what is the lewis structure for OP(N3)3​

Answer:

The correct answer is - option B -  lowest b.p.... isobutane < acetone < isopropanol ...highest b.p.

Explanation:

Isobutane has lowest boiling point due to no hydrogen bonding and no diole to dipole interaction found in them. Isobutane only shows weak dispersion force.

Acetone has dipole dipole interaction but due to lack of Hydrogen bonding they have low boiling point than isopropanol but higher than isobutanol.

Isopropanol is the compound that has ability to form hydrogen bonding with other molecule its boiling point is maximum among all three.

Thus, the correct answer is - option B -  lowest b.p.... isobutane < acetone < isopropanol ...highest b.p.

Answer:

Ca²⁺ and Cl⁻

Explanation:

In a chemical reaction, spectator ions are ions that are not involved in the reaction, that means are the same before and after the reaction.

In water, calcium hydroxide, Ca(OH)₂ is dissociated in Ca²⁺ and OH⁻. Also, hydrochloric acid, HCl, dissociates in H⁺ and Cl⁻. The reaction is:

Ca²⁺ + 2OH⁻ + 2H⁺ + 2Cl⁻ → 2H₂O + Ca²⁺ + 2Cl⁻

The ions that react are H⁺ and OH⁻ (Acid and base producing water)

And the ions that are not reacting, spectator ions, are:

Answer:

The empirical formulae for the compound is CH3.

Answer:

[tex]x_{et}=0.6068[/tex]

Explanation:

Hello,

In this case, since the mole fraction of a compound, in this case ethanol, in a binary mixture, in this constituted by both water and ethanol, is mathematically defined as follows:

[tex]x_{et}=\frac{n_{et}}{n_{et}+n_{w}}[/tex]

Whereas [tex]n[/tex] accounts for the moles in the solution for each species, we must first compute the moles of both ethanol (density: 0.789 g/mL and molar mass: 46.07 g/mol) and water (density: 1g/mL and molar mass: 18.02 g/mol)

[tex]n_{et}=10.00mL\ et*\frac{0.789g\ et}{mL\ et} *\frac{1mol\ et}{46.07g\ et}=0.1713mol\ et\\ \\n_w=2.00mL\ w*\frac{1g\ w}{mL\ w} *\frac{1mol\ w}{18.02g\ w}=0.1110mol\ w[/tex]

Therefore, the mole fraction turns out:

[tex]x_{et}=\frac{0.1713mol}{0.1713mol+0.1110mol}\\\\x_{et}=0.6068[/tex]

Best regards.

Answer:

No precipitate is formed.

Explanation:

Hello,

In this case, given the dissociation reaction of magnesium fluoride:

[tex]MgF_2(s)\rightleftharpoons Mg^{2+}+2F^-[/tex]

And the undergoing chemical reaction:

[tex]MgCl_2+2NaF\rightarrow MgF_2+2NaCl[/tex]

We need to compute the yielded moles of magnesium fluoride, but first we need to identify the limiting reactant for which we compute the available moles of magnesium chloride:

[tex]n_{MgCl_2}=0.3L*1.1x10^{-3}mol/L=3.3x10^{-4}molMgCl_2[/tex]

Next, the moles of magnesium chloride consumed by the sodium fluoride:

[tex]n_{MgCl_2}^{consumed}=0.5L*1.2x10^{-3}molNaF/L*\frac{1molCaCl_2}{2molNaF} =3x10^{-4}molMgCl_2[/tex]

Thus, less moles are consumed by the NaF, for which the moles of formed magnesium fluoride are:

[tex]n_{MgF_2}=3x10^{-4}molMgCl_2*\frac{1molMgF_2}{1molMgCl_2}=3x10^{-4}molMgF_2[/tex]

Next, since the magnesium fluoride to magnesium and fluoride ions is in a 1:1 and 1:2 molar ratio, the concentrations of such ions are:

[tex][Mg^{2+}]=\frac{3x10^{-4}molMg^{+2}}{(0.3+0.5)L} =3.75x10^{-4}M[/tex]

[tex][F^-]=\frac{2*3x10^{-4}molMg^{+2}}{(0.3+0.5)L} =7.5x10^{-4}M[/tex]

Thereby, the reaction quotient is:

[tex]Q=(3.75x10^{-4})(7.5x10^{-4})^2=2.11x10^{-10}[/tex]

In such a way, since Q<Ksp we say that the ions tend to be formed, so no precipitate is formed.

