What is the pOH of a solution at 25.0∘C with [H3O+]=4.8×10−6 M?

Answer:

[tex]1.628 M[/tex]

Explanation:

From the question we were given 0.155 moles of HBr, but Br and H are in ratio 1:1, then there are 0.155 moles of Br- ions.

We were also told that the solution contain NaBr, of 25.9 g. Then it must be converted to moles.

molar mass of NaBr =(22.99g + 79.90 )

= 102.89 g per mol.

the moles of NaBr can be calculated as 25.9 / 102.89

=0.252 moles

But Na and Br are in a ratio 1:1 , then there are 0.252 moles of Br-.

Then to get two Br- mol , we will add the first and second mol of Br- together

= 0.155 + 0.252

=0.407 moles.

The given solution has volume of 250 mL, but we know that there are 1000 ml in a liter, then if we convert to L for unit consistency we have

= 250/1000

= 0.25 L

molarity=0.407 moles/0.25 L

= 1.628 M.

Therefore, Br ion molarity is 1.628 M.

The molarity of the Br ions in the 250 ml solution has been 1.628 M.

Moles can be defined as the mass per unit molecular mass. Moles can be expressed as:

Moles = [tex]\rm \dfrac{weight}{molecular\;weight}[/tex]

Moles of NaBr = [tex]\rm \dfrac{25.9}{102.89}[/tex]

Moles of NaBr = 0.252 mol

Moles of HBr = 0.155 mol.

Since both the compounds have 1:1 ratio of atom: Br, the Br produced has been equal to the concentration of the compound.

Br from NaBr = 0.252 mol

Br from HBr = 0.155 mol.

Total Br ions = 0.407 mol.

Molarity can be expressed as:

Molarity = [tex]\rm moles\;\times\;\dfrac{1000}{Volume\;(ml)}[/tex]

Molarity of Br ions = 0.407 × [tex]\rm \dfrac{1000}{250\;ml}[/tex]

Molarity of Br ions = 1.628 M.

The molarity of the Br ions in the 250 ml solution has been 1.628 M.

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Answer:

Zeros located at the end of significant figures are significant.

Explanation:

Hope it will help :)

Answer:

ΔT=[tex]0.87^{\circ}C[/tex]

Explanation:

Para esta pregunta debemos recordar la ecuación que nos permite calcular el aumento ebulliscopico (aumento del punto de ebullición):

ΔT=[tex]Kb*m[/tex]

Donde ΔT es el valor del aumento del punto de ebullición. Kb es la constante ebulloscopica para el agua ([tex]0.512\frac{Kg~^{\circ}C}{mol}[/tex]) y m es la molalidad ([tex]m=\frac{mol}{Kg~ of~ solvente}[/tex]).

Por lo tanto el primer paso es calcular la molalidad de la solución. Para lo cual tendremos que calcular las moles de sal en los 25 g. Si queremos hacer esto debemos recordar que la formula de sal de mesa es NaCl y que la masa molar de NaCl es 58.44 g/mol. Por lo tanto:

[tex]25~g~NaCl\frac{1~mol~NaCl}{58.44~g~NaCl}~=~0.42~mol~NaCl[/tex]

Ahora bien, también debemos saber los Kg de agua en la solución. Por lo que podemos usar la densidad del agua (1 g/mL) para convertir de mL a g y luego hacer la conversión a Kg:

[tex]250~mL\frac{1~g}{1~mL}\frac{1~Kg}{1000~g}~=~0.25~Kg[/tex]

Finalmente para el calculo de la molalidad podemos dividir los dos valores:

[tex]m=\frac{0.42~mol}{0.25~Kg}=1.68[/tex]

Con el valor de la molalidad se puede calcular ΔT al reemplazar los valores:

ΔT=[tex]1.68~\frac{mol}{Kg}*0.52\frac{Kg~^{\circ}C}{mol}=0.87^{\circ}C[/tex]

Si la temperatura de ebullición normal del agua es 100 ºC. Podemos calcular la temperatura final si adicionamos ΔT:

Temperatura final = 100 + 0.87 = 100.87 ºC

Espero sea de ayuda!

