Consider a galvanic (voltaic) cell that has the generic metals X and Y as electrodes. If X is more reactive than Y (that is, X more readily reacts to form a cation than Y does), classify the following descriptions by whether they apply to the X or Y electrode.
i. anode
ii. cathode
iii. electrons in the wire flow toward
iv. electrons in the wire flow away
v. cations from salt bridge flow toward
vi. anions from salt bridge flow toward
vii. gains mass
viii. loses mass

Answers

Answer 1

Answer:

X

anode

electrons in the wire flow away

anions from salt bridge flow toward

loses mass

Y

cathode

electrons in the wire flow toward

cations from salt bridge flow toward

gains mass

Explanation:

In a galvanic cell, oxidation occurs at the anode while reduction occurs at the cathode. The metal that is more reactive functions as the anode while the less reactive metal functions as the cathode.

Electrons leave the anode and travel via a wire to the cathode. At the anode cations give up electrons and enter into the solution.

At the cathode, cations pick up electrons and are deposited on the cathode leading to a gain in mass at the cathode.

Positive ions from the salt bridge flow towards the cathode while negative ions from the salt bridge flow towards the anode.


Related Questions

Step 7: Measuring the Volume of Air Near 60°C

Answers

Answer:

57------6.9

330------.87

Explanation:

Answer:

The required volume of air is 3.64 L

Explanation:

Ideal gas equation:-

The relation between pressure, volume, and temperature of the gas is known as the ideal gas equation and it is given as,

   PV=RT

Where R=gas constant

Now,

Let the volume of air is, V

According to the question we have

Temperature, T= 60°C= (60+273) K= 333 K

Atmospheric pressure, P= 760 mm

And gas constant, R= 8.314 J/mole K

Substitute values in ideal gas equation and we get

  760V= 8.314(333)

  V= 2768.562/760

  V= 3.642 L

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What is the energy change when 78.0 g of Hg melt at −38.8°C

Answers

Answer:

The correct answer is - 2.557 KJ

Explanation:

In this case, Hg is melting, the process is endothermic, so the energy change will have a positive sign.

we can calculate this energy by the following formula:

Q = met

where, m = mass,

e = specific heat

t = temperature

then,

Q = 78*0.14* (273-38.8)

here 0.14 = C(Hg)

= 2.557 Kj

How can a Bose-Einstein condensate be formed? A. B super-heating a gas. B. By super-cooling certain types of solid. C. By super-cooling certain types of plasma. D. By super-heating a plasma

Answers

Answer:

C. By super-cooling certain types of plasma.

Explanation:

Bose-Einstein condensate is a state of matter whereby atoms or particles become cooled to a very low energy state leading to their condensation to give a single quantum state.

Note that plasma refers to atoms that have had some or even all of its electrons stripped away leaving only positively charged ions. Simply put, plasma is ionized matter.

When certain types of plasma are super cooled, Bose-Einstein condensate are formed.

chemical formula for Ammonia

Answers

Answer:

NH3

Explanation:

It’s a polyatomic ion. You’ll just have to memorize it!

Answer:

NH3

Explanation:

Ammonia is a compound of nitrogen and hydrogen with the formula NH3

Identify the phase of the copper product after each reaction in the copper cycle.

The addition of HNO3 HNOX3 to Cu ______________
The addition of H2SO4 HX2SOX4 to CuO ____________ The addition of Z n Zn to C u S O 4 CuSOX4 Choose... The addition of N a O H NaOH to C u ( N O 3 ) 2 Cu(NOX3)X2 Choose... The heating of C u ( O H )

Answers

Answer:

addition of HNO3 HNOX3 to Cu - Aueous

addition of H2SO4 HX2SOX4 to CuO - Aqueous

addition of Z n Zn to C u S O 4 CuSOX4 - Solid

addition of N a O H NaOH to C u ( N O 3 ) 2 Cu(NOX3)X2 - Solid

heating of C u ( O H ) - Solid

Explanation:

Copper when introduced with acids form an aqueous solution and fumes are released in air during the chemical reaction. When NaOH is added to copper then solid copper product is released. Copper dissolves on HNO but does not dissolves in HCL.

