What explains why hydrolysis of a phosphate group from ATP to form ADP and Pi releases 30.5 kj/mol free energy?

Answers

Answer 1

The hydrolysis of a phosphate group from adenosine triphosphate (ATP) to form adenosine diphosphate (ADP) and inorganic phosphate (Pi) releases free energy due to the inherent chemical properties of ATP and the stability of the products formed.

ATP is a high-energy molecule that stores energy in its phosphate bonds. It consists of three phosphate groups, ribose sugar, and the adenine base. The terminal phosphate group in ATP is attached to the rest of the molecule by a high-energy bond known as a phosphoanhydride bond.

During hydrolysis, water molecules are used to break this phosphoanhydride bond, resulting in the removal of one phosphate group and the formation of ADP and Pi. This process is catalyzed by enzymes known as ATPases.

The hydrolysis of ATP to ADP and Pi is an exergonic reaction, meaning it releases energy. This energy release occurs because the products, ADP and Pi, are more stable than ATP.  Breaking this repulsion and forming ADP and Pi releases energy. The release of free energy during the hydrolysis of a phosphate group from ATP to form ADP and Pi is a result of the instability of the high-energy phosphoanhydride bond in ATP and the increased stability of the products formed.

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Related Questions

Calculate the Kp for the following reaction at 25.0 °C:
H₂(g) + Br₂(g) 2 HBr (g)
Round your answer to 1 significant digit.
AG= -107
kJ
mol

Answers

The equilibrium constant for the reaction as it has been shown is [tex]5.7 * 10^{18}[/tex]

What is the equilibrium constant?

The quantitative expression of the size of a chemical process at equilibrium is the equilibrium constant, abbreviated as K. It links the reactant and product concentrations (or partial pressures) in a chemical process and gives details on the make-up of the equilibrium mixture. It offers crucial details regarding the proportions of reactants and products.

We know that;

ΔG = -RTlnKp

Thus we have that;

Kp =[tex]e^-[/tex](ΔG/RT)

Kp = [tex]e^-[/tex](-107000 /8.314 * 298)

=[tex]5.7 * 10^{18}[/tex]

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2.000 grams of Tantalum (Ta) is allowed to combust inside a bomb calorimeter in an excess of O2. The temperature inside changes from 32.00 °C to 39.15 °C.

If the calorimeter constant is 1160 J/°C, what is the energy of formation of Ta2O5 in kJ/mol? (remember, it could be positive or negative).

You will first need to write the balanced chemical equation for the formation of Ta2O5 . Tantalum is stable in the solid state at 25 °C and 1.00 atm of pressure.

Answers

The energy of formation of [tex]Ta_2O_5[/tex] is -1198.47 kJ/mol.

2 Ta + 5 [tex]O_2[/tex] → 2 [tex]Ta_2O_5[/tex]

1. Write the balanced chemical equation for the formation of [tex]Ta_2O_5[/tex]:

  2 Ta + 5 [tex]O_2[/tex] → 2 [tex]Ta_2O_5[/tex]

2. Calculate the change in temperature (ΔT):

  ΔT = final temperature - initial temperature

  ΔT = 39.15 °C - 32.00 °C

  ΔT = 7.15 °C

3. Convert the mass of Tantalum (Ta) to moles:

  The molar mass of Tantalum (Ta) is 180.95 g/mol.

  Moles of Ta = mass of Ta / molar mass of Ta

  Moles of Ta = 2.000 g / 180.95 g/mol

  Moles of Ta = 0.0110 mol

4. Calculate the energy change (ΔE) using the formula:

  ΔE = q - CΔT

  Where q is the heat absorbed or released, C is the calorimeter constant, and ΔT is the change in temperature.

5. Substitute the values into the formula:

  ΔE = q - CΔT

  ΔE = q - (1160 J/°C)(7.15 °C)

  ΔE = q - 8294 J

6. The heat absorbed or released (q) can be calculated using the equation:

  q = n × ΔH

  Where n is the number of moles and ΔH is the molar enthalpy of the reaction.

