The cylinder shown contains 0.79 moles of nitrogen, 0.19 moles of oxygen and 0.02 moles carbon dioxide, a total of 1.00 mole of molecules in the approximate proportion in which they are present in air. Of the three gases, only carbon dioxide is appreciably soluble in the water in the well at the bottom. Assume an equilibrium between dissolved and undissolved carbon dioxide at the beginning and sufficient time lapse to reestablish that equilibrium after the change described. If 0.02 mole of carbon dioxide is forced into the cylinder, the solubility of carbon dioxide ... a) increases by a factor of about 50. b) increases by a factor of about 2. c) increases by 2%. d) remains unchanged. e) decreases.

A nonmetal and a nonmetal will make molecular compounds like H2O and CO2

Answer:

The element is strontium and the number of neutrons it have is 51.

Explanation:

Based on the given information, the ionic compound is,  

XCl₂ ⇔ X₂⁺ + 2Cl⁻

X2+ is the ion of the mentioned element

As mentioned in the given question, the number of electrons of the element X is 36 and as seen from the reaction the charge present on the ion is +2. Now the atomic number will be,  

No. of electrons = atomic number - charge

36 = atomic number - 2

Atomic number = 38

Based on the periodic table, the atomic number 38 is for strontium element, and the sign of strontium is Sr. Hence, the element X is Sr.  

Now based on the given information, the mass number of the element is 89. Now the no. of neutrons will be,  

No. of neutrons = mass number - atomic number

= 89 - 38

= 51 neutrons.  

Answer:

Mg(s) + 2Ag^+(aq) ---->Mg^2+(aq) + 2Ag(s)

Explanation:

The key to balancing redox reaction equations is this; ensure that the number of electrons lost in the oxidation half reaction equation is equal to the number of electrons gained in the reduction half reaction equation. The both half reaction equations can now be combined to give the overall reaction equation.

For the redox reaction under consideration;

Oxidation half equation;

Mg(s) ------> Mg^2+(aq) + 2e

Reduction half equation;

2Ag^+(aq) +2e ----> 2Ag(s)

Overall balanced redox reaction equation;

Mg(s) + 2Ag^+(aq) ---->Mg^2+(aq) + 2Ag(s)

Answer:

118.6nm

Explanation:

It is possible to calculate wavelength of any energetic process (As an ionization) using:

E = hc / λ (1)

Where E is Energy, h is Planck constant (6.626x10⁻³⁴Js), c speed of light (3x10⁸ms⁻¹) and λ is wavelength In meters.

As the energy to ionize 1 mole of iodine is 1009kJ, one atom requires:

(1009kJ / mol) ₓ (1mol / 6.022x10²³ atoms) = 1.6755x10⁻²¹kJ / atom. = 1.6755x10⁻¹⁸J

Replacing in (1):

λ = hc / E

λ = 6.626x10⁻³⁴Js*3x10⁸ms⁻¹ / 1.6755x10⁻¹⁸J

λ = 1.186x10⁻⁷m

As 1m = 1x10⁹nm:

1.186x10⁻⁷m ₓ (1x10⁹nm / 1m) =

Answer:

B. Valence Electron Pairs

Explanation:

Valence-shell electron-pair repulsion, or VSEPR, describes the shape of molecules by determining the repulsion of valence electrons. Therefore, our answer is B.

Answer:

The correct answer is -  13.33 kJ of heat

Explanation:

To know which one is the limiting reagent, determine the number of moles of each reagent in order .

n(K) = mass/atomic weight = 1.41/39 = 0.036 moles

Density of ICl = Mass/Volume

3.24 = Mass/6.52

Mass of ICl = 21.12 g

n(ICl) = mass/molar mass = 21.12/162.35 = 0.130 moles

2 moles of K reacts with 1 mole of ICl

0.036 moles of K will react with = 0.036/2 = 0.018 moles of ICl

since the amount of moles of ICl is more than 0.018, it is in excess and hence K is the limiting reagent. Now, use the balance equation to determine the amount of heat liberated:

2 moles of K gives out -740.71 kJ of heat

1 mole of K will give out = -740.71/2 = 370.36 kJ of heat

0.036 moles of K will give out = 0.036 × 370.36 = 13.33 kJ of heat

Thus, the correct answer is -  13.33 kJ of heat

The amount of heat liberated at constant pressure is -13.33 kJ

The given parameters are:

Start by calculating the number (n) of moles of each reagent using:

[tex]n = \frac{Mass}{Atomic\ weight }[/tex]

For the potassium metal, we have:

[tex]n_k = \frac{1.41g}{39g/mole}[/tex] ---where 39 is the atomic weight of potassium

[tex]n_k = 0.036\ moles[/tex]

For the liquid iodine monochloride, we start by calculating its mass using:

[tex]Mass = Density \times Volume[/tex]

So, we have:

[tex]Mass = 3.24 \times 6.52[/tex]

[tex]Mass = 21.12g[/tex]

The number of moles is then calculated as:

[tex]n_I=\frac{21.12g}{162.35}[/tex]

[tex]n_I = 0.130\ moles[/tex]

The reaction equation 2K(s) + ICl(l) → KCl(s) + KI(s) means that:

2 moles of potassium reacts with 1 mole of liquid iodine monochloride.

So,  0.036 moles of potassium will react with the following moles of liquid iodine monochloride.

[tex]A = \frac{0.036}{2}[/tex]

[tex]A = 0.018\ moles[/tex]

i.e. 0.036 moles of potassium will react with of liquid iodine monochloride

By comparison: 0.018 moles is less than 0.036 moles

So, the amount of heat liberated at constant pressure is:

[tex]Amount = 0.036 \times -\frac{740.71}{2} kJ[/tex]

[tex]Amount = -13.33 kJ[/tex]

Hence, the amount of heat liberated at constant pressure is -13.33 kJ

Read more about heat at:

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Answer:

Option B. 2096.1 K

Explanation:

Data obtained from the question include the following:

Enthalpy (H) = +1287 kJmol¯¹ = +1287000 Jmol¯¹

Entropy (S) = +614 JK¯¹mol¯¹

Temperature (T) =.?

Entropy is related to enthalphy and temperature by the following equation:

Change in entropy (ΔS) = change in enthalphy (ΔH) / Temperature (T)

ΔS = ΔH / T

With the above formula, we can obtain the temperature at which the reaction will be feasible as follow:

ΔS = ΔH / T

614 = 1287000/ T

Cross multiply

614 x T = 1287000

Divide both side by 614

T = 1287000/614

T = 2096.1 K

Therefore, the temperature at which the reaction will be feasible is 2096.1 K

Answer:

680g of NaNO3.

Explanation:

The balanced equation for the reaction is given below:

2NaNO3 —> 2NaNO2 + O2

Next, we shall determine the mass of NaNO3 that decomposed and the mass of O2 produced from the balanced equation. This is illustrated below:

Molar mass of NaNO3 = 23 + 14 + (16x3) = 85g/mol

Mass of NaNO3 from the balanced equation = 2 x 85 = 170g

Molar mass of O2 = 16x2 = 32g/mol

Mass of O2 from the balanced equation = 1 x 32 = 32g

From the balanced equation above,

170g of NaNO3 decomposed to produce 32g of O2.

Now, we can obtain the mass of NaNO3 needed to produce 128g of O2 as shown below:

From the balanced equation above,

170g of NaNO3 decomposed to produce 32g of O2.

Therefore, Xg of NaNO3 will decompose to produce 128g of O2 i.e

Xg of NaNO3 = (170 x 128)/32

Xg of NaNO3 = 680g

Therefore, 680g of NaNO3 are needed to produce 128g of O2.

Answer:

The new mass/volume percent is 2.0% (m/v)

Explanation:

Dilution is a procedure by which the concentration of a solution is decreased, generally with the addition of a diluent. In other words, dilution is a process in which a concentrated solution is always started, to which a greater volume of solvent is added, causing the concentration and volume of the resulting solution to change. But the amount of solute used to prepare the initial solution remains the same.

The calculation of a dilution is made by:

Cinitial. Vinitial = Cfinal. Vfinal

where C indicates concentration and V indicates volume.