Consider the reaction “2 SO2 (g) + O2 (g) = 2 SO3 was 0.175 M. After 50 s the concentration of SO2 Date: (g)”. Initial concentration of SO2 (g) (g) became 0.0500 M. Calculate rate of the reaction

Answers

Answer:

The answer is "[tex]1.25 \times 10^{-3} \ \frac{m}{s}[/tex]"

Explanation:

Calculating the rate of the equation:

[tex]=-\frac{1}{2} \frac{\Delta [SO_2]}{\Delta t} =-\frac{\Delta [O_2]}{\Delta t}= +\frac{1}{2} \frac{\Delta [SO_3]}{\Delta t}\\\\=\frac{\Delta [SO_2]}{\Delta t}=\frac{0.0500-0.175\ M}{505}= -2.5 \times 10^{-3} \ \frac{m}{s}\\\\[/tex]

Rate:

[tex]=\frac{-2.5 \times 10^{-3}}{2}=1.25 \times 10^{-3} \ \frac{m}{s}[/tex]

Consider the following reaction:

CO(g)+2H2(g)⇌CH3OH(g)

A reaction mixture in a 5.15-L flask at a certain temperature initially contains 26.6 g CO and 2.36 g H2. At equilibrium, the flask contains 8.63 g CH3OH.

Part A
Calculate the equilibrium constant (Kc) for the reaction at this temperature.

Answers

Answer:

26.6

Explanation:

Step 1: Calculate the molar concentrations

We will use the following expression.

M = mass solute / molar mass solute × liters of solution

[CO]i = 26.6 g / (28.01 g/mol) × 5.15 L = 0.184 M

[H₂]i = 2.36 g / (2.02 g/mol) × 5.15 L = 0.227 M

[CH₃OH]e = 8.63 g / (32.04 g/mol) × 5.15 L = 0.0523 M

Step 2: Make an ICE chart

        CO(g) + 2 H₂(g) ⇄ CH₃OH(g)

I        0.184      0.227           0

C         -x           -2x             +x

E     0.184-x   0.227-2x        x

Since [CH₃OH]e = x, x = 0.0523

Step 3: Calculate all the concentrations at equilibrium

[CO]e = 0.184-x = 0.132 M

[H₂]e = 0.227-2x = 0.122 M

[CH₃OH]e = 0.0523 M

Step 4: Calculate the equilibrium constant (Kc)

Kc = [CH₃OH] / [CO] [H₂]²

Kc = 0.0523 / 0.132 × 0.122² = 26.6

How many moles of Al are necessary to form 45.0 g of AlBr, from this
reaction:
2 Al(s) + 3 Br_(1) ► 2 AlBr_(s)?

Answers

Answer:

0.169 mole of Al

Explanation:

We'll begin by by calculating the number of mole in 45 gof AlBr₃. This can be obtained as follow:

Mass of AlBr₃ = 45 g

Molar mass of AlBr₃ = 27 + (3×80)

= 27 + 240

= 267 g/mol

Mole of AlBr₃ =?

Mole = mass /molar mass

Mole of AlBr₃ = 45 / 267

Mole of AlBr₃ = 0.169 mole

Finally, we shall determine the number of mole of Al needed to produce 45 g (i.e 0.169 mole) of AlBr₃. This can be obtained as illustrated below:

2Al + 3Br₂ —> 2AlBr₃

From the balanced equation above,

2 moles of Al reacted to produce 2 moles of AlBr₃.

Therefore, 0.169 mole of Al will also react to produce 0.169 mole of AlBr₃.

Thus, 0.169 mole of Al is needed for the reaction.

How can this product be achieved using the starting material shown?

Answers

Answer:

this product can be achieved using the starting material shown is by use of NaOH as catalyst.

Answer:

By using NaOH as catalyst.

Explanation:

This product can be achieved using the starting material shown is by the use of the NaOH as catalyst.

state the number of protons contained in the atomic number 8

Answers

Answer:

There are 8 protons

Explanation:

[tex]{ \bf{atomic \: number = proton \: number}}[/tex]

What is an emission spectrum?

