How much MnO2(s) should be added to excess HCl(aq) to obtain 195 mL Cl2(g) at 25 °C and 715 Torr g

Answer:

427°C .

Explanation:

Step 1:

Data obtained from the question. This include the following:

Initial temperature (T1) = 77°C

Initial pressure (P1) = P

Final pressure (P2) = 2P

Final temperature (T2) =?

Step 2:

Conversion of celsius temperature to Kelvin temperature.

This is illustrated below:

T(K) = T (°C) + 273

Initial temperature (T1) = 77°C

Initial temperature (T1) = 77°C+ 273 = 350K

Step 3:

Determination of the new temperature. The new temperature can be obtained as follow:

P1/T1 = P2/T2

P/350 = 2P/T2

Cross multiply

P x T2 = 350 x 2P

Divide both side by P

T2 = (350 x 2P ) / P

T2 = 700K

Step 4:

Conversion of Kelvin temperature to celsius temperature.

This can be obtained as follow:

T(°C) = T(K) – 273

T(K) = 700K

T(°C) = 700 – 273

T(°C) = 427°C

Therefore, the new temperature of the gas is 427°C

Answer:

The most common position for an double bond in an unsaturated fatty acid is delta 9 (Δ⁹)

Explanation:

Unsaturated fatty acids are carboxylic acids which contains one or more double bonds. The chain length as well as the number of double bonds is written separated by a colon. The positions of the double bonds are specified starting from the carboxyl carbon, numbered as 1, by superscript numbers following a delta (Δ). For example, an 18-carbon fatty acid containing a  single double bond between carbon number 9 and 10 is written as 18:1(Δ⁹).

In most monounsaturated fatty acids, the double bond is between C-9 and C-10 (Δ⁹), and the other double bonds of polyunsaturated fatty acids are generally Δ¹² and Δ¹⁵. This positioning is due to the nature of the biosynthesis of fatty acids. In the mammalian hepatocytes, double bonds are introduced easily into fatty acids at the Δ⁹ position, but cannot introduce additional double bonds between C-10 and the methyl-terminal end. However, plants are able to introduce these additional double bonds at the  Δ¹² and Δ¹⁵ positions.

Answer:

MgSo4.7H2O = Magnesium sulfate

Cs3PO4.H2O = Cesium Phosphate

Hope this helps!

Answer: 10 moles/kg.

Explanation:

Given, Mass of solute = 24.2 g

Molar mass of solute = 24.2 g/mol

[tex]\text{Moles of solute =}\dfrac{\text{Mass of solute}}{\text{Molar mass of solute}}\\\\=\dfrac{24.2}{24.2}=1[/tex]

Mass of solvent = 100.0g = 0.1 kg  [1 g=0.001 kg]

[tex]\text{Molality}=\dfrac{\text{Moles of solute}}{\text{kilograms of Solvent}}\\\\=\dfrac{1}{0.1}\\\\=10\ moles/kg[/tex]

Hence, the molality of the resulting solution is 10 moles/kg.

Answer:

Ba(NO₃)₂(s) → Ba²⁺ + 2NO₃⁻

Explanation:

A strong electrolyte is a salt (A compound that has an anion and a cation and are neutral) that, in water, dissociates completely in its ions.

In Barium nitrate, Ba(NO₃)₂, the cation is Ba²⁺ (Alkaline earth metal), and the anion is the nitrate ion, NO₃⁻.

Thus, when Ba(NO₃)₂ (s) is dissolved in water, its transformation is:

When solid barium nitrate (Ba(NO₃)₂) dissolves in water, it undergoes a dissociation process where the compound breaks apart into its constituent ions.

Dissociation refers to the process in which a compound breaks apart into its constituent ions when dissolved in a solvent, typically water. In this process, the chemical bonds within the compound are disrupted, resulting in the separation of positive and negative ions.

The dissociation occurs due to the interaction between the solute particles and the solvent molecules, leading to the formation of hydrated ions.

The transformation can be represented as follows:

Ba(NO₃)₂(s) → Ba²⁺(aq) + 2NO₃⁻(aq)

In this process, the barium nitrate compound dissociates into barium ions (Ba²⁺) and nitrate ions (NO₃⁻) in the aqueous solution. The resulting ions are free to move and conduct electricity, indicating that barium nitrate is a strong electrolyte when dissolved in water.

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Answer:

The answer is

Explanation:

Density = mass / volume

mass = density × volume

volume = 2cm³

density = 520g/cm³

mass = 2 × 520

= 1040g

Hope this helps you

3.97×1023 molecules C2H6          1 mol  C2H6  

------------------------------------------ x ------------------------------------   = 0.66 mol C2H6

                                                    6.022 x 1023 molec. C2H6

Answer:

D . Sulphur

Explanation:

the element with a 3p4 valence configuration, look in period 3 and group XVI, and that is ...