In this case:

Replacing:

6.0% (m/v) * 110 mL= Cfinal* 330 mL

Solving:

[tex]Cfinal=\frac{ 6.0 (m/v)*110 mL}{330 mL}[/tex]

Cfinal=   2.0% (m/v)

The new mass/volume percent is 2.0% (m/v)

Answer:

See explanation below

Explanation:

IUPAC came up with the idea of an unambiguous system of nomenclature for organic compounds. This unambiguous system relates the structure of a compound with its name. Thus, IUPAC has established a worldwide standard for the unambiguous naming of organic compounds. Scientists all over the world can now have a uniform system of nomenclature for compounds in order to facilitate easy communication of scientific information.

The systematic names of the following compounds listed in the question are shown below;

(R)-2- butyl bromide has the systematic name (R)-2-bromobutane

(S)-2-butyl bromide has the systematic name (S)-2-bromobutane

This unified system of nomenclature avoids the confusion created by the use of different trivial names in deferent localities and by various scientific academies. This is a major advantage of the systematic nomenclature.

Answer:

[tex]\large \boxed{9.780 \times 10^{4}\text{ u}}[/tex]

Explanation:

The molecular mass of a monomer unit is:

C₂H₃Cl = 2×12.01 + 3×1.008 + 35.45 = 24.02 + 3.024 + 35.45 = 62.494 u

For 1565 units,

[tex]\text{Molecular mass} = \text{1565 units} \times \dfrac{\text{62.494 u}}{\text{1 unit }} = \mathbf{9.780 \times 10^{4}}\textbf{ u}\\\\\text{The molecular mass of the chain is $\large \boxed{\mathbf{9.780 \times 10^{4}}\textbf{ u}}$}[/tex]

Answer:

2.17x10⁻¹⁰ = Kb1

1.69x10⁻¹³ = Kb2

Explanation:

Oxalic acid, C₂O₄H₂, has two intercambiable protons, its equilibriums are:

C₂O₄H₂ ⇄ C₂O₄H⁻ + H⁺ Ka1 = 5.9x10⁻²

C₂O₄H⁻ ⇄ C₂O₄²⁻ + H⁺ Ka2 = 4.6x10⁻⁵

Oxalate ion, C₂O₄²⁻, has as equilibriums:

C₂O₄²⁻ + H₂O ⇄ C₂O₄H⁻ + OH⁻ Kb1

C₂O₄H⁻ + H₂O ⇄ C₂O₄H₂ + OH⁻ Kb2

Also, you can know: KaₓKb = Kw

Where Kw is 1x10⁻¹⁴

Thus:

Kw = Kb2ₓKa1

1x10⁻¹⁴ =Kb2ₓ4.6x10⁻⁵

And:

Kw = Kb1ₓKa2

1x10⁻¹⁴ =Kb1ₓ5.9x10⁻²

That is because the inverse reaction of, for example, Ka1:

C₂O₄H⁻ + H⁺ ⇄ C₂O₄H₂ K = 1 / Ka1

+ H₂O ⇄ H⁺ + OH⁻ K = Kw = 1x10⁻¹⁴

=

C₂O₄H⁻ + H₂O ⇄ C₂O₄H₂ + OH⁻ Kb2 = Kw × 1/Ka1

Answer:

1 mole of CaC₂ will produce 26g of C₂H₂ or 64.1g of CaC₂ will produce 26g of C₂H₂

Explanation:

Hello,

To solve this question, we'll require a balanced chemical equation of reaction between calcium carbide and water.

Equation of reaction

CaC₂ + 2H₂O → Ca(OH)₂ + C₂H₂

Molar mass of calcium carbide (CaC₂) = 64.1g/mol

Molar mass of water (H₂O) = 18g/mol

Molar mass of calcium hydroxide (Ca(OH)₂) = 74g/mol

Molar mass of ethyne (C₂H₂) = 26g/mol

From the equation of reaction, 1 mole of CaC₂ will produce 1 mole of C₂H₂

1 mole of CaC₂ = mass / molar mass

Mass = 1 × 64.1

Mass = 64.1g

1 mole of C₂H₂ = mass / molar mass

Mass = 1 × 26

Mass = 26g

Therefore, 1 mole of CaC₂ will produce 26g of C₂H₂

Note: this is a hypothetical calculation since we were not given the initial mass of CaC₂ that starts the reaction

Answer:

pH at equivalence point is 8.47

Explanation:

Benzoic acid react with NaOH, thus:

HC₇H₅O₂ + NaOH → C₇H₅O₂⁻ + H₂O + Na⁺

You reach equivalence point when moles of the acid = moles of NaOH.

