The ph of a 0.55 m aqueous solution of hypobromous acid, hbro, at 25.0 °c is 4.48. what is the value of ka for hbro?

The percent of lead in Pb(CO3)2 is calculated by dividing the molar mass of the lead by the molar mass of the entire compound and then multiplying by 100, resulting in 63.31% lead content.

To calculate the percent of lead in Pb(CO3)2, first, we need to determine the molar mass of the compound. The molar mass of lead (Pb) is 207.2 u, and the molar mass of carbonate (CO3) is 60.01 u (with 3 oxygens at 16.0 u each plus one carbon at 12.01 u).

The molar mass of the lead carbonate compound is:

Adding these together, the molar mass of Pb(CO3)2 is 207.2 u + 120.02 u = 327.22 u.

To find the percent of lead in the compound, divide the molar mass of lead by the total molar mass of the compound and multiply by 100:

Percent of lead = (207.2 u / 327.22 u) x 100 = 63.31%

Therefore, the percent of lead in lead carbonate is 63.31%.

Final answer:

To find the molecular formula of a compound with an empirical formula of C5H7N and a molar mass of 208.17 g/mol, divide the molar mass by the molar mass of the empirical formula and multiply the empirical formula by the resulting whole number to obtain the molecular formula, which is C15H21N3.

Explanation:

The student's question pertains to finding the molecular formula of a compound given its empirical formula and molar mass. The empirical formula is C5H7N, which has a molar mass of 81.13 g/mol. To find the molecular formula, we divide the given molar mass of the compound, 208.17 g/mol, by the empirical formula mass of 81.13 g/mol to find the multiple (n). This multiple, which should be a whole number, will be used to scale the empirical formula up to the molecular formula.

Performing this calculation, we find that n = 208.17 / 81.13, which simplifies to approximately 2.57. Since n must be a whole number, and this value is close to 2.5, we can deduce that the empirical formula needs to be multiplied by 3 to obtain the molecular formula. Therefore, the molecular formula of the compound is C15H21N3.

The compound formed from sodium and iodine, based on their positions on the periodic table, is Sodium Iodide (NaI). Sodium and Iodine combine in a 1:1 ratio, as Sodium is a metal with a +1 charge and Iodine is a non-metal with a -1 charge.

The compound formed between sodium and iodine based on their positions in the periodic table is Sodium Iodide. In the periodic table, sodium is a metal (from Group 1: alkali metals) and iodine is a non-metal (from Group 17: halogens). When a metal and a non-metal combine, they typically form ionic compounds. Sodium, with a charge of +1, and iodine, with a charge of -1, combine in a 1:1 ratio to form Sodium Iodide, which has the chemical formula NaI.

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The chemical reaction of NaCl with H₂O, leading to the formation of NaOH, H₂, and Cl₂, can be expressed as a balanced molecular equation, a complete ionic equation, and a simplified net ionic equation, where Na⁻ and Cl⁻ act as spectator ions.

The initial equation provided, NaCl(aq) + H₂O(1) → NaOH(aq) + H₂(g) + Cl₂ (g), is an example of a chemical reaction involving the decomposition of sodium chloride (NaCl) in the presence of water to form sodium hydroxide (NaOH), hydrogen gas (H₂), and chlorine gas (Cl₂).

The balanced molecular equation for this reaction is:

2NaCl(aq) + 2H₂O(l) → 2NaOH(aq) + H₂(g) + Cl₂ (g)

The complete ionic equation would be:

2Na+ (aq) + 2Cl + (aq) + 2H₂O(l) → 2Na+ (aq) + 2OH - (aq) + H₂(g) + Cl₂ (g)

And the net ionic equation simplifies to:

2H₂O(l) → H₂(g) + Cl₂ (g), since Na⁺ and Cl ⁻ are spectator ions and do not participate in the reaction.

Answer is: activation energy of this reaction is 212,01975 kJ/mol.
Arrhenius equation: ln(k₁/k₂) = Ea/R (1/T₂ - 1/T₁).
k₁ = 0,000643 1/s.
k₂ = 0,00828 1/s.

T₁ = 622 K.

T₂ = 666 K.

R = 8,3145 J/Kmol.

1/T₁ = 1/622 K = 0,0016 1/K.
1/T₂ = 1/666 K = 0,0015 1/K.
ln(0,000643/0,00828) = Ea/8,3145 J/Kmol · (-0,0001 1/K).
-2,55 = Ea/8,3145 J/Kmol · (-0,0001 1/K).
Ea = 212019,75 J/mol = 212,01975 kJ/mol.

