Given the equation 2KCIO3(s)=2KCI(s) + 3O2(g). A 3.00-g sample of KCIO3 is decomposed and the oxygen at 24 degrees C and 0.982 atm is collected. What volume of oxygen gas will be collected assuming 100% yield?

Answer:

The Avogadro's  number is [tex]N_A = 6.02289 *10^{23}[/tex]

Explanation:

From the question we are told that

   The edge length is  [tex]L = 5.02 * 10^{-8} \ cm= \frac{5.02 * 10^{-8} }{100} = 5.02 * 10^{-10}[/tex]

    The density of the metal is [tex]\rho = 5.30\ g/cm^3 = 5.30 * \frac{g}{cm^3} * \frac{1*10^6}{1*10^3} = 5.30 *10^3kg/m^3[/tex]

     The molar mass of  Ba is  [tex]Z = 137.3 \ g/mol = \frac{137.3}{1000} = 0.1373 \ kg / mol[/tex]

Generally the volume of a unit cell is  

       [tex]V = L^3[/tex]

substituting value

        [tex]V = [5.02 *10^{-10}]^3[/tex]

         [tex]V = 1.265*10^{-28}\ m^3[/tex]  

From the question we are told that 68% of the unit cell is occupied by Ba atoms and that the structure is a metal which implies that the crystalline structure will be  (BCC),

The volume of barium atom is  

        [tex]V_a = \frac{V}{2} * 0.68[/tex]

substituting value

        [tex]V_a = \frac{ 1.265*10^{-28}}{2} * 0.68[/tex]

        [tex]V_a = 4.301 *10^{-29} \ m^3[/tex]

The Molar mass of barium is mathematically represented as

      [tex]Z = N_A V_a * \rho[/tex]

Where [tex]N_A[/tex] is the Avogadro's number

 So  

      [tex]N_A = \frac{ Z}{ V_a * \rho}[/tex]

substituting value

     [tex]N_A = \frac{ 0.1373}{ 4.301*10^{-29} * 5.3*10^{3}}[/tex]

     [tex]N_A = 6.02289 *10^{23}[/tex]

Answer:

simple cations (monatomic cations of elements of only one possible charge)

Explanation:

Simple cations (monatomic cations of elements of only one possible charge)  are named using the unmodified element name and adding the word "ion"

For example, the Na+ is named the sodium ion.

An atom or molecule with a net electric charge as a result of the loss or gain of one or more electrons is known as an ion.

Answer:

2 Hydrogen One oxygen

Explanation:

Answer:

-  Platinum acts as a heterogeneous catalyst in the hydrogenation of ethylene.

- CFCs act as homogeneous catalysts in the conversion of ozone to oxygen gas.

- Magnesium acts as a heterogeneous catalyst in the disproportionantion of hydrogen peroxide.

Explanation:

Hello,

For the given reactions, considering the definition of homogeneous and heterogeneous catalyst, we can identify that is each catalyst behave as follows:

-  Platinum acts as a heterogeneous catalyst in the hydrogenation of ethylene as all the reactants are gaseous but it remains solid.

- CFCs act as homogeneous catalysts in the conversion of ozone to oxygen gas as it remains gaseous as well as both ozone and oxygen.

- Magnesium acts as a heterogeneous catalyst in the disproportionantion of hydrogen peroxide as it is solid whereas the other species are aqueous, liquid and gaseous

Best regards.

Answer:

B

Explanation:

Answer:

Some of the methods in Methods in electrochemistry that can be used for the separation of proteins and enzymes are as follows:

Redox transformations: In this method enzymes or proteins would be adsorbed on the electrode surface and facilitates direct electron transfer that causes denaturation and loss of their electrochemical activities and bioactivities. It is widely used in biosensors and biofuel cells.

Protein electrophoresis: In this process proteins are seperated by placing them in a gel matrix in the presence of an electrical field. In this method a negative charge is applied so that proteins move towards a positive charge.

Answer:

ΔErxn[tex]= -3.90*10^3KJ[/tex]

Explanation:

Given from the question

T1 = 25.87∘C

T2= 38.13∘C.