A. The total amount of energy emitted by an element
B. The products created when an element is burned
C. The energy absorbed when an electron gains energy
D. The colors of light given off when an element loses energy​

Answers

Answer:

D

Explanation:

The electromagnetic radiation is emitted due to a particle moves from a higher to a lower energy state

An emission spectrum is the colors of light given off when an element loses energy​. Therefore, option D is correct.

What is emission spectrum ?

The electromagnetic radiation spectrum produced when an electron changes from a high energy state to a lower energy state is known as the emission spectrum of a chemical element or chemical compound.

An emission spectrum is the range of radiations that are released in different places when electrons jump back and forth between higher and lower energy levels to achieve stability.

Since what you are seeing is the direct radiation produced by the source, this form of spectrum is also known as an emission spectrum. You can see all the colors in the Sun's spectrum because light from the Sun is produced at practically all energies in the visible spectrum.

Thus, option D is correct.

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What direction would equilibrium moves towards based on the following if we increased the volume of the container.

[tex]2A_{(g)} + 5B_{(g)} + 12C_{(g)}[/tex] ↔ [tex]14AC_{(g)} + 5B_{(s)}[/tex]

Answer choices:
a) reactants
b) no change
c) products
d) decrease in volume

Please help!

Answers

To answer this question, we will first find out the number of gaseous moles on each side of the equilibrium

on the left:

we have 2 moles of A, 5 moles of B and 12 moles of C

which gives us a grand total of 19 gaseous moles

on the right:

here, we have 14 moles of AC gas, we will not count the number of moles of B because it's a solid

giving us 14 gaseous moles on the right

Where does the reaction shift?

more gaseous moles means more space taken, because gas likes to fill all the space it can

if we have more volume, more gas can move around without colliding (reacting) with each other

Hence more volume favors the side with more gaseous moles

here, the left has more gaseous moles. So we can say that the reaction will shift towards the left, or the reactants side

Answer:

Explanation:

given reversible chemical reaction:

2A(g) + 5B(g) + 12C(g)  ↔  14AC(g) + 5B(s)

chemicals in solid form do not take up a lot of volume so change in container volume has no effect

look at chemicals in gas form only:

the total no. of moles of reactants in gas form = 2 + 5 + 12 = 19

the total no. of moles of products in gas form = 14

so an increase in volume of the container will favor the reaction direction with higher volume n high volume means higher no. of moles

the ans is the equilibrium will move towards a) reactants

Question In nickel-cadmium batteries: Select the correct answer below: the anodes are nickel-plated and the cathodes are cadmium-plated the anodes are cadmium-plated and the cathodes are nickel-plated both the anodes and cathodes are plated with a nickel-cadmium alloy none of the above

Answers

Answer:

the anodes are cadmium-plated and the cathodes are nickel-plated

Explanation:

Nickel cadmium battery works on the principle as by the other cell. There is anode and a cathode which is separated by a separator (spiral shaped inside the case). The anode is negative and is cadmium plated while the cathode is positive and is nickel plated. An electrolyte is also used.

So the correct answer is : "The anodes are cadmium-plated and the cathodes are nickel-plated."

Generally the vapor pressure of a liquid is related to: I. the amount of liquid II. atmospheric pressure III. temperature IV. intermolecular forces

Answers

Answer:

Atmospheric pressure, because a liquid is said to be boiling when the vapour pressure equals the atmospheric pressure.

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Answers

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Answer:

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What is true of all matter?
A. It pushes or pulls on objects.
B. You can see it.
C. It gives off heat energy.
D. It has mass.

Answers

It would be D (I think)

Read the chemical equation.
N2 + 3H2 - 2NH3
Using the volume ratio, determine how many liters of NH3 is produced if 4.2 liters of H2 reacts with an excess of N2
if all measurements are taken at the same temperature and pressure?
A 2.8 liters
B 3.2 liters
C 5.4 liters
D 6.3 liters

Answers

Answer:

A 2.8 liters

Explanation:

Step 1: Write the balanced equation

N₂ + 3 H₂ ⇄ 2 NH₃

Step 2: Establish the appropriate volume ratio

At the same temperature and pressure, the volume ratio of H₂ to NH₃ is 3:2.