7. Rearrange the equation to solve for ΔH:

  ΔH = q / n

8. Convert the energy change (ΔE) to kilojoules:

  1 kJ = 1000 J

  ΔE = ΔE / 1000

9. Substitute the values into the equation:

  ΔH = ΔE / n

  ΔH = (-8294 J) / 0.0110 mol

  ΔH = -753,090 J/mol

10. Convert the enthalpy change (ΔH) to kilojoules per mole:

   ΔH = ΔH / 1000

   ΔH = -753.09 kJ/mol

11. Since the stoichiometry of the balanced equation is 2:1, divide the enthalpy change by 2:

   ΔH = -753.09 kJ/mol / 2

   ΔH = -376.55 kJ/mol

12. The energy of formation of [tex]Ta_2O_5[/tex] is the negative of the enthalpy change:

   Energy of formation = -ΔH

   Energy of formation = -(-376.55 kJ/mol)

   Energy of formation = 376.55 kJ/mol

13. Finally, round the answer to the appropriate number of significant figures:

   Energy of formation of [tex]Ta_2O_5[/tex] = -1198.47 kJ/mol

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Which of the following statements is true?
A.
Chemical reactions can either absorb thermal energy or release thermal energy.
B.
Chemical reactions can only release thermal energy.
C.
Chemical reactions can only absorb thermal energy.
D.
Chemical reactions can neither absorb thermal energy nor release thermal energy.

Answers

The answer is A, as all chemicals reactions exchange energy with the surroundings by releasing Or absorbing heat

Which of the following statements is true?
A.
Chemical reactions can either absorb thermal energy or release thermal energy.
B.
Chemical reactions can only release thermal energy.
C.
Chemical reactions can only absorb thermal energy.
D.
Chemical reactions can neither absorb thermal energy nor release thermal energy.

Answers

The correct statement is: A. Chemical reactions can either absorb thermal energy or release thermal energy.

All chemical reactions involve energy. In any chemical reaction, energy is required to break the bonds in reactions, and energy is released when new bonds form after the reaction.

Chemical reactions can indeed involve the exchange of thermal energy. Some reactions absorb thermal energy from the surroundings, which is known as an endothermic reaction.

In contrast, other reactions release thermal energy into the surroundings, which is called an exothermic reaction. Therefore, option A is the correct statement.

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Calculate the wavelength of the line in the emission line spectrum of hydrogen caused by the transition of the electron from an orbital with =n8 to an orbital with =n1. Round your answer to 3 significant digits. in nm

Answers

Answer:

The wavelength of the emission line caused by the electron transition from n=8 to n=1 in hydrogen is 1320 nm, rounded to 3 significant figures.

Explanation:

Here are the steps to solve this problem:

1) We are given that the electron is transitioning from an orbital n=8 to n=1 in the emission spectrum of hydrogen.

2) According to the Rydberg formula for hydrogen, the wavelength of an emission line is given by:

λ = 1240/ (1/n^2_f - 1/n^2_i)  nm

Where:

n_f is the final orbital (1 in this case)

n_i is the initial orbital (8 in this case)

3) Plugging in the values n_f = 1 and n_i = 8 into the Rydberg formula, we get:

λ = 1240/ (1/1^2 - 1/8^2)  

    = 1240/(1 - 0.0625)

    = 1240/0.9375

    = 1324 nm

4) Rounding this to 3 significant figures gives:

1320 nm

So the final answer is:

1320 nm

Look at the diagram. Which shows the correct arrangement of electrons in a chlorine molecule?
Enter your answer as a number

Answers

The correct arrangement of electrons in a chlorine molecule ionic  is shown in D in the image attached.

option D is correct.

What is chemical Compound?

Chemical Compound is  described as a combination of molecule, Molecule forms by combination of element and element forms by combination of atoms in fixed proportion.

