AB2AB2 has a molar solubility of 3.72×10−4 M3.72×10−4 M. What is the value of the solubility product constant for AB2AB2? Express your answer numerically.

Answers

Answer 1

Answer:

Ksp = 2.06x10⁻¹⁰

Explanation:

For AB₂. solubility product constant, Ksp, is written as follows:

AB₂(s) ⇄ A²⁺ + 2Br⁻

Ksp = [A²⁺] [Br⁻]²

Molar solubility represents how many moles of AB₂ are soluble per liter of solution. and is obtained from Ksp:

AB₂(s) ⇄ A²⁺ + 2Br⁻

AB₂(s) ⇄ X + 2X

where X are moles that are soluble (Molar solubility)

Ksp = [X] [2X]²

Ksp = 4X³

As molar solubility of the salt is 3.72x10⁻⁴M:

Ksp = 4X³

Ksp = 4(3.72x10⁻⁴)³

Ksp = 2.06x10⁻¹⁰


Related Questions

A strontium hydroxide solution is prepared by dissolving 10.60 gg of Sr(OH)2Sr(OH)2 in water to make 47.00 mLmL of solution.What is the molarity of this solution? Express your answer to four significant figures and include the appropriate units.

Answers

Answer:

Approximately [tex]1.854\; \rm mol\cdot L^{-1}[/tex].

Explanation:

Note that both figures in the question come with four significant figures. Therefore, the answer should also be rounded to four significant figures. Intermediate results should have more significant figures than that.

Formula mass of strontium hydroxide

Look up the relative atomic mass of [tex]\rm Sr[/tex], [tex]\rm O[/tex], and [tex]\rm H[/tex] on a modern periodic table. Keep at least four significant figures in each of these atomic mass data.

[tex]\rm Sr[/tex]: [tex]87.62[/tex].[tex]\rm O[/tex]: [tex]15.999[/tex].[tex]\rm H[/tex]: [tex]1.008[/tex].

Calculate the formula mass of [tex]\rm Sr(OH)_2[/tex]:

[tex]M\left(\rm Sr(OH)_2\right) = 87.62 + 2\times (15.999 + 1.008) = 121.634\; \rm g \cdot mol^{-1}[/tex].

Number of moles of strontium hydroxide in the solution

[tex]M\left(\rm Sr(OH)_2\right) =121.634\; \rm g \cdot mol^{-1}[/tex] means that each mole of [tex]\rm Sr(OH)_2[/tex] formula units have a mass of [tex]121.634\; \rm g[/tex].

The question states that there are [tex]10.60\; \rm g[/tex] of [tex]\rm Sr(OH)_2[/tex] in this solution.

How many moles of [tex]\rm Sr(OH)_2[/tex] formula units would that be?

[tex]\begin{aligned}n\left(\rm Sr(OH)_2\right) &= \frac{m\left(\rm Sr(OH)_2\right)}{M\left(\rm Sr(OH)_2\right)}\\ &= \frac{10.60\; \rm g}{121.634\; \rm g \cdot mol^{-1}} \approx 8.71467\times 10^{-2}\; \rm mol\end{aligned}[/tex].

Molarity of this strontium hydroxide solution

There are [tex]8.71467\times 10^{-2}\; \rm mol[/tex] of [tex]\rm Sr(OH)_2[/tex] formula units in this [tex]47\; \rm mL[/tex] solution. Convert the unit of volume to liter:

[tex]V = 47\; \rm mL = 0.047\; \rm L[/tex].

The molarity of a solution measures its molar concentration. For this solution:

[tex]\begin{aligned}c\left(\rm Sr(OH)_2\right) &= \frac{n\left(\rm Sr(OH)_2\right)}{V}\\ &= \frac{8.71467\times 10^{-2}\; \rm mol}{0.047\; \rm L} \approx 1.854\; \rm mol \cdot L^{-1}\end{aligned}[/tex].

(Rounded to four significant figures.)

