Consider a sample of helium gas in a container fitted with a piston as pictured below. The piston is frictionless, but has a mass of 10.0 kg. How many of the following processes will cause the piston to move away from the base and decrease the pressure of the gas? Assume ideal behavior.
I. Heating the helium.
II. Removing some of the helium from the container.
III. Turning the container on its side.
IV. Decreasing the pressure outside the container.
a) 0
b) 1
c) 2
d) 3
e) 4

Answers

Answer 1

Answer:

I. heating the helium

Explanation:

Answer 2

Final answer:

(I) Heating the helium and (IV) decreasing the pressure outside the container are the two processes that will cause the piston to move away from the base and decrease the pressure of the gas. Hence, (c) is the correct option.

Explanation:

Assuming ideal gas behavior, the question asks which processes will lead the piston to travel away from the base and lower the helium gas's pressure. Let's examine each of the scenarios that are presented:

I. Heating the helium: Heating increases the internal energy of the gas molecules, making them move faster. This increases the force they exert against the piston, pushing it outward and increasing the volume, thus reducing the pressure according to Boyle's Law (P₁V₁ = P₂V₂).

II. Removing some of the helium from the container: If you remove some of the helium, there will be fewer molecules to exert force on the piston, leading to a decrease in pressure. However, the piston will not necessarily move since the external pressure remains the same. Without a corresponding change in external pressure, there is no force to move the piston outward.

III. Turning the container on its side: Turning the container on its side will have no effect on the pressure or volume of the gas if the external conditions remain the same. The position of the piston is influenced by forces and pressures, not by its orientation in space.

IV. Decreasing the pressure outside the container: Decreasing the external pressure acting on the piston will allow the internal pressure of the gas to push the piston outward, increasing the volume and therefore decreasing the gas pressure, as explained by the ideal gas law (PV = nRT).

Therefore, the processes that will cause the piston to move away from the base and decrease the pressure of the gas are heating the helium and decreasing the pressure outside the container. So the correct answer to the question is (c) 2.


Related Questions

A gas has a sample initial pressure of 1.24 atm and an initial volume of 0.671 L. What is the pressure (in torr) if the final volume of the gas is changed to 583. mL? Assume constant temperature and amount of gas.

Answers

Answer:

1086.8 torr

Explanation:

Using Boyle's law  

[tex]{P_1}\times {V_1}={P_2}\times {V_2}[/tex]

Given ,  

V₁ = 0.671 L  

V₂ = 583 mL = 0.583 L ( 1 mL = 0.001 L )

P₁ = 1.24 atm

P₂ = ?

Using above equation as:

[tex]{P_1}\times {V_1}={P_2}\times {V_2}[/tex]

[tex]{1.24\ atm}\times {0.671\ L}={P_2}\times {0.583\ L}[/tex]

[tex]{P_2}=1.43\ atm[/tex]

The conversion of P(atm) to P(torr) is shown below:

[tex]P(atm)={760}\times P(torr)[/tex]

So,  

Pressure = 1.43*760 torr = 1086.8 torr

In the absence of an electric field, a radioactive beam strikes a fluorescent screen at a single point. When an electric field is applied, the radioactive beam is separated into three different components. One of the components is deflected toward the positive electrode because it is negatively charged, one of the components is deflected toward the negative electrode because it is positively charged, and one component is not deflected in any direction; instead, it moves along a straight path.

Identify the charges possessed by the different components of the radioactive beam by observing their behavior under the influence of an electric field.

Sort the appropriate items into Negatively Charged, Neutral, or Positively Charged:

1._____________ ? Rays
2. __________? Rays
3. _____________? Rays

Answers

Answer:

Negatively Charged: Beta - radiation

Neutral: Gamma radiation

Positively Charged: Beta + radiation.

Explanation:

The radioactive beam described is formed by three different types of particules:

The Negatively Charged are formed by electrons and known as Beta - radiationThe Positively Charged are formed by positrons and known as Beta + radiation. The other alternative to a positively charged radiation is Alpha radiation: formed by two positrons and two neutrines-The Neutral ones are formed by Gamma radiation

Of the following reactions occurring at 25ºC, which one involves the greatest increase in entropy?H2(g) + Cl2(g) = 2 HCl(g)H2O(s) = H2O(l)Pb2+(aq) + 2 Cl-(aq) = PbCl2(s)CO2(s) = CO2(g)

Answers

Answer: [tex]CO_2(s)\rightarrow CO_2(g)[/tex]

Explanation:

Entropy is the measure of randomness or disorder of a system. If a system moves from an ordered arrangement to a disordered arrangement, the entropy is said to decrease and vice versa.

[tex]\Delta S[/tex] is positive when randomness increases and [tex]\Delta S[/tex] is negative when randomness decreases.

a) [tex]H_2(g)+Cl_2(g)\rightarrow 2HCl(g)[/tex]

2 molesof gas are converting to 2 moles of another gas , thus [tex]\Delta S[/tex] is zero.

b) [tex]H_2O(s)\rightarrow H_2O(l)[/tex]

1 mole of solid is converting to 1 mole of liquid, the randomness increases and thus [tex]\Delta S[/tex] is positive.

b) [tex]Pb^{2+}(aq)+2Cl^-(aq)\rightarrow PbCl_2(s)[/tex]

2 moles of ions are converting to 1 mole of solid, the randomness decreases and thus [tex]\Delta S[/tex] is negative

d) [tex]CO_2(s)\rightarrow CO_2(g)[/tex]

1 mole of solid is converting to 1 mole of gas, the randomness increases drastically and thus [tex]\Delta S[/tex] is highly positive.

For fatty acids with the same number of carbon atoms, how does the melting point change as the number of double bonds in the fatty acid changes?
a. The melting point of the fatty acid decreases as the number of double bonds increases.
b. There is no relationship between the melting point of a fatty acid and the number of double bonds.
c. The melting point of the fatty acid is unchanged as the number of double bonds changes.
d. The melting point of the fatty acid decreases as the number of double bonds decreases.

