Explain why phosphorus has a low melting point.

Answer:

Explanation:

equilibrium constant

Kc = [ C ]² / [ A ] [ B ]

= .5² / .2  x 3

= .4167

Let moles of A to be added be n

concentration of A unreacted becomes .2 + n M

increase of product C by .2 M will require use of A  and B be .1 M

So unreacted A = .2 + n - .1 = n + .1

Kc = [ C ]² / [ A ] [ B ]

.4167 = .7² / ( n + .1 ) ( 3 - .1 )

n + .1 = .4

n = . 3 moles .

So .3 moles of A to be added .

Answer:

See explanation

Explanation:

For this question, we have to remember the definition of a meso-compound. In a meso-compound, we will have chiral carbons but we don't optical activity. This is due to the symmetry, if we have symmetry in a substance with chiral carbons the optical activity is nullified. So, if we want to find the meso-compounds we have to find symmetry planes in the molecule.

A symmetry plane is an imaginary cut that can divide the molecule in two equal parts. We have to draw the molecule first (see figure 1) and then we can try to find the symmetry planes.

With this in mind, the only compounds with symmetry planes are:

b) cis-1,3-dimethylcyclohexane

d) cis-1,4-dimethylcyclohexane

See figure 2 to more explanations

I hope it helps!

Answer:

Following are the answer to this question:

Explanation:

The structure of the S molecular formula [tex]C_7H_{14}O_2[/tex] defined in the attachment file.

Please find the attachment file.

Answer:

please look at the picture below.

Explanation:

Answer:

The main competing reaction when a primary alkyl halide is treated with alcoholic potassium hydroxide is SN2 substitution.

Explanation:

The relative percentage of products of the reaction between an alkyl halide and alcoholic potassium hydroxide generally depends on the structure of the primary alkylhalide. The attacking nucleophile/base in this reaction is the alkoxide ion. Substitution by SN2 mechanism is a major competing reaction in the elimination reaction intended.

A more branched alkyl halide will yield an alkene product due to steric hindrance, similarly, a good nucleophile such as the alkoxide ion may favour SN2 substitution over the intended elimination (E2) reaction.

Both SN2 and E2 are concerted reaction mechanisms. They do not depend on the formation of a carbocation intermediate. Primary alkyl halides generally experience less steric hindrance in the transition state and do not form stable carbocations hence they cannot undergo E1 or SN1 reactions.

SN2 substitution cannot occur in a tertiary alkyl halides because the stability of tertiary carbocations favours the formation of a carbocation intermediate. The formation of this carbocation intermediate will lead to an SN1 or E1 mechanism. SN2 reactions is never observed for a tertiary alkyl halide due to steric crowding of the transition state. Also, with strong bases such as the alkoxide ion, elimination becomes the main reaction of tertiary alkyl halides.

Answer:

18.308 KJ

Explanation:

From the given above, we obtained the following:

Activation energy for the catalyzed reaction = 4.6 kJ.

Activation energy for the uncatalyzed reaction =..?

Now, a careful observation of the question revealed that the activation energy of the uncatalyzed reaction is about 3.98 times that of the catalyzed reaction.

With this vital information, we can thus, calculate the activation energy of the uncatalyzed reaction as follow:

Activation energy for the uncatalyzed reaction = 3.98 times that of the catalyzed reaction.

Activation energy for the uncatalyzed reaction = 3.98 x 4.6 kJ = 18.308 KJ

Therefore, the activation energy of the uncatalyzed reaction is 18.308 KJ.

Answer:

Explanation:

1 ) Cl₂ + ZnBr₂ = ZnCl₂ + Br₂

In this reaction , oxidation number of Cl decreases  from 0 to -1  so it is reduced  and oxidation number of Br increases from -1 to 0 so it is oxidised . Hence this reaction is oxidation - reduction reaction .

2 )

Pb( ClO₄)₂ + 2KI = PbI₂ + 2KClO₄

In this reaction oxidation number of none is changing so it is not an oxidation - reduction reaction.

