1L of bleach has a mass of 1,100 grams, 7.25% of the mass of bleach is NaClO, 1 mol of NaClO has a mass of 74.44 grams. What is the molarity (mol/L) of NaClO in the bleach? A.0.097 B.0.93 C. 1.07 D.79.75

Answer:

Molarity of sodium acetate you will need to add is 0.0324M

Explanation:

Assuming volume of the buffer is 1L.

The pH of a buffer can be determined using Henderson-Hasselbalch equation:

pH = pKa + log [A⁻] / [HA]

Where pKa is pKa of the weak acid,  [A⁻] molar concentration of conjugate base and [HA] molar concentration of weak acid

Replacing for the acetic buffer (pKa = 4.76):

pH = 4.76 + log [Sodium Acetate] / [Acetic Acid]

As you have 0.010 moles of acetic acid in 1L:

[Acetic Acid] = 0.010mol / 1L = 0.010M

And you require a pH of 5.27:

5.27 = 4.76 + log [Sodium Acetate] / [0.010M]

0.51 = log [Sodium Acetate] / [0.010M]

10^0.51 = [Sodium Acetate] / [0.010M]

3.236 =  [Sodium Acetate] / [0.010M]

3.236 [0.010M] = [Sodium Acetate]

0.0324M = [Sodium Acetate]

Answer:

Trial 1: Moles acetic acid = 0.00686 moles;

Moles of Mg(OH)₂ = 0.00343 moles

Neutralization capacity = 0.00137 mol/g

Trial 2: Moles acetic acid = 0.00722 moles

Moles of Mg(OH)₂ = 0.00361 moles

Neutralization capacity = 0.00144 mol/g

Explanation:

Equation of the reaction: 2CH₃COOH + Mg(OH)₂ ---> Mg(CH₃COO)₂ + 2H₂O

Trial 1:

Moles of acetic acid = concentration * volume in litres

concentration of acetic acid = 0.88 M

volume of acid used = 7.8 mL = (7.8/1000) Litres = 0.0078 L

Moles acetic acid = 0.88 M * 0.0078 L

Moles acetic acid = 0.00686 moles

Moles of Mg(OH)₂:

From the equation of reaction, 2 moles of acetic acid reacts with 1 mole of Mg(OH)₂

Therefore, 0.00686 moles of acetic acid will react with 0.00686/2 moles of Mg(OH)₂ = 0.00343 moles of Mg(OH)₂

Moles of Mg(OH)₂ = 0.00343 moles

Neutralization capacity = moles of Mg(OH)₂/mass of milk of magnesia

Neutralization capacity = 0.00343 mole /2.5 g

Neutralization capacity = 0.00137 mol/g

Trial 2.

Moles of acetic acid = concentration * volume in litres

concentration of acetic acid = 0.88 M

volume of acid used = 8.2 mL = (8.2/1000) Litres = 0.0082 L

Moles acetic acid = 0.88 * 0.0082

Moles acetic acid = 0.00722 moles

Moles of Mg(OH)₂:

From the equation of reaction, 2 moles of acetic acid reacts with 1 mole of Mg(OH)₂

Therefore, 0.00722 moles of acetic acid will react with 0.00722/2 moles of Mg(OH)₂ = 0.00361 moles of Mg(OH)₂

Moles of Mg(OH)₂ = 0.00361 moles

Neutralization capacity = moles of Mg(OH)₂/mass of milk of magnesia

Neutralization capacity = 0.00361 mole /2.5 g

Neutralization capacity = 0.00144 mol/g

Answer:

C

Explanation:

I had this question and C is the right answer

One idea that Dalton taught about atoms was that all atoms of one type were identical in mass and properties.

An atom is defined as the smallest unit of matter which forms an element. Every form of matter whether solid,liquid , gas consists of atoms . Each atom has a nucleus which is composed of protons and neutrons and shells in which the electrons revolve.

The protons are positively charged and neutrons are neutral and hence the nucleus is positively charged. The electrons which revolve around the nucleus are negatively charged and hence the atom as a whole is neutral and stable due to presence of oppositely charged particles.

Atoms of the same element are similar as they have number of sub- atomic particles which on combination do not alter the chemical properties of the substances.

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Answer:please see below for answers in the spaces given.

