (Vo) and the volume of the inner sphere (Vi). Then:

[tex]V_s=V_o-V_i=V_i\\V_o=2*V_i\\[/tex]

If we express the radius of the outer sphere (ro) in function of the radius of the inner sphere (ri), we have (e being the shell thickness):

[tex]r_o=r_i+e[/tex]

The first equation becomes

[tex]\frac{4 \pi}{3}*r_o^{3}  = 2 * \frac{4 \pi}{3}*r_i^{3}  \\\\\frac{4 \pi}{3}*(r_i+e)^{3}  = 2 * \frac{4 \pi}{3}*r_i^{3}  \\\\(r_i+e)^{3}  = 2 *r_i^{3}  \\\\(r_i^{3}+3r_i^{2}e+3r_ie^{2}+e^{3})=2r_i^{3}\\\\-r_i^{3}+3r_i^{2}e+3r_ie^{2}+e^{3}=0\\\\-r_i^{3}+3r_i^{2}+3r_i+1=0[/tex]

To find ri that satisfies this equation we have to find the roots of the polynomial.

Numerically, it could be calculated that ri=3.847 nm satisfies the equation.

So if the radius of the interior sphere is 3.847 nm, the volume of the interior sphere is equal to the volume of the shell of 1nm.

Pyridine is a conjugate base which reacts with H to form pyridine hydrochloride. The hydrochloride dissociates to yield H' with a pKof 5.36. Describe the preparation of a pyridine buffer at pH 5.2 starting with 0.1M pyridine and 1.OM HCI. You may start with one liter of the 0.1 M pyridine.

Answers

Answer:

To one liter of the 0.1 M pyridine you need to add 41 mL of 1,0M HCl to obtain a buffer at 5,2

Explanation:

The reaction is:

pyridine-H⁺ ⇄ pyridine + H⁺ pka = 5,36; k = [tex]10^{-5,36}[/tex]

Using Henderson-Hasselbalch formula:

5,2 = 5,36 + log[tex]\frac{[Py-H^+]}{[Py]}[/tex]

0,692 = [tex]\frac{[Py-H^+]}{[Py]}[/tex] (1)

As total intial moles are 0,1:

0,1 = Py-H⁺ moles + Py moles (2)

Replacing (2) in (1) final moles of both Py-H⁺ and Py are:

Py: 0,059 moles

Thus:

Py-H⁺: 0,041 moles

Moles in reaction are:

Py: 0,1-x moles

H⁺: Y-x moles Y are initial moles of H⁺

Py-H⁺: x moles

Knowing x = 0,041 moles, pyridine volume is 1L and HCl molarity is 1 mol/L and [H⁺] = [tex]10^{-5,2}[/tex]

[tex]10^{-5,2}[/tex] = [tex]\frac{Y-0,041moles}{Y+1L}[/tex]

Y = 0,04100605 moles≡ 41 mL of 1,0M HCl

I hope it helps!

What is the name of the engineer who investigated the efficiency of steam engines? Michael Faraday O Frederic Tudor OSadi Carnot O Robert Boyle

Answers

Answer:

The correct option is: Sadi Carnot

Explanation:

Steam engine is the machine that converts the energy of the steam to mechanical energy in order to perform mechanical work. It is a type of of heat engine.

The efficiency of a steam engine was first given by the French military scientist and physicist, Nicolas Léonard Sadi Carnot. The efficiency of steam engine is the ratio of the output energy of mechanical work and the input energy put provided to engine by burning fuel.

You have ice cubes in your freezer at home. The ice cubes melt into water. This type of change is called a... O a. nuclear change O b.physical change O c. chemical change

Answers

b ) its a physical change

Calculate the number of mol of helium in a 2.01-L balloon at 29°C and 2.71 atm of pressure. Be sure to answer all parts. Imol

Answers

Answer:

0.220 mol He

Explanation:

Hello,

Based on the ideal gas equation of state:

[tex]PV=nRT\\n=\frac{PV}{RT} \\n=\frac{2.71 atm * 2.01 L}{0.082\frac{atm*L}{mol*K}*302K } \\n=0.220 mol[/tex]

Best regards.

Which of the following is not a postulate of the kinetic molecular theory? Select one: a. Gas particles have most of their mass concentrated in the nucleus of the atom. b. The moving particles undergo perfectly elastic collisions with the walls of the container. c. The forces of attraction and repulsion between the particles are insignificant. d. The average kinetic energy of the particles is directly proportional to the absolute temperature. e. All of the above are postulates of the kinetic molecular theory.