Step 3: Calculate the volume of ammonia produced from 4.2 L of hydrogen

4.2 L H₂ × 2 L NH₃/3 L H₂ = 2.8 L

Determine the equilibrium constant, Keq, at 25°C for the reaction
2Br- (aq) + I2(s) <--> Br2(l) + 2I- (aq)



Eocell = (0.0257/n) lnKeq, Calculate Eocell from Use this equation to calculate K value.

Eo (I2/I-) = +0.53, Eo (Br2/Br-) = +1.07,

Answers

Explanation:

The given chemical reaction is:

[tex]2Br^- (aq) + I_2(s) <-> Br_2(l) + 2I^- (aq)[/tex]

[tex]E^ocell=oxidation potential of anode + reduction potential of cathode\\[/tex]

The relation between Eo cell and Keq is shown below:

[tex]deltaG=-RTlnK_e_q\\delta=-nFE^o cell\\=>nFE^o cell=RTlnK_e_q\\lnK_e_q=\frac{nF}{RT} E^o cell[/tex]

The value of Eo cell is:

Br- undergoes oxidation and I2 undergoes reduction.

Reduction takes place at cathode.

Oxidation takes place at anode.

Hence,

[tex]E^ocell= (-1.07+0.53)V\\=-0.54V[/tex]

F=96485 C/mol

n=2 mol

R=8.314 J.K-1.mol-1

T=298K

Substitute all these values in the above formula:

[tex]ln K_e_q=\frac{2mol* 96485 C/mol}{8.314 J.K^-^1.mol^-^1x298K} \\\\lnK_e_q=77.8\\K_e_q=e^7^7^.^8\\=>K_e_q=6.13x10^3^3[/tex]

Answer:

Keq=6.13x10^33

A uniform plastic block floats in water with 50.0 % of its volume above the surface of the water. The block is placed in a second liquid and floats with 23.0 % of its volume above the surface of the liquid.
What is the density of the second liquid?
Express your answer with the appropriate units.

Answers

Answer:

density of second liquid = 650 kg/m³

Explanation:

Given that:

The volume of the plastic block submerged inside the water  = 0.5 V

The force on the plastic block  = [tex]\rho_1V_1g[/tex]

[tex]= 0.5p_1 V_g[/tex]

when the block is floating, the weight supporting the force (buoyancy force) is:

W [tex]= 0.5p_1 V_g[/tex]

[tex]\rho Vg = 0.5p_1 V_g[/tex]

[tex]\rho = 0.5 \rho _1[/tex]

where;

water density [tex]\rho _1[/tex] = 1000

[tex]\rho = 0.5 (1000)[/tex]

[tex]\rho = 500 kg/m^3[/tex]

In the second liquid, the volume of plastic block in the water = (100-23)%

= 77% = 0.7 V

The force on the plastic block is:

[tex]= 0.77p_2 V_g[/tex]

when the block is floating, the weight supporting the force (buoyancy force) is:

[tex]W = 0.77p_2 V_g[/tex]

[tex]\rho Vg = 0.77 \rho_2 V_g \\ \\ \rho = 0.77 \rho_2 \\ \\ 500 = 0.77 \rho_2 \\ \\ \rho_2 = 500/0.77[/tex]

[tex]\mathbf{ \rho_2 \simeq 650 \ kg/m^3}[/tex]

what is speed lathe machine.​

Answers

Answer:

A speed lathe is a type of lathe that is designed to operate much faster than its common counterpart.

Write a net ionic equation for the overall reaction that occurs when aqueous solutions of carbonic acid and sodium hydroxide are combined. Assume excess base.

Answers

Answer:

[tex]H_2CO_3(aq)+2OH^-(aq)\rightarrow (CO_3)^{2-}(aq)+2H_2O(l)[/tex]

Explanation:

Hello there!

