Aqueous sodium bicarbonate was used to wash the crude n-butyl bromide. a. What was the purpose of this wash? Give equations. b. Why would it be undesirable to wash the crude halide with aqueous sodium hydroxide? 5. Look up the density of n-butyl chloride (1-chlorobutane). Assume that this alkyl halide was prepared instead of the bromide. Decide whether the alkyl chloride would appear as the upper or the lower phase at each stage of the separation procedure: after the reflux, after the addition of water, and after the addition of sodium bicarbonate.

Answer:

ΔT = - 2.13°C

Explanation:

Given data:

Mass of NH₄NO₃  = 48.5 g

Specific heat of solution = 4.18 j/g.°C

Initial temperature = 27.5°C

Final temperature = ?

Solution:

First of all we will find the moles of NH₄NO₃.

Number of moles = mass/molar mass

Number of moles =  48.5 g/80 g/mol

Number of moles = 0.6 mol

Now we will find the ΔH when we dissolve the 0.6 mol.

NH₄NO₃ + H₂O  →  NH₄NO₃   ΔH = +25.7 kJ

For 0.6 mol:

0.6 mol × +25.7 kJ/mol = 15.42 kj

15.42kj heat is absorbed by the reaction while -15.42 kj (-1542 j) heat will lost by the water.

The mass of water+ NH₄NO₃ = 125 g + 48.5 g

The mass of water+ NH₄NO₃ = 173.5 g

Q = m.c. ΔT

ΔT = T2 - T1

-1542 j =  173.5 g . 4.18 j/g.°C. ΔT

-1542 j = 725.23 j/°C. ΔT

ΔT = -1542 j / 725.23 j/°C

ΔT = - 2.13°C

Based on the data given, the final temperature in a squeezed cold pack is 6.2° C.

The moles of NH₄NO₃ is determined using the formula:

molar mass of  NH₄NO₃ = 80 g/mol

Number of moles =  48.5/80

Number of moles = 0.6 moles

ΔH for the dissolution of 1 mole NH₄NO₃ = +25.7 kJ

ΔH for the dissolution of 0.6 mol = 0.6 × 25.7 kJ/mol

ΔH for the dissolution of 0.6 mol  = 15.42 kj

Thus, 15.42 kJ heat is absorbed by the reaction while 15.42 kj heat will lost by the water.

Using the formula, Q = mcΔT to calculate the final temperature

mass of solution = 125 g + 48.5 g

mass of solution, m = 173.5 g

ΔT = T2 - T1

Q = 15.42 Kj = -15420 J

-15420 =  173.5 * 4.18 * ΔT

ΔT = -15420 / 725.23

ΔT = - 21.3° C

Final temperature, T2 = ΔT + T1

Final temperature, T2 =  - 21.3° C + 27.5° C

Final temperature, T2 = 6.2° C

Therefore, the final temperature in a squeezed cold pack is 6.2° C.

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Answer : The cell potential for this reaction is 0.50 V

Explanation :

The given cell reactions is:

[tex]Pb^{2+}(aq)+Zn(s)\rightarrow Zn^{2+}(aq)+Pb(s)[/tex]

The half-cell reactions are:

Oxidation half reaction (anode):  [tex]Zn\rightarrow Zn^{2+}+2e^-[/tex]

Reduction half reaction (cathode):  [tex]Pb^{2+}+2e^-\rightarrow Pb[/tex]

First we have to calculate the cell potential for this reaction.

Using Nernest equation :

[tex]E_{cell}=E^o_{cell}-\frac{2.303RT}{nF}\log \frac{[Zn^{2+}]}{[Pb^{2+}]}[/tex]

where,

F = Faraday constant = 96500 C

R = gas constant = 8.314 J/mol.K

T = room temperature = [tex]25^oC=273+25=298K[/tex]

n = number of electrons in oxidation-reduction reaction = 2

[tex]E^o_{cell}[/tex] = standard electrode potential of the cell = +0.63 V

[tex]E_{cell}[/tex] = cell potential for the reaction = ?

[tex][Zn^{2+}][/tex] = 3.5 M

[tex][Pb^{2+}][/tex] = [tex]2.0\times 10^{-4}M[/tex]

Now put all the given values in the above equation, we get:

[tex]E_{cell}=(+0.63)-\frac{2.303\times (8.314)\times (298)}{2\times 96500}\log \frac{3.5}{2.0\times 10^{-4}}[/tex]

[tex]E_{cell}=0.50V[/tex]

Therefore, the cell potential for this reaction is 0.50 V

Answer:

a) Reduction:

Ag⁺(aq) + e⁻ → Ag(s)

Oxidation:

Pb(s) → Pb⁺²(aq) + 2e⁻

Overall reaction:

2Ag⁺(aq) + Pb(s) → 2Ag(s) + Pb²⁺

b) Silver; oxidation;  from the lead electrode to the silver electrode.

Explanation:

a) Ag⁺ had lost 1 electron, so need to gain 1 electron to become Ag(s). Pb needs to lose 2 electrons to become Pb⁺².

Reduction:

Ag⁺(aq) + e⁻ → Ag(s)

Oxidation:

Pb(s) → Pb⁺²(aq) + 2e⁻

Overall reaction:

2Ag⁺(aq) + Pb(s) → 2Ag(s) + Pb²⁺ (it will need 2Ag⁺ to gaind the 2 electrons released by Pb)

b) The cation formed in the redox reaction is Pb²⁺, so, to equilibrate the charges, it will flow towards the silver (Ag) electrode.