Answers

The answer is a. "The melting point of the fatty acid decreases as the number of double bonds increases" because when there are double bonded carbons it cancels the polarity of the carboxyl in the opposite direction, increasing that effect as double bonded carbons add to the molecule, giving the net dipole moment to the molecule. As the net dipole moment of the molecule decrease, the melting point of the fatty acid also decreases.

In addition, when the fatty acid is major than 8 carbons and have a double bond that prevents the fatty acid from crystal lattice formation, so the melting point will be lower also for this reason in this case.

Final answer:

The melting point of a fatty acid decreases as the number of double bonds increases because double bonds create kinks in the fatty acid chains, preventing tight packing and weakening intermolecular forces. The correct answer is option a.

Explanation:

For fatty acids with the same number of carbon atoms, the melting point changes as the number of double bonds in the fatty acid changes. The correct option is (a) - the melting point of the fatty acid decreases as the number of double bonds increases. This decrease in melting point occurs because unsaturated fatty acids, which have one or more double bonds, tend to have kinks or bends at the location of these bonds.

These kinks inhibit the fatty acids from packing closely together, resulting in weaker intermolecular Van der Waals forces (specifically London dispersion forces) and thus a lower melting point. In contrast, saturated fatty acids without double bonds have straight chains and can pack tightly together, leading to stronger intermolecular forces and higher melting points.

A good illustration of this concept is the substantial difference in melting points between palmitoleic acid, which contains a cis-double bond and melts over 60℃ lower than its saturated counterpart, palmitic acid. When considering the physical properties of fatty acids, it's important to note that more saturated fatty acids are typically more solid at room temperature due to their higher melting points, while unsaturated fatty acids are usually liquid.

Almost all amino acids are chiral. The body only uses L amino acids, and proteins are chains of amino acids that in the most simple sense fold up into a single structure by themselves when they are formed. Do you think a cell could live with all D (the opposite of L) amino acids, explain your logic. When I ask 'live', I mean could the organism/cell be composed completely of D amino acids and live (This is just a thought question, don't bother trying to find the answer anywhere…)

Answers

Answer:

An organism that is completely composed D amino acids cannot survive.

Explanation:

Most of amino acids in all organism are present in L conformation.As result  in all organism the enzymes are specific for L amino acids but not for D amino acids.

   Bacterial cell wall contain some D amino acids such as D glutamate,D alanine but not entirely composed of D amino acids.

At a certain temperature, 0.760 mol SO3 is placed in a 1.50 L container. 2SO3(g) = 2SO2(g) + O2(g)At equilibrium, 0.130 mol O2 is present. Calculate Kc.

Answers

Answer:

[tex]K_c=0.0867[/tex]

Explanation:

Moles of SO₃ = 0.760 mol

Volume = 1.50 L

[tex]Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}[/tex]

[tex]Molarity=\frac{0.760}{1.50\ L}[/tex]

[SO₃] = 0.5067 M

Considering the ICE table for the equilibrium as:

[tex]\begin{matrix} & 2SO_3_{(g)} & \rightleftharpoons & 2SO_2_{(g)} & + & O_2_{(g)}\\At\ time, t = 0 & 0.5067 &&0&&0 \\ At\ time, t=t_{eq} & -2x &&2x&&x \\----------------&-----&-&-----&-&-----\\Concentration\ at\ equilibrium:- &0.5067-2x&&2x&&x\end{matrix}[/tex]

Given:    

Equilibrium concentration of  O₂ = 0.130 mol

Volume = 1.50 L

[tex]Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}[/tex]

[tex]Molarity=\frac{0.130}{1.50\ L}[/tex]

[O₂] = x = 0.0867 M

[SO₂] = 2x = 0.1733 M

[SO₃] = 0.5067-2x = 0.3334 M

The expression for the equilibrium constant is:

[tex]K_c=\frac {[SO_2]^2[O_2]}{[SO_3]^2}[/tex]  

[tex]K_c=\frac{(0.3334)^2\times 0.0867}{(0.3334)^2}[/tex]

[tex]K_c=0.0867[/tex]

(a)-Use Lewis symbol store present there action that occurs between Ca and F atoms. (b)-What is the chemical formula of the most likely product? (c)-How many electrons are transferred? (d)-Which atom loses electrons in the reaction?

Answers

Final answer:

The Lewis symbol for Ca is Ca²⁺, while F is F⁻. The Ca atom loses two electrons, and the F atom gains one electron. The most likely product is CaF₂.

Explanation:

(a) The Lewis symbol for Ca is Ca2+, while the Lewis symbol for F is F-. The action that occurs between Ca and F atoms is a redox reaction, where one atom loses electrons (oxidation) and the other gains electrons (reduction).
(b) The chemical formula of the most likely product is CaF2 since one Ca atom can bond with two F atoms.
(c) In the reaction, two electrons are transferred, with Ca losing two electrons and each F atom gaining one electron.
(d) Ca atom loses electrons in the reaction.

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A saturated solution of lead(II) chloride, PbCl2, was prepared by dissolving solid PbCl2 in water. The concentration of Pb2+ ion in the solution was found to be 1.62×10−2 M . Calculate Ksp for PbCl2.

Answers

Answer:

Ksp for PbCl2 is 1.7 *10^-5

Explanation:

Step 1: Data given

The concentration of Pb2+ ion in the solution was found to be 1.62 *10^−2 M

Step 2: The balanced equation

PbCl2 ⇔Pb^2+ + 2Cl -

Step 2: ICE chart

The initial concentration of Pb2+ is 0 M

There will react X M  and 2X of Cl-

At the equillibrium there is X M  of Pb^2+ and 2X M of Cl-

The concentration of Pb2+ ion in the solution was found to be  1.62 *10^−2  M

Step 3: Calculate Ksp

Since PbCl2 is solid, it doesn't aply for Ksp

Ksp = [Pb^2+][Cl-]²

Ksp = X*(2X)²

Ksp = 4X³

  ⇒ X = 1.62 *10^-2 M

Ksp = 4*( 1.62 *10^-2)³

Ksp =1.7 *10^-5

Ksp for PbCl2 is 1.7 *10^-5

In Haber’s process, 30 moles of hydrogen and 30 moles of nitrogen react to make ammonia. If the yield of the product is 50%, what is the mass of nitrogen remaining after the reaction?