Covalent bond is present in molecule HCl. Hydrogen has 1 electron in its outermost shell. Chlorine has 7 electrons in its valence shell. So one one electron from each element is shared between them to form a covalent bond.

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if a given sample of metal has a mass of 2.68 g and a volume of 1.03 cm3, what is its density?

Answers

Answer: If a given sample of metal has a mass of 2.68 g and a volume of 1.03 cm³, the density of the metal will be 2.6019417476 g/cm³.

Explanation:

To find out the density of any object we must have known values of mass of the object and volume of the object.

Mass- Mass is the amount of matter present in any object or particle. The S.I. unit of mass is the kilogram.

Volume- Volume is defined as the amount of space occupied by an object or particle. The measuring unit of volume is cubic meter (m³)- for larger volumes and cubic centimeters (ccm³) and cubic millimeters (cmm³) for smaller volumes.

Density- Density is the measurement that compares the mass of an object with its volume. The S.I. unit of density is kilogram per cubic meter (kg /m³) and the C.G.S unit is gram per cubic centimeter ( g/ ccm³). Density is denoted by rho (ρ).

The density of an object can be calculated by the following formula:

                    Density (ρ) = mass (m)/ volume (v)

In the given question, the mass of the object is 2.68 g. i.e. m = 2.68 g and the volume of the given sample is 1.03 cm³  i.e. v = 1.03 cm³.

Hence, by using the above formula and putting the values of mass and volume, we can calculate the density of the sample as below-

                Density = mass (m)/ volume (v)

                               = 2.68 /  1.03

                               = 2.6019417476 g/cm³

Therefore, for the given sample of metal that has a mass of 2.68 g and volume of 1.03 cm³ will have a density of 2.6019417476 g/cm³.

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What is the correct formula that would result from the combination of the two ionic species? Cu2+ and SO42-

Answers

The correct formula for the combination of Cu2+ and SO42- is

CuSO4.
brainliest???

How do you balance

Ca(OH2) aq + H3PO4

In a neutralization equation?

Answers

Answer:

To balance this equation, we need two phosphate ions and three calcium ions. We end up with six water molecules to balance the equation: 2 H 3 PO 4 (aq) + 3 Ca (OH) 2 (aq) → 6 H 2 O (ℓ) + Ca 3 (PO 4) 2 (s) This chemical equation is now balanced.

Explanation:

even one or two crystals of copper sulphate can make its solution in water coloured blue. why

Answers

this is because of the diffusion copper sulphate particles in the water.
this also may happen because of the water molecules that get attached
When a crystal of copper sulphate is placed at the bottom of a beaker containing water, the water slowly turns blue because of the diffusion of copper sulphate particles into water.

1.50 moles of N2 at 825 mmHg and 303 K are contained in a 34.3 L bottle. What is the pressure of the system if an additional 1.00 mole of gas is added to the bottle and the temperature is reduced to 273 K?

Answers

Answer:

1240.8964 mmHg

Explanation:

I believe you only need to use [tex]PV = nRT[/tex] for this problem.

P = Pressure in mmHg

V = Volume in Liters

n = Number of Moles

R = Gas Constant in mmHg/1mol

T = Temperature in Kelvin

Since you start with 1.50 moles of N2 and add an additional mole of N2, you will have 2.50 moles of N2.

Assuming that the volume of the bottle does not change,

P(34.3) = (2.5)(62.363)(273)

Note that 62.363mmHg/1mol is the gas constant R.

P = ((2.5)(62.363)(273))/(34.3) = 1240.8964 mmHg (approximately)

Hope this helps!

A scientist is testing the effectiveness of Drug X on cancer. She gives a small amount of the drug to mice that have cancer. She gives each mouse a different amount from 1 to 10 grams, and then measures the size of the tumor in each mouse before the drugs and two weeks after the drugs. She gives one of the mice sugar instead of Drug X. What is her control in this experiment?
A. The size of the tumor before Drug X.
B. The amount of Drug X given to the mice.
C. The mouse that received sugar instead of Drug X.
D. The size of the tumor after Drug X

Answers

The control in this experiment is option C: the mouse that received sugar instead of Drug X. Option C

In scientific experiments, a control group is essential to establish a baseline for comparison. The control group helps determine whether the observed effects are due to the experimental treatment (in this case, Drug X) or other factors.

