What can you learn about the pH of a substance with the conductivity test? hint: gives you no info on concentration.

Answers

Answer 1

Answer:

See explanation

Explanation:

So, I'm gonna take a shot at this one and say this:

With a strongly acidic/basic solution, you'll get a high conductivity when preforming a conductivity test.

The more acidic or basic a substance is, the higher the electrical conductivity.

Based on how high or low the conductivity is, it will give you an idea of the substance's pH.

Hope that made since or gave you an idea of what you're looking for. Good luck :)


Related Questions

The solubility of lead(II) iodide is 0.064 g/100 mL at 20ºC. What is the solubility product for lead(II) iodide?

Answers

Answer:

[tex]Ksp=1.07x10^{-8}[/tex]

Explanation:

Hello,

In this case, the dissociation reaction is:

[tex]PbI_2(s)\rightleftharpoons Pb^{2+}(aq)+2I^-(aq)[/tex]

For which the equilibrium expression is:

[tex]Ksp=[Pb^{2+}][I^-]^2[/tex]

Thus, since the saturated solution is 0.064g/100 mL at 20 °C we need to compute the molar solubility by using its molar mass (461.2 g/mol)

[tex]Molar solubility=\frac{0.064g}{100mL}*\frac{1000mL}{1L}*\frac{1mol}{461.2g}=1.39x10^{-3}M[/tex]

In such a way, since the mole ratio between lead (II) iodide to lead (II) and iodide ions is 1:1 and 1:2 respectively, the concentration of each ion turns out:

[tex][Pb^{2+}]=1.39x10^{-3}M[/tex]

[tex][I^-]=1.39x10^{-3}M*2=2.78x10^{-3}M[/tex]

Thereby, the solubility product results:

[tex]Ksp=(1.39x10^{-3}M)(2.78x10^{-3}M)^2\\\\Ksp=1.07x10^{-8}[/tex]

Regards.

Solubility product constant for the product of lead(II) iodide is [tex]\bold { 1.07x 10^-^8}[/tex].

The dissociation reaction for lead (II) iodide

[tex]\bold {Pb I^2 (s) \leftrightharpoons Pb^2^+ + 2I^- }[/tex]

Solubility product constant at equilibrium.

[tex]\bold {Ksp = [Pb^2^++[I^-]^2}[/tex]

The molar solubility of the substance can be calculated by using the molar mass,

[tex]\bold {s = \dfrac {0.064}{100 mL} \times 461.2 g/mol = 1.39x10^-^3}[/tex]

Molar ratio between between PbI to lead and iodide ions is 1:1 and 1:2 respectively.

Thus Ksp will be,

[tex]\bold {Ksp =(1.39x10^-^3)(2.78x10^-^3 )^2}\\\\\bold {Ksp = 1.07x 10^-^8}[/tex]

Therefore, solubility product constant for the product of lead(II) iodide is [tex]\bold { 1.07x 10^-^8}[/tex].

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Calculate the molar solubility of CaF2 at 25°C in a solution that is 0.010 M in Ca(NO3)2. The Ksp for CaF2 is 3.9 x 10-11.

Answers

Answer:

[tex]Molar \ solubility=3.12x10^{-5}M[/tex]

Explanation:

Hello,

In this case, for the dissociation of calcium fluoride:

[tex]CaF_2(s)\rightleftharpoons Ca^{2+}+2F^-[/tex]

The equilibrium expression is:

[tex]Ksp=[Ca^{2+}][F^-]^2[/tex]

In such a way, via the ICE procedure, including an initial concentration of calcium of 0.01 M (due to the calcium nitrate solution), the reaction extent [tex]x[/tex] is computed as follows:

[tex]3.9x10^{-11}=(0.01+x)(2*x)^2\\\\x=0.0000312M[/tex]

Thus, the molar solubility equals the reaction extent [tex]x[/tex], therefore:

[tex]Molar \ solubility=3.12x10^{-5}M[/tex]

Regards.

The molar solubility of Calcium fluoride has been calculated as [tex]3.12\;\times\;10^-^5\;\rm M[/tex].

The dissociation of calcium fluoride has been given by:

[tex]\rm CaF_2\;\rightarrow\;Ca^2^+\;+\;2\;F^-[/tex]

The solubility constant, ksp has been given as:

[tex]ksp=\rm[Mg^2^+]\;[F^-]^2[/tex]

From the dissociation of Calcium nitrate, the concentration of Ca ion in the solution has been 0.01 M.

The dissociation of Calcium fluoride x M has been resulted in x M Ca and 2x M F ions.

The concentration of Ca in the solution has been resulted as x + 0.01 M.

The solubility product can be given as:

[tex]3.9\;\times\;10^-^1^1=[x+0.01]\;[2x]^2\\3.9\;\times\;10^-^1^1=[x+0.01]\;4x^2\\x=3.12\;\times\;10^-^5[/tex]

The molar solubility of Calcium fluoride has been calculated as [tex]3.12\;\times\;10^-^5\;\rm M[/tex].