S, sulfur.

Answer:

The number of moles of Cl₂ present at equilibrium is 3.94x10⁻⁴ moles.

Explanation:

The reaction is:

CO(g) + Cl₂(g) ⇄ COCl₂(g)  

The equilibrium constant of the above reaction is:

K = 1.2x10³

To find the moles of Cl₂ present at equilibrium, let's evaluate the reverse reaction:

COCl₂(g) ⇄ CO(g) + Cl₂(g)  

The equilibrium constant for the reverse reaction is:

[tex] K_{r} = \frac{1}{1.2 \cdot 10^{3}} = 8.3 \cdot 10^{-4} [/tex]

Now, we need to calculate the concentration of CO and COCl₂:

[tex] C_{CO} = \frac{\eta_{CO}}{V} = \frac{0.3500 moles}{3.050 L} = 0.115 M [/tex]

[tex] C_{COCl_{2}} = \frac{\eta_{COCl_{2}}}{V} = \frac{0.05500 moles}{3.050 L} = 0.018 M [/tex]

Now, from the reaction we have:

COCl₂(g) ⇄ CO(g) + Cl₂(g)  

0.018 - x       0.115+x   x    

The concentration of Cl₂ is:

[tex] K_{r} = \frac{[CO][Cl_{2}]}{[COCl_{2}]} [/tex]

[tex] 8.3 \cdot 10^{-4} = \frac{(0.115 + x)(x)}{0.018 - x} [/tex]  

[tex] 8.3 \cdot 10^{-4}*(0.018 - x) - (0.115 + x)(x) = 0 [/tex]  

By solving the above equation for x we have:

x = 1.29x10⁻⁴ M = [Cl₂]

Finally, the number of moles of Cl₂ present at equilibrium is:

[tex] \eta_{Cl_{2}} = C_{Cl_{2}}*V = 1.29 \cdot 10^{-4} mol/L*3.050 L = 3.94 \cdot 10^{-4} moles [/tex]

Therefore, the number of moles of Cl₂ present at equilibrium is 3.94x10⁻⁴ moles.

I hope it helps you!

Answer:

a. boiling

Explanation:

Answer:

Edge length of the unit cell is 4.07x10⁻¹⁰m

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

That means edge lenght is = √8 r

adius

As radius of Silver is 144pm = 144x10⁻¹²m:

a = √8 r

a = √8 ₓ 144x10⁻¹²m

a = 4.07x10⁻¹⁰m

Answer:

0.109 mol/tablespoon

Explanation:

6.37 g/ 58.5 mol = 0.10888888 mol (0.109 significantly)

Answer:

A: 0.109

Explanation:

Edge 2020

Answer:

The reaction is not in equilibrium

Explanation:

For the reaction:

PCl₅ ⇄ PCl₃ + Cl₂

Equilibrium constant, Kp, is defined as:

[tex]Kp = \frac{P_{PCl_3}P_{Cl_2}}{P_{PCl_5}} = 0.497[/tex]

When this ratio is = 0.497, the reaction is in equilibrium. Replacing the pressures of the problem, reaction quotient, Q, is:

[tex]Q =\frac{1.322atm*0.911atm}{0.820atm} = 1.469[/tex]

As Q ≠ Kp, the reaction is not in equilibrium

To reach the equilibrium, the reaction will shift to the left producing more reactant and decreasing amount of products.

Answer:

[tex]\Delta _cH=-1328.3kJ/mol[/tex]

Explanation:

Helllo,

In this case, for the given chemical reaction in gaseous state:

[tex]CH_3OCH_3+3O_2\rightarrow 2CO_2+3H_2O[/tex]

We comoute the combustion enthalpy as the reaction enthalpy for one mole of fuel (dimethyl ether) considering the formation enthalpy of each given substance and whether they are reactants (subtracting) or products (adding), therefore we write:

[tex]\Delta _cH=2*\Delta _fH_{CO_2}+3*\Delta _fH_{H_2O}-\Delta _fH_{CH_3OCH_3}-3*\Delta _fH_{O_2}[/tex]

Whereas the formation enthalpies for carbon dioxide, water, dimethyl ether and oxygen are -393.5, -241.8, -184.1 and 0 kJ/mol respectively, thereby, the combustion enthalpy turns out:

[tex]\Delta _cH=2(-393.5)+3*(-241.8)-(-184.1)-3(0)\\\\\Delta _cH=-1328.3kJ/mol[/tex]

Notice that enthalpy of formation of oxygen is zero since forming an element has no chemical sense, it just exists as it has been early demonstrated.