Un globo lleno de helio tenia un volumen de 8.5 L en el suelo a 20°C y a una presión de 750 torr. Cuando se le soltó, el globo se elevo a una altitud donde la temperatura era de -20°C y la presión de 425 torr, ¿Cuál era el volumen del gas del globo en estas condiciones?

Answers

Answer:

El volumen del gas era 12.95 L

Explanation:

Se relaciona la presión y el volumen mediante la ley de Boyle, que dice:

“El volumen ocupado por una determinada masa gaseosa a temperatura constante, es inversamente proporcional a la presión”

La ley de Boyle se expresa matemáticamente como:  P*V=k

Por otro lado, la Ley de Charles consiste en la relación que existe entre el volumen y la temperatura absoluta de una cierta cantidad de gas ideal, el cual se mantiene a una presión constante. Esta ley dice que cuando la cantidad de gas y de presión se mantienen constantes, el cociente que existe entre el volumen y la temperatura siempre tendrán el mismo valor:  

[tex]\frac{V}{T}=k[/tex]

Por último, la Ley de Gay Lussac dice que la temperatura absoluta y la presión son directamente proporcionales. Es decir, cuando se mantiene todo lo demás constante, mientras suba la temperatura de un gas subirá también su presión. Y mientras la temperatura del gas baje, lo mismo ocurrirá con la presión:

[tex]\frac{P}{T}=k[/tex]

Combinado las mencionadas tres leyes se obtiene:

[tex]\frac{P*V}{T} =k[/tex]

Cuando se desean estudiar dos diferentes estados, uno inicial y una final de un gas, se puede aplicar:

[tex]\frac{P1*V1}{T1} =\frac{P2*V2}{T2}[/tex]

Recordando que la temperatura debe usarse en grados Kelvin, conoces los siguientes datos:

P1: 750 torrV1: 8.5 LT1: 20°C= 293°K (siendo 0°C=273°K)P2: 425 torrV2: ?T2: -20°C= 253 °K

Reemplazando:

[tex]\frac{750 torr*8.5 L}{293K} =\frac{425 torr*V2}{253 K}[/tex]

Resolviendo:

[tex]V2=\frac{750 torr*8.5 L}{293K} *\frac{253 K}{425 torr}[/tex]

V2= 12.95 L

El volumen del gas era 12.95 L

Using the periodic table provided, identify the atomic mass of sodium (Na) . Your answer should have 5 significant figures. Provide your answer below: __ amu

Answers

Answer:

Your answer will either be 22.9897 or 22.990 !!

Explanation:

Identify the elements that have the following abbreviated electron configurations.
A) [Ne] 3s23p5.
B) [Ar] 4s23d7.
C) [Xe] 6s1.

Answers

Answer:

A) Chlorine (Cl)

B) Cobalt (Co)

C) Caesium (Cs)

Hope this helps.

The abbreviated electron configurations that was given in the question belongs to

Chlorine (Cl)

Cobalt (Co)

Caesium (Cs) respectively.

Electronic configurations can be regarded as the  electronic structure, which is the way an electrons is arranged in energy levels towards an atomic nucleus.

The electron configurations is very useful when  describing  the orbitals of an atom in its ground state.

To calculate an electron configuration, we can put the periodic table into sections, and this section will represent the atomic orbitals which is the  regions that house the electrons.

Groups one of the period table and two belongs to s-block, group  3 through 12 belongs to the d-block, while  13 to 18 can be attributed to p-block ,The  rows that is found at bottom are the f-block

Therefore, electron configurations  explain orbitals of an atom when it is in it's ground state.

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If a bottle of olive oil contains 1.2 kg of olive oil, what is the volume, in milliliters (mL), of the olive oil?

Answers

Answer:

1.3 mL

Explanation:

First, get the density of the olive oil, which is 0.917 kg/mL. Then divide the mass by the density:

1.2kg/0.917kg/mL= 1.3086150491 mL. The kg cancel out, leaving us with mL.

It should have 2 significant figures, because 1.2kg has 2 and we are dividing.

The volume of olive oil will be nearly 1300mL or 1.30 L as per the given data.

What is volume?