The lead (Pb) is being oxidized, so oxidation is happening at it.

The electrons flow from the oxidation (anode) to the reduction (cathode), so they flow from the lead electrode to the silver electrode.

In the voltaic cell, the Ag+ is reduced to Ag in the silver half-cell, while Pb is oxidized to Pb2+ in the lead half-cell. The cations flow towards the silver electrode and the electrons flow from the lead to the silver electrode. Hence, the overall reaction is 2Ag+(aq) + Pb(s) -> 2Ag(s) + Pb2+(aq).

In a voltaic cell with an Ag/Ag+ half-cell and a Pb/Pb2+ half-cell, the silver half-cell acts as the cathode or reduction half-cell which gains electrons, while the lead half-cell acts as the anode or oxidation half-cell and loses electrons. Therefore, the balanced half-reactions and overall spontaneous reactions are:

(a) Balanced Half-Reactions and Overall Reaction:

Reduction: Ag+(aq) + 1e- -> Ag(s)

Oxidation: Pb(s) -> Pb2+(aq) + 2e-

Overall Reaction: 2Ag+(aq) + Pb(s) -> 2Ag(s) + Pb2+(aq)

(b) The Cation Flow and Electrons Flow:

The cation flow is towards the silver electrode and the electron flow  is from the lead electrode to the silver electrode. In the voltaic cell, the process that occurs at the lead electrode is oxidation.

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Answer:

The correct answer is 3.36 L

Explanation:

The molecular weight of CO₂ is 44 g/mol (2 O + 1 C= (2 x (16 g/mol )) + 12 g/mol). We have 44 g in 1 mol CO₂, and we want to know how many mol are in 6.6 g:

44 g -------------- 1 mol CO₂

6.6 g -------------- X= 6.6 g x 1 mol / 44 g = 0.15 mol

In normal conditions of temperature and pressure, 1 mol of ideal gas occupies 22.4 L of volume, thus:

1 mol CO₂ ------------ 22.4 L

0.15 mol -------------- X= 3.36 L

So, 6.6 g CO₂ are equivalent to 0.15 mol CO₂ and they occupy 3.36 L.

Answer:

c. Bomb calorimetry

Explanation:

The hydrocarbons are combustibles, it means that they can react in a combustion reaction to release energy. To measure this amount of energy, it's necessary equipment that the reaction can be placed in a controlled way. The bomb calorimeter is this equipment, which is an adiabatic vessel, with water. The heat is calculated based on the increase in the water temperature.

The coffee-cup calorimetry is used to measure the heat of a dissolution reaction and the bomb manometry is used to measure the pressure.

Answer : The element produces lead-206 is, polonium (Po)- 210

Explanation :

Alpha decay : In this process, alpha particles is emitted when a heavier nuclei decays into lighter nuclei. The alpha particle released has a charge of +2 units.

The general representation of alpha decay reaction is:

[tex]_Z^A\textrm{X}\rightarrow _{Z-2}^{A-4}Y+_2^4\alpha[/tex]

As we are given that,

[tex]_{Z-2}^{A-4}Y=_{82}^{206}Pb[/tex]

From the given data we conclude that,

[tex]Z-2=82\\\\Z=82+2=84\\\\A-4=206\\\\A=206+4=210[/tex]

So, the element is polonium (Po) that has atomic number 84 and atomic mass 210.

The complete alpha decay reaction will be,

[tex]_{84}^{210}\textrm{Po}\rightarrow _{82}^{206}\textrm{Pb}+_2^4He[/tex]

Hence, the element produces lead-206 is, polonium (Po)- 210

Answer : The correct option is, (d) 0.632 M

Explanation : Given,

Concentration of HCl = 0.500 M

Volume of HCl = [tex]25.30cm^3=25.30mL=0.2530L[/tex]

conversion used : [tex]1cm^3=1mL\\1mL=0.001L[/tex]

Volume of hypochlorite ion = [tex]20.00cm^3=20.00mL=0.2000L[/tex]

The chemical reaction will be:

[tex]OCl^-+H^+\rightarrow HClO[/tex]

First we have to calculate the moles of [tex]H^+[/tex] ion.

[tex]\text{Moles of }H^+=\text{Concentration of }H^+\times \text{Volume of }H^+[/tex]

[tex]\text{Moles of }H^+=0.500M\times 0.2530L=0.1265mol[/tex]

From the reaction we conclude that,

Moles of [tex]H^+[/tex] = Moles of [tex]ClO^-[/tex] = 0.1265 mol

Now we have to calculate the concentration of hypochlorite ion.

[tex]\text{Moles of }ClO^-=\text{Concentration of }ClO^-\times \text{Volume of }ClO^-[/tex]

[tex]0.1265mol=\text{Concentration of }ClO^-\times 0.2000L[/tex]

[tex]\text{Concentration of }ClO^-=0.632M[/tex]

Therefore, the concentration of hypochlorite ion is 0.632 M.

Answer:

The correct answer is b) ethylmethylamine

Explanation:

This organic compound has the NH group (it is an amine), with 2 substitutions: a methyl group and an ethyl group.

Answer:

ethylmethylamine

Explanation:

Answer:

83 .