3 )

CaCO₃ = CaO + CO₂

In this reaction also oxidation number of none is changing so it is not an oxidation - reduction reaction.

So only first reaction is oxidation - reduction reaction.

2nd option is correct.

Answer:

The answer is Neutralization reaction

It occurs when an acid and a base are present in the same solution and react to form salt and water only

Hope this helps you

Answer:

8.13 g/mol.

Explanation:

The following formula gives us the relationship between the effusion rates of two gases and their molar masses:

[tex]\sqrt{\frac{MM_{x} }{MM_{y} } } = \frac{rate_{y} }{rate_{x} }[/tex]

where x and y are respective sample gases and MM and rate are molar mass and rate of effusion respectively.

⇒[tex]\sqrt{\frac{14}{y} } = \frac{63}{48}[/tex]

[tex]\frac{14}{y} = 1.3125^{2}[/tex]

y= 14 / [tex]1.3125^{2}[/tex]  = 8.13 g/mol.

Answer:

Hey there!

CS2) Carbon Disulfide.

PBr3) Phosphorus Tribromide

NO) Nitric Oxide

CF4) Carbon Tetrafluoride

P2O5) Phosphorus Pentoxide

Let me know if this helps :)

Answer:

[tex]T_2=453.05K[/tex]

Explanation:

Hello,

In this case, the temperature-variable Arrhenius equation is written as:

[tex]\frac{k(T_2)}{k(T_1)}=exp(\frac{Ea}{R}(\frac{1}{T_2}-\frac{1}{T_1} ))[/tex]

Now, for us to solve for the temperature by which the reaction rate constant is 0.0760M/s we proceed as shown below:

[tex]ln(\frac{k(T_2)}{k(T_1)})=\frac{Ea}{R}(\frac{1}{T_2}-\frac{1}{T_1} )\\ln(\frac{0.0760M/s}{0.00000291M/s} )=\frac{183000J/mol}{8.314J/(mol*K)} *(\frac{1}{T_2} -\frac{1}{573K} )\\\frac{1}{T_2} -\frac{1}{573K} =\frac{10.17}{22011.06K^{-1}} \\\\\frac{1}{T_2}=4.62x10^{-4}K^{-1}+\frac{1}{573K}\\\\\frac{1}{T_2}=2.21x10^{-3}K^{-1}\\\\T_2=453.05K[/tex]

Regards.

Answer:

The correct answer is 0.30 M

Explanation:

The molar concentration or molarity of a solution is defined as moles of solute per liter of solution. We found the moles of solute (glucose) by dividing the mass (54.30 g) into the molecular weight (MW) of glucose (C₆H₁₂O₆):

MW(C₆H₁₂O₆)= (12 g/mol x 6) + (1 g/mol x 12) + (16 g/mol x 6) = 180 g/mol

Moles of glucose= mass/MW= 54.30 g/(180 g/mol)= 0.30 mol

There is 0.30 mol of solute per liter of solution, thus the molarity is:

M= moles solute/L solution= 0.30 mol/1 L = 0.30 M

Answer: The molar mass of the unknown compound is 200 g/mol

Explanation:

Depression in freezing point is given by:

[tex]\Delta T_f=i\times K_f\times m[/tex]

[tex]\Delta T_f=2.74^0C[/tex] = Depression in freezing point

i= vant hoff factor = 1 (for molecular compound)

[tex]K_f[/tex] = freezing point constant = [tex]5.12^0C/m[/tex]

m= molality

[tex]\Delta T_f=i\times K_f\times \frac{\text{mass of solute}}{\text{molar mass of solute}\times \text{weight of solvent in kg}}[/tex]

Weight of solvent (benzene)= 0.250 kg  

Molar mass of solute = M g/mol

Mass of solute  = 26.7 g

[tex]2.74^0C=1\times 5.12\times \frac{26.7g}{Mg/mol\times 0.250kg}[/tex]

[tex]M=200g/mol[/tex]

Thus the molar mass of the unknown compound is 200 g/mol

The molar mass of an unknown solute compound in the solution has been 199.626 g/mol.