In this case, according to the given information, it turns out possible for us to set up this net ionic equation, by firstly setting up the complete molecular equation as follows:

[tex]H_2CO_3(aq)+2NaOH(aq)\rightarrow Na_2CO_3(aq)+2H_2O(l)[/tex]

Thus, since carbonic acid is weak it merely ionizes whereas sodium hydroxides ionizes for the 100 % as it is strong; thus, we can write the complete ionic equation:

[tex]H_2CO_3(aq)+2Na^+(aq)+2OH^-(aq)\rightarrow 2Na^+(aq)+(CO_3)^{2-}(aq)+2H_2O(l)[/tex]

Whereas sodium ions act as the spectator ones to be cancelled out for us to obtain:

[tex]H_2CO_3(aq)+2OH^-(aq)\rightarrow (CO_3)^{2-}(aq)+2H_2O(l)[/tex]

Regards!

what is perodic table list all the elements up to 30 with their valencies​

Answers

Answer:

a table of the chemical elements arranged in order of atomic number, usually in rows, so that elements with similar atomic structure (and hence similar chemical properties) appear in vertical columns.

What is the predominant intermolecular force in the liquid state of each of these compounds: ammonia (NH3), methane (CH4), and nitrogen trifluoride (NF3)?

Answers

Answer:

The  predominant intermolecular force in the liquid state of each of these compounds:

ammonia (NH3)

methane (CH4)

and nitrogen trifluoride (NF3)

Explanation:

The types of intermolecular forces:

1.Hydrogen bonding: It is a weak electrostatic force of attraction that exists between the hydrogen atom and a highly electronegative atom like N,O,F.

2.Dipole-dipole interactions: They exist between the oppositely charged dipoles in a polar covalent molecule.

3. London dispersion forces exist between all the atoms and molecules.

NH3 ammonia consists of intermolecular H-bonding.

Methane has London dispersion forces.

Because both carbon and hydrogen has almost similar electronegativity values.

NF3 has dipole-dipole interactions due to the electronegativity variations between nitrogen and fluorine.

A sample of oxygen occupies 1.00 L. If the temperature remains constant, and the pressure on the oxygen is decreased to one third the original pressure, what is the new volume

Answers

Answer:

3.00 L

Explanation:

P₁V₁ = P₂V₂

V₁ = 1.00 L

P₁ = (x) atm

P₂ = [tex]\frac{1}{3}[/tex] · (P₁) = [tex]\frac{x}{3}[/tex]

V₂ = unknown

(x atm)(1.00 L) = ( [tex]\frac{x}{3}[/tex] atm)(V₂)

divide both sides by ( [tex]\frac{x}{3}[/tex] atm)

( 1.00x )( [tex]\frac{3}{x}[/tex] ) = V₂

x cancels out

(1.00)(3) = V₂

V₂ = 3.00 L

Write a balanced half-reaction for the oxidation of liquid water to aqueous hydrogen peroxide in basic aqueous solution. Be sure to add physical state symbols where appropriate.

Answers

Answer:

O2(g) + 2H2O(l) --------> 2H2O2(aq) + 2e

Explanation:

An oxidation reaction reaction refers to a reaction in which electrons are lost. In this case, we are about to see the full balanced half-reaction for the oxidation of liquid water to aqueous hydrogen peroxide in basic aqueous solution.

The full equation is;

O2(g) + 2H2O(l) --------> 2H2O2(aq) + 2e

So, two electrons were lost in the process.

What is "X" in the following reaction?

Answers

x is the answer that we do not know yet!!:)

For each of the following compounds, indicate the pH at which 50% of the compound will be in a form that possesses a charge and at which pH more than 99% of the compound will be in a form that possesses a charge.

ClCH2COOH (pKa = 2.86)
CH3CH2NH+3 (pKa = 10.7)

Express your answer using two decimal places

a. Determine a pH at which 50% of ClCH2COOH will be in a form that possesses a charge.
b. Determine a pH at which pH more than 99% of ClCH2COOH will be in a form that possesses a charge.
c. Determine a pH at which 50% of CH3CH2NH+3 will be in a form that possesses a charge.
d. Determine a pH at which pH more than 99% of CH3CH2NH+3 will be in a form that possesses a charge.

Answers

Answer:

a. 2..86 b. 4.86 c. 10.7 d. 8.7

Explanation:

a. Determine a pH at which 50% of ClCH2COOH will be in a form that possesses a charge.