Explanation:

Radioactive elements: It is defined as the atoms which contains unstable nucleus because of the constantly change and imbalance of energy in the nucleus. Radioactivity of an atom is showing by when the nucleus of an atom loose a neutron, it gives an energy and this process is called to be radioactivity.

Isotopes elements containing different number of neutrons and same number of protons. All isotopes are not considered as radioisotopes. All isotopes of an element with an atomic number greater than 83 are radioactive means, they are having unstable nucleus.

Answer:

Silicon

Explanation:

The formula for the calculation of the average atomic mass is:

[tex]Average\ atomic\ mass=(\frac {\%\ of\ the\ first\ isotope}{100}\times {Mass\ of\ the\ first\ isotope})+(\frac {\%\ of\ the\ second\ isotope}{100}\times {Mass\ of\ the\ second\ isotope})+(\frac {\%\ of\ the\ third\ isotope}{100}\times {Mass\ of\ the\ third\ isotope})[/tex]

Given that:

For first isotope, X-28 :

% = 92.23 %

Mass = 27.977 amu

For second isotope, X-29:

% = 4.67 %

Mass = 28.976 amu

For third isotope, X-30:

% = 3.10 %

Mass = 29.974 amu

Thus,  

[tex]Average\ atomic\ mass=\frac{92.23}{100}\times {27.977}+\frac{4.67}{100}\times {28.976}+\frac{3.10}{100}\times {29.974}[/tex]

[tex]Average\ atomic\ mass=25.8031871+1.3531792+0.929194\ amu[/tex]

Average atomic mass = 28.0855603 amu  ≅ 28.086 amu

This mass corresponds to Silicon, ( Z = 14).

Answer:

it's d

Explanation:

Feldspar, pyroxene, amphibole, mica and the last, made up of the two most abundant elements, is quartz.

An element is a pure material that cannot be transformed into simpler compounds by any physical or chemical process. The atoms that make up an element are all the same kind.

There are three categories for elements are metals, nonmetals, and metalloids. The symbols used to symbolize different elements.

Any compound that cannot be broken down into simpler chemicals by regular chemical processes is referred to as a chemical element or element. The building blocks of which all matter is made are called elements.

Fire, Earth, Water, and Air. Earth, water, air, and fire were the four elements that the ancient Greeks thought made up everything.

Thus, option D is correct.

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The fraction of the volume in a container actually occupied by Ar atoms at 230 atm pressure and 0 C is 92.0%. The van der Waals constant b is 0.0322 L/mol.

The van der Waals constant b is defined as four times the total volume actually occupied by the molecules of a mole of gas 1. To calculate the fraction of the volume in a container actually occupied by Ar atoms at 230 atm pressure and 0 C, we can use the following formula:

V_real = V_ideal - nb

where V_real is the real volume of the gas, V_ideal is the ideal volume of the gas, n is the number of moles of the gas, and b is the van der Waals constant. At 0 C, the ideal volume of one mole of any gas is 22.4 L 2. Therefore, the ideal volume of Ar atoms is 22.4 L/mol.

To calculate the real volume of Ar atoms, we need to know the number of moles of Ar atoms present in the container. We can use the ideal gas law to calculate the number of moles of Ar atoms:

PV = nRT

where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature. Substituting the given values, we get:

n = PV/RT = (230 atm * V)/(0.0821 L atm/mol K * 273 K) = 9.03 V

Substituting the values of n and b into the formula for V_real, we get:

V_real = V_ideal - nb = 22.4 L/mol - 0.0322 L/mol * 4 * 9.03 mol = 20.6 L

Therefore, the fraction of the volume in a container actually occupied by Ar atoms at 230 atm pressure and 0 C is:

(V_real/V) * 100% = (20.6 L/V) * 100% = 92.0%

Answer:family fdhdfgfdsh

Nitric acid aqueous solutions are strong oxidizing agents capable of oxidizing a variety of less active metals, including Cu.

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Answer:

Elements in same column of periodic table have same properties.

Explanation:

The elements in the same group have same number of valance electrons thus have similar properties.

Consider the elements of group two i.e alkaline earth metals. All have two valance electrons and show similar properties.

Magnesium, barium, calcium etc.

All alkaline earth metals form salt with halogens.e.g,

Mg   +   Cl₂    →    MgCl₂

Ba    +   Br₂    →     BaBr₂

Mg   +   Br₂    →     MgBr₂

Ca    +   Br₂    →     CaBr₂

They react with oxygen and form oxides of respective metal.

2Mg   +   O₂   →    2MgO

2Ba   +   O₂   →    2BaO

2Ca   +   O₂   →    2CaO

these oxides form hydroxide when react with water,

MgO  + H₂O   →  Mg(OH)₂

BaO  + H₂O   →  Ba(OH)₂

CaO  + H₂O   →  Ca(OH)₂

With nitrogen it produced nitride,

3Mg + N₂     →  Mg₃N₂

3Ba + N₂     →  Ba₃N₂

3Ca + N₂     →  Ca₃N₂

With acid like HCl,

Mg + 2HCl  →  MgCl₂ + H₂

Ba + 2HCl  →  BaCl₂ + H₂

Ca + 2HCl  →  CaCl₂ + H₂

Elements in the same column of the periodic table have the same number of valence electrons in their outer shell, which leads to similar chemical properties. This includes similarities in bonding with other elements and changes in atomic size as we move down the column.