With the addition of the solute to the solution, there has been a depression in the freezing point. The depression in the freezing point can be expressed as:

Depression in freezing point = Van't Hoff factor × Freezing point constant × molality

The molality can be defined as the moles of solute per kg solvent

Molaity = [tex]\rm \dfrac{Mass\;of\;solute\;(g)}{Molecular\;mass\;of\;solute}\;\times\;\dfrac{1}{Mass\;of\;solvent\;(kg)}[/tex]

The depression in freezing point can be given as:

Depression in freezing point = Van't Hoff factor × Freezing point constant ×  [tex]\rm \dfrac{Mass\;of\;solute\;(g)}{Molecular\;mass\;of\;solute}\;\times\;\dfrac{1}{Mass\;of\;solvent\;(kg)}[/tex] ......(i)

Given, the depression in freezing point = 2.74 [tex]\rm ^\circ C[/tex]

Van't Hoff factor = 1 (Molecular compound)

Freezing point constant (Kf) = 5.12 [tex]\rm ^\circ C[/tex]/m

Mass of solute = 26.7 g

Mass of solvent = 0.250 kg

Substituting the values in equation (i):

2.74 [tex]\rm ^\circ C[/tex] = 1 × 5.12

[tex]\rm \dfrac{2.74}{5.12}[/tex] = [tex]\rm \dfrac{1}{Molecular\;mass\;of\;solute}\;\times\;\dfrac{26.7}{0.250\;kg}[/tex]

0.535 = [tex]\rm \dfrac{1}{Molecular\;mass\;of\;solute}\;\times\;106.8[/tex]

Molecular mass of solute = [tex]\rm \dfrac{106.8}{0.535}[/tex] g/mol

Molecular mass of solute = 199.626 g/mol

The molar mass of an unknown solute compound in the solution has been 199.626 g/mol.

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

https://brainly.com/question/7229194

Answer:

0.83 g

Explanation:

Step 1: Write the balanced equation

Fe + CuSO₄ ⇒ Cu + FeSO₄

Step 2: Calculate the moles corresponding to 0.75 g of Fe

The molar mass of Fe is 55.85 g/mol.

[tex]0.75g \times \frac{1mol}{55.85g} = 0.013 mol[/tex]

Step 3: Calculate the moles of Cu produced from 0.013 moles of Fe

The molar ratio of Fe to Cu is 1:1. The moles of Cu produced are 1/1 × 0.013 mol = 0.013 mol.

Step 4: Calculate the mass corresponding to 0.013 moles of Cu

The molar mass of Cu is 63.55 g/mol.

[tex]0.013mol \times \frac{63.55g}{mol} = 0.83 g[/tex]

Answer:

If 0.75 grams of iron (Fe) react, 0.85 grams of copper (Cu) will be produced.

Explanation:

You know the following balanced reaction:

Fe + CuSO₄ ⇒ Cu + FeSO₄

By stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction), the following quantities react and are produced:

Being:

the molar mass of the compounds participating in the reaction is:

Then, by stoichiometry of the reaction, the amounts of reagent and product that participate in the reaction are:

Then you can apply a rule of three as follows: if 55.85 grams of Fe produces 63.54 grams of Cu, 0.75 grams of Fe how much mass of Cu does it produce?

[tex]mass of Cu=\frac{0.75 grams of Fe*63.54 grams of Cu}{55.85 grams of Fe}[/tex]

mass of Cu= 0.85 grams

If 0.75 grams of iron (Fe) react, 0.85 grams of copper (Cu) will be produced.

Answer:

Rubidium oxide

Explanation:

Answer:

CaF2 + CO3- ----> CaCO3 + 2 F-

Explanation:

The chemical compounds found on the left side of the date are the reagents and those found on the right are the products, where calcium carbonate appears.