Volume is a measurement of three-dimensional space that is occupied. It is frequently numerically quantified using SI derived units or various imperial units. The definition of length is linked to the definition of volume.

Volume is, at its most basic, a measure of space. The units liters (L) and milliliters (mL) are used to measure the volume of a liquid, also known as capacity.

This measurement is done with graduated cylinders, beakers, and Erlenmeyer flasks.

Here, it is given that mass of olive oil is 1.2kg.

We know that,

Density of olive oil = 0.917kg/l.

Volume = mass/density

Volume = 1.2/0.917.

Volume = 1.30 lit.

Volume = 1300mL.

Thus, the volume of olive oil will be 1300 mL.

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cetylene gas is often used in welding torches because of the very high heat produced when it reacts with oxygen gas, producing carbon dioxide gas and water vapor. Calculate the moles of oxygen needed to produce of water.

Answers

Answer:

0.60 mol

Explanation:

There is some info missing. I think this is the original question.

Acetylene gas is often used in welding torches because of the very high heat produced when it reacts with oxygen gas, producing carbon dioxide gas and water vapor. Calculate the moles of oxygen needed to produce 1.5 mol of water.

Step 1: Given data

Moles of water required: 1.5 mol

Step 2: Write the balanced equation

C₂H₂(g) + 2.5 O₂(g) ⇒ 2 CO₂(g) + H₂O(g)

Step 3: Calculate the moles of oxygen needed to produce 1.5 mol of water

The molar ratio of O₂ to H₂O is 2.5:1. The moles of oxygen needed to produce 1.5 mol of water are (1/2.5) × 1.5 mol = 0.60 mol

Complete ionic equation K2CO3(aq)+2CuF(aq) → Cu2CO3(s)+2KF(aq) Examine each of the chemical species involved to determine the ions that would be present in solution. Be sure to consider both the coefficients and subscripts of the molecular equation, and then write this precipitation reaction in the form of a balanced complete ionic equation. Express your answer as a chemical equation including phases.

Answers

Answer:

2K+(aq) + CO3²¯(aq) + Ca^2+(aq) + 2F¯(aq) —› Cu2CO3(s) + 2K+(aq) + 2F¯(aq)

Explanation:

K2CO3(aq) + 2CuF(aq) → Cu2CO3(s) + 2KF(aq)

The complete ionic equation for the above equation can be written as follow:

In solution, K2CO3 and CuF will dissociate as follow:

K2CO3(aq) —› 2K+(aq) + CO3²¯(aq)

CuF(aq) —› Ca^2+(aq) + 2F¯(aq)

Thus, we can write the complete ionic equation for the reaction as shown below:

K2CO3(aq) + 2CuF(aq) —›

2K+(aq) + CO3²¯(aq) + Ca^2+(aq) + 2F¯(aq) —› Cu2CO3(s) + 2K+(aq) + 2F¯(aq)

11. How did the solubility product constant Ksp of KHT in pure water compare to its solubility product constant Ksp of KHT in KCl solution? Are these results what you would expect? Why?

Answers

Answer:

Explanation:

KHT is a salt which ionises in water as follows

KHT ⇄ K⁺ + HT⁻

Solubility product Kw= [ K⁺ ] [ HT⁻ ]

product of concentration of K⁺ and HT⁻ in water

In KCl solution , the solubility product of KHT will be decreased .

In KCl solution , there is already presence of K⁺  ion in the solution . So

in the equation  

[ K⁺ ] [ HT⁻ ]  = constant

when K⁺ increases [ HT⁻ ] decreases . Hence less of KHT dissociates due to which its  solubility decreases . It is called common ion effect . It is so because here the presence of common ion that is K⁺ in both salt to be dissolved and in solvent , results in decrease of solubility of the salt .

Carbon-14 has a half-life of 5720 years and this is a fast-order reaction. If a piece of wood has converted 75 % of the carbon-14, then how old is it?