C= 5.73Kj/C

CHECK THE ATTACHMENT FOR DETAILED EXPLATION

Answer:

It's ferric oxide Fe2O3

Explanation:

I don't say u must have to mark my ans as brainliest but if it has really helped u plz don't forget to thank me plz...

Answer: The lower temperature reduces molecule speeds, reducing the number of effective collisions.

Explanation:

Answer:

(a) [tex]m=2.69m[/tex]

(b) [tex]x_{LiBr}=0.099[/tex]

(c) [tex]\% LiBr=18.9\%[/tex]

Explanation:

Hello,

In this case, given the molality in mol/L, we can compute the required units of concentration assuming a 1-L solution of acetonitrile and lithium bromide that has 1.80 moles of lithium bromide:

(a) For the molality, we first compute the grams of lithium bromide in 1.80 moles by using its molar mass:

[tex]m_{LiBr}=1.80mol*\frac{86.845 g}{1mol}=156.32g[/tex]

Next, we compute the mass of the solution:

[tex]m_{solution}=1L*0.826\frac{g}{mL}*\frac{1000mL}{1L}=826g[/tex]

Then, the mass of the solvent (acetonitrile) in kg:

[tex]m_{solvent}=(826g-156.32g)*\frac{1kg}{1000g}=0.670kg[/tex]

Finally, the molality:

[tex]m=\frac{1.80mol}{0.670kg} \\\\m=2.69m[/tex]

(b) For the mole fraction, we first compute the moles of solvent (acetonitrile):

[tex]n_{solvent}=669.68g*\frac{1mol}{41.05 g} =16.31mol[/tex]

Then, the mole fraction of lithium bromide:

[tex]x_{LiBr}=\frac{1.80mol}{1.80mol+16.31mol}\\ \\x_{LiBr}=0.099[/tex]

(c) Finally, the mass percentage with the previously computed masses:

[tex]\% LiBr=\frac{156.32g}{826g}*100\%\\ \\\% LiBr=18.9\%[/tex]

Regards.

Answer:

Since the tape has extra electrons, it has a negative charge. When you move your finger close to the tape, electrons in your skin are repelled and move away. This makes the skin on your finger tip have a slight positive charge. Since positive and negative attract, the tape moves toward your finger.

Answer:

[tex]Q=-361.56kJ[/tex]

Explanation:

Hello,

In this case, the decomposition of hydrogen peroxide is given by:

[tex]2H_2O_2\rightarrow 2H_2O+O_2[/tex]

Which occurs in gaseous phase, therefore the enthalpy of reaction is:

[tex]\Delta _rH=2\Delta _fH_{H_2O}-2\Delta _fH_{H_2O_2}[/tex]

Oxygen is not included as it is a pure element. The enthalpies of formation for both hydrogen peroxide and water are -136.11 and -241.83 kJ/mol respectively, so we compute the enthalpy of reaction:

[tex]\Delta _rH=2(-241.83kJ/mol)-2(-136.11kJ/mol)=-211.44kJ/mol[/tex]

Then, the total heat that is released for 1.71 mol of hydrogen peroxide is:

[tex]Q=n*\Delta _rH=1.71mol*-211.44kJ/mol\\\\Q=-361.56kJ[/tex]

Whose sign means a released heat.

Regards.

Answer:

The correct answer is 5.447 × 10⁻⁵ vacancies per atom.

Explanation:

Based on the given question, the at 750 degree C the number of vacancies or Nv is 2.8 × 10²⁴ m⁻³. The density of the metal is 5.60 g/cm³ or 5.60 × 10⁶ g/m³. The atomic weight of the metal given is 65.6 gram per mole. In order to determine the fraction of vacancies, the formula to be used is,  

Fv = Nv/N------ (i)  

Here Nv is the number of vacancies and N is the number of atomic sites per unit volume. To find N, the formula to be used is,  

N = NA×P/A, here NA is the Avogadro's number, which is equivalent to 6.022 × 10²³ atoms per mol, P is the density and A is the atomic weight. Now putting the values we get,  

N = 6.022 × 10²³ atoms/mol × 5.60 × 10⁶ g/m³ / 65.6 g/mol

N = 5.14073 × 10²⁸ atoms/m³

Now putting the values of Nv and N in the equation (i) we get,  

Fv = 2.8 × 10²⁴ m⁻³ / 5.14073 × 10²⁸ atoms/m^3

Fv = 5.44669 × 10⁻⁵ vacancies per atom or 5.447 × 10⁻⁵ vacancies/atom.  