Elements in the same column of the periodic table, also known as a chemical group, share similar chemical properties. This arises because they have the same number of valence electrons in their outer shells, which determines how they interact and bond with other elements. For example, Lithium (Li) and Sodium (Na), both in the first column, exhibit comparable chemical behaviors because they each have one valence electron in their outer shell.

The similarity in behavior extends to other attributes as well. As we move down any given group, the atomic size tends to increase, which also impacts the chemical properties of the elements. Each column in the periodic table is a unique group that shares these properties, making the table a critical tool for understanding elemental behaviors and relationships.

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Answer:

A)Mass of  gallium plated out is 0.3440 grams

B) For 0.67 hours current of 5.79 A must to be applied to plate out 8.70 g of tin.

Explanation:

To calculate the total charge, we use the equation:

[tex]C=I\times t[/tex]

where,

C = Charge

I = Current in time t (seconds)

To calculate the moles of electrons, we use the equation:

[tex]\text{Moles of electrons}=\frac{C}{F}[/tex]

where,

F = Faraday's constant = 96500

A) The equation for the deposition of Ga(s) from Ga(III) solution follows:

[tex]Ga^{3+}(aq.)+3e^-\rightarrow Ga(s)[/tex]

I = 0.790 A, t = 30.0 min = 1800 seconds

[tex]C=I\times t[/tex]

[tex]C=0.790 A\times 1800 s=1422 C[/tex]

Moles of electron transferred:

[tex]=\frac{1422 C}{96500 F}=0.01474 mol[/tex]

Now, to calculate the moles of gallium, we use the equation:

[tex]\text{Moles of Gallium}=\frac{\text{Moles of electrons}}{n}[/tex]

n = number of electrons transferred = 3

[tex]\text{Moles of Gallium}=\frac{0.01474 mol}{3}=0.004913 mol[/tex]

Mass of 0.004913 moles of gallium = 0.004913 mol × 70 g/mol=0.3440 g

B) The equation for the deposition of Sn(s) from Sn(II) solution follows:

[tex]Sn^{2+}(aq.)+2e^-\rightarrow Sn(s)[/tex]

Moles of tin = [tex]\frac{8.70 g}{119 g/mol}=0.07311 mol[/tex]

n = number of electrons transferred = 2

[tex]\text{Moles of tin}=\frac{\text{Moles of electrons}}{n}[/tex]

Moles of electron =  [tex]n\times \text{Moles of tin}[/tex]

[tex]=2\times 0.07311 mol=0.14622 mol[/tex]

Charge transferred during time t :

[tex]\text{Moles of electrons}=\frac{C}{F}[/tex]

[tex]C=96500 F\times 0.14622 mol=14,110.23 C[/tex]

Current applied for t time = I = 5.79 A

[tex]t=\frac{C}{I}=\frac{14,110.23 C}{5.79 A}=2,437 s=0.67 hrs[/tex]

For 0.67 hours current of 5.79 A must to be applied to plate out 8.70 g of tin.

Answer:

Qtotal = 90.004 kJ

Explanation:

To start resolving the problem we need to first convert the kJ/mol units from the thermodynamic values to J/g, so that we can work with the units of the heat capacity values. We know that the molar mass of water is 18.015 g/mol, so with this we do the respective conversion:

ΔH°fus of H2O = (6.02 kJ/mol) (1 mol/18.015g) (1000J/kJ) = 334.165 J/g

ΔH°vap of H2O = (40.7 kJ/mol) (1mol/18.015g) (1000J/kJ) = 2259.228 J/g

Now we need to find out the heat energy required to rise the temperature (specific heat capacity) and the energy required for each change of phase (specific latent heat), and add everything up. For this we will require the specific heat capacity and latent heat equations:

Q = mCΔT ; where m = mass, C = Hear capacity, ΔT = change of temperature

Q = mL ; where m = mass, L = specific latent heat

First change of phase (solid to liquid - fusion)

Q1 = (25g) (2.09 J/g°C) (0°C - (-129°C) = 6740.25 J

Q2 = (25g) (334.165 J/g) = 8354.125 J

Second change of phase (liquid to gas - vaporization)

Q3 = (25g) (4.18 J/g°C) (100°C - 0°C = 10450 J

Q4 = (25g) (2259.228 J/g) = 56480.7 J

Rise of temperature of the gaseous water

Q5 = (25g) (1.97 J/g°C) (262°C - 100°C = 7978.5 J

Finally we add everything up:

Qtotal = Q1 + Q2 + Q3 + Q4 + Q5 = 6740.25 J + 8354.125 J + 10450 J + 56480.7 J + 7978.5 J = 90003.575 J = 90.004 kJ

Change of state has to do with the movement from one state of matter to another.