Regards.

The question is incomplete; the complete question is: Classify each molecule by whether its real bond angles are the same as or different than its model (ideal) bond angles. In other words, do the bond angles change when you switch between Real and Model mode at the top of the page? Same (angles do not change) Different (angles change) Answer Bank | H2O | CO2, SO2, XeF2, BF3 CIF3, NH3, CH4, SF4, XeF4, BrF5, PCI5,SF6

Answer:

Compounds whose real bond angle are the same as ideal bond angle;

SF6, BF3, CH4, PCI5

Compounds whose real bond angles differ from ideal bond angles;

H2O, CO2, SO2, XeF2, CIF3, NH3, SF4, XeF4, BrF5

Explanation:

According to the valence shell electron pair repulsion theory (VSEPR), molecules adopt various shapes based on the number of electron pairs on the valence shell of the central atom of the molecule. The electron pairs usually orient themselves as far apart in space as possible leading to various observed bond angles.

The extent of repulsion of lone pairs is greater than that of bond pairs. Hence, the presence of lone pairs on the valence shell of the central atom in the molecule distorts the bond angles of molecules away from the ideal bond angles predicted on the basis of valence shell electron pair repulsion theory.

For instance, methane is a perfect tetrahedron having an ideal bond angle of 109°28'. Both methane and ammonia are based on a tetrahedron, however, the presence of a lone pair of electrons on nitrogen distorts the bond angle of ammonia to about 107°. The distortion of lone pairs in water is even more as the bond angles of water is about 104°.

Answer:

C. An electron at this electrode has a higher potential energy than it has at a standard hydrogen electrode.

Explanation:

The standard hydrogen electrode (SHE) is used to measure the electrode potential of substances. The standard hydrogen electrode is arbitrarily assigned an electrode potential of zero. Recall that electrode potentials are always measured as reduction potentials in electrochemical systems.

For an electrode that has a negative electrode potential, electrons at this electrode have a higher potential energy compared to electrons at the standard hydrogen electrode. Electrons flow from this electrode to the hydrogen electrode.

On the other hand, a positive electrode potential implies that an electron at this electrode has a lower potential energy than it has at a standard hydrogen electrode. Hence electrons will flow from the standard hydrogen electrode to this electrode.

Answer:

See explanation

Explanation:

Hydration of alkenes is a common reaction in organic chemistry. Hydration is simply the addition of water to an alkene. This is an acid catalysed reaction as we can see from the mechanism attached.

Recall that our task is to carry out the synthesis of 2-butanol using an alkene starting material in which there will be no rearrangement of the intermediate carbocation. If we start with the compound shown in the image (but-2-ene), the first step is the formation of the secondary carbocation. This is followed by the addition of water. Subsequently, the added water is deprotonated by another water molecule to yield 2-butanol and the acid catalyst. All these steps have been clearly outlined in the image attached.

Answer:

i. n = 5

ii. ΔE = 7.61 × [tex]10^{-46}[/tex] KJ/mole

Explanation:

1. ΔE = (1/λ) = -2.178 × [tex]10^{-18}[/tex]([tex]\frac{1}{n^{2}_{final} }[/tex] - [tex]\frac{1}{n^{2}_{initial} }[/tex])

    (1/434 × [tex]10^{-9}[/tex]) = -2.178 × [tex]10^{-18}[/tex] ([tex]\frac{n^{2}_{initial} - n^{2}_{final} }{n^{2}_{final} n^{2}_{initial} }[/tex])

⇒ 434 × [tex]10^{-9}[/tex] = (1/-2.178 × [tex]10^{-18}[/tex])[tex]\frac{n^{2}_{final} *n^{2}_{initial} }{n^{2}_{initial} - n^{2}_{final} }[/tex]

But, [tex]n_{final}[/tex] = 2

434 × [tex]10^{-9}[/tex] = (1/2.178 × [tex]10^{-18}[/tex])[tex]\frac{2^{2} n^{2}_{initial} }{n^{2}_{initial} - 2^{2} }[/tex]

434 × [tex]10^{-9}[/tex]  × 2.178 × [tex]10^{-18}[/tex] = [tex](\frac{4n^{2}_{initial} }{n^{2}_{initial} - 4 })[/tex]

⇒ [tex]n_{initial}[/tex] = 5

Therefore, the initial energy level where transition occurred is from 5.