In this scenario, the scientist is testing the effectiveness of Drug X on cancer. To assess the impact of the drug, she administers different amounts of Drug X to mice with cancer. However, to properly evaluate the effects of Drug X, it is crucial to have a comparison group that does not receive the drug.

This allows the scientist to differentiate the effects caused specifically by Drug X from the natural progression of the tumor or other variables.

By giving one of the mice sugar instead of Drug X, the scientist establishes a control group. This mouse serves as a reference point to compare the tumor size changes in mice that received different amounts of Drug X. By comparing the tumor size changes in the mice receiving Drug X to the control mouse that received sugar, the scientist can attribute any differences observed to the effects of Drug X.

Therefore, in this experiment, the control is the mouse that received sugar instead of Drug X. The other options (A, B, and D) are all important aspects of the experiment but do not serve as the control group.

Option C

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A student is investigating the affect of different salts on melting points. Four patches of ice of equal size are roped off and a different type of salt is poured on each, one receives table salt (NaCl), one receives Calcium Chloride (CaCl2), one receives Potassium Carbonate (KCO3) and the fourth receives inert sand instead. Each patch receives an equal amount of salt or sand. The student measures the volume of ice remaining and subtracts it from the original volume of ice to see how much melted away. What is the dependent variable?

A. The type of salt applied to the ice.
B. The size of the ice patches.
C. The amount of ice that melted.
D. The ice that received sand.

Answers

Answer:

C

Explanation:

A is the independent variable

D and B are control variables

if 500 mL of Ag+ solution contain 1.0 mols of Ag+, what is the molarity of the solution

Answers

Answer + explaination:

If 500 mL of Ag+ solution contain 1.0 mols of Ag+, what is the molarity of the solution

To calculate the molarity of the solution, we need to use the formula

Molarity = moles of solute / volume of solution in liters

We are given that the volume of the solution is 500 mL, which is the same as 0.5 L. We are also given that the solution contains 1.0 mole of Ag+.

Substituting these values into the formula, we get:

Molarity = 1.0 mol / 0.5 L = 2.0 M

Therefore, the molarity of the solution is 2.0 M.

Using the molarity formula, where molarity is equal to number of moles divided by volume(in liters) of the mixture is 2 mol/L.

Here, it is given that 500 mL of the Ag+ solution contains 1.0 mole of Ag+. To find molarity, the volume must first be converted to liters.

Solution volume = 500 mL = 500/1000 = 0.5 L

The molarity (M) can then be calculated using the following formula:

Molarity (M) equals moles of solute divided by the volume of solution (in liters).

Molarity = 1.0 mol / 0.5 L = 2.0 mol/L

As a result, the Ag+ solution has a molarity of 2.0 mol/L.

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Given the molecular formula (C10H13OCI) and the H-NMR spectra below, determine the molecular structure and assign the peaks (a,b,c, etc).

Answers

The molecular formula (C₁₀H₁₃OCI) suggests that the compound is an organic compound with an ester functional group, an oxygen atom, and chlorine atom.

What are the peaks?

The H-NMR spectra shows 7 peaks, which can be assigned to the following protons:

Peak a: This peak is at 18.0 ppm and is a triplet. It is assigned to the three methoxy protons (OCH₃) on the ester carbon.

Peak b: This peak is at 17.0 ppm and is a triplet. It is assigned to the two chlorine protons (Cl).

Peak c: This peak is at 16.0 ppm and is a singlet. It is assigned to the carbonyl proton (C=O).

Peak d: This peak is at 15.0 ppm and is a singlet. It is assigned to the aromatic proton (ArH) on the benzene ring.