Change of state has to do with the movement from one state of matter to another. We need to first change the kJ/mol units  to J/g,knowing that the molar mass of water is 18.015 g/mol hence:

ΔH°fus of H2O = (6.02 kJ/mol) (1 mol/18.015g) (1000J/kJ) = 334.165 J/g

ΔH°vap of H2O = (40.7 kJ/mol) (1mol/18.015g) (1000J/kJ) = 2259.228 J/g

For each phase change;

Q = mCΔT

where

And

Q = mL

where

First,  solid to liquid change

Q1 = (25g) (2.09 J/g°C) (0°C - (-129°C) = 6740.25 J

Q2 = (25g) (334.165 J/g) = 8354.125 J

Second liquid to gas change

Q3 = (25g) (4.18 J/g°C) (100°C - 0°C) = 10450 J

Q4 = (25g) (2259.228 J/g) = 56480.7 J

Then,

Q5 = (25g) (1.97 J/g°C) (262°C - 100°C) = 7978.5 J

Finally:

Qtotal = Q1 + Q2 + Q3 + Q4 + Q5 = 6740.25 J + 8354.125 J + 10450 J + 56480.7 J + 7978.5 J = 90003.575 J = 90.004 kJ

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Answer:

The coefficients should be: 2, 5, 2, 5

Explanation:

Given redox reaction: MnO₄⁻ + SO₃²⁻ → Mn²⁺+ SO₄²⁻

To balance the given redox reaction in acidic medium, the oxidation and the reduction half-reactions should be balanced first.

Reduction half-reaction: MnO₄⁻ → Mn²⁺

Oxidation state of Mn in MnO₄⁻ is +7 and the oxidation state of Mn in Mn²⁺ is +2. Therefore, Mn accepts 5e⁻ to get reduced from +7 to +2 oxidation state.

⇒ MnO₄⁻ + 5e⁻ → Mn²⁺

Now the total charge on reactant side is (-6) and the total charge on product side is +2. Therefore, to balance the total charge, 8H⁺ must be added to the reactant side.

⇒ MnO₄⁻ + 5e⁻ + 8H⁺ → Mn²⁺

To balance the number of hydrogen and oxygen atoms, 4H₂O must be added to the product side.

⇒ MnO₄⁻ + 5e⁻ + 8H⁺ → Mn²⁺ + 4H₂O               .....equation 1

Oxidation half-reaction: SO₃²⁻ → SO₄²⁻

Oxidation state of S in SO₃²⁻ is +4 and the oxidation state of S in SO₄²⁻ is +6. Therefore, S loses 2e⁻ to get oxidized from +4 to +6 oxidation state.

⇒ SO₃²⁻ → SO₄²⁻ + 2e⁻

Now the total charge on reactant side is (-2) and the total charge on product side is (-4). Therefore, to balance the total charge, 2H⁺ must be added to the product side.

⇒ SO₃²⁻ → SO₄²⁻ + 2e⁻ + 2H⁺

To balance the number of hydrogen and oxygen atoms, 1 H₂O must be added to the reactant side.

⇒ SO₃²⁻ + H₂O → SO₄²⁻ + 2e⁻ + 2H⁺                .....equation 2

Now, to cancel the electrons transferred, equation (1) is multiplied by 2 and equation (2) is multiplied by 5.

Balanced Reduction half-reaction:

MnO₄⁻ + 5e⁻ + 8H⁺ → Mn²⁺ + 4H₂O ] × 2

⇒ 2MnO₄⁻ + 10e⁻ + 16H⁺ → 2Mn²⁺ + 8H₂O                .....equation 3

Balanced Oxidation half-reaction:

SO₃²⁻ + H₂O → SO₄²⁻ + 2e⁻ + 2H⁺ ] × 5

⇒ 5SO₃²⁻ + 5H₂O → 5SO₄²⁻ + 10e⁻ + 10H⁺                  .....equation 4  

Now adding equation 3 and 4, to obtain the overall balanced redox reaction:

2MnO₄⁻ + 5SO₃²⁻ + 6H⁺ → 2Mn²⁺ + 5SO₄²⁻ + 3H₂O

Therefore, the coefficients should be: 2, 5, 2, 5

Answer:

The molal boiling point elevation constant is 1.59 ≈  1.6 [tex]Kkgmol^{-1}[/tex]

Explanation:

To solve this question , we will make use of the equation ,

Δ[tex]T_{b} = i*K_{b} *m[/tex]

where ,

           ∴ [tex]m = \frac{\frac{24.6}{60} }{\frac{650}{1000} }[/tex]

    4. [tex]K_{b}[/tex] ⇒ ?

∴

[tex]1=1*K_{b} *\frac{\frac{24.6}{60} }{\frac{650}{1000} }[/tex]

⇒ [tex]K_{b}[/tex] = [tex]1.59[/tex] ≈ 1.6 [tex]Kkgmol^{-1}[/tex]

The molal boiling point elevation constant (Kb) of substance X is 4.1.

The molal boiling point elevation constant (Kb) can be calculated using the formula: ΔT = Kb × m

Where ΔT is the change in boiling point, Kb is the molal boiling point elevation constant, and m is the molality of the solution.

In this case, the change in boiling point (ΔT) is 1 degree C (124.3 - 123.3), the molality (m) can be calculated by dividing the molal mass of urea by the mass of the solvent water, which gives a value of 0.0246 kg urea / 0.100 kg water = 0.246 mol/kg, and the formula becomes: 1 = Kb × 0.246

Now, rearrange the equation to solve for Kb: Kb = 1 / 0.246 = 4.07

Rounding to 2 significant digits, the molal boiling point elevation constant Kb of substance X is 4.1.

Answer:

The sphere on the left has the most inertia because it has more mass.

Explanation:

Inertia is a property of matter of a substance.

According to Newton's first law of motion, a body continues to stay in the state of rest or constant velocity unless acted upon a external force.

The amount of inertia that an object possess is proportional to the mass of the object.

The sphere on the left is of 300 kg and that on the right is of 30 kg.