Explanation:

There are three types of cubic-unit cells of a cubic system which include Simple cubic unit cell, body-centered cubic unit cell and face-centered cubic-unit cell and Thier characteristics are completed below.

1) In the simple cubic unit cell, the centers of _______eight _____ identical particles define the _________corners___ of a cube. The particles do touch along the cube's _______edges_____ but do not touch along the cube's ____diagonal_______ or through the center. There is/are _______one_____ particle per unit cell and the coordination number is

__six______ .

2. In the body-centered cubic unit cell, the centers of _______eight _____ identical particles define the _______corners_____ of the cube plus ______one______ particle at the _______center_____ of ______the cube______ . The particles do not touch along the cube's _______edges_____ or faces but do touch along the cube's ____diagonal________ . There is/are _____two_______ particles per unit cell and the coordination number is _____eight_______ .

3. In the face-centered cubic cell, the centers of ______eight______ identical particles define the _______corner____ of the cube plus ________one____ particle in the _____center_______ of ______each face______ . The particles on the _____corners_______ do not touch each other but do touch those on the ______faces____ . There is/are ________four___ particles per unit cell and the coordination number is _____twelve_______ .

Answer:

2MnO4^- (aq) + 3C2O4^2- (aq) + 2H2O (l) --> 2MnO2(s) +6CO3^2 -(aq) + 4H^+ (aq)

Explanation:

First, write the half equations for the reduction of MnO4^- and the oxidation of C2O4^2- respectively. Balance it.

Reduction requires H+ ions and e- and gives out water, vice versa for oxidation.

Reduction:

MnO4^- (aq) + 4H^+ (aq) + 3e- ---> MnO2(s) + 2H2O (l)

Oxidation:

C2O4^2- (aq) + 2H2O (l) ---> 2CO3^2 -(aq) + 4H^+ (aq) + 2e-

Balance the no. of electrons on both equations so that electrons can be eliminated. we can do so by multiplying the reduction eq by 2, and oxidation eq by 3.

2MnO4^- (aq) + 8H^+ (aq) + 6e- ---> 2MnO2(s) + 4H2O (l)

3C2O4^2- (aq) + 6H2O (l) ---> 6CO3^2 -(aq) + 12H^+ (aq) + 6e-

Now combine both equations and eliminate repeating H+ and H2O.

2MnO4^- (aq) + 8H^+ (aq) + 3C2O4^2- (aq) + 6H2O (l) --> 2MnO2(s) + 4H2O (l) +6CO3^2 -(aq) + 12H^+ (aq)

turns into:

2MnO4^- (aq) + 3C2O4^2- (aq) + 2H2O (l) --> 2MnO2(s) +6CO3^2 -(aq) + 4H^+ (aq)

Answer:

The equilbrium concentration of NH₃(g) is 2.4 M

Explanation:

The balanced reaction is:

4 NH₃(g) + 7 O₂(g) ⇔ 2 N₂O₄(g) + 6 H₂O(g)

By stoichiometry of the reaction,  2 moles of N₂O₄ are formed from 4 moles of NH₃.

Considering that the concentration is [tex]concentration=\frac{number of moles}{volume}[/tex] and with a volume of 1 liter, it is possible to apply the following rule of three: if 2 M of N₂O₄ are formed from 4 M of NH₃, 0.6 M of N₂O₄ from what concentration  of NH₃ are formed?

[tex]concentration of NH_{3}=\frac{0.6 M of N_{2}O_{3} *4MofNH_{3} }{2 M of N_{2}O_{3} }[/tex]

concentration of NH₃= 1.2 M

By subtracting the moles of NH3 in equilibrium from the moles of NH₃ initially, you will see how many moles of NH₃ were converted and remain in equilibrium: 3.6 M - 1.2 M= 2.4 M

The equilbrium concentration of NH₃(g) is 2.4 M

Answer:

[HI] = 0.264M

Explanation:

Based on the equilibrium:

2HI(g) ⇄ H₂(g) + I₂(g)

It is possible to define Kc of the reaction as the ratio between concentration of products and reactants using coefficients of each compound, thus:

Kc = 0.0156 = [H₂] [I₂] / [HI]²

As initial concentration of HI is 0.660mol / 2.00L = 0.330M, the equlibrium concentrations will be:

[HI] = 0.330M - 2X

[H₂] = X

[I₂] = X

Where X is reaction coefficient.