Answers

Answer:

Therewill be produced 170.6 grams NH3, there will remain 25 moles of N2, this is 700 grams

Explanation:

Step 1: Data given

Number of moles hydrogen = 30 moles

Number of moles nitrogen = 30 moles

Yield = 50 %

Molar mass of N2 = 28 g/mol

Molar mass of H2 = 2.02 g/mol

Molar mass of NH3 = 17.03 g/mol

Step 2: The balanced equation

N2 + 3H2 → 2NH3

Step 3: Calculate limiting reactant

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

Hydrogen is the limiting reactant.

The 30 moles will be completely be consumed.

N2 is in excess. There will react 30/3 =10 moles

There will remain 30 -10 = 20 moles (this in the case of a 100% yield)

In a 50 % yield, there will remain 20 + 0,5*10 = 25 moles. there will react 5 moles.

Step 4: Calculate moles of NH3

There will be produced, 30/ (3/2) = 20 moles of NH3 (In case of 100% yield)

For a 50% yield there will be produced, 10 moles of NH3

Step 5: Calculate the mass of NH3

Mass of NH3 = mol NH3 * Molar mass NH3

Mass of NH3 = 20 moles * 17.03

Mass of NH3 = 340.6 grams = theoretical yield ( 100% yield)

Step 6: Calculate actual mass

50% yield = actual mass / theoretical mass

actual mass = 0.5 * 340.6

actual mass = 170.3 grams

Step 7: The mass of nitrogen remaining

There remain 20 moles of nitrogen + 50% of 10 moles = 25 moles remain

Mass of nitrogen = 25 moles * 28 g/mol

Mass of nitrogen = 700 grams

Answer:

25 moles

Explanation:

1 mole of nitrogen reacts with 3 moles of hydrogen to give 2 moles of ammonia. 30 moles of hydrogen will react with 10 moles of nitrogen to give 20 moles of ammonia. As the actual yield is 50%, ammonia formed is 10 moles, the amount of nitrogen reacted is 5 moles, and the amount of hydrogen reacted is 15 moles. The mass of the remaining hydrogen is 15 moles and of the remaining nitrogen is 25 moles.

You have two solutions separated by a semipermeable membrane that only allows water to pass through it. The right side of the membrane has 2.0 moles of sucrose (MW 324g/mol), and the left side has 2.0 moles of fructose (MW 180g/mol). Will the solution rise on the right, rise on the left, or remain at a matching level on both sides? What reason do you have for the answer you have selected?

Answers

Answer:

The solution remain at a matching level on both sides, because they have the same molarity.

Explanation:

The osmosis is the spontaneous passage of water by a membrane, from a less concentrated solution to a higher concentration solution, to made them reach an equilibrium.

We know that the number of moles of the compound is the same in both sides of the membrane, without knowing the volume it's impossible to identify the molarity, to identify which one is more concentrated. Let's suppose that the volumes are the same.

Because of that, the molarity is the same on both sides of the membrane, so, the solutions are already in equilibrium, then the solution remains at a matching level on both sides.

What mass of FeSO4^2- x 6H20 (Molar Mass=260g/mol) is required to produce 500 mL of a .10M iron (II) sulfate solution.
A.) 9g
B.) 13g
C.) 36g
D.) 72g

Answers

Answer:

The correct option is: B. 13g

Explanation:

Given: Molar mass of iron (II) sulfate: m = 260g/mol,

Molarity of iron (II) sulfate solution: M =  0.1 M,

Volume of iron (II) sulfate solution: V = 500 mL = 500 × 10⁻³ = 0.5 L           (∵ 1L = 1000mL)

Mass of iron (II) sulfate taken: w = ? g

Molarity: [tex]M = \frac{n}{V (L)} = \frac{w}{m\times V(L)}[/tex]

Here, n- total number of moles of solute, w - given mass of solute, m- molar mass of solute, V- total volume of solution in L

Molarity of iron (II) sulfate solution:  [tex]M = \frac{w}{m\times V(L)}[/tex]

⇒  [tex]w = M\times m\times V(L)[/tex]

⇒  [tex]w = (0.1 M)\times (260g/mol)\times (0.5L) [/tex]

⇒  mass of iron (II) sulfate taken: [tex]w = 13 g[/tex]

Therefore, the mass of iron (II) sulfate taken for preparing the given solution is 13 g.

Which of the following statements is TRUE? Vapor pressure increases with temperature. Dispersion forces are generally stronger than dipole-dipole forces. Hydrogen bonds are chemical bonds. Intermolecular forces hold the atoms in molecules together. None of the above are true.

Answers

Vapor pressure increases with temperature” is the true statement from the given statements.

Option: A

Explanation:

Vapor pressure exerted in “thermodynamic equilibrium” with solid or liquid phase called “condensed phase” at a given temperature in packed or closed system. “Liquid particles” are arranged with more inter molecular space than solid and not as fixed as solid. Therefore when temperature is increased “kinetic energy” of the molecules also increases and thereby molecules transitioning into vapor also increases and in this way whole process is responsible for increase in “vapor pressure”.

Final answer:

The statement that vapor pressure increases with temperature is true because higher temperatures lead to increased kinetic energies, causing more molecules to escape into the vapor phase.

Explanation:

The correct statement among those listed is that vapor pressure increases with temperature. As the temperature rises, the kinetic energy of the molecules increases, leading to a greater number of molecules having enough energy to escape the liquid phase and enter the vapor phase, thereby increasing the vapor pressure. On the other hand, hydrogen bonds are not chemical bonds; they are strong intermolecular forces that occur between molecules, not within them as chemical bonds do.

Dispersion forces are generally weaker than dipole-dipole forces, and they increase with the mass and size of the molecules involved due to increased polarizability. Finally, intermolecular forces are responsible for the attractions between molecules, not for holding the atoms within a molecule together; that role is filled by intramolecular forces or chemical bonds.