Clearly, the sphere on the left has more mass.

Therefore, the sphere on the left has the most inertia.

Answer:

it would be the sphere on the left that has more mass

Explanation:

because the more weight that is applied to a force with being a small object appling force upon it it will be harder to move

Answer:

A reaction mechanism is a series of elementary reactions that describe how an overall reaction occurs and explain the experimentally determined rate law.

Explanation:

A reaction machanism is a succession of steps to go from the reactants to the products of a reaction.

Each of this steps is a different elementary reaction and produces an intermediary product. This intermediaries not always can be seen in real life due to the high rate of reaction of following steps.

This intermediary reactions where created to explain the experimental determined rate law of some reactions, that doesn't fit in a one-step reaction.

In conclusion we can say that: A reaction mechanism is a series of elementary reactions that describe how an overall reaction occurs and explain the experimentally determined rate law

Final answer:

A reaction mechanism explains the step-by-step sequence in which reactants are converted into products, detailing elementary reactions that make up the entire process, including the rate-determining step that dictates the overall rate of the reaction.

Explanation:

A reaction mechanism is the detailed process by which a chemical reaction occurs, broken down into a series of elementary steps. Each of these steps involves a certain number of reactant species, as detailed by the molecularity (unimolecular, bimolecular, or termolecular) of the reaction. The overall rate of the chemical reaction is determined by the rate-determining step, which is the slowest step within the reaction mechanism. The rate laws for the elementary reactions must align with the experimentally determined rate law to confirm the proposed reaction mechanism as plausible.

Catalysts are substances that alter the reaction mechanism, offering an alternative pathway with a lower activation energy, which in turn affects the reaction rate without impacting the chemical equilibrium of the reaction.

A balanced chemical equation alone does not convey the complexity of the reaction mechanism. For example, the decomposition of ozone is a multi-step process, not evident from the simple overall equation. The mechanism provides a detailed description of these steps, much like a roadmap of a journey illustrating every turn and stop, rather than just showing the start and end points.

Answer:

ΔH° = -521,6kJ

Explanation:

It is possible to obtain the ΔH° of a reaction using Hess's law that consist in the algebraic sum of the ΔH° of semireactions.

For the semireactions:

(1) H₂(g) + F₂(g) ⟶ 2HF(g) ΔHrxn° = −546.6kJ

(2) 2H₂(g) + O₂(g) ⟶ 2H₂O(l) Δ Hrxn° = −571.6kJ

The sum of 2×(1) - (2) gives:

2F₂(g) + 2H₂O(l) ⟶ 4HF(g) + O₂(g)

The ΔH° for this reaction is:

ΔH° = -546,6kJ×2 - (-571,6kJ)

ΔH° = -521,6kJ

I hope it helps!

ΔHrxn ° for 2 F₂ (g) + 2 H₂O (l) ⟶ 4HF (g) + O₂ (g): -521.6 kJ

The change in enthalpy in the formation of 1 mole of the elements is called enthalpy of formation

The enthalpy of formation measured in standard conditions (25 ° C, 1 atm) is called the standard enthalpy of formation (ΔHf °)

Determination of the enthalpy of formation of a reaction can be through a calorimetric experiment, based on the principle of Hess's Law, enthalpy of formation table, or from bond energy data

Delta H reaction (ΔH) is the amount of heat / heat change between the system and its environment

(ΔH) can be positive (endothermic = requires heat) or negative (exothermic = releasing heat)

The value of ° H ° can be calculated from the change in enthalpy of standard formation:

ΔH∘rxn = ΔH∘f of the product (s) if ∆Hf ° (reactants) = 0

Based on the principle of Hess's Law, the change in enthalpy of a reaction will be the same even though it is through several stages or ways

Known ΔHrxn from reaction:

H₂ (g) + F₂ (g) ⟶2HF (g) ΔHrxn ° = −546.6 kJ reaction 1 (R1)

2H₂ (g) + O₂ (g) ⟶2H₂O (l) ΔHrxn ° = −571.6 kJ reaction 2 (R2)

ΔHrxn from reaction:

2 F₂ (g) + 2 H₂O (l) ⟶ 4HF (g) + O₂ (g) reaction 3 (R3)

Can be searched from ΔHrxn ° R1 and R2

From R3 it is known that the reaction coefficient of F₂ is 2, so we can multiply R1 by 2 (include ΔHrxn °)

2H₂ (g) + 2F₂ (g) ⟶4HF (g) ΔHrxn ° = −1093.2 kJ reaction 4 (R4)

R3 H₂O lies in the reactants, so that we can reverse R2,and so ΔHrxn ° is marked +

2H₂O (l) ⟶2H₂(g) + O₂ (g) ΔHrxn ° = + 571.6 kJ reaction 5 (R5)

We add R4 to R5 to get R3, by removing 2H₂ (g) because it is located in the reactants and products

2H₂ (g) + 2F₂ (g) ⟶4HF (g) ΔHrxn ° = −1093.2 kJ

2H₂O (l) ⟶2H₂ (g) + O₂ (g) ΔHrxn ° = + 571.6 kJ

-------------------------------------------------- --------------------  +

2F₂ (g) + 2H₂O (l) ⟶ 4HF (g) + O₂ (g) ΔHrxn ° = -521.6 kJ

Delta H solution

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an exothermic reaction

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as endothermic or exothermic

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an exothermic dissolving process

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Answer:

272.2 grams is the mass of calcium carbonate needed to produce 61.0L of carbon dioxide at STP

Explanation:

If the gas is produced at STP, by the volume and the Ideal Gas Law, we can know the moles.