Moles of benzoic acid are:

0.025L ₓ (0.0988mol / L) = 0.00247 moles

To have 0.00247 moles of NaOH in solution and reach equivalence point you need to add:

0.00247 moles NaOH ₓ (1L / 0.115mol) = 0.0215L of NaOH solution.

Total volume is 0.0465L.

There are produced 0.00247 moles of C₇H₅O₂⁻ and its molarity will be:

0.00247 mol C₇H₅O₂⁻ / 0.0465L = 0.0531M C₇H₅O₂⁻

C₇H₅O₂⁻ is in equilibrium with water, thus:

C₇H₅O₂⁻(aq) + H₂O ⇄ HC₇H₅O₂(aq) + OH⁻(aq)

Where Kb = Kw / Ka = 1x10⁻¹⁴ / 6.5x10⁻⁵ = 1.54x10⁻¹⁰ is:

Kb = 1.54x10⁻¹⁰ = [HC₇H₅O₂] [OH⁻] / [C₇H₅O₂⁻]

The concentrations in equilibrium of the species are:

[HC₇H₅O₂] = X

[OH⁻] = X

[C₇H₅O₂⁻] = 0.0531M - X

Where X represents how much C₇H₅O₂⁻ react, X is reaction coordinate

Replacing in Kb expression:

1.54x10⁻¹⁰ = [HC₇H₅O₂] [OH⁻] / [C₇H₅O₂⁻]

1.54x10⁻¹⁰ = [X] [X] / [0.0531 - X]

8.169x10⁻¹² - 1.54x10⁻¹⁰X = X²

8.169x10⁻¹² - 1.54x10⁻¹⁰X - X² = 0

Solving for X:

X = -2.858x10⁻⁶M → False solution, there is no negative concentrations

X = 2.858x10⁻⁶M → Right solution

As [OH⁻] = X

[OH⁻] = 2.858x10⁻⁶M

pOH is -log [OH⁻]

pOH = 5.54

pH = 14 - pOH

pH = 8.46

Answer:

Polarity in chemistry referred to physical properties of compounds related to solubility, melting and boiling properties.

Polarity of black pepper can be seen when black pepper is sprinkled on water. The balck pepper float on water and get displaced if touched.

It means black pepper is non-polar and have no difference in electronegativity between bonded atoms. Black pepper is so light in weight and non-polar, the surface tension of water keep it floating in the water.

Answer:

THE MOLAR MASS OF THE UNKNOWN COMPOUND IS 242.02 g/mol.

Explanation:

First:

Calculate the change in freezing point:

          Freezing point of pure benzene = 5.5°C

Change in temperature = 5.5 - 3.06 = 2.44 °C

Second:

Using the formula:

Δt = i Kf m

Let's assume i = 1

Kf = 5.12 °C/m

M = x / 0.250 kg of benzene

Then we can calculate x which is the molarity

Re-arranging the formula, we have:

m = Δt / i Kf

x / 0.250 = 2.44 / 1 * 5.12

x = 2.44 * 0.250 / 5.12

x = 0.61 / 5.12

x = 0.119 M

Since it is well known that molarity is the mass of a substance divided by its molar mass. We can then calculate the molar mass.

Molar mass = Mass / molarity

Molar mass = 28.8 g / 0.119 M

Molar mass =242.02 g/mol

Hence, the molar mass of the unknown molecular compound is 242.02 g/mol.

Answer: The amount of sample left after 8323 years is 4.32g

Explanation:

Expression for rate law for first order kinetics is given by:

[tex]t=\frac{2.303}{k}\log\frac{a}{a-x}[/tex]

where,

k = rate constant

t = age of sample

a = let initial amount of the reactant

a - x = amount left after decay process  

a) for completion of half life:

Half life is the amount of time taken by a radioactive material to decay to half of its original value.