A solution of the primary standard potassium hydrogen phthalate, (204.22 g/mol), was prepared by dissolving 0.4877 g of in about 50 mL of water. The solution was titrated with an solution and mL were needed to reach the phenolphthalein end point. What is the molarity of the solution?

Answers

Answer: The molarity of KOH solution is 0.0663 M.

Explanation:

To calculate the number of moles, we use the equation:

[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]

Given mass of KHP = 0.4877 g

Molar mass of KHP = 204.22 g/mol

Putting values in equation 1, we get:

[tex]\text{Moles of KHP}=\frac{0.4877g}{204.22g/mol}=0.0024mol[/tex]

The chemical reaction for the formation of chromium oxide follows the equation:

[tex]KHC_8H_4O_4(aq.)+KOH\rightarrow K_2C_8H_4O_4(aq.)+H_2O(l)[/tex]

By Stoichiometry of the reaction:

1 mole of KHP reacts with 1 mole of KOH.

So, 0.0024 moles of KHP will react with = [tex]\frac{1}{1}\times 0.0024=0.0024mol[/tex] of KOH.

To calculate the molarity of KOH, we use the equation:

[tex]\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}[/tex]

We are given:

Moles of KOH = 0.0024 moles

Volume of solution = 36.21 mL = 0.03621 L (Assuming)      (Conversion factor:  1L = 1000 mL)

Putting values in above equation, we get:

[tex]\text{Molarity of KOH }=\frac{0.0024mol}{0.03621L}=0.0663M[/tex]

Hence, the molarity of KOH solution is 0.0663 M.

The purpose of the salt bridge in an electrochemical cell is to ____.
A. maintain electrical neutrality in the half-cells via migration of ions
B. provide a source of ions to react at the anode and cathode
C. provide oxygen to facilitate oxidation at the anode
D. provide a means for electrons to travel from the anode to the cathode
E. provide a means for electrons to travel from the cathode to the anode

Answers

Answer:

A. maintain electrical neutrality in the half-cells via migration of ions

Explanation:

Salt bridge -

For an electrochemical reaction , involving an anode and a cathode , both the electrodes are connect via a salt bridge to complete the circuit for the reaction .

One of the very important use of a salt bridge is to maintain the electrical neutrality of the respective half cells , which is achieved by the movement of ions .

Hence , from the given options , the correct option is ( a ) .

Final answer:

A salt bridge in an electrochemical cell helps maintain electrical neutrality. It does so by providing a pathway for ions to flow between the half-cells, balancing out the charge imbalance resulting from the reactions, thereby allowing the cell to continue to operate.

Explanation:

The purpose of the salt bridge in an electrochemical cell is to maintain electrical neutrality in the half-cells via the migration of ions. A salt bridge, often saturated with a salt solution, provides a pathway for ions to flow between the two half-cells. During the operation of the cell, the reactions produce or use ions in the solutions, which could result in a charge imbalance. The salt bridge helps to balance out these charges, keeping the solutions electrically neutral. This flow of ions helps complete the circuit, allowing the cell to continue to operate.

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"In the absence of an adequate supply of oxygen, yeasts obtains metabolic energy by fermentation of glucose to produce ethanol. C6H12O6(s) LaTeX: \longrightarrow⟶ 2 C2H5OH(l) + 2 CO2(g) Use the standard enthalpies of formation to calculate ΔH for this reaction" Substance ΔHo glucose(s) -304.5 kcal/mol CO2(g) -93.9 kcal/mol C2H5OH(l) -66.4 kcal/mol

Answers

Answer: [tex]\Delta H=-16.5 kcal[/tex]

Explanation:

The balanced chemical reaction is,

[tex]C_6H_{12}O_6(s)\longrightarrow 2C_2H_5OH(l)+2CO_2(g)[/tex]

The expression for enthalpy change is,

[tex]\Delta H=\sum [n\times \Delta H_f(product)]-\sum [n\times \Delta H_f(reactant)][/tex]

[tex]\Delta H=[(n_{C_2H_5OH}\times \Delta H_{C_2H_5OH})+(n_{CO_2}\times \Delta H_{CO_2})]-[(n_{C_6H_{12}O_6}\times \Delta H_{C_6H_{12}O_6})][/tex]

where,

n = number of moles

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

[tex]\Delta H=[(2\times -66.4)+(2\times -93.9)]-[(1\times -304.5)][/tex]

[tex]\Delta H=-16.5kcal[/tex]

Therefore, the enthalpy change for this reaction is -16.5 kcal

Draw the major organic substitution product(s) for (2R,3S)-2-bromo-3-methylpentane reacting with the given nucleophile. Indicate the stereochemistry, including H\'s, at each stereogenic center. Omit any byproducts.

Answers

Answer:

The final product of the reaction is (2S,3S)-2-ethoxy-3-methylpentane.

Explanation:

The given reaction undergoes [tex]S_{N}2[/tex]  mechanism in which the nucleophile attacks the backside and it is substituted by the elimination of bromine.

Due to the backside attack of nucleophile , the inverse in stereo-chemistry is observed.

After the substitution of ethoxy group, the configuration is assigned according to the priority it shows clock wise direction(R) - configuration.

When hydrogen faces the front side , it results shows inverse configuration i.e, S- configuration.

The chemical reaction is as follows.

 

Part A Add single electrons and/or electron pairs as needed to complete the electron-dot symbol for astatine, At. To return the atom to its original state, use the reset button. Click to select an electron-pair or single electron, then click on the element symbol to add electrons. Click the reset button to clear all electrons and start over. View Available Hint(s) AANNN

Answers

Answer:

7 valence electrons

Explanation:

Astatine has the atomic number 85. Thus, its electron configuration is:

[Xe] 4f¹⁴ 5d¹⁰ 6s² 6p⁵

As we can see, in the last level (6) it has 2 + 5 = 7 electrons, that is, astatine has 7 electrons in its valence shell. In the Lewis dot structure (attached) we write 3 pairs of electrons and 1 unpaired electron around the symbol of At.