Calcium carbonate is a quaternary salt

yes yes yes yes

yes

yes

yes

yes

yes

Answer:

NaClO + CH₃COOH ----> HClO + CH3CO- + Na

Explanation:

This reaction occurs between the combination of a salt and an acid, that is, an oxide-reduction reaction

Answer:

because it is

Answer:

4.2g of gasoline

Explanation:

In the problem, you need to give energy to the cup from the combustion of gasoline. The energy you need to give is:

Qcup + QWater + QCopper

As you need to increase (110ºC - 21ºC = 89º = Increase 89K) 89K, the Qcup is:

Qcup = 89K × (47J/K) = 4183J.

You can find Qwater using its specific heat, C (4.18Jg⁻¹K⁻¹), its mass (450mL = 450g) and the change of temperature, 89K:

QWater = CₓmₓΔT

QWater = 4.184Jg⁻¹K⁻¹ ₓ 450g×89K

QWater = 167569J

And Q of Copper, QCu, could be obtained in the same way (Specific heat Cu: 0.387 J/g⁻¹K⁻¹:

QCu = CₓmₓΔT

QCu = 0.387 J/g⁻¹K⁻¹ₓ850gₓ89K

QCu = 29277J

Thus, total heat you need is:

Q = Qcup + QWater + QCopper

Q = 4183J + 167569J + 29277J

Q = 201029J = 201kJ

The combustion of gasoline (Octane) produce 47.8kJ/g (Its heat of combustion). that means to produce 201kJ of energy you require:

201kJ × (1g / 47.8kJ) =

Answer:

Constant-volume molar heat capacity of argon is 12.47 J K ⁻¹mol⁻¹

Explanation:

Argon is a monoatomic gas that behaves as an ideal gas at 298K.

Using the first law of thermodinamics you can obtain:

Work, Q, for constant pressure molar heat capacity,CP:

CP = (5/2)R

For constant-volume molar heat capacity,CV:

CV = (3/2)R

That means:

2CP/5 = 2CV/3

3/5 = CV / CP

As CP of Argon is 20.79 J K ⁻¹mol⁻¹, CV will be:

3/5 = CV / CP

3/5 = CV / 20.79 J K ⁻¹mol⁻¹

12.47 J K ⁻¹mol⁻¹ = CV

Answer:

the molecular mass of hydrogen sulphide, which contains two atoms of hydrogen and one atom of sulphur is = 2 — 1 + 1 — 32 = 34 a.m.u.

Answer:

[Ar] 4s²

Explanation:

Ca is the symbol for Calcium. It is the 20th element and it has 20 electrons.

The full electronic configuration for calcium is given as;

1s²2s²2p⁶3s²3p⁶4s²

The condensed electronic configuration is given as;

[Ar] 4s²

Answer:

Δ [tex]G_{rxn}[/tex] = −51. 4 kJ/mol

However, since Δ [tex]G_{rxn}[/tex] is negative. The hydrolysis of ATP for this reaction is said to be spontaneous

Explanation:

From the question; The equation for this reaction can be represented as :

[tex]ATP_{(aq)} + H_2O_{(l)} \to ADP_{(aq)}+ HPO_4^{2-}} _{(aq)}[/tex]

where:

[tex]\Delta G ^0 _{rxn} =[/tex]-30.5 kJ/mol

= -30.5 kJ/mol × 1000 J/ 1 kJ

= -30.5 × 10 ⁻³ J/mol

Temperature T = 37 ° C

= (37+273)

= 310 K

pH = 7.0

[ATP] = 5.0 mM

= 5.0mM × 1M/1000mM

= 0.005 M

[ADP] = 0.30 mM

= 0.30 mM × 1M/1000mM

= 0.0003 M

[tex][HPO_4^{2-}}][/tex] = 5.0 mM

= 5.0mM × 1M/1000mM

= 0.005 M

The objective is to calculate the value for Δ [tex]G_{rxn}[/tex] in the biological cell and to determine if the hydrolysis  of  ATP is spontaneous under these conditions.