Answer:

True

Explanation:

The percentage by mass of a substance in a solution can be calculated by dividing the mass of the substance dissolved in the solution by the total mass of the solution. This can be expressed mathematically as:

Percentage by mass = mass of substance in solution/mass of solution x 100

In this case;

mass of KBr = 4.34 grams

mass of water = 17.4 grams

mass of solution = mass of KBr + mass of water = 4.34 + 17.4 = 21.74

Percentage by mass of KBr = 4.34/21.74 x 100

                                              = 19.96 %

19.96 is approximately 20%.

Hence, the statement is true.

Answer:

A. Increasing the temperature will favor forward reaction and more CaCo3 formed.

B. More CaCo3 will be formed.

C. CaCo3 will decrease and more react ants formed.

D. Less CaCo3 will be formed.

E. Iridium is a catalyst so there is no effect

Explanation:

A. Temperature will increase because it's an endothermic reaction.

B. Adding Cao will favor forward reaction and more CaCo3 formed.

C. Removing methane, more react ants are formed and CaCo3 decreases.

D. Irridi is a catalyst so it has no effect on the CaCo3 but only speeds its rate of reaction.

Answer:

1. In the compound, H3C-NH-CH2-CH3, there are no chiral centers present, chiral centers refer to the configuration in which carbon is attached with four different groups. In the molecules, as there are no chiral centers, therefore the molecule is optically inactive, that is, it will not demonstrate pair of an enantiomer is one of the essential characteristics of optically active compounds is the possession of enantiomeric pairs.  

2. On the nitrogen of aniline, the lone pair of electrons can produce hydrogen bonds, play an essential function in basicity, play an essential role in dipole moment or polarity, and wit the increase in solubility there is an increase in the formation of the hydrogen bond, eventually increasing to boiling point. However, the melting point is not affected. As the melting point is the characteristic of the packing efficacy of a molecule and does not rely upon the anilinic nitrogen's lone pairs.  

3. With the increase in the tendency to donate an electron, basicity increases. However, if the electron is taking part in resonance, the donation will not take place easily, and the compound will be the least basic. Apart from cyclohexyl amine, in all the other given compounds, the lone pair of nitrogen takes part in the process of delocalization or conjugation. Thus, cyclohexyl amine will be most basic as the lone pairs are easily available for donation.  

Answer:

21 KJ/mol

Explanation:

For this question, we have to start with the linear structure of 2-methylbutane. With the linear structure, we can start to propose all the Newman projections keep it in mind that the point of view is between carbons 2 and 3 (see figure 1).

Additionally, we have several energy values for each interaction present in the Newman structures:

-) Methyl-methyl gauche: 3.8 KJ/mol

-) Methyl-H eclipse: 6.0 KJ/mol

-) Methyl-methyl eclipse: 11.0 KJ/mol

-) H-H eclipse: 4.0 KJ/mol

Now, we can calculate the energy for each molecule.

Molecule A

In this molecule, we have 2 Methyl-methyl gauche interactions only, so:

(3.8x2) = 7.6 KJ/mol

Molecule B

In this molecule, we have a Methyl-methyl eclipse interaction a Methyl-H eclipse interaction and an H-H eclipse interaction, so:

(11)+(6)+(4) = 21 KJ/mol

Molecule C

In this molecule, we have 1 Methyl-methyl gauche interaction only, so:

3.8 KJ/mol

Molecule D

In this molecule, we have three Methyl-H eclipse interaction, so:

(6*3) = 18 KJ/mol

Molecule E

In this molecule, we have 1 Methyl-methyl gauche interaction only, so:

3.8 KJ/mol

Molecule F

In this molecule, we have a Methyl-methyl eclipse interaction a Methyl-H eclipse interaction and an H-H eclipse interaction, so:

(11)+(6)+(4) = 21 KJ/mol

The structures with higher energies would be less stable. In this case, structures B and F with an energy value of 21 KJ/mol (see figure 2).

I hope it helps!