Answers

Answer:

11445.8years

Explanation:

Half-life of carbon-14 = 5720 years

First we have to calculate the rate constant, we use the formula :

below are three reactions showing how chlorine from CFCs (chlorofluorocarbons) destroy ozone (O3) in the stratosphere. Ozone blocks harmful ultraviolet radiation from reaching earth’s surface. Show how these 3 equations sum to produce the net equation for the decomposition of two moles of ozone to make three moles of diatomic oxygen (2 O3→ 3 O2), and calculate the enthalpy change. (6 points) R1 O2 (g) → 2 O (g) ΔH1°= 449.2 kJ R2 O3 (g) + Cl (g) → O2 (g) + ClO (g) ΔH2° = -126 kJ R3 ClO (g) + O (g) → O2 (g) + Cl (g) ΔH3°= -268 kJ

Answers

Answer:

ΔH = -338.8kJ

Explanation:

it is possible to sum the enthalpy changes of some reactions to obtain the enthalpy change of the whole reaction (Hess's law).

Using the reactions:

R₁ O₂(g) → 2O(g) ΔH₁°= 449.2 kJ

R₂ O₃(g) + Cl(g) → O₂(g) + ClO(g) ΔH₂° = -126 kJ

R₃ ClO (g) + O (g) → O₂ (g) + Cl (g) ΔH₃°= -268 kJ

By the sum 2R₂ + 2R₃:

(2R₂ + 2R₃) = 2O(g) + 2O₃(g) → 4O₂(g)

ΔH = 2ₓ(-126kJ) + (2ₓ-268kJ) = -788kJ

Now, this reaction + R₁

2O₃(g) → 3O₂(g)

ΔH = -768kJ + 449.2kJ

ΔH = -338.8kJ

The equilibrium between carbon dioxide gas and carbonic acid is very important in biology and environmental science. CO2 ( aq) + H2O ( l) H2CO3 ( aq) Which one of the following is the correct equilibrium constant expression (K c) for this reaction?
a) K =[H2CO3]/ [CO2]
b) K=[CO2]/ [H2CO3]
c) K=[H2CO3]/ [CO2][H2O]
d) K=[CO2][H2O]/ [H2CO3]
e) K=1/[H2CO3]

Answers

Answer:

Kc = [H₂CO₃] / [CO₂]

Explanation:

Equilibrium constant expression (Kc) of any reaction is defined as the ratio between molar concentrations in equilibrium of products over reactants.

Pure solids and liquids don't affect the equilibrium and you don't have to take its concentrations in the equilibrium.

Also, each specie must be powered to its reactant coefficient.

For example, for the reaction:

aA(s) + bB(aq) ⇄ cC(l) + nD(g) + xE(aq)

The equilibrium constant, kc is:

Kc = [D]ⁿ / [B]ᵇ[E]ˣ

You don't take A nor C species because are pure solids and liquids. b, n and x are the reactant coefficients of each substance. Ratio of products over reactants

Thus, for the reaction:

CO₂(aq) + H₂O(l) ⇄ H₂CO₃(aq)

The Kc is:

Kc = [H₂CO₃] / [CO₂]

Draw a Lewis structure for one important resonance form of HBrO4 (HOBrO3). Include all lone pair electrons in your structure. Do not include formal charges in your structure.

Answers

Answer:

The Lewis structure is attached with the answer -

Explanation:

Lewis structure or Lewis dot diagram are diagrams or representation of showing the bonding between different or same atoms of a molecule in any and also shows lone pairs of electrons that may exist in the molecule as dots.

HBrO₄ is bromine oxoacid which is also known as perbromic acid. It is a unstable inorganic compound.

The Lewis structure is attached in form of image with representation of lone pairs of electrons.

A base solution contains 0.400 mol of OH–. The base solution is neutralized by 43.4 mL of sulfuric acid. What is the molarity of the sulfuric acid solution?

Answers

Answer:

Molarity of the sulfuric acid solution is 4.61M

Explanation:

The neutralization of a base of OH⁻ with sulfuric acid, H₂SO₄, occurs as follows:

2 OH⁻ + H₂SO₄ → 2H₂O + SO₄²⁻

That means, 2 moles of base react with 1 mole of sulfuric acid.