P . V = n. R .T

1 atm . 61 L = n . 0.082L.atm/mol.K . 273K

61 L.atm / (0.082 mol.K/L.atm . 273K) = 2.72 moles

As the relation is 1:1

1 mol of CO₂ cames from 1 mol of CaCO₃, so 2.72 moles of CO₂ comes from 2.72 moles of CaCO₃

Molar mass CaCO₃ = 100.08 g/m

Moles . molar mass = grams

2.72 m . 100.08 g/m = 272.7 g

Answer: 35.4 grams

Explanation:

Molarity of a solution is defined as the number of moles of solute dissolved per Liter of the solution.

[tex]Molarity=\frac{n\times 1000}{V_s}[/tex]

where,

Molality = 2.65

n= moles of solute =?

 [tex]V_s[/tex] = volume of solution in ml = 445 ml

Putting in the values we get:

[tex]2.65=\frac{n\times 1000}{445ml}[/tex]

[tex]n=1.18[/tex]

Mass of solute in g=[tex]moles\times {\text {molar mass}}=1.18mol\times 30.02g/mol=35.4g[/tex]

Thus 35.4 grams of [tex]CH_2O[/tex] is needed to prepare 445 ml of a 2.65 m solution of [tex]CH_2O[/tex].

Final answer:

To prepare 445 mL of a 2.65 M solution of CH2O, 35.42 grams of CH2O are needed. The calculation involves converting volume to liters, calculating moles required using molarity, and then finding the mass needed by multiplying with the molar mass of CH2O.

Explanation:

The question asks, how much CH2O is needed to prepare 445 ml of a 2.65 M solution of CH2O? In chemistry, the concentration of a solution is typically expressed in molarity (M), which is defined as moles of solute per liter of solution. To calculate the amount of CH2O needed, we use the formula:

Moles of solute = Molarity (M) × Volume of solution (L)

First, convert the volume of the solution from milliliters to liters:

445 mL = 0.445 L

Then, use the molarity and the volume of the solution to find the moles of CH2O needed:

Moles of CH2O = 2.65 M × 0.445 L = 1.17925 moles

To find the mass of CH2O required, multiply the moles by the molar mass of CH2O (approximately 30.03 g/mol):

Mass of CH2O = 1.17925 moles × 30.03 g/mol = 35.42 grams

Therefore, to prepare 445 mL of a 2.65 M solution of CH2O, 35.42 grams of CH2O are needed.

Answer:

Electron shielding

Explanation:

Ionization energy decreases moving down a group in the periodic table because of a phenomen known as Electron shielding, in which valence electrons do not interact with the positively charged nucleus as strongly as inner electrons do, because these inner electrons shield the valence electrons. This means it's easier for these valence electrons to leave the atom the more inner electrons are between them and the nucleus, this translates into a decreased ionization energy value.

Ionization energy decreases moving down a group due to the increased distance of the valence electrons from the nucleus, the greater shielding effect of inner electrons, and the higher principal quantum number of the valence electrons, which reduces the effective nuclear charge.

Ionization energy refers to the amount of energy required to remove an electron from an atom in the gaseous state. A key periodic trend observed is that ionization energy decreases as we move down a group in the periodic table. There are several reasons for this trend:

For example, within Group 1 of the periodic table, which contains the alkali metals, we notice a significant drop in ionization energy from lithium to cesium as each succeeding element has more filled inner electron shells that shield the outer electron. Furthermore, the increase in atomic number down a group does not proportionately increase the nuclear charge's effect on the valence electrons due to the reasons mentioned above.

In summary, the larger atomic radius and greater shielding by inner electrons result in a reduced attraction between the valence electrons and the nucleus, leading to a decrease in ionization energy down a group.

George should include two main points: First, the hypothesis of the existence of neutrons arose from the need to explain the remaining mass in an atom's nucleus not accounted for by protons. Second, the detection of neutrons was especially difficult due to their lack of charge, which required more advanced techniques to observe.

In his essay, George should include the following statements:

Understanding the nature of neutrons was a significant step in the development of atomic theory, as it helped explain isotopes and variants of a particular chemical element that differ in neutron number.

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#SPJ12

Answer:

Explanation:"Watch the video and identify which of the following statements are correct." Excuse no video

Kinetic energy

Energy is something that is very inherent in every activity of life. Simply stated, energy can be interpreted as the ability of an object to make an effort. An object is said to have energy when it is able to produce power that can work.

Kinetic, Potential, and Mechanical Energy is a type of energy that is almost present in every part of human life. Its existence cannot be eliminated because of its enormous benefits for the development of human technology, especially those relating to the movement of objects, the position of objects, or a combination of the two.

Kinetic energy is the energy of motion, the energy possessed by objects or objects due to their motion. Kinetic energy comes from the Greek word kinetics which means to move. So, you certainly know that every object that moves then that object has kinetic energy.

The kinetic energy of an object is defined as the effort needed to move an object with a certain mass from rest to a certain speed.

The kinetic energy of an object is equal to the amount of effort required to express its speed and rotation, starting from the others.