Now;

From the equation given; the equilibrium constant [tex]K_{eq}[/tex] can be expressed as:

[tex]K_{eq} = \dfrac{[ADP][ HPO_4^{2-}]} {[ATP]}[/tex]

[tex]K_{eq} = \dfrac{(0.0003 \ M)(0.005 \ M)} {(0.005 \ M)}[/tex]

[tex]K_{eq} = 3*10^{-4}[/tex]

The  Δ [tex]G_{rxn}[/tex] in the biological cell can now be calculated as:

Δ [tex]G_{rxn}[/tex] = [tex](-30.5 * 10 ^3 \ J/mol) + (8.314 \ J/mol.K)(310 K ) In ( 3*10^{-4})[/tex]

Δ [tex]G_{rxn}[/tex] = [tex](-30.5 * 10 ^3 \ J/mol) + (-20906.68126)[/tex]

Δ [tex]G_{rxn}[/tex] = −51406.68 J/mol

Δ [tex]G_{rxn}[/tex] = −51. 4 × 10³ J/mol

Δ [tex]G_{rxn}[/tex] = −51. 4 kJ/mol

Thus since Δ [tex]G_{rxn}[/tex] is negative. The hydrolysis for this reaction is said to be spontaneous

Answer:

Thin Layer Chromatography (TLC) can be used to analyze chemical reactions. During this reaction monitoring, a typical TLC plate would have three spots: the reactant lane, the reaction mixture lane, and a "co-spot" where reaction product would be spotted directly on top of reactant.

The co-spot serves as a reference point and is vital for reactions where reactant and product have similar Rfs, and many other variations of eluent tracking.

To indicate completion of the reaction, the disappearance of a spot (usually the starting reactant) is observed.

Answer:  13.9 g of [tex]H_2O[/tex] will be produced from the given mass of oxygen

Explanation:

To calculate the moles :

[tex]\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}[/tex]    

[tex]\text{Moles of} O_2=\frac{28.8g}{32.00g/mol}=0.900moles[/tex]

The balanced chemical reaction is:

[tex]4NIO_2(g)+7O_2(g)\rightarrow 4NO_2(g)+6H_2O(g)[/tex]

According to stoichiometry :

7 moles of [tex]O_2[/tex] produce =  6 moles of [tex]H_2O[/tex]

Thus 0.900 moles of [tex]O_2[/tex] will produce =[tex]\frac{6}{7}\times 0.900=0.771moles[/tex]  of [tex]H_2O[/tex]

Mass of [tex]H_2O=moles\times {\text {Molar mass}}=0.771moles\times 18.02g/mol=13.9g[/tex]

Thus 13.9 g of [tex]H_2O[/tex] will be produced from the given mass of oxygen

Answer:

RbOH

Explanation:

For this question, we have to remember what is the definition of a base. A base is a compound that has the ability to produce hydroxyl ions [tex]OH^-[/tex], so:

[tex]AOH~->~A^+~+~OH^-[/tex]

With this in mind we can write the reaction for each substance:

[tex]HCOOH~->~HCOO^-~+~H^+[/tex]

[tex]RbOH~->~Rb^+~+~OH^-[/tex]

[tex]H_2CO_3~->~CO_3^-^2~+~2H^+[/tex]

[tex]NaNO_3~->~Na^+~+~NO_3^-[/tex]

The only compound that fits with the definition is [tex]RbOH[/tex], so this is our base.

I hope it helps!

RbOH or rubidium hydroxide is a base.  

• RbOH or rubidium hydroxide is the inorganic compound.  

• It comprises hydroxide anions and an equal number of rubidium cations.  

• Rubidium hydroxide, like other strong bases is highly corrosive.  