Final answer:

To find the activation energy for 1-bromopropane's decomposition, use the Arrhenius equation with given rate constants and temperatures, resulting in a calculated activation energy.

Explanation:

To calculate the activation energy for the gas-phase decomposition of 1-bromopropane, we can use the Arrhenius equation which relates the rate constant (k) of a reaction to the temperature (T) and activation energy (Ea). The equation in its logarithmic form is:

ln(k2/k1) = (Ea/R) * (1/T1 - 1/T2)

Where:

k1 = 6.43×10-4 /s and k2 = 8.28×10-3 /s are the rate constants at temperatures T1 and T2, respectively

Using the Arrhenius equation:

ln(8.28×10-3 / 6.43×10-4) = (Ea/8.314) * (1/622 - 1/666)

After calculating:

Ea = ((ln(8.28×10-3 / 6.43×10-4)) * 8.314) / (1/622 - 1/666)

This gives the activation energy for the gas-phase decomposition of 1-bromopropane.

Answer:

Container will rupture at temperature of 375 K.

Explanation:

Initial pressure of the  nitrogen gas =[tex]P_1= 1200 torr = 1.572 atm[/tex]

(1 torr = 0.00131 atm)

Initial temperature of nitrogen gas =[tex]T_1= 250 K[/tex]

Final pressure of the nitrogen gas =[tex]P_2=1800 torr=2.358 atm[/tex]

Final temperature of nitrogen gas =[tex]T_2=?[/tex]

Since, the container is inflexible that is volume remains constant we can apply Gay Lussac's law:

[tex]\frac{P_1}{T_1}=\frac{P_2}{T_2}[/tex]

[tex]\frac{1.572 atm}{250 K}=\frac{2.358 atm}{T_2}[/tex]

[tex]T_2=375 K[/tex]

Container will rupture at temperature of 375 K.

A barometer reading of 29.8 inches of mercury is equivalent to 1012.8 millibars of atmospheric pressure.

A barometer displaying a reading of 29.8 inches of mercury would be equivalent to 1012.8 millibars of atmospheric pressure.

To convert inches of mercury to millibars, you can use the conversion factor of 1 inch of mercury = 33.864 millibars. Multiply the given reading by the conversion factor to find the equivalent pressure in millibars.

Therefore, 29.8 inches of mercury = 29.8 inches * 33.864 millibars/inch = 1012.8 millibars.

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Answer : The value of [tex]k_a[/tex] for benzoic acid is, [tex]6.4\times 10^{-5}[/tex]

Solution :

The balanced equilibrium reaction will be,

                             [tex]C_6H_5COOH\rightleftharpoons H^++C_6H_5COO^-[/tex]

initial conc.          0.2 M                  0                0

at eqm.  [tex](0.2-0.00355)M[/tex]       [tex]0.00355M[/tex] [tex]0.00355M[/tex]

The expression for dissociation constant for a benzoic acid will be,

[tex]k_a=\frac{[H^+]\times [C_6H_5COO^-]}{[C_6H_5COOH]}[/tex]

Now put all the given values in this formula, we get the value of [tex]k_a[/tex]

[tex]k_a=\frac{(3.55\times 10^{-3})\times (3.55\times 10^{-3})}{(0.2-3.55\times 10^{-3})}=6.4\times 10^{-5}[/tex]

Therefore, the value of [tex]k_a[/tex] for benzoic acid is, [tex]6.4\times 10^{-5}[/tex]

The concentration of X2 in a 0.150 M solution of the diprotic acid H2X is calculated by using equilibrium constants Ka1 and Ka2, which are specific to each step in the ionization process. This involves solving for the concentration of HX- first, and then X2-. The assumption that the change in concentration (x) is negligible compared to initial concentrations is valid if x is less than 5%.

The concentration of X2 in a 0.150 M solution of the diprotic acid H2X can be calculated using the given equilibrium constants Ka1 and Ka2. It is important to remember that diprotic acids undergo ionization in two steps and each step has its own equilibrium constant.

Ka1 = 4.5×10-6 is the equilibrium constant for the first dissociation and Ka2 = 1.2×10-11 is for the second.

In the first step, H2X dissociates into HX- and H+. From the value of Ka1 and the initial concentration of H2X, one can solve for the concentration of HX-. The next step is the dissociation of HX- into X2- and H+. Similarly, by using Ka2 and the concentration of HX-, the concentration of X2- can be calculated. The calculation usually assumes that x is small compared to initial concentrations and this assumption is valid if the concentration of x is less than 5% of initial concentrations.