Replacing in Kc:

0.0156 = [X] [X] / [0.330M - 2X]²

0.0156 = X² / [0.1089 - 1.32X + 4X² ]

0.00169884 - 0.020592 X + 0.0624 X² = X²

0.00169884 - 0.020592 X - 0.9376 X² = 0

Solving for X:

X = - 0.055 → False solution, there is no negative concentrations

X = 0.0330 → Right solution.

Replacing in HI formula:

[HI] = 0.330M - 2×0.033M

Answer:

D

Explanation:

Double Displacement reaction

Both sides are balanced with option D

The balanced form of the chemical equation shown below is [tex]\rm Ca(OH)_2(aq) + Na_2CO_3(aq) \rightarrow CaCO_3(s) + 2NaOH(aq).[/tex] The correct option is D.

A balanced equation is where the reactant and the product have the number of moles of elements. According to the law, the reaction, and the product have the same number of moles after the reaction, so balancing an equation is important.

To balance an equation, it is significant to see the number of moles of reactant and the same number of moles is in the product side. Here the moles of sodium has to be balanced.

Thus, the correct option is D, [tex]\rm Ca(OH)_2(aq) + Na_2CO_3(aq) \rightarrow CaCO_3(s) + 2NaOH(aq).[/tex]

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Answer:

25.99mL is the volume internal volume of the flask

Explanation:

To complete the question:

The temperature of the water was measured to be 21ºC. Use this data to find the internal volume of the stoppered flask

The flask was filled with water, that means the internal volume of the flask is equal to the volume that the water occupies.

To find the volume of the water you need to find the mass and by the use of density of water at 21ºC (0.997992g/mL), you can find the volume of the flask, thus:

Mass water = Mass filled flask - Mass of clean flask

Mass water = 60.167g - 34.232g

Mass water = 25.935g of water.

To convert this mass to volume:

25.935g × (1mL / 0.997992g) =

Answer: When a solute is added to water, the vapor pressure will decrease and the boiling point will increase.

Explanation:

Answer:

24.15%

Explanation:

According to the given situation the computation of the percent yield of the reaction is shown below:-

PV = NRT = N = [tex]\frac{PV}{RT}[/tex]

Mole of [tex]N_2[/tex] = [tex]\frac{PV}{RT}[/tex]

= [tex]\frac{1\times 0.450}{0.0821\times 295}[/tex]

= [tex]\frac{0.450}{24.2195}[/tex]

= 0.0186

Mole of [tex]N_2H_4 = \frac{2.45}{32}[/tex]

= 0.077

Now, the percentage of yield is

= [tex]\frac{Practical\ yield}{Theoretical\ yield}\times 100[/tex]

= [tex]\frac{0.0186}{0.077}\times 100[/tex]

= 24.15%

Therefore for computing the percentage of yield we simply divide the practical yield by theoretical yield and multiply with 100 so that we can get the result into the percentage form.

Answer:

- The molar mass of the solute, in order to convert from moles of solute to grams of solute.

- The density of solution, to convert from volume of solution to mass of solution.

Explanation:

Hello,

In this case, since molarity is mathematically defined as the moles of solute divided by the volume of solution and the weight/weight percent as the mass of solute divided by the mass of solution, we need:

- The molar mass of the solute, in order to convert from moles of solute to grams of solute.

- The density of solution, to convert from volume of solution to mass of solution.

For instance, if a 1-M solution of HCl has a density of 1.125 g/mL, we can compute the w/w% as follows:

[tex]w/w\%=1\frac{molHCl}{L\ sln}*\frac{36.45gHCl}{1molHCl}*\frac{1L\ sln}{1000mL\ sln}*\frac{1mL\ sln}{1.125g\ sln} *100\%\\\\w/w\%=3.15\%[/tex]

Whereas the first factor corresponds to the molar mass of HCl, the second one the conversion from L to mL of solution and the third one the density to express in terms of grams of solution.

Regards.

For the w/w% of the solution, information about the molecular mass of the solute, and density of the solution has been required.

Molarity can be defined as the moles of the solute per liter of the solution. The molarity can be used for the determination of the weight of the solute, by the information about the molecular weight of the compound.

Thus, for the w/w% of the solution, the weight of the solute has been determined with information about the molecular mass of the solute.

