If a salt is formed by combining NH3 (Kb=1.8×10−5) and CH3COOH (Ka=1.8×10−5), an aqueous solution of this salt would be:

Answer:

CHC12

Explanation:

i am not really sure i am onna do a quick research 4 u tho

3 2 1 is the set of quantum numbers.

The set of numbers used to describe the position and energy of the electron in an atom is called quantum numbers. There are four quantum numbers, namely, principal, azimuthal, magnetic, and spin quantum numbers.

The rules for quantum numbers are: (n) can be any positive, nonzero integral value. (l) can be zero or any positive integer but not larger than (n-1). l = 0, 1, 2, 3, 4, …. (n-1) (ml) values follow the equation.

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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:

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:

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.

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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.

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

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:

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:

The rms speed of its molecules becomes. (T) has become four times. Therefore, v_(rms) will become two times,...

Answer:

b.) Put on protective gloves

Answer:

2020 Put on protective gloves.

Explanation:

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:

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.

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]

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:

The empirical formula of the compound is NO2.

Explanation:

The following data were obtained from the question:

Mass of compound = 30.5 g

Mass of nitrogen (N) = 9.29 g

Empirical formula of compound =?

Next, we shall determine the mass of oxygen in the compound. This can be obtained as follow:

Mass of compound = 30.5 g

Mass of nitrogen (N) = 9.29 g

Mass of oxygen (O) =?

Mass of O = mass of compound – mass of N.

Mass of O = 30.5 – 9.29

Mass of O = 21.21 g

Finally, we shall determine the empirical formula of the compound as follow:

Mass of nitrogen (N) = 9.29 g

Mass of oxygen (O) = 21.21 g

Divide by their molar mass.

N = 9.29 / 14 = 0.664

O = 21.21 / 16 = 1.326

Divide by the smallest

N = 0.664/ 0.664 = 1

O = 1.326/ 0.664 = 2

Therefore the empirical formula of the compound is NO2.

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.

Answer:

1.543 pounds = 700 grams

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:

Yes

Explanation:

The balanced equation of the reaction is;

FeBr2 (aq) + Cl2 (g) → FeCl2 (aq) + Br2 (aq)

This reaction is possible because chlorine is more electronegative than bromine and can displace it from its salt.

In group seventeen, electro negativity decreases down the group. Hence as we move down the group, elements become less electronegative and can be displaced from their salt by more electronegative elements found earlier in the group.

Hence chlorine can displace bromine in FeBr2 to form FeCl2.

Answer:

Yes, because Cl2 has higher activity than Br2

Explanation:

Answer: C) [tex]CO^{2-}_{3}[/tex]

Explanation: The Valence Shell Electron Pair Repulsion Model (VSEPR Model) shows bonding and nonbonding electron pairs present in the valence, outermost, shell of an atom connecting to other atoms. It also gives the molecular geometric shape of a molecule.

To determine molecular geometry:

1) Draw Lewis Structure, i.e., a simplified representation of the valence shell electrons;

2) Count the number of electron pairs (count multiple bonds as 1 pair);

3) Arrange electron pairs to minimise repulsion;

4) Position the atoms to minimise the lone pair;

5) Name the molecular geometry from the atom position;

Trigonal planar molecular geometry is a model which molecule's shape is triangular and in one plane. Such molecule has three regions of electron density extending out from the central atom and the repulsion will be at minimum when angle between any two is 120°.

The Lewis structure of each molecule is shown in the attachment.

Analysing each one, it can be concluded that molecule with trigonal planar geometry is [tex]CO^{2-}_{3}[/tex]

Answer:

C) 237.96682 amu

Explanation:

The symbol for neptunium-236 is given as;

²³⁶₉₃Np

This element has 93 protons and (236 - 93 = 143) neutrons.

Mass Number =Total mass of Protons + Total mass of neutrons

Total Mass pf protons = 93 * 1.007825 amu, = 93.727725 amu

Total mass of Neutrons = 143 * 1.008665 amu = 144.239095 amu

Mass = 144.239095 + 93.727725  = 237.96682 amu

Correct option is option C.

Answer:

[tex]1.242 \times 10^{25}\text{ atoms H}[/tex]

Explanation:

You must convert the mass of B₄H₁₀ to moles of B₄H₁₀, then to molecules of B₄H₁₀, and finally to atoms of H.

1. Moles of B₄H₁₀

[tex]\text{Moles of B$_{4}$H}_{10} = \text{110.0 g B$_{4}$H}_{10} \times \dfrac{\text{1 mol B$_{4}$H}_{10}}{\text{53.32 g B$_{4}$H}_{10}} = \text{2.063 mol B$_{4}$H}_{10}[/tex]

2. Molecules of B₄H₁₀

[tex]\text{No. of molecules} = \text{2.063 mol B$_{4}$H}_{10} \times \dfrac{6.022 \times 10^{23}\text{ molecules B$_{4}$H}_{10}}{\text{1 mol B$_{4}$H}_{10}}\\\\=1.242 \times 10^{24}\text{ molecules B$_{4}$H}_{10}[/tex]

3. Atoms of H

[tex]\text{Atoms of H} = 1.242 \times 10^{24}\text{ molecules B$_{4}$H}_{10} \times \dfrac{\text{10 atoms H}}{\text{1 molecule B$_{4}$H}_{10}}\\\\= \mathbf{1.242 \times 10^{25}}\textbf{ atoms H}[/tex]

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:

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:

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:

[tex]M_{acid}=0.141M[/tex]

Explanation:

Hello,

In this case, the reaction between sulfuric acid and hydroxide is:

[tex]H_2SO_4+2NaOH\rightarrow Na_2SO_4+2H_2O[/tex]

We can notice a 1:2 molar ratio between the acid and the base respectively, therefore, at the equivalence point we have:

[tex]2*n_{acid}=n_{base}[/tex]

And in terms of volumes and concentrations:

[tex]2*M_{acid}V_{acid}=M_{base}V_{base}[/tex]

So we compute the molarity of sulfuric acid as shown below:

[tex]M_{acid}=\frac{M_{base}V_{base}}{2*V_{acid}} =\frac{0.100M*28.15mL}{2*10.0mL}\\ \\M_{acid}=0.141M[/tex]

Best regards.

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:

[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.