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Answer:

1. 1/2

2. 36

3. 18

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The ratio of the central angle to the entire circle measure is 1:2. The area of the sector is equal to the area of the entire circle.

The ratio of the measure of the central angle MNL to the entire circle measure is 1:2. This is because the measure of the central angle is given as π radians, which is half the measure of the entire circle, which is 2π radians.

The area of the entire circle is given as π units², and the area of the sector (or the sector is the region enclosed by the central angle MNL in the circle) is also given as π units². This means that the sector occupies the entire area of the circle.

The pH of the 0.005 M K₂O solution is 12. This is calculate using K₂O dissociation in water and then dealing with concentrations.

To calculate the pH of a 0.005 M solution of potassium oxide (K₂O), first, we need to understand how K₂O dissociates in water. Potassium oxide reacts with water to form potassium hydroxide (KOH), which completely dissociates in water:

K₂O + H₂O → 2 KOH

Given the 0.005 M concentration of K₂O, it will produce an equivalent concentration of 0.01 M KOH because one K₂O produces two KOH molecules.

Next, KOH fully dissociates into K⁺ and OH⁻ ions:

KOH → K⁺ + OH⁻

This means the concentration of OH⁻ is also 0.01 M. To find the pOH of the solution, use the following formula:

pOH = -log[OH⁻]

So,

pOH = -log(0.01) = 2

Now, we can calculate the pH using the formula:

pH = 14 - pOH

So,

pH = 14 - 2 = 12

Therefore, the pH of the 0.005 M K₂O solution is 12.

[tex]\boxed{{\text{No precipitate will be formed}}}[/tex] in [tex]2{\text{NaCl}} + {\text{Ba}}{\left( {{\text{OH}}} \right)_2} \to {\text{BaC}}{{\text{l}}_2} + 2{\text{NaOH}}[/tex]

[tex]\boxed{{\text{AgI}}}[/tex] is the precipitate formed in [tex]{\text{AgCl}}{{\text{O}}_3} + {\text{Mg}}{{\text{I}}_2} \to {\text{AgI}} + {\text{Mg}}{\left( {{\text{Cl}}{{\text{O}}_3}} \right)_2}[/tex]

Further Explanation:

Precipitation reaction:

It is the type of reaction in which an insoluble salt is formed by the combination of two solutions containing soluble salts. That insoluble salt is known as precipitate and therefore such reactions are named precipitation reactions. An example of precipitation reaction is,

[tex]{\text{AgN}}{{\text{O}}_3}\left( {aq} \right) + {\text{KBr}}\left( {aq} \right) \to {\text{AgBr}}\left( s \right) + {\text{KN}}{{\text{O}}_3}\left( {aq} \right)[/tex]

Here, AgBr is a precipitate.

The solubility rules to determine the solubility of the compound are as follows:  

1. The common compounds of group 1A are soluble.

2. All the common compounds of ammonium ion and all acetates, chlorides, nitrates, bromides, iodides, and perchlorates are soluble in nature. Only the chlorides, bromides, and iodides of [tex]{\text{A}}{{\text{g}}^ + }[/tex], [tex]{\text{P}}{{\text{b}}^{2 + }}[/tex], [tex]{\text{C}}{{\text{u}}^ + }[/tex] and [tex]{\text{Hg}}_2^{2 + }[/tex] are not soluble.

3. All common fluorides, except for [tex]{\text{Pb}}{{\text{F}}_{\text{2}}}[/tex] and group 2A fluorides, are soluble. Moreover, sulfates except [tex]{\text{CaS}}{{\text{O}}_{\text{4}}}[/tex], [tex]{\text{SrS}}{{\text{O}}_{\text{4}}}[/tex],  [tex]{\text{BaS}}{{\text{O}}_{\text{4}}}[/tex], [tex]{\text{A}}{{\text{g}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}[/tex] and [tex]{\text{PbS}}{{\text{O}}_{\text{4}}}[/tex] are soluble.

4. All common metal hydroxides except [tex]{\text{Ca}}{\left( {{\text{OH}}} \right)_{\text{2}}}[/tex], [tex]{\text{Sr}}{\left( {{\text{OH}}} \right)_{\text{2}}}[/tex], [tex]{\text{Ba}}{\left( {{\text{OH}}} \right)_{\text{2}}}[/tex] and hydroxides of group 1A, are insoluble.