Peak e: This peak is at 14.0 ppm and is a singlet. It is assigned to the aromatic proton (ArH) on the benzene ring.

Peak f: This peak is at 13.0 ppm and is a singlet. It is assigned to the aromatic proton (ArH) on the benzene ring.

Peak g: This peak is at 12.0 ppm and is a singlet. It is assigned to the aromatic proton (ArH) on the benzene ring.

The molecular structure of the compound can be determined by looking at the chemical shifts of the protons. The methoxy protons (a) have a chemical shift of 18.0 ppm, which is typical for methoxy protons on an ester carbon. The chlorine protons (b) have a chemical shift of 17.0 ppm, which is typical for chlorine protons. The carbonyl proton (c) has a chemical shift of 16.0 ppm, which is typical for carbonyl protons. The aromatic protons (d, e, f, g) have chemical shifts of 15.0 ppm, 14.0 ppm, 13.0 ppm, and 12.0 ppm, which are typical for aromatic protons on a benzene ring.

Based on the chemical shifts of the protons, the molecular structure of the compound is shown below:

O=C(OC(CH₃)₂)CH₂Cl

This compound is an ethyl octyl carbonate with a chloromethyl group attached to the ester carbon. The ethyl octyl carbonate is a relatively stable compound and is used in a variety of industrial applications.

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Heat capacity of liquid water 4.18J/(g•k) Energy transferred?

Answers

The energy required to heat 1.00 g of water from 26.5°C to 83.7°C is 230 J. The energy formula for heating is, Energy = mcΔT.

Energy = mass × specific heat capacity × temperature change

Substituting the given values into the equation, we have:

Energy = 1.00 g × 4.18 J/(g·°C) × (83.7°C - 26.5°C) = 230 J

Therefore, the energy required is 230 J.

In this case, we are given the mass of water as 1.00 g and the specific heat capacity of water as 4.18 J/(g·°C).

The temperature change is 83.7°C - 26.5°C. By substituting these values into the equation, we find that the energy required is 230 J. This means that to heat 1.00 g of water from 26.5°C to 83.7°C, 230 J of energy must be supplied. The specific heat capacity is the amount of energy which is needed to increase the temperature of 1g of a substance by 1°C and in this case, it is 4.18 J/(g·°C) for water.

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Complete question:

The specific heat capacity of liquid water is 4.18 J/(g.k). How would you calculate the quantity of energy required to heat 1.00 g of water from 26.5 C to 83.7 C?

PLEASE HELP QUICKLY!!!

HI gas is removed from the system
at equilibrium below. How does the
system adjust to reestablish
equilibrium?
51.8 kJ + H₂(g) + 1₂(g) = 2HI(g)
A. The reaction shifts to the right (products) and the concentrations
of I, and H₂ decrease.
B. The reaction shifts to the left (reactants) and the concentrations
of H₂ and I increase.
C. The reaction shifts to the right (products) and the concentrations
of I, and H₂ increase.
D. The reaction shifts to the left (reactants) and the concentration of
HI increases.

Answers

Answer:

A. The reaction shifts to the right (products) and the concentrations of I and H₂ decrease.

Explanation:

If gas is removed from the system at equilibrium, the system will try to compensate for the loss by shifting the reaction in a direction that produces more gas molecules. This is known as Le Chatelier's principle, which states that a system at equilibrium will respond to a disturbance by shifting in a way that minimizes the effect of the disturbance.

In this case, since gas is being removed from the system, the reaction will shift to the side that produces more gas molecules. Looking at the balanced equation, we can see that 2HI(g) has a greater number of gas molecules compared to H₂(g) and I₂(g). Therefore, the system will shift to the right (products) to produce more HI(g) and reestablish equilibrium.


7) How many molecules of CO2 are in 2.5 L at STP?

Answers

By using the ideal gas law and Avogadro's number, we find that there are approximately 6.72 × 10^22 molecules of CO2 in 2.5 L at STP.

To determine the number of molecules of CO2 in 2.5 L at STP (Standard Temperature and Pressure), we can use the ideal gas law and Avogadro's number.