Using the Henderson-Hasselbalch equation,

pH = pKa + log[A⁻]/[HA]

where [A⁻] = concentration of conjugate base (or charged form) and [HA] = concentration of acid.

At 50% concentration, [A⁻] = [HA] ⇒ [A⁻]/[HA] = 1

So, pH = pKa + log[A⁻]/[HA]

pH = pKa + log1

pH = pKa = 2.86

b. Determine a pH at which pH more than 99% of ClCH2COOH will be in a form that possesses a charge.

Let x be the concentration of the acid. Since 99% of it should possess a charge, the basic concentration is 0.99x while the acidic concentration is remaining 1 % (1 - 0.99)x = 0.01x

Using the Henderson-Hasselbalch equation,

pH = pKa + log[A⁻]/[HA] where [A⁻] = concentration of conjugate base (or charged form) = 0.99x and [HA] = concentration of acid = 0.01x.

pH = pKa + log0.99x/0.01x

pH = pKa + log0.99/0.01

pH = 2.86 + log99

pH = 2.86 + 1.996

pH = 4.856

pH ≅ 4.86

c. Determine a pH at which 50% of CH3CH2NH+3 will be in a form that possesses a charge.

Using the Henderson-Hasselbalch equation,

pH = pKa + log[A⁻]/[HA]

where [A⁻] = concentration of conjugate base and [HA] = concentration of acid.

At 50% concentration, [A⁻] = [HA] ⇒ [A⁻]/[HA] = 1

So, pH = pKa + log[A⁻]/[HA]

pH = pKa + log1

pH = pKa = 10.7

d. Determine a pH at which pH more than 99% of CH3CH2NH+3 will be in a form that possesses a charge.

Let x be the concentration of the acid. Since 99% of it should possess a charge, the basic concentration is 0.01x while the acidic concentration is remaining 99 % (1 - 0.01)x = 0.99x (which possesses the charge).

Using the Henderson-Hasselbalch equation,

pH = pKa + log[A⁻]/[HA] where [A⁻] = concentration of conjugate base = 0.01x and [HA] = concentration of acid = 0.99x.

pH = pKa + log0.01x/0.99x

pH = pKa + log1/99

pH = 10.7 - log99

pH = 10.7 - 1.996

pH = 8.704

pH ≅ 8.7

How many grams of NaF should be added to 500 mL of a 0.100 M solution of HF to make a buffer with a pH of 3.2

Answers

Answer:

2.25g of NaF are needed to prepare the buffer of pH = 3.2

Explanation:

The mixture of a weak acid (HF) with its conjugate base (NaF), produce a buffer. To find the pH of a buffer we must use H-H equation:

pH = pKa + log [A-] / [HA]

Where pH is the pH of the buffer that you want = 3.2, pKa is the pKa of HF = 3.17, and [] could be taken as the moles of A-, the conjugate base (NaF) and the weak acid, HA, (HF).

The moles of HF are:

500mL = 0.500L * (0.100mol/L) = 0.0500 moles HF

Replacing:

3.2 = 3.17 + log [A-] / [0.0500moles]

0.03 = log [A-] / [0.0500moles]

1.017152 = [A-] / [0.0500moles]

[A-] = 0.0500mol * 1.017152

[A-] = 0.0536 moles NaF

The mass could be obtained using the molar mass of NaF (41.99g/mol):

0.0536 moles NaF * (41.99g/mol) =

2.25g of NaF are needed to prepare the buffer of pH = 3.2

a chemist combines 4.9 g of nitrogen gas with 9.4 grams of nitrogen gas to form 11.4 g of ammonia 2.9 g of nitrogen is remaining

Answers

The results agree with the law of conservation of mass Explanation: The law of conservation of mass states that mass is neither created nor destroyed in a chemical reaction. On the reactant side, the total mass of reactants is 14.3g and the total product masses is also 14.3g. That implies that no mass was !most in the reaction. The sum of masses on the left hand side corresponds with sum of masses on the right hand side of the reaction equation.

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How much heat capacity, in joules and in calories, must be added to a 75.0-g iron block with a specific heat of 0.499J/g °C to increase its temperature from 25 °C to its melting temperature of 1535 °C?