Answer:

1188.976 mmHg

Explanation:

Initial pressure P1= 755 mmHg

Initial volume V1 = 6.00 litres

Final volume V2 = 3.81 litres

Final pressure P2= the unknown

Now applying Boyle's Law,we have;

P1V1 = P2V2

Since P2 is the unknown then it has to be made the subject of the formula.

P2=P1V1/ V2

P2= 755 × 6.00/ 3.81

P2= 1188.976 mmHg

Therefore, the new pressure is; 1188.976 mmHg

Answer:

The solubility is  [tex]S = 0.0014 \ g[/tex]

Explanation:

From the question we are told that

   The volume of the solution is  [tex]V = 600 mL[/tex]

    The initial temperature is  [tex]T_i = 37 ^oC[/tex]

     The final temperature is [tex]T_f = 21^oC[/tex]

      The additional precipitate is [tex]m = 0.084 \ kg = 84 \ g[/tex]

Yes because the solubility of the substance X is the amount of X needed to saturate a unit volume of the solvent  (for solubility of a solute to be calculated the solute must be able to saturate the solvent)

now we see that the substance X saturated the solvent because a precipitate was formed which the student threw away

   The solubility at  21 ° C is mathematically represented as

            [tex]S = \frac{m}{m_w * 100 g \ of water }[/tex]

Mass of water([tex]m_w[/tex]) in the solution is mathematically represented as

       [tex]m_w = V * \rho_w[/tex]

Where  [tex]\rho = 1 \frac{g}{mL}[/tex]

So  

      [tex]m_w =600 * 1[/tex]

       [tex]m_w =600g[/tex]

So  

    [tex]S = \frac{84}{600 * 100 g \ of water }[/tex]

    [tex]S = 0.0014 \ g[/tex]

Answer:

  C

Explanation:

The molecule has 8 carbon atoms joined by 7 C-C bonds.

The first two diagrams show 6 carbon atoms, not 8.

The last two diagrams show line segments representing C-C bonds. Only choice C shows 7 such segments.

The appropriate choice is C.

Answer:

C.

Explanation:

Answer:

25,50,25

Explanation:

Answer:

H2CO3 = 2H+ + CO3-

Explanation:

It is simply what carbonic acid breaks down into when placed in water. Since carbonic acid is made up of H and CO3, these are the products.

Explanation:

P(aq)+Q(aq)⇌3R(aq)

This problem involves applying LeChatelier's principle.

LeChatelier's principle states that whenever a system in equilibrium is disturbed, the equilibrium position would change in order to annul that change.

1) Increase [P]

This would cause the equilibrium position to shift to the right. This is because more reactions have been added, to annul that change more products have to be formed.

2) Increase [Q]

This would cause the equilibrium position to shift to the right. This is because more reactions have been added, to annul that change more products have to be formed.

3) Increase [R]

This would cause the equlibrium position to shift to the left. This is because more products have been formed, to annul that change more reactants have to be formed.

4) Decrease [P]

This would cause the equlibrium position to shift to the left. This is because there are now less reactants, to annul that change more reactants have to be formed.

5) Decrease [Q]

This would cause the equilibrium position to shift to the left. This is because there are now less reactants, to annul that change more reactants have to be formed.

6) Decrease [R]

This would cause the equilibrium position to shift to the right. This is because there are now less products, to annul that change more products have to be formed.

7) Triple [P] and reduce [Q] to one third

No shift in the direction of the net reaction because both changes cancels each other.

8) Triple both [Q] and [R]

No shift in the direction of the net reaction because both changes cancels each other.