Kinetic energy is influenced by the mass and velocity of an object as it moves. The mass is symbolized by the letter m, while the speed is symbolized by the letter v. The amount of energy is directly proportional to the amount of mass and magnitude of the speed of an object as it moves.

Objects with large mass and velocity must have large kinetic energy when moving. And vice versa, objects whose mass and velocity are small, their kinetic energy is also small.

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Details

Grade:  High School

Subject:  Chemistry

keywords: Energy, kinetic.

Answer:

The amount of hydrogen gas collected will be 0.5468 g

Explanation:

We are given:

Vapor pressure of water = 44.6 torr = 44.6 mm Hg

Total vapor pressure = 855 mm Hg

Vapor pressure of hydrogen gas = Total vapor pressure - Vapor pressure of water = (855 - 44.6) mmHg = 810.4 mmHg

To calculate the amount of hydrogen gas collected, we use the equation given by ideal gas which follows:

[tex]PV=nRT[/tex]

where,

P = pressure of the gas = 810.4 mmHg

V = Volume of the gas = 6.50 L

T = Temperature of the gas = [tex]36^oC=[36+273]K=309K[/tex]

R = Gas constant = [tex]62.3637\text{ L.mmHg }mol^{-1}K^{-1}[/tex]

n = number of moles of hydrogen gas = ?

Putting values in above equation, we get:

[tex]810.4mmHg\times 6.50L=n\times 62.3637\text{ L.mmHg }mol^{-1}K^{-1}\times 309K\\\\n=\frac{810.4\times 6.50}{62.3637\times 309}=0.2734\ mol[/tex]

To calculate the mass from given number of moles, we use the equation:

[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]

Moles of hydrogen gas = 0.2734 moles

Molar mass of hydrogen gas = 2 g/mol

Putting values in above equation, we get:

[tex]0.2734mol=\frac{\text{Mass of hydrogen gas}}{2g/mol}\\\\\text{Mass of hydrogen gas}=(0.2734mol\times 2g/mol)=0.5468g[/tex]

Hence, the amount of hydrogen gas collected will be 0.5468 g

Answer: The mass of sulfur dioxide gas at STP for given amount is 16.8 g

Explanation:

At STP conditions:

22.4 L of volume is occupied by 1 mole of a gas.

So, 5.9 L of volume will be occupied by = [tex]\frac{1mol}{22.4L}\times 5.9L=0.263mol[/tex]

Now, to calculate the mass of a substance, we use the equation:

[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]

Moles of sulfur dioxide gas = 0.263 mol

Molar mass of sulfur dioxide gas = 64 g/mol

Putting values in above equation, we get:

[tex]0.263mol=\frac{\text{Mass of sulfur dioxide gas}}{64g/mol}\\\\\text{Mass of sulfur dioxide gas}=(0.263mol\times 64g/mol)=16.8g[/tex]

Hence, the mass of sulfur dioxide gas at STP for given amount is 16.8 g

Answer:

ATP

Explanation:

The main purpose for cellular respiration, is to finally obtain ATP (Adenosine Triphosphate), this process occurres through the electron transport chain: this is the final step of the aerobic respiration, and takes place when energy from NADH and FADH₂ (both products from Krebs Cycle) is transferred to ATP  

This process occurres within the inner membrane of the mitochondria: while protons (H⁺) pass through the ATP synthase (this protein acts as a “tunnel” where H⁺ go through), which uses the difference of protons (H⁺) concentration between the matrix (between 2 mitochondrial membranes) and the inner matrix of mitochondria.  

The ATP synthase also acts as an enzyme, creating ATP using ADP + Pi (inorganic phosphorus)

The electrons used to help with this process, finally attach to O₂ (oxygen) to form H₂O

Answer:  The correct answer is :  ATP

Explanation:  The phosphorylation reaction is a type of metabolic reaction that results in the formation of (ATP) or (GTP) by the direct transfer and donation of a phosphoryl (PO3) group to (ADP) or (GDP) from a phosphorylated reactive intermediate. The breath is an ATP generating process in which an inorganic compound serves as the ultimate e-acceptor. O2 is delivered by blood flow.

Answer:

Final temperature will be T = 67.68°C

Explanation:

The heat evolved by the copper tubing will be absrobed by both water and the vessel used.

The heat evolved by the copper tubing will be:

Heat = [tex]Q1=massXspecificheatX(changeintemperature)[/tex]

Mass = 670 g

Specific heat = 0.387 J/g · K

Change in temperature = Initial - Final

[tex]Q1=670X0.387X(ChangeinTemperature)[/tex]

The heat absorbed by water will be

[tex]Q2=massXspecificheatXchangeintemperature[/tex]

mass = 52.5

Specific heat = 4.184 J/g · K

the heat absorbed by vessel will be:

[tex]Q3=heatcapacityXchange intemperature[/tex]

Heat capacity = 10J/K

Final temperature of all the three will be same (say T)

[tex]Q1=Q2+Q3[/tex]

[tex]670X0.387X(ChangeinTemperature)=massXspecificheatXchangeintemperature+heatcapacityXchange intemperature[/tex]

[tex]670X0.387X(95.3-T)=(52.5X4.184X(T-36.5))+(10X(T-36.5)[/tex]

[tex]259.29(95.3-T)=219.66(T-36.5)+10(T-36.5)[/tex]

[tex]24710.337-259.29T=219.66T-8017.59+10T-365[/tex]

[tex]33092.59=488.95T[/tex]

T = 67.68°C