5. All carbonates and phosphates, except those formed by group 1A and ammonium ion, are insoluble.

6. All sulfides, except those formed by group 1A, 2A, and ammonium ion are insoluble.

7. Salts that contain [tex]{\text{C}}{{\text{l}}^ - }[/tex], [tex]{\text{B}}{{\text{r}}^ - }[/tex] or [tex]{{\text{I}}^ - }[/tex] are usually soluble except for the halide salts of [tex]{\text{A}}{{\text{g}}^ + }[/tex], [tex]{\text{P}}{{\text{b}}^{2 + }}[/tex] and [tex]{\left( {{\text{H}}{{\text{g}}_2}} \right)^{{\text{2 + }}}}[/tex].

8. The chlorides, bromides, and iodides of all the metals are soluble in water, except for silver, lead, and mercury (II). Mercury (II) iodide is water insoluble. Lead halides are soluble in hot water.

9. The perchlorates of group 1A and group 2A are soluble in nature.

(1) The given reaction is as follows:

 [tex]2{\text{NaCl}} + {\text{Ba}}{\left( {{\text{OH}}} \right)_2} \to {\text{BaC}}{{\text{l}}_2} + 2{\text{NaOH}}[/tex]

This is an example of a double displacement reaction in which two ionic compounds are exchanged with each other and two new compounds are formed. [tex]{\text{BaC}}{{\text{l}}_2}[/tex] and NaOH are soluble salts according to the solubility rules. So no precipitate will be formed in this reaction.

(2) The given reaction is as follows:

 [tex]{\text{AgCl}}{{\text{O}}_3} + {\text{Mg}}{{\text{I}}_2} \to {\text{AgI}} + {\text{Mg}}{\left( {{\text{Cl}}{{\text{O}}_3}} \right)_2}[/tex]

According to the solubility rules, AgI is an insoluble salt. The perchlorates of group 2 are soluble in nature and therefore [tex]{\text{Mg}}{\left( {{\text{Cl}}{{\text{O}}_3}} \right)_2}[/tex] is soluble in water. So AgI forms the precipitate in the above reaction.

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Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Chemical reaction and equation

Keywords: precipitation reaction, precipitate, insoluble, soluble, AgI, AgClO3, Mg(ClO3)2, MgI2, NaCl, Ba(OH)2, BaCl2, NaOH, solubility rules, halides, sulfides.

The volume of in milliliters of the HClO₄ needed to neutralize the NaOH solution is 24.9mL

From the question, we are to determine the volume of HClO₄ needed to neutralize the NaOH solution

First, we will write the balanced chemical equation for the reaction

The balanced chemical equation for the reaction is

HClO₄ + NaOH → NaClO₄ + H₂O

This means

1 mole of HClO₄ is required to neutralize 1 mole of NaOH

Now, we will determine the number of moles of NaOH present in the solution

From the given information

Volume of NaOH = 50.00 mL = 0.05 L

Concentration of NaOH = 0.0832 M

Using the formula,

Number of moles = Concentration × Volume

Then,

Number of moles of NaOH present = 0.0832 × 0.05

Number of moles of NaOH present = 0.00416 mole

Now,

Since 1 mole of HClO₄ is required to neutralize 1 mole of NaOH

Then,

0.00416 mole of HClO₄ will be required to neutralize the 0.00416 mole of NaOH

Thus, the number of moles of HClO₄ required is 0.00416 mole

Now, for the volume of HClO₄ required

From the formula,

[tex]Volume = \frac{Number\ of\ moles }{Concentration}[/tex]

Then,

Volume of HClO₄ needed = [tex]\frac{0.00416}{0.167}[/tex]

Volume of HClO₄ needed = 0.0249101 L

Volume of HClO₄ needed = 24.9101 mL

Volume of HClO₄ needed ≅ 24.9 mL

Hence, the volume of in milliliters of the HClO₄ needed to neutralize the NaOH solution is 24.9mL

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Answer:

295.3744 grams of NaBr will produce 244 grams of [tex]NaNO_3[/tex].

Explanation:

[tex]Pb(NO_3)_2(aq)+2 NaBr(aq)\rightarrow PbBr_2(s)+2 NaNO_3(aq)[/tex]

Moles of sodium nitarte= [tex]\frac{244 g}{85 g/mol}=2.8705 mol[/tex]

According to reaction, 2 moles of sodium nitrate is obtained from 2 moles of sodium bromide.