Answer:

A. [tex]\mathbf{\Delta G^0 = -72.6 \ kJ/mol}[/tex] ; as such the reaction is said to be spontaneous since the value of [tex]\mathbf{\Delta G^0 }[/tex] is negative.

B.  [tex]\mathbf{\Delta G^0_{702 \ K} = -13.29 \ kJ/mol.K}}[/tex] and the reaction is spontaneous

Explanation:

The equation for this chemical reaction is :

[tex]2NO_{(g)} +O_{2(g)} \to 2NO_{2(g)}[/tex]

Using the following relation to calculate [tex]\Delta G^0[/tex];

[tex]\Delta G^0 = [2(\Delta G^0_{NO_{2(g)}}] - [1(\Delta G^0_{O_{2(g)}})+ 2(\Delta G^0_{NO_{g}})][/tex]

At 298 K; the standard Gibbs Free Energy for the formation are as follows:

[tex]\Delta G^0_{NO_{2(g)}} = 51.2 \ kJ/mol[/tex]

[tex]\Delta G^0_{O_{2(g)}} = 0[/tex]

[tex]\Delta G^0_{NO_{g}}= 87.6 \ kJ/mol[/tex]

Replacing them into the above equation;

[tex]\Delta G^0 = [2(51.2 \ kJ/mol}] - [1(0)+ 2(87.6 \ kJ/mol})][/tex]

[tex]\Delta G^0 = [102.4 \ kJ/mol}] - [175.2 \ kJ/mol})][/tex]

[tex]\mathbf{\Delta G^0 = -72.6 \ kJ/mol}[/tex]

Thus; [tex]\mathbf{\Delta G^0 = -72.6 \ kJ/mol}[/tex] ; as such the reaction is said to be spontaneous since the value of [tex]\mathbf{\Delta G^0 }[/tex] is negative.

B.

Using the same above chemical equation;

The relation used for calculating [tex]\mathbf{\Delta G^0}[/tex] of the reaction when the temperature is 702 K is:

[tex]\Delta G^0_{702 \ K} = \Delta H^0_{xn} - T \Delta S^0_{rxn}[/tex]

where;

[tex]\Delta G^0_{702 \ K} =[/tex] Gibbs free energy of the reaction at 702 K

[tex]\Delta H^0_{xn}[/tex] = standard enthalpy of the reaction = -116.2 kJ/mol

[tex]\Delta S^0_{rxn}[/tex] = standard entropy of the reaction = -146.6 J/mol/K

Temperature T = 702 K

[tex]\Delta G^0_{702 \ K} = -1162. \ kJ/mol - 702 \ K ( -146.6 \ J/mol. K (\dfrac{1 \ kJ }{1000 \ J})[/tex]

[tex]\Delta G^0_{702 \ K} = -1162. \ kJ/mol - 702 \ K ( 0.1466 \ kJ/mol.K})[/tex]

[tex]\Delta G^0_{702 \ K} = -13.2868 \ kJ/mol.K}[/tex]

[tex]\mathbf{\Delta G^0_{702 \ K} = -13.29 \ kJ/mol.K}}[/tex]

Thus [tex]\mathbf{\Delta G^0_{702 \ K} = -13.29 \ kJ/mol.K}}[/tex] and the reaction is spontaneous

Answer:

AT STOICHIOMETRIC POINT, THE VOLUME OF ACID ADDED IS 0.01 L

AT HALF-WAY POINT, THE VOLUME OF ACID IS 0.0050 L

Explanation:

In solving titration problems, you must remember this formula;

MaVa = MbVb

Since M a= 0.005 M

Mb = 0.010 M

Vb = 5 mL = 5 /1000 = 0.005 L

Va = unknown.

Solving for Va, we have:

Va = MbVb / Ma

Va = 0.010 * 0.005 / 0.005

Va = 0.01 L

So therefore, the volume of acid added at:

1. the stoichiometric point is 0.01 L

2. half-way point of titration is 0.01 /2 = 0.0050 L

For the pH:

Since HCl is a strong acid, it dissociate into {H30}+ ion.