The weight of the solvent has been determined with the density of the solution. The density has been defined as the mass per unit volume.

Thus, for the w/w% of the solution, the weight of the solvent has been determined by the density of the solution.

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Answer:

C. rate = k[F₂] [Cl₂O]

Explanation:

Based on the reaction, rate law can be obtained from the initial concentration of reactants thus:

rate = k[F₂]ᵃ [Cl₂O]ᵇ

Where the exponents a and b can be finded doing a experiment changing initial concentrations and seeing how a variation contribute in rate law.

If you analize experiments 1 and 2, the only change is [Cl₂O] (From 0.010 to 0.040, four times more) that changes its concentration in four times. This change produce rate law change from 5x10⁻⁴ to 2.0x10⁻³, also four times. That means the exponent b of [Cl₂O] is 1.

rate = k[F₂]ᵃ [Cl₂O]ᵇ

rate = k[F₂]ᵃ [Cl₂O]¹

Now, comparing experiments 1 and 3, the [F₂] change from 0.05 to 0.10, (Twice), and initial rate change from 5x10⁻⁴ to 1x10⁻³ (Also, twice). That means a = 1 and rate law is:

rate = k[F₂]¹ [Cl₂O]

rate = k[F₂] [Cl₂O]

Thus, right answer is:

Answer:

1.5 mol of CO₂

Explanation:

Use the mole ratio to find how many moles of CO₂ are produced from the reaction.

For every 5 moles of O₂, three moles of CO₂ is produced.

2.5 mol O₂ × 3 mol CO₂ ÷ 5 mol O₂

= 2.5 mol O₂ × 0.6

= 1.5 mol CO₂

When 2.5 mol of O₂ is consumed in the reaction, 1.5 mol of CO₂ is produced.

Hope that helps.

Answer:

filtration, drying, and weighing

Explanation:

The procedures that would facilitate the measurement of the actual yield of the solid would be filtration of the precipitate, drying of the precipitate, and weighing of the precipitate respectively.

The liquid/solid mixture resulting from the reaction can be separated by the process of filtration using a filter paper. The residue in the filter paper would be the solid while the filtrate would be the liquid portion of the reaction's product.

The residue can then be allowed to dry, and then weighed using a laboratory-grade weighing balance. The weight of the solid represents the actual yield of the solid.

Answer:

Kc = 166.7

[Fe³⁺] =  0.18 M

[SCN⁻] = 2×10⁻⁴ M

Explanation:

In the reaction of Fe³⁺ and SCN⁻, it is formed a complex:

Fe³⁺  +  SCN⁻  ⇄  FeSCN²⁺             Kc

Let's make the expression for Kc →  [FeSCN²⁺] / [Fe³⁺] . [SCN⁻]

5.7×10⁻⁵ / 9.5×10⁻⁴. 3.6×10⁻⁴  = 166.7

We determine the mmoles, we add from each reactant:

18 ml . 0.2M = 3.6 mmoles of Fe³⁺

2 ml . 0.002M = 4×10⁻³ mmoles of SCN⁻

General form of the dilution equation is:

Concentrated [C] . Concentrated Volume = Diluted [C] . Diluted Volume

Total volume = 20mL

[Fe³⁺]: 3.6 mmoles /20mL = 0.18 M

[SCN⁻]: 4×10⁻³ mmoles /20mL = 2×10⁻⁴ M

The value should be 1.67 x 10^2

The initial concentration should be 0.18 M and 2.0 x 10^(-4) M

The value is

= 5.7 x 10^(-5)/(9.5 x 10^(-4) x 3.6 x 10^(-4))

= 167

= 1.67 x 10^2

we know that

Initial moles  = volume x concentration

So,

= 18/1000 x 0.200

= 0.0036 mol

Now

Initial moles  = volume x concentration

= 2/1000 x 0.0020

= 4.0 x 10^(-6) mol

So,

Total volume should be

= 18 + 2

= 20 mL

= 0.02 L

Now

Initial concentration   

= moles /total volume

= 0.0036/0.02

= 0.18 M

Now

Initial concentration

= moles  /total volume

= 4.0 x 10^(-6)/0.02

= 2.0 x 10^(-4) M

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Answer: C. Metals tend to easily lose their valence electrons.