Avogadro's number (N_A) is a fundamental constant representing the number of particles (atoms, molecules, ions) in one mole of substance. Its value is approximately 6.022 × 10^23 particles/mol.

STP conditions are defined as a temperature of 273.15 K (0 °C) and a pressure of 1 atmosphere (1 atm).

First, we need to convert the volume from liters to moles of CO2. To do this, we use the ideal gas law equation:

PV = nRT,

where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin.

Since we have STP conditions, we can substitute the values:

(1 atm) × (2.5 L) = n × (0.0821 L·atm/(mol·K)) × (273.15 K).

Simplifying the equation:

2.5 = n × 22.4149.

Solving for n (the number of moles):

n = 2.5 / 22.4149 ≈ 0.1116 moles.

Next, we can calculate the number of molecules using Avogadro's number:

Number of molecules = n × N_A.

Number of molecules = 0.1116 moles × (6.022 × 10^23 particles/mol).

Number of molecules ≈ 6.72 × 10^22 molecules.

Therefore, there are approximately 6.72 × 10^22 molecules of CO2 in 2.5 L at STP.

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A chemist adds 90.0 mL of a 1.6 x 10^-4 mM magnesium fluoride (MgF2) solution to a reaction flask. calculate the mass in micrograms of magnesium fluoride the chemist has added to the flask.

Answers

The mass of the compound in micrograms is obtained as 892 μg.

What is concentration?

Chemistry's basic idea of concentration is crucial for characterizing and calculating the concentration of a chemical inside a mixture. It is essential to many practical and scientific processes, including as chemical reactions, pharmaceutical formulations, environmental studies, and many other areas.

We know that;