2. ΔE = hf

     = (hc) ÷ λ

    = (6.626 × 10−34 × 3.0 × [tex]10^{8}[/tex] ) ÷ (434 × [tex]10^{-9}[/tex])

    = (1.9878 × [tex]10^{-25}[/tex]) ÷ (434 × [tex]10^{-9}[/tex])

    = 4.58 × [tex]10^{-19}[/tex] J

    = 4.58 × [tex]10^{-22}[/tex] KJ

But 1 mole = 6.02×[tex]10^{23}[/tex], then;

energy in KJ/mole = (4.58 × [tex]10^{-22}[/tex] KJ) ÷ (6.02×[tex]10^{23}[/tex])

         = 7.61 × [tex]10^{-46}[/tex] KJ/mole

The initial energy level is 5  and the energy of this transition in units of kJ/mole is 7.57 * 10^-43 kJ/mole

We must first calculate ΔE as follows;

ΔE = hc/λ

h = Plank's constant = 6.6 * 10^-34 Js

c = speed of light = 3 * 10^8 m/s

λ = wavelength = 434 * 10^-9

ΔE =  6.6 * 10^-34 * 3 * 10^8/434 * 10^-9

ΔE = 0.0456 * 10^-17 J

ΔE = [tex]ΔE = -2.178 x10^-18 (\frac{1}{n^2final} - \frac{1}{n^2initial}) \\ΔE = -2.178 x10^-18 (\frac{1}{2^2} - \frac{1}{n^2initial} )\\\\4.56 * 10^-19/2.178 x10^-18 = (\frac{1}{2^2} - \frac{1}{n^2initial})\\0.210 = (\frac{1}{2^2} - \frac{1}{n^2initial})\\\frac{1}{n^2initial} = 0.25 - 0.210\\\frac{1}{n^2final} = 0.04\\n = (\sqrt{(0.04)^-1} \\n = 5[/tex]

Energy of this transition in units of kJ/mole = 4.56 * 10^-19/ 6.02 * 10^23

= 7.57 * 10^-43 kJ/mole

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Answer:

conductivity increasing order CH₃-CH₂-OH < HF< NaCl

Explanation:

NaCl is the better conductor comparing with remaining two. it is strong electrolyte. dissociation percent always nearly eqaul to 100% but HF is weaker acid than NaCl and dissociation percent <100% . So, the no of ions furnished by HF less than that of NaCl

CH₃-CH₂-OH organic compound . in general it is not treated as an electrolyte and it cannot carry any charge. If it  carries, it is very very less compared to remaining two

Answer:

254.23 g/mol

Explanation:

Atomic mass for Osmium tetroxide would be 254.23 g/ml

Answer:254.2276

Explanation:

Answer:

here, as we have known the elements of group 3A(13) such as aluminium , boron has three valance electron and in perodic table the elements are kept with similar proterties in same place so, their valance electron is 3.

hope it helps...

The number of valence electrons are in the electron dot structures for the elements in group 3A(13) is three.

The periodic table is organized into groups (vertical columns), periods (horizontal rows), and families (groups of elements that are similar). Elements in the same group have the same number of valence electrons.

Groups are the columns of the periodic table, and periods are the rows. There are 18 groups, and there are 7 periods plus the lanthanides and actinides.

There are two different numbering systems that are commonly used to designate groups, and you should be familiar with both.

The traditional system used in the United States involves the use of the letters A and B. The first two groups are 1A and 2A, while the last six groups are 3A through 8A. The middle groups use B in their titles.

Therefore, The number of valence electrons are in the electron dot structures for the elements in group 3A(13) is three.

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Answer:

1) HCl (Hydrochloric acid reacts with NH3 and forms dense fumes)

2) Methane (It is from the group of alkanes and is a green house gas)

3) Deliquescent substance (It is a solid which when kept in open forms a solution after sometime)

4) Brass (It is a solid-in-solid solution used to make electrical fittings)

5) Aluminium

Question 2:

1)    Al₂O₃ + 2NaOH    ⇒   2NaAlO₂  (Sodium Aluminate) + H₂O

2)   Zn +  H₂SO₄   (dilute)  =>  ZnSO₄  (Zinc Sulphate) + H₂

Answer:

equations

Aluminium oxide and Sodium hydroxide react to form water and sodium aluminate

Zinc reacts with dilute sulphuric acid to form zinc sulphate and hydrogen gas

Answer:

Li

Explanation:

The phenomenon of wave particle duality was well established by Louis deBroglie. The wavelength associated with matter waves was related to its mass and velocity as shown below;

λ= h/mv

Where;

λ= wavelength of matter waves

m= mass of the particle

v= velocity of the particle

This implies that if the velocities of all particles are the same, the wavelength of matter waves will now depend on the mass of the particle. Hence; the wavelength of a matter wave associated with a particle is inversely proportional to the magnitude of the particle's linear momentum. The longest wavelength will then be obtained from the smallest mass of matter. Hence lithium which has the smallest mass will exhibit the longest DeBroglie wavelength

The atom that have the longest de Broglie wavelength is ; ( A ) Li

Wave particle duality is a phenomenon by de Broglie. that shows that The wavelength associated with matter waves is related to its mass and velocity .