Answers

Answer:

a. Gas particles have most of their mass concentrated in the nucleus of the atom

Explanation:

The main postulates of the molecular kinetic theory are:

A gas consists of a set of small particles that move with rectilinear motion and obey Newton's laws (this means that this theory doesn't divide the atom in a nucleus and electrons, this is why a. is not correct)

The molecules of a gas do not occupy volume (this is assumed because between the molecules of a gas is a lot of space).

Collisions between the molecules are perfectly elastic (this means that energy is not gained or lost during the crash).

There're no forces of attraction or repulsion between the molecules (because each molecule is "far" of the others).

The average kinetic energy of a molecule is [tex]E_k=\frac{3}{2} \times k\times T[/tex] (where T is the absolute temperature and k the Boltzmann constant. This means that kinetic energy is directly proportional to temperature).

Final answer:

The kinetic molecular theory of gases has several postulates that describe the behavior of gas particles.

Explanation:

The kinetic molecular theory of gases has several postulates that describe the behavior of gas particles. The postulates are as follows:

Gases consist of very large numbers of tiny spherical particles that are far apart from one another compared to their size. The particles of a gas may be either atoms or molecules.Gas particles are in constant rapid motion in random directions. The fast motion of gas particles gives them a relatively large amount of kinetic energy.Collisions between gas particles and between particles and the container walls are elastic collisions. An elastic collision is one in which there is no overall loss of kinetic energy.There are no forces of attraction or repulsion between gas particles. It is assumed that the particles of an ideal gas have no such attractive forces.The average kinetic energy of gas particles is dependent upon the temperature of the gas. As the temperature of a sample of gas is increased, the speeds of the particles are increased, resulting in an increase in the kinetic energy of the particles.

Based on these postulates, the answer to the question is:

a. Gas particles have most of their mass concentrated in the nucleus of the atom


Calculate the answer to the following expression to the correct number of significant figures.

4 X 10^8/2.0 X 10^-5

Answers

Answer :  The answer will be [tex]\Rightarrow 2\times 10^{13}[/tex]

Explanation :

Significant figures : The figures in a number which express the value -the magnitude of a quantity to a specific degree of accuracy is known as significant digits.

The rule apply for the multiplication and division is :

The least number of significant figures in any number of the problem determines the number of significant figures in the answer.

As we are given the expression :

[tex]\frac{4\times 10^8}{2.0\times 10^{-5}}[/tex]

[tex]\Rightarrow 2\times 10^{13}[/tex]

In the given expression, [tex]4\times 10^8[/tex] has 1 significant figure and [tex]2.0\times 10^{-5}[/tex] has 2 significant figures. From this we conclude that 1 is the least significant figures in this problem. So, the answer should be in 1 significant figures.

Thus, the answer will be [tex]\Rightarrow 2\times 10^{13}[/tex]

Two pipes with 1 meter diameters join to become one pipe with 2 meter diameter. Water was flowing in first pipe at 500 kg/s and oil was flowing in the second pipe with average velocity of 0.8 m/s. Oil mass density is 1500 kg/s. What will be the mass flow rate at the exit of the 2 meter diameter pipe?

Answers

Answer:

The mass flow rate at the exit of the 2 meter diameter pipe is 1442 kg/s

Explanation:

If the oil is flowing in the 1m diameter pipe at 0.8 m/s, we can calculate the flow as

[tex]Q=\bar{v}*A=\bar{v}*(\pi/4*D^{2} )=0.8m/s*0.785m^{2} =0.628m^{3}/s[/tex]

The mass flow is

[tex]M=\rho*Q=1500 kg/m3*0.628m3/s=942kg/s[/tex]

The mass flow rate at the exit of the 2 meter diameter pipe is

[tex]M=M_w+M_o=500 kg/s+942kg/s=1442kg/s[/tex]

A runner runs 4339 ft in 7.45 minutes. What is the runner's average speed in miles per hou 6.62 mi/hr O 0.110 mi/hr 6.618 mi/h e 582 mi/hr

Answers

Answer: The average speed of the runner is 6.618 miles/hr

Explanation:

Average speed is defined as the ratio of total distance traveled to the total time taken.