If you add 0.400 moles of OH⁻, moles of sulfuric acid you need to neutralize this amount of OH⁻ are:

0.400 moles OH⁻ ₓ (1 mole H₂SO₄ / 2 moles OH⁻) = 0.200 moles of H₂SO₄

As you add 43.4mL = 0.0434L of sulfuric acid to neutralize this solution, molarity (Ratio between moles and liters) is:

0.200 moles H₂SO₄ / 0.0434L = 4.61M

Molarity of the sulfuric acid solution is 4.61M

What states can electrons exist in? A. Electron clouds or energy levels B. Positive and negative C. Up and down spin D. In phase and out of phase

Answers

Answer:

A. Electron clouds or energy levels

Explanation:

Electrons can exist in two states:

Stablized in electronic orbitalsFreely moving

Electrons can exist in an electron cloud or energy level. Electron in an atoms have ability to change energy levels either by emitting or absorbing a photon that form the energy equal to the energy difference between the two levels.

Hence, the correct answer is A.

Answer:

Up and DOWN spin

Explanation:

Draw the Lewis structure for methane (CH4) and ethane (C2H6) in the box below. Then predict which would have the higher boiling point. Finally, explain how you came to that conclusion.

Answers

Answer:

Ethane would have a higher boiling point.

Explanation:

In this case, for the lewis structures, we have to keep in mind that all atoms must have 8 electrons (except hydrogen). Additionally, each carbon would have 4 valence electrons, with this in mind, for methane we have to put the hydrogens around the carbon, and with this structure, we will have 8 electrons for the carbon. In ethane, we will have a bond between the carbons, therefore we have to put three hydrogens around each carbon to obtain 8 electrons for each carbon.

Now, the main difference between methane and ethane is an additional carbon. In ethane, we have an additional carbon, therefore due to this additional carbon, we will have more area of interaction for ethane. If we have more area of interaction we have to give more energy to the molecule to convert from liquid to gas, so, the ethane will have a higher boiling point.

I hope it helps!

The Lewis structure shows the valence electrons in a molecule. Ethane will have a higher boiling point than methane.

We can deduce the number of valence electrons in a molecule by drawing the Lewis structure of the molecule. The Lewis structure consists of the symbols of elements in the compound and the valence electrons in the compound.

We know that the higher the molar mass of a compound the greater its boiling point. Looking at the Lewis structures of methane and ethane, we cam see that ethane has a higher molecular mass (more atoms) and consequently a higher boiling point than methane.

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Determine the volume occupied by 10 mol of helium at 27 ° C and 82 atm

please.

Answers

Answer:

3.00 L

Explanation:

Convert the pressure to Pascals.

P = 82 atm × (101325 Pa/atm)

P = 8,308,650 Pa

Convert temperature to Kelvins.

T = 27°C + 273

T = 300 K

Use ideal gas law:

PV = nRT

(8,308,650 Pa) V = (10 mol) (8.314 J/mol/K) (300 K)

V = 0.00300 m³

If desired, convert to liters.

V = (0.00300 m³) (1000 L/m³)

V = 3.00 L

Answer:

[tex]\large \boxed{\text{3.0 L}}[/tex]

Explanation:

[tex]\begin{array}{rcl}pV &=& nRT\\\text{82 atm} \times V & = & \text{10 mol} \times \text{0.082 06 L}\cdot\text{atm}\cdot\text{K}^{-1}\text{mol}^{-1} \times \text{300.15 K}\\82V & = & \text{246 L}\\V & = & \textbf{3.0 L} \\\end{array}\\\text{The volume of the balloon is $\large \boxed{\textbf{3.0 L}}$}[/tex]

Q1. Calculate the amount of copper produced in 1.0 hour when aqueous CuBr2 solution was electrolyzed by using a current of 4.50 A. Q2. In another electroplating experiment, if electric current was passed for 3 hours and 2.00 g of silver was deposited from a AgNO3 solution, what was the current used in amperes

Answers

Answer:

[tex]\boxed{\text{Q1. 3.6 g; Q2. 0.2 A}}[/tex]

Explanation:

Q1. Mass of Cu

(a) Write the equation for the half-reaction.