Complete and balance the following reaction by filling in the missing coefficients. Assume the reaction is occurring in a basic, aqueous solution.CH3CH2OH(aq)+MnO−4(aq)⟶CH3COO−(aq)+MnO2(s) Include all coefficients, even those equal to 1.

Answers

Answer:

4 MnO₄⁻(aq) + 3 CH₃CH₂OH(aq) ⟶ 4 MnO₂(s) + 1 OH⁻(aq) + 3 CH₃COO⁻(aq) + 4 H₂O(l)

Explanation:

To balance a redox reaction we use the ion-electron method.

Step 1: Identify both half-reactions

Reduction: MnO₄⁻(aq) ⟶ MnO₂(s)

Oxidation: CH₃CH₂OH(aq) ⟶ CH₃COO⁻(aq)

Step 2: Balance the mass adding H₂O and OH⁻ where necessary.

2 H₂O(l) + MnO₄⁻(aq) ⟶ MnO₂(s) + 4 OH⁻(aq)

5 OH⁻(aq) + CH₃CH₂OH(aq) ⟶ CH₃COO⁻(aq) + 4 H₂O(l)

Step 3: Balance the charge adding eelctrons where necessary.

2 H₂O(l) + MnO₄⁻(aq) + 3 e⁻ ⟶ MnO₂(s) + 4 OH⁻(aq)

5 OH⁻(aq) + CH₃CH₂OH(aq) ⟶ CH₃COO⁻(aq) + 4 H₂O(l) + 4 e⁻

Step 4: Multiply both half-reactions by numbers that assure that the number of electrons gained and lost are the same.

4 × (2 H₂O(l) + MnO₄⁻(aq) + 3 e⁻ ⟶ MnO₂(s) + 4 OH⁻(aq))

3 × (5 OH⁻(aq) + CH₃CH₂OH(aq) ⟶ CH₃COO⁻(aq) + 4 H₂O(l) + 4 e⁻)

Step 5: Add both half-reactions and cancel what is repeated.

8 H₂O(l) + 4 MnO₄⁻(aq) + 12 e⁻ + 15 OH⁻(aq) + 3 CH₃CH₂OH(aq) ⟶ 4 MnO₂(s) + 16 OH⁻(aq) + 3 CH₃COO⁻(aq) + 12 H₂O(l) + 12 e⁻

4 MnO₄⁻(aq) + 3 CH₃CH₂OH(aq) ⟶ 4 MnO₂(s) + 1 OH⁻(aq) + 3 CH₃COO⁻(aq) + 4 H₂O(l)

Final answer:

To balance the given reaction: CH3CH2OH(aq)+MnO−4(aq)⟶CH3COO−(aq)+MnO2(s), in a basic, aqueous solution, the balanced equation is 8CH3CH2OH(aq) + MnO−4(aq) + 8OH-(aq) ⟶ 8CH3COO-(aq) + MnO2(s) + 4H2O(l).

Explanation:

To complete and balance the given reaction: CH3CH2OH(aq)+MnO−4(aq)⟶CH3COO−(aq)+MnO2(s) in a basic, aqueous solution, we need to assign appropriate coefficients to each compound to ensure that the reaction is balanced. The balanced equation is:

8CH3CH2OH(aq) + MnO−4(aq) + 8OH-(aq) ⟶ 8CH3COO-(aq) + MnO2(s) + 4H2O(l)

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Describe the reactions during the electrolysis of water :

A) Oxygen is reduced and hydrogen is oxidized.

B) Oxygen and hydrogen are both oxidized.

C) Oxygen and hydrogen are both reduced.

D) Oxygen is oxidized and hydrogen is reduced.

E) Neither oxygen or hydrogen are oxidized or reduced.

Please give a reason why you chose that answer.

Answers

Answer:

D) Oxygen is oxidized and hydrogen is reduced.

Explanation:

The electrolysis of water is the decomposition of water ( as the name suggests) of water into hydrogen and oxygen. The general equation is

2 H₂O(l)  ⇒ 2 H₂ (g) + O₂ (g)

The hydrogen atom in the water molecule has been reduced since its oxidation snumber goes from  1⁺ to 0, and the oxygen has oxidized from 2⁻ to 0  , and in the balanced equation. The overall exchange of electron is 4.

Answer: D

Explanation:

Electrolysis of water which is also called water splitting, is the decomposition of water into oxygen and hydrogen gas when an electric current is passed through it, from a platinum electrode.

Pure water (H2O) is used to produce hydrogen gas (H2), which is fuel, and breathable oxygen gas(O2).

Hydrogen is collected at cathode and oxygen is collected at anode.

Chemical reaction

2 H2O(l) → 2 H2(g) + O2(g)

In the reactant H20, the oxidation state of Hydrogen is (+1) the oxidation state of Oxygen is (-2)

In the products, the oxidation number of elements in the uncombine state is zero so, the oxidation state of Hydrogen is 0 and Oxygen is 0.

Therefore Hydrogen is reduced (from +1 to 0)and oxygen is oxidized (from -2 to 0)

The half reaction are:

Reduction: 2 H+(aq) + 2e− → H2(g)

Oxidation: 2 H2O(l) → O2(g) + 4 H+(aq) + 4e−

An air/gasoline vapor mix in an automobile cylinder has an initial temperature of 180 ∘C and a volume of 13 cm3 . If the mixture is heated to 587 ∘C with the pressure and amount held constant, what will be the final volume of the gas in cubic centimeters? Express your answer in cubic centimeters to three significant figures.

Answers

Answer:

The final volume will be 24.7 cm³

Explanation:

Step 1: Data given:

Initial temperature = 180 °C

initial volume = 13 cm³ = 13 mL

The mixture is heated to a fina,l temperature of 587 °C

Pressure and amount = constant

Step 2: Calculate final volume

V1/T1 = V2/T2

with V1 = the initial volume V1 = 13 mL = 13*10^-3

with T1 = the initial temperature = 180 °C = 453 Kelvin

with V2 = the final volume = TO BE DETERMINED

with T2 = the final temperature = 587 °C = 860 Kelvin

V2 = (V1*T2)/T1

V2 = (13 mL *860 Kelvin) /453 Kelvin

V2 = 24.68 mL = 24.7 cm³

The final volume will be 24.7 cm³

An element with the valence electron configuration 4s1 would form a monatomic ion with a charge of_______ . In order to form this ion, the element will _________ (lose/gain)______ electron(s) from/into the ________subshell(s).
If an element with the valence configuration 4s23d6 loses 3 electron(s), these electron(s) would be removed from the ________ subshell(s).