Answer:

[tex]MM_{acid}=140.1g/mol[/tex]

Explanation:

Hello,

In this case, since the acid is monoprotic, we can notice a 1:1 molar ratio between, therefore, for the titration at the equivalence point, we have:

[tex]n_{acid}=n_{base} \\\\V_{acid}M_{acid}=V_{base}M_{base}\\\\n_{acid}=V_{base}M_{base}[/tex]

Thus, solving for the moles of the acid, we obtain:

[tex]n_{acid}=0.0215L*0.250\frac{mol}{L}=5.375x10^{-3}mol[/tex]

Then, by using the mass of the acid, we compute its molar mass:

[tex]MM_{acid}=\frac{0.753g}{5.375x10^{-5}mol} \\\\MM_{acid}=140.1g/mol[/tex]

Regards.

Answer:

2.81 g of H2O.

Explanation:

We'll begin by calculating mass of O2 that contains 1.25×10²³ molecules O2.

This can be obtained as follow:

From Avogadro's hypothesis, we understood that 1 mole of any substance contains 6.022×10²³ molecules. This implies that 1 mole of O2 also contains 6.022×10²³ molecules.

1 mole of O2 = 16x2 = 32 g.

Thus 6.022×10²³ molecules is present in 32 g of O2,

Therefore, 1.25×10²³ molecules will be present in =

(1.25×10²³ × 32) / 6.022×10²³ = 6.64 g of O2.

Therefore, 1.25×10²³ molecules present in 6.64 g of O2.

Next, the balanced equation for the reaction. This is given below:

CH4(g) + 2O2(g) —> CO2(g) + 2H2O(g)

Next, we shall determine the masses of CH4 and O2 that reacted and the mass of H2O produced from the balanced equation.

This can be obtained as follow:

Molar mass of CH4 = 12 + (4x1) = 16 g/mol.

Mass of CH4 from the balanced equation = 1 x 16 = 16 g

Molar mass of O2 = 16x2 = 32 g/mol.

Mass of O2 from the balanced equation = 2 x 32 = 64 g

Molar mass of H2O = (2x1) + 16 = 18 g/mol.

Mass of H2O from the balanced equation = 2 x 18 = 36 g

From the balanced equation above,

16 g of CH4 reacted with 64 g of O2 to produce 36 g if H2O.

Next, we shall determine the limiting reactant.

This can be obtained as follow:

From the balanced equation above,

16 g of CH4 reacted with 64 g of O2.

Therefore, 1.25 g of CH4 will react with = (1.25 x 64)/16 = 5 g of O2.

From the above calculations, we can see that only 5 g out of 6.64 g of O2 is needed to react completely with 1.25 g of CH4.

Therefore, CH4 is the limiting reactant.

Finally, we shall determine the mass of H2O produced from the reaction.

In this case, the limiting reactant will be used because it will give the maximum yield of H2O.

The limiting reactant is CH4 and the mass of H2O produced from the reaction can be obtained as follow:

From the balanced equation above,

16 g of CH4 reacted to produce produce 36 g if H2O.

Therefore, 1.25 g of CH4 will react to produce = (1.25 x 36)/16 = 2.81 g of H2O.

Therefore, 2.81 g of H2O were obtained from the reaction.

The mass in grams of H₂O which can be formed when 1.25g CH₄ are combined with 1.25×10²³ molecules O₂ is 2.8 grams.

Stoichiometry of any reaction tells about the amount of species present before and after the completion of the reaction.

Given chemical reaction is:

CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(g)

Moles of CH₄ will b calculate as:

n = W/M, where

W = given mass = 1.25g

M = molar mass = 16g/mol

n = 1.25/16 = 0.078 moles

Molecues of CH₄ in 0.078 moles = 0.078×6.022×10²³ = 0.46×10²³

Given molecules of O₂ = 1.25×10²³

Required molecules of CH₄ is less as compared to the molecules of O₂, so here CH₄ is the limiting reagent and formation of water is depends on it only.