Answers

Answer:

56511.75 J

13506.3 Calories

Explanation:

Applying,

Q = cm(t₂-t₁).................. Equation 1

Where Q = amount of heat, m = mass of the iron, c = specific heat capacity of the iron, t₁ = initial temperature, t₂ = final temperature.

From the question,

Given: m = 75 g, c = 0.499 J/g.°C, t₂ = 1535°C, t₁ = 25°C

Substitute these values into equation 1

Q = 75(0.499)(1535-25)

Q = 75(0.499)(1510)

Q = 56511.75 J

Q in Calories is

Q = (56511.75×0.239)

Q = 13506.3 Calories

Other Questions
Use the interactive to solve. What is the difference between 5 and 3? how do the lack of health education,health services and facilities affect the date rate? write in brief PLEASE HELP SOON AND TRY TO EXPLAIN.Prove that if a, b, c, d are non zero real numbers, not all equal, and if a+1/b=b+1/c=c+1/d=d+1/a then |abcd|=1. Thx! 1. Which of the following does law enforcement NOT do to enforce the law? A. Apprehend and detain individuals who violate lawsB. Discover law violationsC. Determine how the law applies to a situationD. Deter law violations Solve : 3X + 3x + 1 + 3x + 2 = 39 how to deal with your addictions???...? H qu tch cc ca tnh t tr l: how many CH4 molecules are in 14.8 g of CH4 If f(x) = 5x squared -3 and g(x) = x squared - 4x -8, find (f-g)(x) The ovals above represent mangrove islands of various sizes and distances from shore. According to the MacArthur/Wilson Island Biogeography Model, which island is expected to have the LOWEST number of insect species at equilibrium 2. A photon of wavelength for 4 x10 -7 m strikes on the metal surface the workfunction of the metal being 2.13 eV.Calculate (i)the energy of the photon ineV(ii) the kinetic energy of the emission(iii) velocity of the photoelectron[1ev=1.602x10-19 J] Find the surface area of the composite figure, 2 cm 6 cm 14 cm 8 cm 8 cm 4 cm 6 cm 2 cm SA = [?] cm A ball is launched from the ground with a horizontal speed of 30 m/s and a vertical speed of 30 m/s. How far horizontally will it travel in 2 seconds?A. 30 mB. 90 mC. 45 mD. 60 m Help !!!!!!!!!!!!!!! 2, Nine people fit comfortably in a 3 ft. by 3 ft. area. Use this value toestimate the size of a crowd that is 8 yards deep and 1 mile long.Determine the Ratio? ARatio = 9 x (3x3). BO Ratio = 9/3CRatio = 9/(3x3)D Ratio = 8/5280 What about Iceland's location makes it particularly well-suited to produce electricity from geothermal energy Please answer ASAP I would like to check my answer with someone. Thank youHow did Japanese authoritarian governmentcompare to Fascism in Germany and Russia?A. Japan lacked a charismatic leadersB. Japan resocialized dissenters, rather than killing themC. Both A and BD. None of the above Which is true of plantation owners in the nineteenth century? Group of answer choices They frequently broke the law by knowingly buying slaves imported from Africa. They were often first-generation British or French immigrants. They typically supported the Republican Party. They were very public about their ambivalence toward slavery. They insisted that slavery was required in order for whites to be truly free. Below are works cited entries for a book in print by one author. Select the one that is completely correct.Entry A:Fleet, Megan L. A Biography of Jimmy Carter. Chicago: Lippincott, 2007. Print.Entry B:Fleet, Megan L. A Biography of Jimmy Carter. Lippincott: Chicago, 2007. Print.Entry C.Fleet, Megan L. "A Biography of Jimmy Carter." Chicago: Lippincott, 2007.Entry D:Fleet, Megan L. A Biography of Jimmy Carter. Chicago, Lippincott. 2007. Print In a lottery game, a single ball is drawn at random from a container that contains 25 identical balls numbered from 1 through 25. Use the equation P(AUB)=P(A) + P(B) - P(ANB), where A and B are any events, to compute the probability that the number drawn is prime or greater than 12. The probability that the number drawn is prime or greater than 12 is : ___________