Regards.

Answer:

I hope it works

Explanation:

As we know that

w=m*g

given m=0.145 , g=9.8

hence we get

w= (9.8)*(0.145)

w=1.421 m/sec 2

if its help-full thank hit the stars and brain-list it thank you

Answer:

Molar solubility is 1.12x10⁻⁴M

Explanation:

The dissolution of magnesium hydroxide is:

Mg(OH)₂(s) ⇄ Mg²⁺ + 2OH⁻

The molar solubility represents the moles of the solid that the solution can dissolve, that could be written as:

Mg(OH)₂(s) ⇄ X + 2X

Where X is solubility.

If you obtained a [OH⁻] = 2.24x10⁻⁴M and you know [OH⁻] = 2X:

2X = 2.24x10⁻⁴M

X = 2.24x10⁻⁴M/2

X =1.12x10⁻⁴M

Answer:

[tex]r_A=-1\frac{M}{s}[/tex]

Explanation:

Hello,

In this case, given the rate of production of C, we can compute the rate of consumption of A by using the rate relationships which include the stoichiometric coefficients at the denominators (-1 for A and 2 for C) as follows:

[tex]\frac{1}{-1} r_A=\frac{1}{2}r_C[/tex]

In such a way, solving the rate of consumption of A, we obtain:

[tex]r_A=-\frac{1}{2} r_C=-\frac{1}{2}*2\frac{M}{s}\\ \\r_A=-1\frac{M}{s}[/tex]

Clearly, such rate is negative which account for consumption process.

Regards.

Answer:

Oxidation half equation;

3Cd(s) -------> 3Cd^2+(aq) + 6e

Reduction half equation;

2In^3+(aq) + 6e -----> 2In(s)

Explanation:

Since the reduction potentials of Indium and Cadmium are -0.34 V and - 0.40 V respectively, we can see that cadmium will be oxidized while indium will the reduced.

We arrived at this conclusion by examining the reduction potential of both species. The specie with more negative reduction potential is oxidized in the process.

Oxidation half equation;

3Cd(s) -------> 3Cd^2+(aq) + 6e

Reduction half equation;

2In^3+(aq) + 6e -----> 2In(s)

Answer:

b) the Ka of NH₄⁺ to find the hydronium concentration

Explanation:

The equilbrium of NH₄⁺ (The conjugate acid of NH₃, a weak base), is:

NH₄⁺ ⇄ NH₃ + H⁺

Where Ka of the conjugate acid is:

Ka = [NH₃] [H⁺] / [NH₄⁺]

Thus, if you know Ka of NH₄⁺ and its molar concentration you can calculate  [H⁺], the hydronium concentration, to find pH (Because pH =  -log [H⁺])

Thus, right option is:

Answer:

Difference of the enthalpy (of a system) minus the product of the entropy and absolute temperature

Explanation:

The basis of spontaneity in a chemical reaction is that ∆G must be negative. ¡∆G is known as the change in free energy of a system. If ∆G is negative, then the reaction will occur without any external help (the reaction is spontaneous at room temperature).

∆G is given by;

∆G= ∆H -T∆S

Where;

∆H= change in enthalpy of the system

T= absolute temperature of the system

∆S= change in entropy

Hence; when ∆H -T∆S gives a negative result, the reaction proceeds without any external help.

The advantage of a resource person would be that it will provide a hands-on activity that will allow the students to experience spacing between organs and on the body of the person.

It will also allow them to identify challenges when doing this and will engage them more in the activity and lesson.

Answer:A resource person add knowledge to the course

Explanation:

Answer:

The rate would have doubled

Explanation:

Answer:

[tex]1.25~mol~H_2O[/tex] and [tex]0.627~mol~N_2[/tex]

Explanation:

Our goal for this question is the calculation of the number of moles of the molecules produced by the reaction of hydrazine ([tex]N_2H_4[/tex]) and oxygen ([tex]O_2[/tex]). So, we can start with the reaction between these compounds:

[tex]N_2H_4~+~O_2~->~N_2~+~H_2O[/tex]

Now we can balance the reaction:

[tex]N_2H_4~+~O_2~->~N_2~+~2H_2O[/tex]