From the stoichiometry of the reaction it is clear that:

1 mole of CH₄ = will produce 2 moles of H₂O

0.078 moles of CH₄ = will produce 2×0.078=0.156 moles of H₂O

Mass of H₂O will be calculated by using its moles as:

W = (0.156)(18) = 2.8g

Hence required mass of H₂O is 2.8g.

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Answer:

ethanolamine ( H2N-CH2-CH2-OH ) and acetamide ( CH3-CO-NH2 )

Explanation:

Isomers are compounds that have the same molecular formula but different structural formulas. Hence any pair of compounds that can be represented by exactly the same molecular formula are isomers of each other.

If we look at the pair of compounds; ethanolamine ( H2N-CH2-CH2-OH ) and acetamide ( CH3-CO-NH2 ), one compound has molecular formula, C2H7ON while the other has a molecular formula, C2H5ON, hence they are not isomers of each other.

Answer: [tex]\Delta G^{0}[/tex] = 168.12 kJ

Explanation: Gibbs Free Energy, at any time, is defined as the enthalpy of the system minus product of temperature and entropy of the reaction, i.e.:

[tex]\Delta G^{0} = \Delta H^{0} - T.\Delta S^{0}[/tex]

Enthalpy is defined as internal heat existent in the system. It is calculated as:

[tex]\Delta H^{0} = \Sigma H^{0}_{product} - \Sigma H^{0}_{reagent}[/tex]

Using Enthalpy Formation Table:

[tex]\Delta H^{0} = [3*(-299.9)+(-524.7)] - (-1277)[/tex]

[tex]\Delta H^{0} = 62,6 kJ[/tex]

Entropy is the degree of disorder in the system. It is found by:

[tex]\Delta S^{0} = \Sigma S^{0}_{products} - \Sigma S^{0}_{reagents}[/tex]

Calculating:

[tex]\Delta S^{0} = (-321.7) + 3(-10.8) - 0[/tex]

[tex]\Delta S^{0} = -354.1J[/tex]

And so, Gibbs Free energy will be:

[tex]\Delta G^{0} = \Delta H^{0} - T.\Delta S^{0}[/tex]

[tex]\Delta G^{0} = 62600 - [298.(-354.1)][/tex]

[tex]\Delta G^{0} = 168121.8 J[/tex]

Rounding to the nearest kJ:

[tex]\Delta G^{0}[/tex] = 168.12 kJ

Answer:

The number of moles of excess reagent NO₂ that are present after the reaction has completed is 7 moles.

Explanation:

The balanced reaction is:

3 NO₂ + H₂O → 2 HNO₃ + NO

By stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of reactants and products participate in the reaction:

The limiting reagent is one that is consumed in its entirety first, determining the amount of product in the reaction. When the limiting reagent ends, the chemical reaction will stop.

In other words, the limiting reagent is that reagent that is consumed first in a chemical reaction, determining the amount of products obtained. The reaction depends on the limiting reagent, because the other reagents will not react when one is consumed.

You can apply the following rule of three: if by stoichiometry of the reaction 3 moles of NO₂ react with 1 mole of H₂O, 13 moles of NO₂ react with how many moles of H₂O?

[tex]moles of H_{2}O=\frac{13 moles of NO_{2}*1 mole of H_{2}O }{3 moles of NO_{2}}[/tex]

moles of H₂O= 4.33 moles

But 4.33 moles of H₂O are not available, 3 moles are available. Since you have less moles than you need to react with 13 moles of NO₂, water H₂O will be the limiting reagent.

To determine the number of moles of excess reagent NO2 that are present after the reaction is complete, you can apply the following rule of three: if by stoichiometry of the reaction 1 moles of H₂O react with 3 mole of NO₂, 3 moles of H₂O react with how many moles of NO₂?

[tex]moles of NO_{2}=\frac{3 moles of NO_{2}*3 mole of H_{2}O }{1 mole of H_{2}O}[/tex]

moles of NO₂= 6 moles

If 6 moles of NO₂ react and 13 moles of the compound are present, the amount that remains in excess is calculated as: 13 moles - 6 moles= 7 moles

The number of moles of excess reagent NO₂ that are present after the reaction has completed is 7 moles.