Explanation:

Metals are those substances which have tendency to loose their valence electrons to attain noble gas configuration and forms positive ions called as cations.

Example: Gold, potassium etc

[tex]M\rightarrow M^++e^-[/tex]

Non metals are those substances which have tendency to gain valence electrons to attain noble gas configuration and form negative ions called as anions.

Example: Sulphur, Chlorine

[tex]N+e^-\rightarrow N^-[/tex]

Answer:

a) for adiabatic reversible, ΔU(internal energy is constant) = 0, ΔH = 0(no heat is entering or leaving the surrounding)

workdone (w) = -8442.6 J  ≈ -8.443 KJ

heat transferred (q) of the ideal gas = - w

q = 8.443 KJ

b) For ideal gas at adiabatic reversible, Internal energy (ΔU) = 0 and enthalpy (ΔH) = 0

the workdone(w) in the ideal gas= - 4567.5 J  ≈ - 4.57 KJ

the heat transfer (q) of an ideal gas = 4.5675 KJ

Explanation:

given

mole of an ideal gas(n) = 2.5 mol

Temperature (T) = 20°C

= (20°C + 273) K  = 293 K

Initial pressure of the ideal gas(P₁) = 20 atm

Final pressure of the ideal gas(P₂) = 5 atm.

2) (a)for adiabatic reversible process,

note: adiabatic process is a process by which no heat or mass is transferred between the system and its surrounding.

Work done (w) = nRT ln[tex]\frac{P_{1} }{P_{2} }[/tex]

= 2.5 mol × 8.314 J/mol K × 293 K × ln[tex]\frac{5atm}{20atm}[/tex]

= 6090.01 J × [-1.3863]

= -8442.6 J  ≈ -8.443 KJ

So, the work done (w) of ideal gas = -8.443 KJ

For ideal gas at adiabatic reversible, Internal energy (U) = 0 and Enthalpy (H) = 0

From first law of thermodynamics:-

U = q + w

0 = q + w

q = - w

q = - (-8.443 KJ)

q = 8.443 KJ

heat transfer (q) of the ideal gas = 8.443 KJ

(b) For adiabatic irreversible, the temperature T remains constant because the internal energy U depends only on temperature T. Since at constant temperature, the entropy is proportional to the volume, therefore, entropy will increase.

Work done (w) = -nRT(1 - ln[tex]\frac{P_{1} }{P_{2} }[/tex] )

= - 2.5 mol × 8.314 J / mol K× 293 K × [1- (5 atm /20 atm)]

= - 6090.01 J × 0.75

= - 4567.5 J  ≈ - 4.57 KJ

∴work done(w) of an ideal gas = - 4.57 KJ

For ideal gas at adiabatic Irreversible, Internal energy (U) = 0 and Enthalpy (H) = 0

From first law of thermodynamics:-

U = q + w

0 = q + w

q = - w

q = - (-4.5675 KJ)

q = 4.5675 KJ

the heat transfer (q) of an ideal gas = 4.5675 KJ

Answer:

Explanation:

Given that:

mass of the antacid tablet = 5.8400 g

required mass of the antacid tablet = 4.2800 g was added to 200. mL of simulated stomach acid.

The amount of the  original 200. mL of stomach acid (in mL) needed to  neutralize the 4.2800 g crushed sample of the tablet can be calculated as:

= 11.6 mL of NaOH × 25.00 mL /29.0 mL NaOH

= 10.00 mL original stomach acid

Now; since it requires 11.6  mL of  NaOH o neutralize 10.00 mL of  original acid , then:

the antacid neutralized = 200 mL - 10.00 mL

the antacid neutralized = 190.00 mL

Answer:

They are:

H2, N2, O2, F2, Cl2, Br2, and I2.

Note: whether the element At molecule is monoatomic or diatomic is incredibly arguable. While some say it exists as diatomic because it is a halogen like bromine, iodine etc, At is in fact extremely unstable and no one has ever really studied the molecules on it, so, when others say it is monoatomic, this is also based on calculations. But the other 7 elements listen above is for sure diatomic.

Hydrogen (H2) , Nitrogen (N2) , Oxygen (O2) , Fluorine (F2) , Chlorine (Cl2) , Iodine (I2) , carbonmonoxide (CO) and Bromine (Br2).