In the problem, we have the values for both reagents. Therefore we have to calculate the limiting reagent. Our first step, is to calculate the moles of each compound using the molar masses values (32.04 g/mol for [tex]N_2H_4[/tex] and 31.99 g/mol for [tex]O_2[/tex]):

[tex]20.1~g~N_2H_4\frac{1~mol~N_2H_4}{32.04~g~N_2H_4}=0.627~mol~N_2H_4[/tex]

[tex]20.1~g~O_2\frac{1~mol~O_2}{31.99~g~O_2}=0.628~mol~O_2[/tex]

In the balanced reaction we have 1 mol for each reagent (the numbers in front of [tex]O_2[/tex] and [tex]N_2H_4[/tex] are 1). Therefore the smallest value would be the limiting reagent, in this case, the limiting reagent is [tex]N_2H_4[/tex].

With this in mind, we can calculate the number of moles for each product. In the case of [tex]N_2[/tex] we have a 1:1 molar ratio (1 mol of [tex]N_2[/tex] is produced by 1 mol of [tex]N_2H_4[/tex]), so:

[tex]0.627~mol~N_2H_4\frac{1~mol~N_2}{1~mol~N_2H_4}=~0.627~mol~N_2[/tex]

We can follow the same logic for the other compound. In the case of [tex]H_2O[/tex] we have a 1:2 molar ratio (2 mol of [tex]H_2O[/tex] is produced by 1 mol of [tex]N_2H_4[/tex]), so:

[tex]0.627~mol~N_2H_4\frac{2~mol~H_2O}{1~mol~N_2H_4}=~1.25~mol~H_2O[/tex]

I hope it helps!

Answer: The number of moles of a solute per kilogram of solvent

Explanation:

Answer:

1.543 pounds = 700 grams

Answer: Reaction (1) , (3) and (4) are accompanied by an increase in entropy.

Explanation:

Entropy is the measure of randomness or disorder of a system. If a system moves from  an ordered arrangement to a disordered arrangement, the entropy is said to decrease and vice versa.

(1) [tex]2SO_2(g)+O_2(g)\rightarrow SO_3(g)[/tex]

3 moles of reactant are changing to 1 mole of product , thus the randomness is increasing. Thus the entropy also increases.

2) [tex]H_2O(l)\rightarrow H_2O(s)[/tex]

1 mole of Liquid reactant is changing to 1 mole of solid product , thus the randomness is decreasing. Thus the entropy also decreases.

3) [tex]Br_2(l)\rightarrow Br_2(g)[/tex]

1 mole of Liquid reactant is changing to 1 mole of gaseous product , thus the randomness is increasing. Thus the entropy also increases.

4)  [tex]H_2O_2(l)\rightarrow H_2O(l)+\frac{1}{2}O_2(g)[/tex]

1 mole of Liquid reactant is changing to half mole of gaseous product and 1 mole of liquid product, thus the randomness is increasing. Thus the entropy also increases.

Answer:

Mass = 11.688g

Explanation:

Volume = 2.00L

Molar concentration = 0.100M

Mass = ?

These quantities are relatted by the following equation;

Conc = Number of moles / volume

Number of moles = Conc * Volume = 2 * 0.100 = 0.2 mol

Number of moles = Mass / Molar mass

Mass = Number of moles * Molar mass

Mass = 0.2mol * 58.44g/mol

Mass = 11.688g

Answer:

b. the amount of heat given off or absorbed

Explanation:

Hello,

In this case, we should take into account a formal definition of enthalpy change such as an energetic change that occurs in a system when matter is transformed by a given chemical reaction from reactants to products. Thus, such energetic change is macroscopically exhibited and it is related with either a temperature increase or decrease; it means that if a reaction exhibits a temperature increase, we say that heat was given off and if the temperature exhibits a decrease, we say that heat is absorbed. For that reason, answer is b. the amount of heat given off or absorbed.

Regards.

Answer:

[tex]Cd(s)|Cd^{2+}(aq) || Cu^{2+}(aq)|Cu(s)[/tex]

Explanation:

A galvanic cell is composed of two electrodes immersed in a suitable electrolyte and connected via a salt bridge. One of the electrodes serves as a cathode where reduction or gain of electrons takes place. The other half cell functions as an anode where oxidation or loss of electrons occurs.

The representation is given by writing the anode on left hand side followed by its ion with its molar concentration. It is followed by a salt bridge. Then the cathodic ion with its molar concentration is written and then the cathode.