Cu²⁺ + 2e⁻ ⟶ Cu

The number of electrons transferred (z) is 2 mol per mole of Cu.

(b) Calculate the number of coulombs

q  = It  

[tex]\text{t} = \text{1.0 h} \times \dfrac{\text{3600 s}}{\text{1 h}} = \text{3600 s}\\\\q = \text{3 C/s} \times \text{ 3600 s} = \textbf{10 800 C}[/tex]

(c) Mass of Cu

We can summarize Faraday's laws of electrolysis as

[tex]\begin{array}{rcl}m &=& \dfrac{qM}{zF}\\\\& = &\dfrac{10 800 \times 63.55}{2 \times 96 485}\\\\& = & \textbf{3.6 g}\\\end{array}\\\text{The mass of Cu produced is $\boxed{\textbf{3.6 g}}$}[/tex]

Note: The answer can have only two significant figures because that is all you gave for the time.

Q2. Current used

(a) Write the equation for the half-reaction.

Ag⁺ + e⁻ ⟶ Ag

The number of electrons transferred (z) is 1 mol per mole of Ag.  

(a) Calculate q

[tex]\begin{array}{rcl}m &=& \dfrac{qM}{zF}\\\\2.00& = &\dfrac{q \times 107.87}{1 \times 96 485}\\\\q &=& \dfrac{2.00 \times 96485}{107.87}\\\\& = & \textbf{1789 C}\\\end{array}[/tex]

(b) Calculate the current

t = 3 h = 3 × 3600 s = 10 800 s

[tex]\begin{array}{rcl}q&=& It\\1789 & = & I \times 10800\\I & = & \dfrac{1789}{10800}\\\\& = & \textbf{0.2 A}\\\end{array}\\\text{The current used was $\large \boxed{\textbf{0.2 A}}$}[/tex]

Note: The answer can have only one significant figure because that is all you gave for the time.

A sample of neon gas at a pressure of 0.609 atm and a temperature of 25.0 °C, occupies a volume of 19.9 liters. If the gas is compressed at constant temperature to a
volume of 12.7 liters, the pressure of the gas sample will be
atm.​

Answers

Answer:

The pressure of the gas sample will be  0.954 atm.​

Explanation:

Boyle's law states that the pressure of a gas in a closed container is inversely proportional to the volume of the container, when the temperature is constant. That is, if the pressure increases, the volume decreases; conversely if the pressure decreases, the volume increases.

Boyle's law is expressed mathematically as:

Pressure * Volume = constant

o P * V = k

To determine the change in pressure or volume during a transformation at constant temperature, the following is true:

P1 · V1 = P2 · V2

That is, the product between the initial pressure and the initial volume is equal to the product of the final pressure times the final volume.

In this case:

P1= 0.609 atmV1= 19.9 LP2=?V2= 12.7 L

Replacing:

0.609 atm* 19.9 L= P2* 12.7 L

Solving:

[tex]P2=\frac{0.609 atm* 19.9 L}{12.7 L}[/tex]

P2= 0.954 atm

The pressure of the gas sample will be  0.954 atm.​

The volume of a sample of oxygen is 300mL when the pressure is 1 atm and the temperature is 27 C . At what temperature is the volume 1.00 L and the pressure.500 atm?

Answers

Answer:

T2 = 500K

Explanation:

Given data:

P1 = 1atm

V1 = 300ml

T1= 27 + 273 = 300K

T2 = ?

V2 = 1.00ml

P2 = 500atm

Apply combined law:

P1xV1//T1 = P2xV2/T2 ...eq1

Substituting values into eq1:

1 x 300/300 = 500 x 1/T2

Solve for T2:

300T2 = 500 x 300

300T2 = 150000

Divide both sides by the coefficient of T2:

300T2/300 = 150000/300

T2 = 500K

Which of the following elements is in the same family as fluorine?
a. silicon
b. antimony
O c. iodine
O d. arsenic
e. None of these.