To calculate the average speed of the runner, we use the equation:

[tex]\text{Average speed}=\frac{\text{Total distance traveled}}{\text{Total time taken}}[/tex]

We are given:

Distance traveled = 4339 ft

Time taken = 7.45 mins

Putting values in above equation, we get:

[tex]\text{Average speed of runner}=\frac{4330ft}{7.45min}=582.42ft/min[/tex]

To convert the speed into miles per hour, we use the conversion factors:

1 mile = 5280 ft

1 hr = 60 mins

Converting the speed into miles per hour, we get:

[tex]\Rightarrow \frac{528.42ft}{min}\times (\frac{1miles}{5280ft})\times (\frac{60min}{1hr})\\\\\Rightarrow 6.618mil/hr[/tex]

Hence, the average speed of the runner is 6.618 miles/hr

An astronaut uses a laboratory balance and weighs an object on earth and again on the moon Which statement below about the weight and mass of the object is true? A) The mass and weight will be identical on the earth and the moon. B) The mass will be the same on earth and moon but the weight will be less on the moon C) Both the mass and weight will be different on earth and moon. D) The weight will be the same on earth and moon but the mass will be less on the moon.

Answers

Answer:

B) The mass will be the same on earth and moon but the weight will be less on the moon

Explanation:

Mass of a substance is a intrinsic property. It is same throughout the universe.  On the other hand,

Weight = Mass × Gravitational acceleration

Weight depends on the gravitational acceleration of the planet on which the substance is present.

Thus, the astronaut weight on Moon will be approximately six times less than the weight on Earth as gravitational acceleration on Moon is six times less than the gravitational acceleration on Earth the but number of the atoms in body of the astronaut has not changed and thus, mass is same same at two places.

Convert 3.15 x 10 m to the equivalent length in nanometers. 3.15 x 10-ºm =

Answers

Answer:

∴ 3.15 * 10 m = 3.15 E10 nm

Explanation:

1 m ≡ 1 E9 nm

⇒ 3.15 * 10 m = 31.5 m * ( 1 E9 nm /m ) = 3.15 E10 nm

A 25 liter gas cylinder contains 10 mole of carbon dioxide. Calculate the pressure, in atm, of the gas at 325K.

Answers

Answer: The pressure of carbon dioxide gas is 11 atm

Explanation:

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

PV = nRT

where,

P = pressure of the gas = ?

V = Volume of gas = 25 L

n = number of moles of gas = 10 mole

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

T = temperature of the gas = 325 K

Putting values in above equation, we get:

[tex]P\times 25L=10mol\times 0.0821\text{ L atm }mol^{-1}K^{-1}\times 325K\\\\P=11atm[/tex]

Hence, the pressure of carbon dioxide gas is 11 atm

There is inverse relation between pressure and diffusion in liquid phase?
•True
•False

Answers

Answer:

True

Explanation:

There is inverse relation between pressure and diffusion in liquid phase is true statement because ,with the increase in pressure of the liquid phase, the material in the pores of the find it difficult to penetrate and therefore it is difficult for them to diffuse. Hence, there is an inverse relation between pressure and diffusion.

A liquid mixture of density 0.85 g/mL flowing at a rate of 700 lbm/h is formed from mixing a stream of benzene S = 0.879 and another of n-hexane S = 0.659. Determine the volumetric flow rates of all three streams as well as the mass flow rate of benzene and n-hexane.

Answers

Answer:

Volumetric flow:

Mixture = 373.55 l/h; Benzene = 310.05 l/h; n-hexane = 63.5 l/h

Mass flow

Mixture = 317.52 kg/h; Benzene = 275.66 kg/h; n-hexane = 41.85 kg/h

Explanation:

First, we can express the mixture mass flow as

[tex]M=700\frac{lbm}{h}*\frac{453.6g}{1lbm} = 317,520g/h=317,52kg/h[/tex]

The volumetric flow of the mixture is

[tex]V=M/\rho = \frac{317520g/h}{0.85g/ml} =373,553ml/h=373.55l/h[/tex]

The mass balance can be written as

[tex]M_1+M_2=M\\[/tex]

And the volumetric flow as

[tex]V_1+V_2=V\\\\\rho_1*M_1+\rho_2*M_2=\rho*M\\\\\rho_1(M-M_2)+\rho_2*M_2=\rho*M\\\\(\rho_2-\rho_1)*M_2=(\rho-\rho_1)*M\\\\M_2=\frac{(\rho-\rho_1)}{(\rho_2-\rho_1)} *M[/tex]

If fluid 2 is n-hexane, we have

[tex]M_2=\frac{(\rho-\rho_1)}{(\rho_2-\rho_1)} *M\\\\M_2=\frac{(0.85-0.879)}{(0.659-0.879)} *317.52kg/h\\\\M_2=0.1318*317.52kg/h=41.85kg/h[/tex]

The mass flow of n-hexane is 41.85 kg/h.