First calculate the number of moles of hydronium ion

number of mole = concentration of hydronium ion {H30}+ * Volume

n = 0.005 * 0.01  = 0.00005 moles

A. At initial point of the titration, the volume of base added is 0 L

{H30]+ = n(H+)/ V = 0.00005 / 0.01 = 0.005 M

pH  = - log {0.005}

pH = 2.3

B. At the final point, since the volumes and concentrations of acid and base are the same, the pH is equal to 7.

n(H+) = n(OH^-)

pH = 7

Answer: -2.373  x 10^-24J/K(particles

Explanation: Entropy is defined as the degree of randomness of a system which is a function of the state of a system and depends on the number of the random microstates present.

The entropy change for a particle in a system  depends on the initial and final states of a system and is given by Boltzmann equation as  

S = k ln(W) .

where S =Entropy

K IS Boltzmann constant ==1.38 x 10 ^-23J/K

W is the number of microstates available to the system.

 The  change in entropy is given as

S2 -S1 = kln W2 - klnW1

dS = k ln (W2/W1)

where w1 and w2 are initial and final microstates

from the question, W2(final) = 0.842 x W1(initial), so:

= 1.38*10-23 ln (0.842)

=1.38*10-23  x -0.1719

= -2.373  x 10^-24J/K(particles)

Answer:

THE HEAT REQUIRED TO CHANGE 2 KG OF ETHANOL FROM 26 °C TO THE BOILING POINT AND TO VAPOR AT THAT TEMPERATURE IS 1965.175 KJ.

Explanation:

Boiling point of ethanol = 78.29 °C = 78.29 + 273 K = 351.29 K

Mass = 2 kg = 2000 g

Final temp. = 26.0 °C = 26 + 273 K= 299 K

Change in temperature = (78.29 - 26) °C = 52.29 °C

1. Heat required to raise the temperature from 26 °C to the boiling point?

Heat = mass * specific heat * change in temperature

Heat = 2000 * 2.44 * 52.29

Heat = 255 175.2 J

2. Heat required to change the liquid to vapor at that temperature?

Heat = mass * enthalphy of vaporization

Heat = 2000 * 855

Heat =1 710000 J

The total heat required to raise the temperature of 2 kg of ethanol from 26 °C to the boiling point and then to change the liquid to vapor at that temperature will be:

Heat = mcT + m Lv

Heat = 255 175.2 J + 1710000 J

Heat = 1965175.2 J

Heat = 1965.175 kJ of heat.

Answer:

Time taken = 174 days

Explanation:

Half life is the time take for a subsrtance taken to decay to half of it's origial or initial concentration.

In this probel, the haf life is 87 days, this means that after evry 87 days, the concentration or mass of sulfur-35 decreases by half.

If the starting mass is 1.00g, then we have;

1.00g --> 0.5g (First Half life)

0.5g --> 0.25g (Second half life)

This means that sulphur-335 would undergo two half lives for 0.25g to remain.

Total time taken = Number of half lives  * Half life

Time taken = 2 * 87

Time taken = 174 days

Answer:

an atom with a half-filled subshell - hydrogen

an atom with a completely filled outer shell - argon

an atom with its outer electrons occupying a half-filled subshell and a filled subshell- copper

Explanation:

The outermost shell or the valence shell of the atom is the last shell in the atom. Chemical reactions occur at this outer most shell. The number of electrons on the outermost shell of an atom determines the group to which it belongs in the periodic table as well as its chemical properties.

Hydrogen has a half filled 1s sublevel. Only one electron is present in this sublevel.

Let us consider argon

1s2 2s2 2p6 3s2 3p6

The outermost ns and np levels are completely filled. Thus the outermost shell is completely filled.

In the last case; let us look at the electronic configuration of nitrogen;

1s2 2s2 2p3

The outermost 2p subshell is exactly half filled while the 2s sublevel is fully filled. The outermost shell of nitrogen is made up of 2s2 and 2p3 sublevels.