[tex]t_{\frac{1}{2}}=\frac{0.693}{k}[/tex]

[tex]k=\frac{0.693}{8694years}=7.97\times 10^{-5}years^{-1}[/tex]

b) amount left after 8323 years

[tex]t=\frac{2.303}{7.97\times 10^{-5}}\log\frac{8.30g}{a-x}[/tex]

[tex]8323=\frac{2.303}{7.97\times 10^{-5}}\log\frac{8.30g}{a-x}[/tex]

[tex]0.285=\log\frac{8.30}{a-x}[/tex]

[tex]\frac{8.30}{a-x}=1.92[/tex]

[tex](a-x)=4.32g[/tex]

The amount of sample left after 8323 years is 4.32g

Answer:

211.4g/mol.

Explanation:

Data obtained from the question includes:

Mass of unknown compound = 0.98g

Mass of solvent = 10.30g

Molality = 0.45 M

Next, we shall determine the number of mole of the unknown compound present in the solution.

This can be obtained as follow:

Molality = mole /kg of solvent

Mole of the unknown compound =.?

Mass of solvent = 10.30g = 10.30/1000 = 0.0103Kg

Molality = 0.45 M

Molality = mole /kg of solvent

0.45 = mole /0.0103

Cross multiply

Mole = 0.45 x 0.0103

Mole = 4.635×10¯³ mole

Therefore the mole of the unknown compound that dissolve in solution is 4.635×10¯³ mole

Now, we can obtain the molar mass of the unknown compound as follow:

Mole of the unknown compound = 4.635×10¯³ mole

Mass of unknown compound = 0.98g

Molar mass of the unknown compound =?

Mole = mass /Molar mass

4.635×10¯³ = 0.98 /Molar mass

Cross multiply

4.635×10¯³ x molar mass = 0.98

Divide both side by 4.635×10¯³

Molar mass = 0.98 / 4.635×10¯³

Molar mass = 211.4g/mol.

Therefore, the molar mass of the unknown compound is 211.4g/mol.

The molecular mass of the unknown has been 211.66 g/mol.

Molality can be defined as the moles of the solute per kg of solvent.

Molality can be expressed as:

Molality = [tex]\rm \dfrac{Mass\;of\;solute\;(g)}{molecular\;mass\;of\;solute}\;\times\;\dfrac{1000}{Mass\;of\;solvent\;(g)}[/tex] ......(i)

The given unknown has been the solute.

The mass of solute = 0.98 g.

The mass of solvent = 10.30 g.

The molality of the solution formed has been = 0.45 m.

Substituting the values in equation (i):

0.45 m = [tex]\rm \dfrac{0.98\;g}{molecular\;mass\;of\;solute}\;\times\;\dfrac{1000}{10.30\;g}[/tex]

0.45 m = [tex]\rm \dfrac{0.98\;g}{molecular\;mass\;of\;solute}\;\times\;97.087[/tex]

[tex]\rm \dfrac{0.98\;g}{molecular\;mass\;of\;solute}[/tex]  =  [tex]\rm \dfrac{0.45}{97.087}[/tex]

[tex]\rm \dfrac{0.98\;g}{molecular\;mass\;of\;solute}[/tex]  = 0.00463

Molecular mass of solute = [tex]\rm \dfrac{0.98}{0.00463}[/tex]

Molecular mass of solute = 211.66 g/mol.

The molecular mass of the unknown has been 211.66 g/mol.

For more information about the molality of the solution, refer to the link:

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Answer:

3,4,1 and 6,5,2

Explanation:

In the electromagnetic spectrum the arrangement of the waves in increasing frequencies and decreasing wavelengths are as follows;

Radio waves

Microwaves

Infrared waves

Visible light rays

Ultraviolet rays

X-rays

Gamma rays

(a simple mnemonic is RMIVUXG)

Answer:

You can take 1mL of your stock solution in a 100mL volumetric flask and complete to volume.

Explanation:

You need to create a 0.00200M solution of Fe(NO₃)₃. First, you have to obtain the concentration of the first solution you made. That is:

0.8g Fe(NO₃)₃.9H₂O × (1mol / 403.9972g) =

0.0020 moles of Fe(NO₃)₃.9H₂O = Moles of Fe(NO₃)₃

In 10mL = 0.010L:

0.0020 moles of Fe(NO₃)₃ / 0.010L = 0.20M Fe(NO₃)₃

This is the concentration of your stock solution, as you want to obtain a 0.0020M solution, you dilution factor must be:

0.20M / 0.0020M = 100

That means you need to dilute your stock solution 100 times.