Its volumetric flow is 63.5 litre/h

[tex]V=M/\rho=41.85\frac{kg}{h}*\frac{1}{0.659g/ml} *\frac{1000g}{1kg}\\\\V = 63505 ml/h=63.5 l/h[/tex]

The mass flow of benzene can be deducted by difference:

[tex]M_1= M-M_2= 317.52-41.85=275.66 kg/h\\\\V_1=V-V_2= 373.55 - 63.5=310.05 l/h[/tex]

267 moles of NADH are produced by the Citric Acid Cycle (CAC). How many moles of glucose needed to be broken down during glycolysis in order to produce enough pyruvate for the CAC to produce this many moles of NADH?

Answers

Answer:

45 moles

Explanation:

From glycolysis, 1 mole of glucose gives 2 moles of pyruvate which undergoes citric acid cycle.

1 mole of pyruvate undergoes citric acid cycle (After conversion to acetyl-CoA) gives 3 moles of NADH.

Also,

2 moles of pyruvate undergoes citric acid cycle (After conversion to acetyl-CoA) gives 6 moles of NADH.

Thus,

6 moles of NADH are produced from 2 moles of pyruvate or 1 mole of glucose.

1 mole of NADH is produced from 1/6 mole of glucose

267 moles of NADH are produced from [tex]\frac {1}{6}\times 267[/tex] moles of glucose.

Thus, moles of glucose needed to be broken ≅ 45 moles

The chemical formula for sucrose (also known as table sugar) is C12H22011. What is the molar mass of sucrose? O a. 144 g/mole O b. 166 g/mole O c. 22 g/mole O d. 29 g/mole O e. 342 g/mole

Answers

e ) it must be 342 g/mol

Two moles of a monatomic ideal gas are contained at a pressure of 1 atm and a temperature of 300 K; 34,166 J of heat are transferred to the gas, as a result of which the gas expands and does 1216 J of work against its surroundings. The process is reversible. Calculate the final temperature of the gas.

Answers

Answer:

Final temperature is 302 K

Explanation:

You can now initial volume with ideal gas law, thus:

V = [tex]\frac{n.R.T}{P}[/tex]

Where:

n are moles: 2 moles

R is gas constant: 0,082 [tex]\frac{atm.L}{mol.K}[/tex]

T is temperature: 300 K

P is pressure: 1 atm

V is volume, with these values: 49,2 L

The work in the expansion of the gas, W, is: 1216 J - 34166 J = -32950 J

This work is:

W = P (Vf- Vi)

Where P is constant pressure, 1 atm

And Vf and Vi are final and initial volume in the expansion

-32950 J = -1 atm (Vf-49,2L) × [tex]\frac{101325 J}{1 atm.L}[/tex]

Solving: Vf = 49,52 L

Thus, final temperature could be obtained from ideal gas law, again:

T = [tex]\frac{P.V}{n.R}[/tex]

Where:

n are moles: 2 moles

R is gas constant: 0,082 [tex]\frac{atm.L}{mol.K}[/tex]

P is pressure: 1 atm

V is volume: 49,52 L

T is final temperature: 302 K

I hope it helps!

using the thermodynamic information in the ALEKS Data tab, calculate the standard reaction entropy of the following chemical reaction:
P4O10 + 6H2O → 4H3PO4
Round your answer to zero decimal places.

Answers

Answer: The value of [tex]\Delta S^o[/tex] for the reaction is -206 J/K

Explanation:

Entropy change is defined as the difference in entropy of all the product and the reactants each multiplied with their respective number of moles.