• The formation of rubidium hydroxide takes place when the metal rubidium reacts with water.  

• The bases refers to the substance, which gets dissociate in an aqueous solution to produce OH- or hydroxide ions.  

• Rubidium hydroxide is a base, when it is dissolved in water it give rise to an alkali, when reacts with acid it generates a rubidium salt.  

Thus, RbOH is a base.

To know more about:

https://brainly.com/question/18616956

Answer:

a change in temperature would be observed(ΔH is -ve)

Explanation:

Hydrochloric acid react with sodium hydroxide to give salt(sodium chloride) and water

HCl(aq) + NaOH(aq) =====> NaCl(aq) + H2O(l)

There would be no notable change since sodium chloride dissolved in water but there would be a change in temperature.

Since neutralization is exothermic(heat is evolved), therefore ΔH is negative

Answer:

pH ≅ 4.80

Explanation:

Given that:

the volume of HN₃ = 25 mL = 0.025 L

Molarity of HN₃ = 0.150 M

number of moles of HN₃ = 0.025 × 0.150

number of moles of HN₃ =  0.00375  mol

Molarity of NaOH = 0.150 M

the volume of NaOH = 13.3 mL = 0.0133

number of moles of NaOH = 0.0133× 0.150

number of moles of NaOH = 0.001995 mol

The chemical equation for the reaction of this process can be written as:

[tex]HN_3 + OH- ---> N^-_{3} + H_2O[/tex]

1 mole of hydrazoic acid react with 1 mole of hydroxide to give nitride ion and water

thus the new number of moles of HN₃ = 0.00375 - 0.001995 = 0.001755 mol

Total volume used in the reaction =  0.025 +  0.0133 = 0.0383  L

Concentration of [tex]HN_3[/tex] = [tex]\dfrac{0.001755}{0.0383}[/tex] = 0.0458 M

Concentration of [tex]N^{-}_3[/tex] = [tex]\dfrac{ 0.001995 }{0.0383}[/tex] = 0.0521 M

GIven that :

Ka = [tex]1.9 x 10^{-5}[/tex]

Thus; it's pKa = 4.72

[tex]pH =4.72 + log(\dfrac{ \ 0.0521}{0.0458})[/tex]

[tex]pH =4.72 + log(1.1376)[/tex]

[tex]pH =4.72 + 0.05598[/tex]

[tex]pH =4.77598[/tex]

pH ≅ 4.80

The pH of the solution 0.150 M hydrazoic acid after 13.3 mL of NaOH base is added is 4.80.

pH of the given solution will be used by using the following equation:

pH = pKa + log[conjugate base] / [weak acid]

Given chemical reaction will be represented as:

HN₃ + OH⁻ → N₃⁻ + H₂O

Moles will be calculated as:

n = M×V, where

M = molarity

V = volume

Moles of 0.150 M hydrazoic acid = (0.150M)(0.025L) = 0.00375  mol

Moles of 0.150 M NaOH = (0.0133)(0.150) = 0.001995 mol

From the above calculation it is clear that moles of hydrazoic acid is present in excess and it will be:

0.00375 - 0.001995 = 0.001755 mol

And 0.001995 mol of N₃⁻ is preduced by the reaction.

Total volume of the solution = 0.025 +  0.0133 = 0.0383  L

To calculate the pH after titration, first we have to calculate the concentration in terms of molarity of N₃⁻ and HN₃ as:

[N₃⁻] = 0.001995 mol / 0.0383  L = 0.0521 M

[HN₃] = 0.001755 mol / 0.0383  L = 0.0458 M

Ka for HN₃ = 1.9 × 10⁻⁵

pKa = -log( 1.9 × 10⁻⁵ ) = 4.72

On putting all these values on the above equation, we get

pH = 4.72 + log (0.0521) / (0.0458)

pH = 4.80

Hence, pH of the solution is 4.80.

To know more about pH, visit the below link:
https://brainly.com/question/10313314