You can make this dilution, for example,

Answer:

THE SPECIFIC HEAT OF THE METAL IS 0.4843 J/G °C

Explanation:

Heat involved in the reaction is 169.5 J

mass of the metal = 10 g

Change in temperature = 60 °C - 25 °C = 35 °C

Specific heat of the metal = unknown

Specific heat of a metal or substance is the amount of heat per unit mass required to raise the temperature by one degree Celsius.

Using the formula for heat of a reaction, which is:

Heat = mass * specific heat * change in temperature

Heat = m C ΔT

Re-arranging the formula by making C the subject of the formula, we get:

C = Heat / m ΔT

C = 169.5 J / 10 g * 35 C

C = 169.5 J / 350 g °C

C = 0.4843 J/g°C

So therefore, the specific heat of the metal is 0.4843 J/ g°C

Answer:

Option E. 2.04 L

Explanation:

Data obtained from the question include:

Molarity of NaCl = 2.25 M

Mole of NaCl = 4.58 moles

Volume =..?

Molarity is simply defined as the mole of solute per unit litre of the solution. It is represented mathematically as:

Molarity = mole /Volume

With the above formula, we can obtain the volume of the solution as follow:

Molarity = mole /Volume

2.25 = 4.58/volume

Cross multiply

2.25 x volume = 4.58

Divide both side by 2.25

Volume = 4.58/2.25

Volume = 2.04 L

Therefore, the volume of the solution is 2.04 L

Answer:

10.96 grams of compound C will be produced from 18.24 grams of compound A and excess compound B.

Explanation:

3A + 2B ⇒ 5C

By stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction) the following amounts of reagent and products participate in the reaction:

The excess reagent will be that which is not completely depleted during the reaction.

The amount of product obtained from the reaction will always depend on the amount of limiting reagent in the reaction. Then, being B the excess reagent and therefore A the limiting reagent and knowing that compound A has a molar mass of 159.7 g/mole and compound C has a molar mass of 57.6 g/mole, by stoichiometry the following mass amounts of A and C participate in the reaction:

Then it is possible to apply the following rule of three: if by stoichiometry of the reaction 479.1 grams of A produce 288 grams of C, 18.24 grams of A, how much mass of C does it produce?

[tex]mass of C=\frac{18.24 grams of A*288 grams of C}{479.1 grams of A}[/tex]

mass of C= 10.96 grams

10.96 grams of compound C will be produced from 18.24 grams of compound A and excess compound B.

Answer:

true

Explanation:

This is true because elements aim to have a full octet of electrons in their outermost (also called valence) shell. Noble gases already have a full valence shell which is why the elements that are not noble gases aim to be like them.

Answer:

a) true

Explanation:

this is the answer coz elements aim to have a full octet of electrons in their outermost (also called valence) shell.

Answer:

0.48 dm3  (or 480 cm3)

Explanation:

First find the original no. of moles existing in the sulphuric acid:

no. of moles = volume (in dm3) x concentration

                     = 120/1000 x 10

                     = 1.2 mol

Then let the total volume of the diluted acid be v dm3.

Since

Concentration = no. of moles / volume,

so by substituting the given information,

2 = 1.2 / v

v = 0.6 dm3

Hence, the volume of water required

= 0.6 - 120/1000

= 0.48 dm3  (or 480 cm3)

Considering the definition of dilution, 600 cm³ of water is required to dilute 120 cm³ of 10 [tex]\frac{mol}{dm^{3} }[/tex] sulphuric acid to a concentration of 2 [tex]\frac{mol}{dm^{3} }[/tex].

First of all, you have to know that when it is desired to prepare a less concentrated solution from a more concentrated one, it is called dilution.

Dilution is the procedure followed to prepare a less concentrated solution from a more concentrated one and consists of reducing the amount of solute per unit volume of solution. This is accomplished simply by adding more solvent to the solution in the same amount of solute.

In a dilution the amount of solute does not change, but as more solvent is added, the concentration of the solute decreases, as the volume of the solution increases.