Hydrogen (H2) , Nitrogen (N2) , Oxygen (O2) , Fluorine (F2) , Chlorine (Cl2) , Iodine (I2) , carbonmonoxide (CO) and Bromine (Br2) are the formulas of the elements that is present as diatomic molecules under normal environmental conditions. Diatomic molecules refers to those molecules that is composed of only two atoms of the same or different elements. There are large number of diatomic molecules which is made up of two similar elements or different elements.

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Explanation:

system at equilibrium, will the reaction shift towards reactants ~

--?'

2. (2 Pts) Consider the reaction N2(g) + 3H2(g) =; 2NH3(g). The production of ammonia is an

exothermic reaction. Will heating the equilibrium system increase o~e amount of

ammonia produced? . .co:(

3. (2 Pts) Consider the reaction N2(g) + 3H2(g) =; 2NH3(g). Ifwe use a catalyst, which way will

the reaction shift? ':'\

.1.+- w~t s~,H (o')l r'eo.c. e~ ei~i"liht-,·u.fn\ P~~,

4. (3 Pts) ff 1ven th e o £ 11 owmg d t a a £ or th ere action: A(g) + 2B(s) =; AB2(g)

Temperature (K) Kc

300 1.5x104

600 55 k ' pr, cl l<..J~

e- ~ r fee, ct o. ~ 1<

900 3.4 X 10-3

Is the reaction endothermic or exothermic (explain your answer)?

t d- IS o.,;r-. \4\a..i~1f't~ °the te.Y'il(lf1,:J'u.r-a a•~S. j lrvdu..c,,.) +~H~to{' '\

exothe-rnh't.-- ,.. ..,. (/.., ,~.

5. (4 Pts) Consider the reaction, N2(g) + 3H2(g) =; 2NH3(g). Kc= 4.2 at 600 K.

What is the value of Kc for 4 NH3(g) =; 2N2(g) + 6H2(g)

N ... ~l + 3 H~(ri ~ ~Nli3~) kl,= ~:s;H,J3 # 4. J..

~ ;)N~~) ~ ~ H ~) ~\-_ == [A!;J:t D~~Jb

J. [,v 1+3] ~

I

4,:i.~ = 0,05

5.00 mol of ammonia are introduced into a 5.00 L reactor vessel and when the equilibrium is reached, 1.00 mole remains. The concentration equilibrium constant is 17.3.

Initially, there are 5.00 mol of ammonia in a 5.00 L reactor vessel. The initial concentration of ammonia is:

[tex][NH_3]_i = \frac{5.00mol}{5.00L} = 1.00 M[/tex]

At equilibrium, there is 1.00 mole of ammonia in the 5.00 L vessel. The concentration of ammonia at equilibrium is:

[tex][NH_3]eq = \frac{1.00mol}{5.00L} = 0.200 M[/tex]

We can calculate the concentrations of all the species at equilibrium using an ICE chart.

       2 NH₃(g) ⇄ 3 H₂(g) + N₂(g)

I          1.00              0          0

C         -2x              +3x        +x

E       1.00-2x          3x          x

Since the concentration of ammonia at equilibrium is 0.200 M,

[tex]1.00-2x = 0.200\\\\x = 0.400 M[/tex]

The concentrations of all the species at equilibrium are:

[tex][NH_3] = 0.200 M\\[H_2] = 3x = 1.20 M\\[N_2] = x = 0.400 M[/tex]

The concentration equilibrium constant (Kc) is:

[tex]Kc = \frac{[H_2]^{2} [N_2]}{[NH_3]^{2} } = \frac{(1.20^{3})(0.400) }{0.200^{2} } = 17.3[/tex]

5.00 mol of ammonia are introduced into a 5.00 L reactor vessel and when the equilibrium is reached, 1.00 mole remains. The concentration equilibrium constant is 17.3.

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Answer:

11.6mL of the 0.1400M NaOH solution

Explanation:

0.154g of chloroacetic acid diluted to 250mL. Titrated wit 0.1400M NaOH solution.

The reaction of chloroacetic acid, ClCH₂COOH (Molar mass: 94.5g/mol) with NaOH is:

ClCH₂COOH + NaOH → ClCH₂COO⁻ + Na⁺ + H₂O

Where 1 mole of the acid reacts per mole of the base.