The equation used to calculate entropy change of a reaction is:

[tex]\Delta S^o_{rxn}=\sum [n\times \Delta S^o_{(product)}]-\sum [n\times \Delta S^o_{(reactant)}][/tex]

For the given chemical reaction:

[tex]P_4H_{10}(s)+6H_2O(l)\rightarrow 4H_3PO_4(s)[/tex]

The equation for the entropy change of the above reaction is:

[tex]\Delta S^o_{rxn}=[(4\times \Delta S^o_{(H_3PO_4)})]-[(1\times \Delta S^o_{(P_4O_{10})})+(6\times \Delta S^o_{(H_2O)})][/tex]

We are given:

[tex]\Delta S^o_{(H_3PO_4)}=110.5J/K.mol\\\Delta S^o_{(H_2O(l))}=69.91J/K.mol\\\Delta S^o_{(P_4H_{10})}=228.86J/K.mol[/tex]

Putting values in above equation, we get:

[tex]\Delta S^o_{rxn}=[(4\times (110.5))]-[(1\times (228.86))+(6\times (69.91))]\\\\\Delta S^o_{rxn}=-206.32J/K=-206J/K[/tex]

Hence, the value of [tex]\Delta S^o[/tex] for the reaction is -206 J/K

Final answer:

To calculate the standard reaction entropy, ΔS°, find the Standard Molar Entropies of the products and reactants from the ALEKS Data tab, multiply them by their respective coefficients, and then subtract the sum of products from the sum of reactants. The result should be rounded to zero decimal places.


Explanation:The standard reaction entropy (ΔS°) can be calculated by subtracting the sum of the entropy values of reactants from the sum of entropy values of products, as per the formula: ΔS° = ΣS°(products) - ΣS°(reactants). First, find the standard molar entropies of P4O10, H2O, and H3PO4 in the ALEKS Data tab. For example, suppose they are 200, 50, and 80 J/mol•K respectively. Then, apply the formula: ΔS° = [4*(80) - (200 + 6*50)], which gives -20 J/K/mol. Rounded to zero decimal places, ΔS° for the reaction would be -20 J/K/mol.
Learn more about Standard Reaction Entropy here:https://brainly.com/question/31046571
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Hexane and octane are mixed to form a 45 mol% hexane solution at 25 deg C. The densities of hexane and octane are 0.655 g/cm3 and 0.703 g/cm3, respectively. Assume you have 1.0 L of octane. Calculate the required volume of hexane. Report your answer in liters.

Answers

Answer:

The required volume of hexane is 0.66245 Liters.

Explanation:

Volume of octane = v=1.0 L=[tex]1000 cm^3[/tex]

Density of octane= d = [tex]0.703 g/cm^3[/tex]

Mass of octane ,m= [tex]d\times v=0.703 g/cm^3\times 1000 cm^3=703 g[/tex]

Moles of octane =[tex]\frac{m}{114 g/mol}=\frac{703 g}{114 g/mol}=6.166 mol[/tex]

Mole percentage of Hexane = 45%

Mole percentage of octane = 100% - 45% = 55%

[tex]55\%=\frac{6.166 mol}{\text{Total moles}}\times 100[/tex]

Total moles = 11.212 mol

Moles of hexane :

[tex]45%=\frac{\text{moles of hexane }}{\text{Total moles}}\times 100[/tex]

Moles of hexane = 5.0454 mol

Mass of 5.0454 moles of hexane,M = 5.0454 mol × 86 g/mol=433.9044 g

Density of the hexane,D = [tex]0.655 g/cm^3[/tex]

Volume of hexane = V

[tex]V=\frac{M}{D}=\frac{433.9044 g}{0.655 g/cm^3}=662.4494 cm^3\approx 0.66245 L[/tex]

(1 cm^3= 0.001 L)

The required volume of hexane is 0.66245 Liters.

How many grams of F are in 12.56 g of SF6? h.

Answers

Answer:

9.80 g

Explanation:

The molecular mass of the atoms mentioned in the question is as follows -

S = 32 g / mol

F = 19 g / mol

The molecular mass of the compound , SF₆ = 32 + ( 6 * 19 ) = 146 g / mol

The mass of 6 F = 6 * 19 = 114 g /mol .

The percentage of F in the compound =

mass of 6 F / total mass of the compound * 100

Hence ,  

The percentage of F in the compound = 114 g /mol  / 146 g / mol * 100

78.08 %

Hence , from the question ,

In 12.56 g of the compound ,

The grams of F = 0.7808 * 12.56 = 9.80 g

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