Answer:

A. NaHCO₃

Explanation:

NaHCO₃ ⇒ NaOH + H₂CO₃

NaOH is a strong base and H₂CO₃ is a weak acid. Therefore, NaHCO₃ is a salt of a strong base-weak acid reaction. The salt is basic because carbonic acid (H₂CO₃) is a weak acid so it remains undissociated. So, there is a presence of additional OH⁻ ions that makes the solution basic.

Hope that helps.

Answer:

acetic acid and phosphoric acid

Explanation:

After refluxing the reaction mixture ( synthesis of isoamyl acetate ) what is likely present in the solution is acetic acid and phosphoric acid, this due to the fact that if the reaction time between the reactants was less than the refluxing time which is 25 minutes,

there will be no reactant ( 3-methylbutanol )left in the reaction mixture

Answer:

d. the conjugate base.

Explanation:

The general reaction of a weak acid, HA, with a strong base YOH, is:

HA + YOH → A⁻ + H₂O + Y⁻

Where A⁻ is the conjugate base of the weak acid and Y⁻ usually is a strong electrolyte.

That means after he complete reaction you don't have weak acid nor strong base, just conjugate base that will be in equilibrium with water, thus (Strong electrolyte doesn't change pH:

A⁻ + H₂O ⇄ HA + OH⁻

As the equilibrium is producing OH⁻, the pH of the solution is being affected for the conjugate base

Right option:

Answer:

It is a hydrogen bond but if I had to coose one of thee answers it is b. chemical bonding

Explanation:

Answer:

C. electronegativity

Explanation:

THE ELEMENT WITH THE GREATE ELECTRONEGATIVITY IS ASSIGNED A NEGATIVE OXIDATION NUMBER EQUAL TIO ITS CHARGE

Answer:

A)coupling between 1H nuclei attached to adjacent C atoms.

Explanation:

The word ‘vicinal’ in chemistry means three bonds from the functional groups. The two functional groups are in a relationship with the atoms in adjacent position to them.

The 1H nuclei consists of two Hydrogen nucleus which acts as the functional groups. They are however attached and in a relationship with the adjacent C atoms. This makes option A the right choice.

Answer:

The 3rd one

Explanation:

Answer: 6.31×10⁻³ moles Fe

Explanation:

To calculate moles when given atoms, we need to use Avogadro's number.

Avogadro's number: 6.022×10²³ atoms/mol

[tex](3.8*10^2^1 atoms)*\frac{mol}{6.022*10^2^3 atoms} =6.31*10^-^3 mols[/tex]

The atoms cancel out, and we are left with moles. There are 6.31×10⁻³ moles Fe.

Answer:

Explanation:

lewis structure can be defined as a process of how the valence shell electrons of a molecule is being arranged, the pattern of it arrangement and the relationship between the bonding atoms and the lone pairs present in the molecule.

In order to draw the Lewis structure for Arsenate ion  [tex]\mathsf{AsO \ _4^{3-}}[/tex], first thing is to count the valence electrons in the molecule. Once we determine the valence electrons, then we distribute them around the central atom. The Arsenate ion structure is tetrahedral in nature with a bond angle of 109.5° and it is sp³ hybridized.

Answer:

25.7 mL

Explanation:

Step 1: Given data

Step 2: Calculate the volume of the initial solution

We have a concentrated solution and we want to prepare a diluted one. We can calculate the initial volume using the dilution rule.

C₁ × V₁ = C₂ × V₂

V₁ = C₂ × V₂ / C₁

V₁ = 0.150 M × 600 mL / 0.350 M

V₁ = 25.7 mL

Answer:

4.22

Explanation:

We know from the question, that the pOH of the solution is 9.78. Now the pOH is defined as -log [OH^-].

If the pOH of a solution is given, one may obtain the pH of such solution from the formula;

pH + pOH =14

Hence we can write;

pH = 14-pOH

pH = 14 - 9.78 = 4.22

Hence the pH of the solution is 4.22.

Answer: Gamma Radiation

Explanation:

The emission of Gamma rays does not cause a change in both the atomic and mass number. They are  electromagnetic radiation.