Answer:

327.89g/mol

Explanation:

Step 1:

The following data were obtained from the question:

Van 't Hoff factor (i) = 1 (since the molecule is non-electrolyte)

Temperature (T) = 25°C = 25°C + 273 = 298K

Gas constant (R) = 0.0821 atm.L/Kmol

Mass of molecule = 241mg

Volume of water = 250mL

Molarity (M) =?

Osmotic pressure (Π) = 0.072 atm

Step 2:

Determination of the molarity of the molecule.

This can be obtained as follow:

Π = iMRT

0.072 = 1 x M x 0.0821 x 298

Divide both side by 0.0821 x 298

M = 0.072 / (0.0821 x 298)

M = 2.94×10¯³ mol/L

Step 3:

Determination of the number of mole of the molecule. This can be obtained as follow:

Molarity = 2.94×10¯³ mol/L

Volume of water = 250mL = 250/1000 = 0.25L

Mole of molecule =..?

Molarity = mole /Volume

2.94×10¯³ = mole / 0.25

Cross multiply

Mole of molecule = 2.94×10¯³ x 0.25

Mole of molecule = 7.35×10¯⁴ mole.

Step 4:

Determination of the molar mass of the molecule.

Mole of molecule = 7.35×10¯⁴ mole.

Mass of molecule = 241mg = 241×10¯³g

Molar mass of molecule =..?

Mole = Mass /Molar Mass

7.35×10¯⁴ = 241×10¯³/ Molar Mass

Cross multiply

7.35×10¯⁴ x molar mass = 241×10¯³

Divide both side by 7.35×10¯⁴

Molar Mass = 241×10¯³/7.35×10¯⁴

Molar Mass = 327.89g/mol

Therefore, the molar mass of the molecule is 327.89g/mol

Answer:

[tex]2.11\ * 10^{-2}[/tex]  is the correct answer to the given question.

Explanation:

Given k=6.40 x 10-3 min-1.

According to the first order reaction .

The concentration of time can be written as

[tex][\ A\ ]\ = \ [\ A_{0}\ ] * e \ ^\ {-kt}[/tex]

Here [tex][\ A\ ]_{0}[/tex] = Initial concentration.

So  [tex][\ A\ ]_{0}= 0.0314 M[/tex]

Putting this value into the above equation.

[tex]0.0314 \ *\ e^{6.40 x 10^{-3} \ * \ 62.0 }[/tex]

=0.211 M

This can be written as

[tex]=\ 2.11 *\ 10^{-2}[/tex]

Answer:

1. Born Haber cycle is used to calculate enthalpy of formation of an ionic solid

2. Resonance structures are used to represent the bonding in some chemical species.

Explanation:

The Born–Haber cycle is a method popularly known in chemistry used in computing enthalpy. The enthalpy of formation of an ionic solid cannot be measured directly. The lattice enthalpy refers to the enthalpy change involved in the formation of an ionic compound from gaseous ions the process is exothermic process. A Born–Haber cycle works on the principle of Hess's law. It can be used to calculate the lattice enthalpy by comparing the standard enthalpy change of formation of the ionic compound from the elements to the enthalpy required to make gaseous ions from the elements.

Resonance is an idea introduced by Linus Pauling to explain chemical bonding from the valence bond perspective. The idea of resonance affords us the opportunity to describe the bonding in certain molecules by combining several structures called chemical or canonical structures. The real structure of the specie lie somewhere between the structures indicated by the resonance structures. The resonance structures of the nitrate ion are shown in the image attached.

Answer:

Molecular:

Ba(NO3)2 + K2CO3 -> BaCO3 + 2KNO3

Complete ionic:

Ba2+ + 2NO3- + 2K+ + CO3 2- -> BaCO3 + 2K+ + 2NO3-

Net ionic:

Ba2+ + CO3 2- - > BaCO3

Explanation:

Molecular consists of all species reacting.

Complete ionic consists of all ionic species (ex. K+, NO3-) separated, as well as any compounds that didn't dissociate into ions (BaCO3 doesn't dissolve).

Net ionic doesn't include spectator ions (in this case, nitrate and potassium) and only species that aren't present on both sides of the arrow (barium and carbonate become a solid precipitate, so the ions aren't present as products, making them appear in the net ionic equation).