Wave particle duality is represented as ;  λ = h / mv

λ= wavelength of matter waves

m= mass of the particle

v= velocity of the particle

Given that the elements have the same velocity the atom that would have the longest de Broglie wavelength is Li

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Answer:

C. Tetrahedral

Explanation:

Tetrahedral would be the correct choice because the central atom has 4 domains (1 bond counts as 1 domain so 4 bonds =4) and no lone pairs which means it has tetra (which translates to four) domains hence tetrahedral.

Tetrahedral is the shape of a molecule that has 4 atoms bonded to a central atom and no lone pairs of electrons. Hence, option C is correct.

An atom consists of a central nucleus that is usually surrounded by one or more electrons.

Tetrahedral would be the correct choice because the central atom has 4 domains (1 bond counts as 1 domain so 4 bonds =4) and no lone pairs which mean it has tetra (which translates to four) domains hence tetrahedral.

Hence, option C is correct.

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Hey! :)

__________ ☆ ☆__________________________________

Answer:

The answer is Electronegativity increases up and to the right

Explanation:

When you move from left to right it increases ( in the periodic table )

But when you move down the table electronegativity decreases.

So “ Electronegativity increases up and to the right” describes the trends the best.

Hope this helps! :)

____________☆ ☆________________________________

By, BrainlyMember ^-^

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Answer:

Explanation: M(PCL5)= 31 + 5(35.5)

=208.5g/mol

M(H20)= 18g/mol

n(PCL5) = 75.5÷208.5

= 0.362mol

n(HCl)/n(PCL5)= 5/1

n(HCl)= 5×0.362

=1.81mol of HCl

Answer:

the mass of HBr that would react is 25.41 g of HBr

Explanation:

attached is the calculations.

Answer:

Acid, Horizontally, Drain off the acid by opening the stopcock

Explanation:

A buret is a calibrated glass appratus used to measure and deliver accurate volume of liquid, usually acids, in acid-base titrations.

Before usually the buret to carry out your titration reaction, it is advised to condition or prepare your buret for use.

To condition the buret, add a small volume of the acid to the buret and rotate the barrel horizontally to ensure that the liquid ( acid) makes contact with the inner surface of the barrel o the buret.

This procedure is done to wash off any previous acid or liquid that the buret had been used for in previous titrations.

Complete this action by opening the stopcock of the buret to drain off the acid and discarding each volume in a designated waste container.

Note that acids could be corrosive and dangerous to the skin and so should be handled with great care

Answer:

The calculator add the CO2 released from the use of electricity, released from driving and the CO2 from the waste that we disposed.

Explanation:

The carbon dioxide, CO2 is what the human body does not need, therefore, we breathe it out, hence taking in oxygen(respiration process). The plants need oxygen for the production of their own food.

The carbon calculator estimate the amount of CO2 that each individual releases into the atmosphere through the consideration of several factors such as the kind of food that we eat.

Therefore, if we are to use the carbon calculator to determine the amount of CO2 that each individual releases into the atmosphere we will have:

The amount of CO2 that each individual releases into the atmosphere =( CO2 released from the use of electricity) + (CO2 released from driving) + (the CO2 from the waste that we disposed).

Answer:

The percent composition of fluorine is 65.67%

Explanation:

Percent Composition is a measure of the amount of mass an element occupies in a compound. It is measured in percentage of mass.

That is, the percentage composition is the percentage by mass of each of the elements present in a compound.

The calculation of the percentage composition of an element is made by:

[tex]percent composition element A=\frac{total mass of element A}{mass of compound} *100[/tex]

In this case, the percent composition of fluorine is:

[tex]percent composition of fluorine=\frac{39.6 g}{60.3 g} *100[/tex]

percent composition of fluorine= 65.67%

The percent composition of fluorine is 65.67%

Answer:

The percent composition of fluorine is 65.67%

Explanation:

Percent Composition is a measure of the amount of mass an element occupies in a compound. It is measured in percentage of mass.

That is, the percentage composition is the percentage by mass of each of the elements present in a compound.

The calculation of the percentage composition of an element is made by:

In this case, the percent composition of fluorine is:

percent composition of fluorine= 65.67%

The percent composition of fluorine is 65.67%