Answers

Answer:

+1, lose, 1, 4s, 4s and 3d

Explanation:

An element with the valence electron configuration 4s¹ would form a monatomic ion with a charge of +1. In order to form this ion, the element will lose 1 electron from the 4s subshell.

The corresponding oxidation reaction is:

K ⇒ K¹⁺ + 1 e⁻

[Ar] 4s¹ ⇒ [Ar]

If an element with the valence configuration 4s² 3d⁶ loses 3 electrons, these electrons would be removed from the 4s and 3d subshell(s).

The corresponding oxidation reaction is:

Fe ⇒ Fe³⁺ + 3 e⁻

[Ar] 4s² 3d⁶ ⇒ [Ar] 4s⁰ 3d⁵

Final answer:

An element with 4s1 valence configuration will form a +1 ion by losing 1 electron from 4s subshell. An element with 4s2 3d6 configuration will lose electrons firstly from 4s subshell and then from 3d, if it loses 3 electrons.

Explanation:

An element with the valence electron configuration 4s1 would form a monatomic ion with a charge of +1. In order to form this ion, the element will lose 1 electron from the 4s subshell.
Meanwhile, an element with the valence configuration 4s2 3d6 that loses 3 electrons will remove these electrons first from the 4s subshell (2 electrons), and then one from the 3d subshell. This is due to the fact that 4s electrons are generally removed before 3d electrons in transitional metals.

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At 518°C, the rate of decomposition of a sample of gaseousacetaldehyde, initially at a pressure of 363 Torr, was 1.07 Torr s−1when 5.0 percent had reacted and 0.76 Torr s−1when 20.0 per cent had reacted. Determinethe order of the reaction.web.bumc.bu.edu/otlt/MPH-Modules/BS/BS704_Probability/BS704_Probability4.html.

Answers

Answer:

2nd order reaction

Explanation:

Let us assume the reaction to be:

                         R → P

Where R is the reactant and P is the product.

So here, say initially we have "a" amount of reactant.

                        R → P

At t=0:             a      0       (initial condition)

At t=t:            a-x      x

Say x be the amount of reactant which forms the product in time t.

So from the rate law, we have

                 rate of decomposition = k (R)ⁿ

Where k is rate constant , R is amount of reactant at time t and n is the order of the reaction

From the question, at the instant when 5% and 20% have reacted, we will be left with 95% and 80% of the reactant respectively. So writing the rate law equation:

1.07 = k ( 95a / 100)ⁿ

0.76 = k ( 80a/100)ⁿ

Dividing these two equations, we get:

(1.07 / 0.76 ) = ( 95/80 )ⁿ

Taking logarithm on both sides we get

n = ( ㏒ (1.07 / 0.76) ) ÷ ㏒(95/80) = 2.0067 ≈ 2

Therefore the reaction is of order 2

Identify the type of interactions involved in each of the following processes taking place during the dissolution of sodium chloride (NaCl) in water.

I. Solvent-solvent interactions
II. Solute-solute interactions
III. Solute-solvent interactions

A. Interactions between the ions of sodium chloride
B. Interactions involving dipole-dipole attractions
C. Interactions formed during hydration
D. Interactions involving ion-ion attractions
E. Interactions associated with an exothermic process during the dissolution of sodium chloride
F. Interactions between the water molecules
G. Interactions formed between the sodium ions and the oxygen atoms of water molecules

Answers

Answer:

I) Solvent-solvent interactions involves B & F.

B. Interactions involving dipole-dipole attractions.

F. Interactions between the water molecules.

II) Solute-solute interactions involves A & D.

A. Interactions between the ions of sodium chloride.

D. Interactions involving ion-ion attractions.

III. Solute-solvent interactions involves G & C

G. Interactions formed between the sodium ions and the oxygen atoms of water molecules.

C. Interactions formed during hydration.

The melting of an ice sculpture of BEVO at room temperature requires 10 kJ of energy. Calculate the change in entropy of the surroundings. Please report your answer one point past the decimal with the unit J/K.

Answers

Final answer:

The change in entropy, given the 10 kJ of energy needed to melt the ice sculpture at room temperature, works out to be 1.22x10³ J/K. This underscores the substantial amounts of energy required for phase changes.

Explanation:

Given the problem at hand, the energy required to melt the ice sculpture is given by Q = 10 kJ, and the melting temperature of ice (T) is 273 K. The change in entropy can be calculated using the formula ΔS = Q/T. Substituting the provided values, the change in entropy works out to be 1.22x10³ J/K. This calculation emphasizes the concept that phase changes require significant energy. For instance, the energy needed to turn ice into liquid water, as in this example, is immense compared to the energy associated with mere temperature changes.

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Final answer:

The change in entropy for the melting ice sculpture at room temperature is calculated to be 34.1 J/K. This represents an increase in disorder due to the melting, an endothermic process that breaks bonds in the ice, allowing the molecules to move freely.

Explanation:

The calculation of the change in entropy for the melting ice sculpture is based on the formula ΔS = Q/T, where Q is the heat required for the phase change (melting), and T is the absolute temperature in Kelvin during the process. Given that Q is 10 kJ (or 10,000 J) and T is room temperature, which is approximately 293 K, we substitute these values into the formula.

So, ΔS = Q/T = 10000 J / 293 K = 34.1 J/K.

This represents the change in entropy of the surroundings when the ice sculpture melts at room temperature. It's important to note that the process of melting is an endothermic process requiring energy to break the bonds in the ice, which contributes to this increase in entropy: the system becomes more disordered as the structured ice lattice becomes freely moving water molecules.