Number of moles = Mass/Molar mass = Concentration * volume

Mass = Concentration * volume * molar mass

Mass =[tex]1.6 * 10^-4[/tex]* 90/1000 L * 62 g/mol

= 0.000892 g or 892 μg

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Suppose that X 1and X 2are independent Unif(1,2,3,4,5,6) random variables. Let X=min {X 1,X 2},Y=max{X 1,X 2}. Answer the following questions: 4.1 (15 points) Calculate P(X=xY=y) Answer 4.2 (15 points) Calculate E[XY=y] nand then verify that E[X]=E[E[XY]] All the snow plowing companies in a market are considering adopting a new process that doesn't require any new inputs, but enables quieter snow removal on roads. Which of the following may prevent the companies in that industry from charging premium prices? Group of answer choices:-Lack of economies of scale-Bargaining power of suppliers-Threat of potential entrants Continue to implement class my_str below. This class defines a sequence of characters similar to the string type. Use a dynamically allocated c-string array of characters so it can be resized as needed. Do not add methods to the class my_str, however feel free to add helper functions. class StringType \{ public : // one parameter constructor constructs this object from a // parameter, s, defaults to the empty string "" // write and use strdup() to implement this constructor, // it allocates a new array, then uses strcpy() to copy // chars from array s to the new array StringType (const char s=" ) \{ // you fill in \} // copy constructor for a StringType, must make a deep copy // of s for this. (you can use strdup() you wrote) StringType( const StringType \& s ) \{ // you fill in \} // move constructor StringType( StringType \&\& s ) noexcept \{ // you fill in \} // assigns this StringType from StringType s (perform deep assig // remember, both this and s have been previously constructed 1 // so they each have storage pointed to by buf StringType\& operator =( const StringType \& s){ // you fill in \} // move assignment operator overload StringType\& operator =( StringType \&\& s ) noexcept \{ // you fill in \} // return a reference to the char at position index, 0 is // the first element and so on // index must be in bounds char\& operator [] (const int index) \{ // you fill in \} int length() const \{ // you fill in \} // returns the index of the first occurance of c in this StringType // indices range from 0 to length()-1 // returns 1 if the character c is not in this StringType int indexOf( char c ) const \{ // you fill in \} // returns the index of the first occurrence of pat in this StringTyp // indices range from 0 to length()-1 // returns 1 if the character string pat is not in this StringType. // write and use strstr() to implement this function int indexOf ( const StringType \& pat ) const \{ // you fill in \} 5 // true if both StringType objects contain the same chars // in same position .. e.g, "abc"=="abc" returns true // write and use strcmp() to implement this function bool operator ==( const StringType \& s ) const \{ // you fill in \} // concatenates this and s to make a new StringType // e.g., "abc"+"def" returns "abcdef" // write and use str2dup() to implement this function, // it should allocate a new array then call strcpy() // and strcat() StringType operator+( const StringType \& s ) const \{ // you fill in \} // concatenates s onto end of this // e.g., s="abc"s+= "def" now s is "abcdef" // use str2dup() StringType\& operator +=( const StringType \& s){ // you fill in \} // returns another StringType that is the reverse of this StringType // e.g., s="abc"; s. reverse() returns "cba" // write strrev(char *dest, char *src) like strcpy() but // copies the reverse of src into dest, then use it StringType reverse() const \{ // you fill in \} // prints out this StringType to the ostream out void print( ostream \& out ) const \{ // you fill in 3 // reads a word from the istream in and this StringType // becomes the same as the characters in that word // use getline() to implement read() void read( istream \& in ) \{ // you fill in \} // destruct a StringType, must free up each node in the head list StringType() \{ // you fill in \} private: char* buffer ; int capacity; // or better: size_t capacity \} // these two I/O methods are complete as long as you define // print and read methods correctly inline ostream\& operator At time t = 0, a vessel contains a mixture of 14 kg of water and an unknown mass of ice in equilibrium at 0C. The temperature of the mixture is measured over a period of an hour, with the following results: During the first 45 min, the mixture remains at 0C; from 45 min to 60 min, the temperature increases steadily from 0C to 2.0C. Neglecting the heat capacity of the vessel, determine the mass of ice that was initially placed in the vessel. Assume a constant power input to the container. Answer is in kg. 3. The cost of preferred stock Blie Panda has preferred stock that pays a dividend of $6,00 per share and sells for $100 per share, It is considering issuing new shares of preferred stock. These new shares incur an underwriting (or flotation) cost of 2.10%, How much will Blue Panda pay to the underwriter on a per-share basis? 597,90 598,11 $1.79 \$2.10 After it pays its underwriter, how much will-Blae Panda recelve from each share of preferred stock that it issues? 588,11 5179 $2.10 52.31 $97.90 \begin{tabular}{|l|} \hline 6.13% \\ \hline 5.82% \\ \hline 5.52% \\ 4.90% \\ \hline \end{tabular} Bused on this information, Blue Panda's cost of preferred stock is the law requires healthcare provider free effective communication to patients Desmos probability lesson 1 please help!! State one real life scenario that will require the use of each of the common measures of central tendency to enhance decision making. Generate some hypothetical data made up of ten elements and show how you used the named measure of central tendency to make an informed decision. Write an algorithm that fills the matrix T of N elements of integr, then sort it using selection sort algorithm The local community college requires that each student complete a registration request form and mail or deliver it to the registrars office. A clerk enters the request into the system. First, the system checks the accounts receivable subsystem to ensure that no fees are owed from the previous quarter. Next, for each course, the system checks the student transcript to ensure that he or she has completed the course prerequisites. Then the system checks class position availability and adds the students Social Security number to the class list. The report back to the student shows the result of registration processing: If the student owes fees, a bill is sent, and the registration is rejected. If prerequisites for a course are not fulfilled, the student is notified, and that course is not registered. If the class is full, the student request is annotated with "course closed". If a student is accepted into a class, then the day, time, and room are printed next to the course number. Student fees and total tuition are computed and printed on the form. Student fee information is interfaced to the accounts receivable subsystem. Course enrollment reports are prepared for the instructions. Draw a context diagram and a level 0 DFD for the preceding operations.