Answers

Answer:

c iodine

Explanation:

fluorine is a halogen group element like Bromine, Iodine,Astatine,Chloride

How many mL of calcium hydroxide are required to neutralize 25.0 mL of 0.50 M
nitric acid?

Answers

Answer:

6.5 mL

Explanation:

Step 1: Write the balanced reaction

Ca(OH)₂ + 2 HNO₃ ⇒ Ca(NO₃)₂ + 2 H₂O

Step 2: Calculate the reacting moles of nitric acid

25.0 mL of 0.50 M  nitric acid react.

[tex]0.0250L \times \frac{0.50mol}{L} = 0.013 mol[/tex]

Step 3: Calculate the reacting moles of calcium hydroxide

The molar ratio of Ca(OH)₂ to HNO₃ is 1:2. The reacting moles of Ca(OH)₂ are 1/2 × 0.013 mol = 6.5 × 10⁻³ mol

Step 4: Calculate the volume of calcium hydroxide

To answer this, we need the concentration of calcium hydroxide. Since the data is missing, let's suppose it is 1.0 M.

[tex]6.5 \times 10^{-3} mol \times \frac{1,000mL}{1.0mol} = 6.5 mL[/tex]

When the following molecular equation is balanced using the smallest possible integer coefficients, the values of these coefficients are:
P2O5 (s) + H2O (l) =H3PO4 (aq)
The balanced chemical equation for the reaction between hydrogen sulfide and oxygen is:
2H2S(g) + 3O2(g) =2H2O(l) + 2SO2(g)
We can interpret this to mean:
3moles of oxygen and_______moles of hydrogen sulfide react to produce______moles of water and_______ moles of sulfur dioxide.

Answers

Answer:

1. The coefficients are: 1, 3, 2

2. From the balanced equation, we obtained the following:

3 moles oxygen, O2 reacted.

2 moles of Hydrogen sulfide, H2S reacted.

2 moles of water were produced.

2 moles of sulphur dioxide, SO2 were produced.

Explanation:

1. Determination of the coefficients of the equation.

This is illustrated below:

P2O5(s) + H2O(l) <==> H3PO4(aq)

There are 2 atoms of P on the left side and 1 atom on the right side. It can be balance by putting 2 in front of H3PO4 as shown below:

P2O5(s) + H2O(l) <==> 2H3PO4(aq)

There are 2 atoms of H on the left side and 6 atoms on the right side. It can be balance by putting 3 in front of H2O as shown below:

P2O5(s) + 3H2O(l) <==> 2H3PO4(aq)

Now the equation is balanced.

The coefficients are: 1, 3, 2.

2. We'll begin by writing the balanced equation for the reaction. This is given below:

2H2S(g) + 3O2(g) => 2H2O(l) + 2SO2(g)

From the balanced equation above,

3 moles of oxygen, O2 reacted with 2 moles of Hydrogen sulfide, H2S to produce 2 moles of water, H2O and 2 moles of sulphur dioxide, SO2.

When a solution is diluted with water, the ratio of the initial to final
volumes of solution is equal to the ratio of final to initial molarities
Select one:
True
-​

Answers

Hello!!

The correct answer for this problem would be TRUE.

Explanation: it is true that when a solution is diluted with water, the ratio of the initial to final volumes of solution is equal to the ratio of final to initial molarities.

When a solution is diluted with water, the ratio of the initial to final volumes of solution is equal to the ratio of final to initial molarities. The statement is True.

Concentration refers to the amount of a substance in a defined space. Another definition is that concentration is the ratio of solute in a solution to either solvent or total solution.

There are various methods of expressing the concentration of a solution.

Concentrations are usually expressed in terms of molarity, defined as the number of moles of solute in 1 L of solution.

Solutions of known concentration can be prepared either by dissolving a known mass of solute in a solvent and diluting to a desired final volume or by diluting the appropriate volume of a more concentrated solution (a stock solution) to the desired final volume.

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