Can someone please help me!!!

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

Answer:

[tex]\alpha=17.7[/tex]

Explanation:

From the question we are told that:

Mass [tex]m=13g[/tex]

Volume [tex]V=10mL[/tex]

Angle [tex]\theta=23[/tex]

Sample Tube=10cm

Generally the equation for concentration is mathematically given by

 [tex]C=m/v[/tex]

 [tex]C=\frac{13}{10}\\C=1.3g/mL[/tex]

Therefore the Specific Rotation

 [tex]\alpha=frac{\theta }{m*l}[/tex]

 [tex]\alpha=frac{23 }{1.3*1.0}[/tex]

 [tex]\alpha=17.7[/tex]

Answer:

Mass of C = 47.37g

Mass of H = 10.59g

Mass of O = 42.04g

The total mass of these elements is 100g, taking a proportion of their molar masses.

C = 47.37/12= 3.95

H = 10.59/1 = 10.59

O = 42.04/16= 2.63.

Dividing through with the smallest proportion which is 2.63

C=3.95/2.63 = 1.5

H =10.59/2.63 =4

O = 2.63/2.63= 1

Multiplying through by 2 to get a whole number.

C = 1.5x2 = 3

H= 4x2 = 8

O = 1x2= 2

The empirical formula is C3H6O2

(Empirical formula)n= molecular mass

(C3H8O2)n =228.276

(12x3 +8+16x2)n= 228.276

76n = 228.276

n = 228.276/76

n = 3

Molecular formula = Empirical formula

=(C3H8O2)3 = C9H24O6

The molecular formula is C9H24O6

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.

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.

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.

[tex] \boxed{\boxed{\mathfrak{ 1\: mole \:of \:Na_2Cr_2O_2\: = \:it's \:Gram\: Mol. \: mass}} }[/tex]

[tex]\underline{ \mathfrak{ Gram \:molecular \:mass \:of \: \red{ Na_2Cr_2O_2}}}[/tex]

= 2 × 23 + 2 × 52 + 2 × 16

= 182 grams

1 mole of [tex]Na_2Cr_2O_2[/tex] weighs = 182 g

8 moles weigh = 8× 182

=[tex] \mathfrak{\blue {\boxed{\underline {1456 \: grams}}}} [/tex]

or

[tex] \mathfrak{\blue {\boxed{\underline {1. 46 \:kg }}}} [/tex]

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

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]

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

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

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.

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

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.

Answer:

10.71%

Explanation:

The dissociation of acetic acid can be well expressed as follow:

CH₃COOH ⇄   CH₃COO⁻  + H⁺

Let assume that the prepared amount of the aqueous solution is 14mM since it is not given:

Then:

The I.C.E Table is expressed as follows:

                     CH₃COOH       ⇄   CH₃COO⁻        +           H⁺  

Initial              0.0014                       0                                0

Change            - x                           +x                               +x

Equilibrium   (0.0014 - x)                 x                                 x

Recall that:

Ka for acetic acid CH₃COOH  = 1.8×10⁻⁵

∴

[tex]K_a = \dfrac{[x][x]]}{[0.0014-x]}[/tex]

[tex]1.8*10^{-5} = \dfrac{[x][x]]}{[0.0014-x]}[/tex]

[tex]1.8*10^{-5} = \dfrac{[x]^2}{[0.0014-x]}[/tex]

[tex]1.8*10^{-5}(0.0014-x) = x^2[/tex]

[tex]2.52*10^{-8} -1.8*10^{-5}x = x^2[/tex]

[tex]2.52*10^{-8} -1.8*10^{-5}x - x^2 =0[/tex]

By rearrangement:

[tex]- x^2 -1.8*10^{-5}x +2.52*10^{-8}= 0[/tex]

Multiplying through  by (-) and solving the quadratic equation:

[tex]x^2 +1.8*10^{-5}x-2.52*10^{-8}= 0[/tex]

[tex](-0.00015 + x) (0.000168 + x) =0[/tex]

x = 0.00015 or x = -0.000168

We will only consider the positive value;

so x=[CH₃COO⁻] = [H⁺] = 0.00015

CH₃COOH = (0.0014 - 0.00015) = 0.00125

However, the percentage fraction of the dissociated acetic acid is:

[tex]= \dfrac{ 0.00015}{0.0014}\times 100[/tex]

= 10.71%

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

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."

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]

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.

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!

QUESTION:- Which particle has a mass of 9.11 x 10^-28g and charge of -1?

OPTIONS:-

A. electron

B. proton

C. neutron

ANSWER:-

CHARGE ON PROTRON IS +1 AND IT HAS MASS OF [tex]1.6 \times 10 {}^{ - 27} [/tex] SO IT CANNOT BE URE ANSWER

THERE IS NO CHARGE ON NEUTRON AND HAS MASS ALMOST EQUAL TO THE PROTON SO IT ALSO CANNOT BE URE ANSWER

MASS OF THE ELECTRON:- [tex]9.11 \times 10^{ - 28} [/tex]

CHARGE ON ELECTRON:- [tex] -1[/tex]

SO URE ANSWER IS ELECTRON

Answer:

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

Answer:

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

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

Answer:

hydrogen bonding

Explanation:

just took the test :D

Answer:

[tex]\boxed {\boxed {\sf {One \ atom \ donates \ both \ electrons \ in \ a \ pair}}}[/tex]

Explanation:

A covalent bond involves the sharing of electrons to make the atoms more stable, and so they satisfy the Octet Rule (8 valence electrons).

Typically each atom contributes an electron to form an electron pair. This is a single bond. There are also double bonds (two pairs of electrons), triple bonds (three pairs of electrons), and coordinate covalent bonds.

Sometimes, to satisfy the Octet Rule and achieve stability, one atom contributes both of the electrons in an electron pair. This is different from other covalent bonds because usually each of the 2 atoms contributes an electron to make a pair.

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.

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.

To learn more about emission spectrum, follow the link;

https://brainly.com/question/27268130

#SPJ2