Then 2.8705 mol of sodium nitrate will be obtained from :

[tex]\frac{2}{2}\times 2.8705mol=2.8705 mol[/tex] of sodium nitrate

Mass of 2.8705 moles of sodium nitrate:

[tex]2.8705 mol\times 102.9 g/mol=295.3744 g[/tex]

295.3744 grams of NaBr will produce 244 grams of [tex]NaNO_3[/tex].

Our planet is formed of elements and the combinations of elements known as compounds. An element is a pure component formed of atoms, which are all of a similar kind. Till now, 116 elements are identified, and of these only 90 occur naturally.  

At the time of universe formation, which is, about 14 billion years ago known as Big Bang, the creation of only the lightest elements took place, that is, hydrogen and helium along with minute concentrations of beryllium and lithium. The rest of the 86 elements found in nature were formed in the nuclear reactions, which took place in the stars and in huge stellar explosions called supernovae.  

To determine the molar mass, you need to get the atomic mass of the molecule. To do this, check the periodic table for the atomic mass or average atomic weight of each element.

Mg = 24.305 x 1 = 24.305 amu

O = 15.9994 x 2 =31.9988 amu

H = 1.0079 x 2 = 2.0158 amu

Then, add all the components to get the atomic mass of the molecule.

24.305 amu + 31.9988 amu + 2.0158 amu = 58.3196 amu

The atomic mass is just equivalent to its molar mass.

So, the molar mass of Magnesium hydroxide (Mg(OH)2) is 58.3196 g/mol.

The three alkenes that will form 2-methylbutane upon hydrogenation are 1-pentene, 2-pentene, and 3-pentene.

To find the three alkenes that will form 2-methylbutane upon hydrogenation, we need to consider the number of carbon atoms in the molecule. In this case, the formula is C5H10, which means it has 5 carbon atoms. Since we want to form 2-methylbutane upon hydrogenation, we know that the starting molecule must have a double bond on one of the outer carbon atoms and a methyl group attached to one of the middle carbon atoms.

The three alkenes that satisfy these conditions are:

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Answer:   c. potassium nitrate

Among the given substances the solubility of carbon dioxide in liquids is more when compared with potassium nitrate, sugar and sodium chloride.

Explanation:

Increasing pressure doesn’t change the amount of solid or liquid particles dissolved in a solution. Thus pressure doesn’t have an effect on the solubility of solids and liquids. But the case of gases is different.

Increasing pressure above a liquid causes more gas molecules to get dissolved in the liquid. Thus pressure of the system directly affects the solubility of gases in a liquid.

In this question sugar, potassium nitrate and sodium chloride re solids and the pressure cannot change the solubility of these in a liquid.

But carbon dioxide being a gas can be dissolved more in a liquid if the pressure is increased.

The pressure of the enclosed gas is likely equivalent to the given atmospheric pressure, so in this scenario, it would be 0.975 atm.

The pressure of the enclosed gas in this scenario would be equivalent to the given atmospheric pressure, in this case being 0.975 atm. Atmospheric pressure is defined by the sum of all the partial pressures of the atmospheric gases added together. In a closed system, typically if no other factors are impacting the system, the pressure of a gas would be equal to the atmospheric pressure. More precise measurement might require the consideration of factors like temperature and volume as per the ideal gas law, but with the information provided, the pressure of the enclosed gas can be assumed to be the same as the atmospheric pressure which is 0.975 atm in this case.

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The pressure of an enclosed gas can differ from atmospheric pressure based on factors like volume, temperature and amount of gas. In the absence of these variables, we assume equality with atmospheric pressure. For the given atmospheric pressure of 0.975 atm, the assumed pressure of the enclosed gas is also 0.975 atm.

The question refers to understanding the pressure of an enclosed gas when the atmospheric pressure is 0.975 atm. The pressure of a gas enclosed within a closed system might not necessarily be the same as the atmospheric pressure, as it depends on several factors such as the volume of the gas, the temperature, and the number of gas molecules present.

Without additional information, we must assume that the pressure inside is equal to the atmospheric pressure according to the principles of equilibrium. So, if the atmospheric pressure is 0.975 atm, then the pressure of the enclosed gas should be assumed to be also 0.975 atm, unless stated otherwise. This is the same principle as the pressure inside and outside of a properly inflated tyre, or the pressure inside an unopened soda can and the atmospheric pressure.

If this is a manometer style problem where there is an additional pressure from a column of fluid (like mercury), we would need the height of the column to calculate the additional pressure exerted by the gas. For most homework problems, the system is at equilibrium and the pressure inside the container is the same as outside.

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