That means the student will reach equivalence point when moles of chloroacetic acid = moles NaOH.

You will titrate the 0.154g of ClCH₂COOH. In moles (Using molar mass) are:

0.154g ₓ (1mol / 94.5g) = 1.63x10⁻³ moles of ClCH₂COOH

To reach equivalence point, the student must add 1.63x10⁻³ moles of NaOH. These moles comes from:

1.63x10⁻³ moles of NaOH ₓ (1L / 0.1400moles NaOH) = 0.0116L of the 0.1400M NaOH =

Answer:

0.1110 mol

Explanation:

Mass = 2g

Molar mass = 18.016 g/mol

moles = ?

These quantities are realted by the following equation;

Moles = Mass / Molar mass

Substituting the values of the quantities and solving for moles, we have;

Moles = 2 / 18.016 = 0.1110 mol

Answer:

Total momentum of both player after collision =93  Kg m/s

Explanation:

According to law of conservation of momentum

For an isolated system of bodies , momentum of bodies before and after collision remains same.

momentum is given by mass* velocity

_________________________________________

Here the isolated system of bodies are

two football players.

Momentum of player before collision

Momentum of player 1 = 105*8.6 = 903 Kg m/s

Momentum of player 2 = 90*-9 = -810 Kg m/s

Total momentum of both player before collision = 903 + (-810) = 93 Kg m/s

as by conservation of

Total momentum of both player before collision = Total momentum of both player after collision

Total momentum of both player after collision =93  Kg m/s

Answer:A is the Answer

Explanation:

Answer: The Clean Air Act was important because it emphasized cost-effective methods to protect the air; encouraged people to study the effects of dirty air on human health; and created a regulation that makes any activities that pollute the air illegal.

Explanation:

Answer:

Clean Air Act (CAA), U.S. federal law, passed in 1970 and later amended, to prevent air pollution and thereby protect the ozone layer and promote public health. The Clean Air Act (CAA) gave the federal Environmental Protection Agency (EPA) the power it needed to take effective action to fight environmental pollution.

Answer:

The initial temperature was  [tex]36.4^\circ \:C[/tex]

Explanation:

[tex]\Delta t=\frac{q}{m\cdot C_s}=\frac{5.83\times10^5}{2190\times 4.184}\\\\=63.6^\circ\:C[/tex]

The temperature difference [tex]=100-63.6=36.4^\circ\:C[/tex]

Best Regards!

Answer:

D

Explanation:

Answer:

Explanation:

count given by old sample = .97 disintegrations per minute per gram

count given by fresh sample = 6.68 disintegrations per minute per gram

Half life of radioactive carbon = 5568 years

rate of disintegration

dN / dt = λ N

In other words rate of disintegration is proportional to no of radioactive atoms present . As number reduces rate also reduces .

Let initial no of radioactive be N₀ and after time t , number reduces to N

N₀ / N = 6.68 / .97

Now

[tex]N=N_0e^{-\lambda t}[/tex]

[tex]\frac{N}{N_0} =e^{-\lambda t}[/tex]

[tex]\frac{6.68}{.97} = e^{\lambda t}[/tex]

λ is disintegration constant

λ = .693 / half life

= .693 / 5568

= .00012446 year⁻¹

Putting the values in the equation above

[tex]\frac{6.68}{.97} = e^{.00012446\times t}[/tex]

[tex]6.8866 = e^{.00012446\times t}[/tex]

1.929577 = .00012446 t

t = 15503.6 years .  

Answer:

Fluorine, Chlorine, Bromine, or Iodine

Explanation:

These all have an ALMOST full valence shell. And they need one more electron so they'd react with hydrogen

Answer:

its chlorine

Explanation:

just trust me do i look like i would lie too you ;-)

btw i just took the test :-)

Answer:

3.97

Explanation:

pH of buffer solution = pKa+Log(Cb/Ca)

pH of buffer solution = -log(Ka)+log(Cb/Ca)............... Equation 1

Where Ca = concentration of acid, Cb = concentration of base.

Given: Ka = 6.5×10⁻⁵, Ca = 0.25 M, Cb = 0.15 M

Substitute into equation 1

pH of buffer solution = -log(6.5×10⁻⁵)+log(0.15/0.25)

pH of buffer solution = 4.19+(0.22)

pH of buffer solution = 3.97.