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What is the electron pair arrangement around the central atom in the molecule IF5?
A.) Trigonal pyramidal
B.) Square Planar
C.) Octahedral
D.) Square Pyramidal

Answers

Answer:

The correct option is: C.) Octahedral

Explanation:

Iodine pentafluoride (IF₅) is an inorganic interhalogen compound in which the oxidation state of iodine is +5 and the oxidation state of fluorine is (-1).

In this molecule, iodine is sp³d² hybridized and covalently bonded to five fluorine atoms. So there are 5 bond pair of electrons and 1 lone pair of electron around iodine.

Thus the steric number = 6

According to the VSEPR theory, the electron pair arrangement of a molecule with steric number 6 is octahedral.

Therefore, electron pair arrangement around iodine in IF₅ molecule is octahedral.

Final answer:

The molecule IF5 (Iodine Pentafluoride) has a square pyramidal electron pair arrangement around the central atom, Iodine, resulting from its electron configuration. In this arrangement, five fluorine atoms surround iodine in the shape of a pyramid with a square base. Therefore, the answer would be D) Square Pyramidal.

Explanation:

The molecule IF5 (Iodine Pentafluoride) follows the square pyramidal electron pair arrangement around the central atom (Iodine). In this arrangement, five Fluorine atoms are arranged around Iodine in the shape of a pyramid with a square base. This structure comes from the electron configuration of the central atom. Iodine has seven valence electrons, five of which are shared with the Fluorine atoms (forming covalent bonds), and the remaining two occupy a lone pair of non-bonding electrons. Therefore, the answer would be D) Square Pyramidal.

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Nitric oxide (NO) can be formed from nitrogen, hydrogen and oxygen in two steps. In the first step, nitrogen and hydrogen react to form ammonia:N2(g) + 3H2(g) →2NH3(g)ΔH=−92.kJIn the second step, ammonia and oxygen react to form nitric oxide and water:4NH3(g) + 5O2(g) → 4NO(g) +6H2O(g)ΔH=−905.kJCalculate the net change in enthalpy for the formation of one mole of nitric oxide from nitrogen, hydrogen and oxygen from these reactions.Round your answer to the nearest kJ.

Answers

Answer:

ΔH  = - 272 kJ

Explanation:

We are going to use the fact that Hess law allows us to calculate the enthalpy change of a reaction no matter if the reaction takes place in one step or in several steps. To do this problem we wll add two times the first step to second step as follows:

N2(g) +         3H2(g) →          2NH3(g)                 ΔH=−92.kJ  Multiplying by 2:      

2N2(g) +       6H2(g) →          4NH3(g)                        ΔH=− 184 kK

plus

4NH3(g) +     5O2(g) →          4NO(g) +6H2O(g)        ΔH=−905.kJ

__________________________________________________

2N2(g) +   6H2(g) + 5O2(g)→  4NO(g)  + 6H2O(g)      ΔH = (-184 +(-905 )) kJ

                                                                                     ΔH =    -1089 kJ

Notice how the intermediate NH3 cancels out.

As we can see this equation is for the formation of 4 mol NO, and we are asked to calculate the ΔH  for the formation  of one mol NO:

-1089 kJ/4 mol NO  x 1 mol NO =  -272 kJ (rounded to nearest kJ)

Final answer:

The net enthalpy change for the formation of one mole of nitric oxide from nitrogen, hydrogen, and oxygen is -318.25 kJ, which is obtained by summing up the enthalpy changes of the two relevant reactions and adjusting for the number of moles of NO produced.

Explanation:

To calculate the net change in enthalpy for the formation of one mole of nitric oxide from nitrogen, hydrogen and oxygen, we necessitate to add the enthalpy changes of the two reactions. For the initial reaction, nitrogen and hydrogen react to produce ammonia, with ΔH = -92 kJ. In the second stage, ammonia reacts with oxygen to form nitric oxide and water, with ΔH = -905 kJ.

Since the second equation leads to formation of 4 moles of nitric oxide, we divide the enthalpy change for this reaction by 4 to get the enthalpy change for one mole of nitric oxide, which is -905/4 = -226.25 kJ.

The total enthalpy change for the formation of a mole of nitric oxide hence will be the sum of ΔH for the both reactions, i.e., -92 kJ + -226.25 kJ = -318.25 kJ.

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Arizona was the site of a 400,000-acre wildfire in June 2002. How much carbon dioxide (CO2) was produced into the atmosphere by that fire? [Hints: Assume that the density of carbon on the acreage was 10 kg/m2 and that 50% of the biomass burned. In addition, 10,000 m2 = 2.47 acre].

Answers

Answer:

2.97 × 10¹³ g

Explanation:

First, we have to calculate the biomass the is burned. We can establish the following relations:

2.47 acre = 10,000 m² 10 kg of C occupy an area of 1 m²50% of the biomass is burned

The biomass burned in the site of 400,000 acre is:

[tex]400,000acre\times\frac{10,000m^{2} }{2.47acre} \times \frac{10kgC}{m^{2} } \times 50\% = 8.10 \times 10^{9} kgC[/tex]

Let's consider the combustion of carbon.

C(s) + O₂(g) ⇒ CO₂(g)

We can establish the following relations:

The molar mass of C is 12.01 g/mol1 mole of C produces 1 mole of CO₂The molar mass of CO₂ is 44.01 g/mol

The mass of  produced is CO₂:

[tex]8.10 \times 10^{12}gC \times \frac{1molC}{12.01gC} \times \frac{1molCO_{2}}{1molC} \times \frac{44.01gCO_{2}}{1molCO_{2}} =2.97 \times 10^{13} gCO_{2}[/tex]

A chemist carefully measures the amount of heat needed to raise the temperature of a 894.0 g sample of a pure substance from -5.8C to 17.5C. The experiment shows that 4.90kJs of heat are needed. What can the chemist report for the specific heat capacity of the substance? (Round your answer to 3 significant digits.)