As it is given that cadmium acts as anode, it must be on the left hand side and copper must be on right hand side.

[tex]Cd(s)|Cd^{2+}(aq) || Cu^{2+}(aq)|Cu(s)[/tex]

Answer:

7.4 × 10⁻⁹ M

Explanation:

Step 1: Given data

Solubility product constant (Ksp) for Li₃PO₄: 5.9 × 10⁻¹⁷

Concentration of lithium ion: 0.0020 M

Step 2: Write the reaction for the solution of Li₃PO₄

Li₃PO₄(s) ⇄ 3 Li⁺(aq) + PO₄³⁻(aq)

Step 3: Calculate the phosphate concentration required for a precipitate to occur

The solubility product constant is:

Ksp = 5.9 × 10⁻¹⁷ = [Li⁺]³ × [PO₄³⁻]

[PO₄³⁻] = 5.9 × 10⁻¹⁷ / [Li⁺]³

[PO₄³⁻] = 5.9 × 10⁻¹⁷ / 0.0020³

[PO₄³⁻] = 7.4 × 10⁻⁹ M

Answer:

That means that if you are calculating entropy change, you must multiply the enthalpy change value by 1000. So if, say, you have an enthalpy change of -92.2 kJ mol-1, the value you must put into the equation is -92200 J mol-1

The entropy change in the surroundings associated with this reaction occurring at 25 degree C is calculated as ΔS = -ΔH/T J/K.

Entropy is a quantity which gives idea about the randomness or arrangement of atoms or molecules present in any sample.

Entropy change will be calculated as:
ΔS = -ΔH/T, where

ΔH = chnage in enthalpy (J/mole)

T = temperature (K)

So to calculate the entropy change first we have to know about the value of enthalpy in joules and then divide it by the temperature.

Hence the unit of entropy is joule per kelvin.

To know more about entropy, visit the below link:
https://brainly.com/question/6364271

Answer:

Al(s)<NaF(s)<HF(g)<SiH4(g)<SiF4(g)

Explanation:

Hello,

In this case, we can arrange the increasing order of entropy at 25 \°C by taking into account, at first, that since solids are more molecularly organized than gases, the first we have solid sodium fluoride and solid aluminium, but in this case, as the higher the molar mass, the higher the entropy, the molar mass of aluminium is 27 g/mol and 42 g/mol for sodium fluoride, therefore, we first have:

Al(s)<NaF(s)

Afterwards, since the molar mass of hydrogen fluoride (HF), silicon fluoride (SiF4) and silane (SiH4) are 20, 104 and 32 g/mol respctively, since silicon fluoride has the greater molar mass, it also has the higher entropy. In such a way, the overall order turns out:

Al(s)<NaF(s)<HF(g)<SiH4(g)<SiF4(g)

Best regards.

Answer:

Major product does not undergo oxidation since it is a tertiary alcohol whereas minor product undergoes oxidation to ketone as it is  secondary alcohol.

Explanation:

Hello,

In this case, given the attached picture, the hydration of the 1 methylcyclohexene yields to alcohols; 1-methylcyclohexan-1-ol and 1-methylcyclohexan-2-ol. Thus, since the OH in the 1-methylcyclohexan-1-ol (major product) is bonded to a tertiary carbon (bonded with other three carbon atoms) it is not able to increase the number of oxygen bonds (oxidation) as it already attained the octet whereas the 1-methylcyclohexan-2-ol (minor product) is able to undergo oxidation to ketone as the carbon bonded to it is secondary (bonded with other two carbon atoms), so one extra bond the oxygen is allowed to be formed to carbonyl.

Best regards.

Answer:

NaN3 = 47.2 g

Explanation:

Given:

2 NaN3 ⇒ 2 Na  + 3 N2

Find:

Amount of NaN3

Computation:

N2 moles = Product of N2 / molar mass of N2

N2 moles =30.5/28

N2 moles = 1.0893

2NaN3 makes 3(N2 )

So,

NaN3 moles = (2/3) moles of N2  

NaN3 moles = ( 2/3) × 1.0893

NaN3 moles = = 0.7262

NaN3 mass = 0.7262 x 65

NaN3 = 47.2 g

Answer:

NaN3 = 47.2 g

Explanation:

Given:

2 NaN3 ⇒ 2 Na  + 3 N2

Find:

Amount of NaN3

Computation:

N2 moles = Product of N2 / molar mass of N2

N2 moles =30.5/28

N2 moles = 1.0893

2NaN3 makes 3(N2 )

So,

NaN3 moles = (2/3) moles of N2  

NaN3 moles = ( 2/3) × 1.0893

NaN3 moles = = 0.7262

NaN3 mass = 0.7262 x 65

NaN3 = 47.2 g

Explanation:

Answer:

n = 0.207 mole

Explanation:

We have,

P = 1 atm

V = 5 liter

R = 0.0821 L.atm/mol.K

T = 293 K

We need to find the value of n. The relation is as follows :

PV = nRT

Solving for n,

[tex]n=\dfrac{PV}{RT}\\\\n=\dfrac{1\times 5}{0.0821 \times 293}\\\\n=0.207\ \text{mol}[/tex]

So, the value of n is 0.207 mol.

The given question is incomplete. The complete question is:

Compound X has a molar mass of 153.05 g/mol and the following composition:

element mass %

carbon 47.09%

hydrogen 6.59%

chlorine 46.33%

Write the molecular formula of X.

Answer: The molecular formula of X is [tex]C_6H_{10}Cl_2[/tex]

Explanation:

If percentage are given then we are taking total mass is 100 grams.

So, the mass of each element is equal to the percentage given.

Mass of C= 47.09 g

Mass of H = 6.59 g

Mass of Cl = 46.33 g

Step 1 : convert given masses into moles.

Moles of C =[tex]\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{47.09g}{12g/mole}=3.92moles[/tex]

Moles of H =[tex]\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{6.59g}{1g/mole}=6.59moles[/tex]

Moles of Cl =[tex]\frac{\text{ given mass of Cl}}{\text{ molar mass of Cl}}= \frac{46.33g}{35.5g/mole}=1.30moles[/tex]

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For C = [tex]\frac{3.92}{1.30}=3[/tex]

For H = [tex]\frac{6.59}{1.30}=5[/tex]

For Cl =[tex]\frac{1.30}{1.30}=1[/tex]

The ratio of C : H: Cl= 3: 5 :1

Hence the empirical formula is [tex]C_3H_5Cl[/tex]

The empirical weight of [tex]C_3H_5Cl[/tex] = 3(12)+5(1)+1(35.5)= 76.5g.

The molecular weight = 153.05 g/mole

Now we have to calculate the molecular formula.

[tex]n=\frac{\text{Molecular weight }}{\text{Equivalent weight}}=\frac{153.05}{76.5}=2[/tex]

The molecular formula will be=[tex]2\times C_3H_5Cl=C_6H_{10}Cl_2[/tex]

The element which can not loose electron easily and having electronagtive character is called non-metal it has following property-

1. it can not conduct heat and electricity

2. it is netiher ductile not malleable

3. it is not lsuturous and also not sonorous

Explanation:

a nonmetal (or non-metal) is a chemical element that mostly lacks the characteristics of a metal. Physically, a nonmetal tends to have a relatively low melting point, boiling point, and density. A nonmetal is typically brittle when solid and usually has poor thermal conductivity and electrical conductivity. Chemically, nonmetals tend to have relatively high ionization energy, electron affinity, and electronegativity. They gain or share electrons when they react with other elements and chemical compounds. Seventeen elements are generally classified as nonmetals: most are gases (hydrogen, helium, nitrogen, oxygen, fluorine, neon, chlorine, argon, krypton, xenon and radon); one is a liquid (bromine); and a few are solids (carbon, phosphorus, sulfur, selenium, and iodine). Metalloids such as boron, silicon, and germanium are sometimes counted as nonmetals.

Explanation:

The spontaneity of a system is deduced by the sign of the gibbs free energy value. If it is negative, it means the process / reaction is spontaneous however a positive value indicates the such process is not spontaneous.

Gibbs free energy, enthalpy and entropy are related by the following equation;

ΔG = ΔH - TΔS

A positive value of enthalpy, H and entropy, S means that G would always be a negative value at all temperatures.

Answer:

226Ra88→222Rn86+4He2

Explanation:

An α-particle usually consists of a helium nucleus which indicates the type of decay that was undergone in this radioactive process.

During α-decay(alpha decay), an atomic nucleus emits an alpha particle.