A dilution is mathematically expressed as:

Ci×Vi = Cf×Vf

where

In this case, you know:

Replacing in the definition of dilution:

10[tex]\frac{mol}{dm^{3} }[/tex]× 120 cm³= 2 [tex]\frac{mol}{dm^{3} }[/tex]× Vf

Solving:

Vf= (10[tex]\frac{mol}{dm^{3} }[/tex]× 120 cm³) ÷2 [tex]\frac{mol}{dm^{3} }[/tex]

Vf= 600 cm³

In summary, 600 cm³ of water is required to dilute 120 cm³ of 10 [tex]\frac{mol}{dm^{3} }[/tex] sulphuric acid to a concentration of 2 [tex]\frac{mol}{dm^{3} }[/tex].

Learn more about dilution:

Answer:

This reaction isn't yet at an equilibrium. It must shift in the direction of the reactant (namely [tex]\rm NOBr\; (g)[/tex]) in order to reach an equilibrium.

For this mixture, the reaction quotient is [tex]Q_c = 0.0126[/tex].

Explanation:

A reversible reaction is at equilibrium if and only if its reaction quotient [tex]Q_c[/tex] is equal to the equilibrium constant [tex]K_c[/tex].

Start by calculating the equilibrium quotient [tex]Q_c[/tex] of this reaction. Given the reaction:

[tex]\rm 2\; NOBr\; (g) \rightleftharpoons 2\; NO\; (g) + Br_2\; (g)[/tex].

Let [tex][\mathrm{NOBr\; (g)}][/tex], [tex][\mathrm{NO\; (g)}][/tex], and [tex][\mathrm{Br_2\; (g)}][/tex] denote the concentration of the three species. The formula for the reaction quotient of this system will be:

[tex]\displaystyle Q_c = \frac{[\mathrm{NO\; (g)}]^2 \cdot [\mathrm{Br_2\; (g)}]}{[\mathrm{NOBr\; (g)}]^2}[/tex].

(Note, that in this formula, both [tex][\mathrm{NO\; (g)}][/tex] and [tex][\mathrm{NOBr\; (g)}][/tex] are raised to a power of two. That corresponds to the coefficients in the balanced reaction.)

Calculate the reaction quotient given the concentration of each species:

[tex]\displaystyle Q_c = \frac{[\mathrm{NO\; (g)}]^2 \cdot [\mathrm{Br_2\; (g)}]}{[\mathrm{NOBr\; (g)}]^2} \approx 1.26\times 10^{-2} = 0.0126[/tex].

(Note that the unit is ignored.)

Apparently, [tex]Q_c > K_c[/tex]. Since [tex]Q_c[/tex] and [tex]K_c[/tex] are not equal, this reaction is not at an equilibrium. If external factors like temperature stays the same,

Keep in mind that [tex]Q_c[/tex] denotes a quotient. To reduce the value of a quotient, one may:

In [tex]Q_c[/tex], that means reducing the concentration of the products while increasing the concentration of the reactants. In other words, the system needs to shift in the direction of the reactants before it could reach an equilibrium.

Answer:

Atomic radius of Strontium is 27.38pm

Explanation:

In a face-centered cubic structure, the edge, a, could be obtained using pythagoras theorem knowing the hypotenuse of the unit cell, b, is equal to 4r:

a² + a² = b² = (4r)²

2a² = 16r²

a = √8 r

As edge length of Strontium is 77.43pm:

77.43pm / √8 = r

27.38pm = r

Answer:

B

Explanation:

Heat capacity = mass x specific heat capacity.

(C = mc)

87.2 = 368 x c

= 0.237 J/g.°C

Answer:

Equilibrium constant Kc = [x]² / [A - x] [B - x]

Explanation:

The equilibrium constant is defined as the ratio of the concentration of the products to that of the reactants at equilibrium

ie Kc = [products] / [reactants].

The balanced equation of the reaction is given as : A + B ⇄ AB

At the beginning of the reaction,

Initial concentration I = A = 1M

                                       B = 1M

                                      AB = 0M

After a period of time and assuming 'x' to be the concentration of product AB formed, the concentrations become

                                         C = reactant A = [A - x] M

                                                 rectant B =   [B - x] M

                                              Product AB =  [x] [x] M

At equilibrium, the concentrations are,

                                            E  = rectant A = [A - x] M

                                                   reactant B = [B - x] M

                                                   product AB = [x]² M

therefore , the equilibrium constant, Kc  = [products]/[reactants]

                                                                   = [x]² / [A - x] [B - x]