Answers

Answer : The specific heat of substance is [tex]0.235J/g^oC[/tex]

Explanation :

Formula used :

[tex]Q=m\times c\times \Delta T[/tex]

or,

[tex]Q=m\times c\times (T_2-T_1)[/tex]

where,

Q = heat needed = 4.90 kJ  = 4900 J

m = mass of sample = 894.0 g

c = specific heat of substance = ?

[tex]T_1[/tex] = initial temperature  = [tex]-5.8^oC[/tex]

[tex]T_2[/tex] = final temperature  = [tex]17.5^oC[/tex]

Now put all the given value in the above formula, we get:

[tex]4900J=894.0g\times c\times [17.5-(-5.8)]^oC[/tex]

[tex]c=0.235J/g^oC[/tex]

Therefore, the specific heat of substance is [tex]0.235J/g^oC[/tex]

Consider a reaction that has a positive ΔH and a positive ΔS. Which of the following statements is TRUE? Consider a reaction that has a positive ΔH and a positive ΔS. Which of the following statements is TRUE? This reaction will be spontaneous only at low temperatures. This reaction will be spontaneous at all temperatures. This reaction will be nonspontaneous only at low temperatures. This reaction will be nonspontaneous at all temperatures. It is not possible to determine without more information.

Answers

Answer:

This reaction will be nonspontaneous only at low temperatures.

Explanation:

An equation that helps us determine the spontaneity of a reaction is:

ΔG = ΔH - TΔS

A reaction will be spontaneous when ΔG is negative.

A reaction will be nonspontaneous when ΔG is positive.

With a positive ΔS and ΔH, ΔG will only be positive when the multiplication TxΔS is lower than ΔH. That happens when T is a low value. That's why the answer is This reaction will be nonspontaneous only at low temperatures.

The reaction 2NO(g)+O2(g)−→−2NO2(g) is second order in NO and first order in O2. When [NO]=0.040M, and [O2]=0.035M, the observed rate of disappearance of NO is 9.3×10−5M/s.
(a) What is the rate of disappearance of O2 at this moment?
(b) What is the value of the rate constant?
(c) What are the units of the rate constant?
(d) What would happen to the rate if the concentration of NO were increased by a factor of 1.8?

Answers

Answer:

(a) The rate of disappearance of [tex]O_{2}[/tex] is: [tex]4.65*10^{-5}[/tex] M/s

(b) The value of rate constant is: 0.83036 [tex]M^{-2}s^{-1}[/tex]

(c) The units of rate constant is:  [tex]M^{-2}s^{-1}[/tex]

(d) The rate will increase by a factor of 3.24

Explanation:

The rate of a reaction can be expressed in terms of the concentrations of the reactants and products in accordance with the balanced equation.

For the given reaction:

[tex]2NO(g)+O_{2}->2NO_{2}[/tex]

rate = [tex]-\frac{1}{2} \frac{d}{dt}[NO][/tex] = [tex]-\frac{d}{dt}[O_{2}][/tex] = [tex]\frac{1}{2}\frac{d}{dt}[NO_{2}][/tex] -----(1)

According to the question, the reaction is second order in NO and first order in  [tex]O_{2}[/tex].

Then we can say that, rate = k[tex][NO]^{2}[O_{2}][/tex] -----(2)

where k is the rate constant.

The rate of disappearance of NO is given:

[tex]-\frac{d}{dt}[NO][/tex] = [tex]9.3*10^{-5}[/tex] M/s.

(a) From (1), we can get the rate of disappearance of [tex]O_{2}[/tex].

    Rate of disappearance of  [tex]O_{2}[/tex] = [tex]-\frac{d}{dt}[O_{2}][/tex] = (0.5)*([tex]9.3*10^{-5}[/tex]) M/s = [tex]4.65*10^{-5}[/tex] M/s.

(b) The rate of the reaction can be obtained from (1).

    rate = [tex]-\frac{1}{2} \frac{d}{dt}[NO][/tex] = (0.5)*([tex]9.3*10^{-5}[/tex])

    rate = [tex]4.65*10^{-5}[/tex] M/s

   The value of rate constant can be obtained by using (2).

    rate constant = k = [tex]\frac{rate}{[NO]^{2}[O_{2}]}[/tex]

    k = [tex]\frac{4.65*10^{-5}}{(0.040)^{2}(0.035)}[/tex] = 0.83036 [tex]M^{-2}s^{-1}[/tex]

(c) The units of the rate constant can be obtained from (2).

    k = [tex]\frac{rate}{[NO]^{2}[O_{2}]}[/tex]

    Substituting the units of rate as M/s and concentrations as M, we get:

[tex]\frac{Ms^{-1} }{M^{3}}[/tex] = [tex]M^{-2}s^{-1}[/tex]

(d) The reaction is second order in NO. Rate is proportional to square of the concentration of NO.

     [tex]rate\alpha [NO]^{2}[/tex]

If the concentration of NO increases by a factor of 1.8, the rate will increase by a factor of [tex](1.8)^{2}[/tex] = 3.24

     

Final answer:

The rate of disappearance of O2 is 4.65×10−5 M/s, the rate constant is 0.19875 M^-1s^-1. The units of the rate constant for a third-order reaction are M^-1s^-1. If NO concentration were to increase by a factor 1.8, the reaction rate would increase by a factor of 3.24.

Explanation:

In a chemical reaction, the rate of disappearance of a reactant matches the stoichiometric ratio. Here, the ratio of NO to O2 is 2:1, therefore, the rate of disappearance of O2 is half that of NO. So, (a) the rate of disappearance of O2 is 9.3×10−5 M/s ÷ 2 = 4.65×10−5 M/s.

From the rate law, rate = k[NO]^2[O2], we can determine the rate constant (k) by substituting the known values into the equation. (b) the value of the rate constant k is rate ÷ ([NO]^2[O2]) = 9.3×10−5 M/s ÷ ((0.040 M)^2(0.035 M)) = 0.19875 M^-1s^-1.

As for (c) the units of the rate constant for a third-order reaction are M^-1s^-1. (d) If the concentration of NO were increased by a factor of 1.8, the rate would increase by a factor of (1.8)^2 due to the reaction being second order in NO. Hence the rate would increase by a factor of 3.24 times.

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