The radiations that leaves without changing the atomic mass and atomic number of the particle have been gamma radiations. Thus, option D is correct.

Radiations have been the energy that has been evolved by the particles during energy transitions. The nuclear decay results with the release of the energy from the particle resulting in the change in the atomic mass.

The electromagnetic radiations have been capable of emitting the radiation without changing the mass and atomic number of the element. The gamma radiations have been the electromagnetic radiations. Thus, option D is correct.

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Explanation:

The whole process of titration involves finding the concentration of a solution (usually an acid or base) by adding (titrating) it to a solution(acid or base) with a known concentration.

The solution of unknown concentration (the analyte) is usually placed in an flask, while the solution of known concentration (titrant) is placed in a burette and slowly added to the flask.

Answer:

Explanation:

a) 4-nitrobenzoic acid         pKa= 3.41

    benzoic acid                   pKa= 4.19

    4-chlorobenzoic acid     pKa= 3.98

b) benzoic acid                    pKa= 4.19    

   cyclohexanol                   pKa= 18.0

   phenol                              pKa= 9.95

c) 4-Nitrobenzoic acid             pKa= 3.41

   4-nitrophenol                       pKa= 7.15

   4-nitrophenylacetic acid     pKa= 3.85

Answer:

ΔHrxn = -635.14kJ/mol

Explanation:

We can make algebraic operations of reactions until obtain the desire reaction and, ΔH of the reaction must be operated in the same way to obtain the ΔH of the desire reaction (Hess's law). Using the reactions:

(1)Ca(s) + 2 H+(aq) → Ca2+(aq) + H2(g) ΔH = 1925.9 kJ/mol

(2) 2H2(g) + O2 g) → 2 H2O(l) ΔH = −571.68 kJ/mole

(3) CaO(s) + 2 H+(aq) → Ca2+(aq) + H2O(l) ΔH = 2275.2 kJ/mole

Reaction (1) - (3) produce:

Ca(s) + H2O(l) → H2(g) + CaO(s)

ΔH = 1925.9kJ/mol - 2275.2kJ/mol = -349.3kJ/mol

Now this reaction + 1/2(2):

Ca(s) + ½ O2(g) → CaO(s)

ΔH = -349.3kJ/mol + 1/2 (-571.68kJ/mol)

Answer:

5.74M or 5.74 mol/L (to 3 sign. fig.)

Explanation:

The molar mass of NaNO3 is 85g/mol, which means that:

1 mole of NaNO3 - 85g

? moles - 122.0g

= 122/85 = 1.44 moles

Concentration in mol/L = no. of moles (moles) ÷ volume (L)

[tex]\frac{1.44}{0.250}[/tex] = 5.74M or 5.74 mol/L (to 3 sign. fig.)

I hope the steps are clear and easy to follow.

Answer:

ΔH∘rxn of the reaction is -521.6kJ

Explanation:

Complete question: "Calculate the value of ΔH°rxn for 2F2(g)+2H2O(l)⟶4HF(g)+O2(g)"

You can find the ΔH of a reaction by the algebraic sum of similar reactions (Hess's law) as follows:

(1) H₂(g) + F₂(g) ⟶ 2HF(g) ΔH∘rxn=−546.6 kJ

(2) 2H₂(g)+O₂(g)⟶2H₂O(l) ΔH∘rxn=−571.6 kJ

Subtracting 2ₓ(1) - (2)

2ₓ(1) 2H₂(g) + 2F₂(g) ⟶ 4HF(g) ΔH∘rxn=2ₓ−546.6 kJ = -1093.2kJ

-(2) 2H₂O(l) ⟶ 2H₂(g)+O₂(g) ΔH∘rxn=- (-571.6 kJ) = 571.6kJ

2ₓ(1) - (2) 2F₂(g)+ 2H₂O(l) ⟶ 4HF(g) + O₂(g)

H₂(g) are cancelled because are the same in products and reactants

ΔH∘rxn = -1093.2kJ + 571.6kJ

ΔH∘rxn = -521.6