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Advantages of using a resource person in handling the first aid lesson

Answers

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:

A soft drink contains 63 g of sugar in 378 g of H2O. What is the concentration of sugar in the soft drink in mass percent

Answers

Answer:

[tex]\% m/m= 14.3\%[/tex]

Explanation:

Hello,

In this case, the by mass percent is computed as shown below:

[tex]\% m/m=\frac{m_{solute}}{m_{solute}+m_{solvent}} *100\%[/tex]

Whereas the solute is the sugar and the solvent the water, therefore, the concentration results:

[tex]\% m/m=\frac{63g}{63g+378g} *100\%\\\\\% m/m= 14.3\%[/tex]

Best regards.

Click on ALL of the following that use sound waves to communicate with their surroundings and find their way! bats cars dolphins whales birds submarines buses school

Answers

Bats, dolphins, whales, ( some ) birds and submarines all use echolocation ( or sound waves ) to navigate through their surroundings.!

Why will the conjugate base of a weak acid affect pH? Select the correct answer below: it will react with hydroxide

Answers

Answer:

It will react with water

Explanation:

I know this is an older question, but I just wanted to provide the correct answer.

Since we are dealing with a weak base, and the acid is somewhat stronger, it will react with the water molecules to produce hydronium, which will affect the pH of the solution.

The conjugate base of a weak acid affect pH because it will react with hydronium ion.

A weak acid is an acid that does not dissociate completely in water. On the other hand, a strong acid achieves almost 100% dissociation in water.

Acids dissociate in water to yield the hydronium ion and a conjugate base. For instance, the weak acid, acetic acid is dissociated as follows;

CH3COOH(aq) + H2O(l) ⇄ CH3COO-(aq) + H3O+(aq)

We can see that the conjugate base( CH3COO-) could react with the hydronium ions thereby moving the equilibrium position to the left hand side and affecting the pH by decreasing the hydronium ion concentration.

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Missing parts;

Why will the conjugate base of a weak acid affect pH? Select the correct answer below: O it will react with hydroxide O it will react with water O it will react with hydronium O none of the above

The direction of the functional group is called?

Answers

Explanation:

they are called hydrocarbyls

pls mark me brainliest

Answer:

The first carbon atom that attaches to the functional group is referred to as the alpha carbon.

What is advertising used for? Check all that apply. influencing consumer tastes tracking product popularity increasing product awareness promoting company branding gathering data about potential consumers

Answers

Answer:

influencing consumer tastes

increasing product awareness

promoting company branding

Explanation:

Advertising is basically a form of communication using creative ideas and communicating benefits of the products. Advertising plays a very crucial role in product business and some of the important uses of advertising are as follows:

Creative advertisements, influence customers or consumers to buy the product.Advertisings involve information regarding the product and so increases product awareness.Advertising on social media platforms, TVs, radio and newspapers, promotes company branding.

Hence, the correct options are:

influencing consumer tastesincreasing product awarenesspromoting company branding

Answer:

1,3,4

Explanation:

I took the test

I add a 50. g piece of Al (c = 0.88 J/g-deg) that is at 225°C to 100. mL of water at 20°C. What is the final temperature of the water in °C? The density of water is approximately 1g/mL.

Answers

Answer:

THE FINAL TEMPERATURE OF WATER IS -4.117 °C

Explanation:

Mass of the aluminium = 50 g

c = 0.88 J/g C

Initial temperature of aluminium = 225 °C

Volume of water = 100 ml

Density of water = 1 g/ml

Mass of water = density * volume of water

Mass of water = 1 * 100 = 100 g of water

Initial temperature of water = 20 C

It is worthy to note that the heat of a system is constant and conserved as no heat is lost or gained by a closed system,

So therefore,

heat lost by aluminium = heat gained by water

H = mass * specific heat capacity * temeprature change

So:

m c ( T2- T1) = m c (T2-T1)

50 * 0.88 * ( T2 - 225) = 100 * 4.18 *( T2 - 20)

44 ( T2 - 225 ) = 418 ( T2 - 20)

44 T2 - 9900 = 418 T2 - 8360

-9900 + 8360 = 418 T2 - 44 T2

-1540 = 374 T2

T2 = - 4.117

So therefore the final temperature of water is -4.117 °C

Which of the following types of electromagnetic radiation have higher frequencies than visible light and which have shorter frequencies than visible light? Sort them accordingly. ltems (6 items) (Drag and drop into the appropriate area below)
a. Gamma rays
b. Infrared radiation
c. Ultraviolet liht
d. X-rays
e. Microwaves
f. Radio waves

Answers

Answer:

Higher frequency than visible light -  Ultraviolet light, X-rays, and Gamma rays

Lower frequency than visible light - Infrared radiation, microwaves, and Radio waves

Explanation:

The frequencies of electromagnetic radiations vary according to their wavelengths. The relationship between the frequency and wavelength of the waves is expressed such that:

      λ = c/f, where λ = wavelength, c = speed of light, and f = frequency.

Thus, there is an inverse relationship between the wavelength and the frequency of electromagnetic waves.

The order of the electromagnetic waves based on their frequency from the lowest to the highest is radio waves, microwaves, infrared, visible light, ultraviolet, X-rays, and gamma-rays

Hence, electromagnetic waves with higher frequencies than visible light include ultraviolet light, X-rays, and Gamma rays while those with lower frequencies include Infrared radiation, microwaves, and Radio waves.

Answer:

need points

Explanation:

i) Briefly discuss the strengths and weaknesses of the four spectroscopy techniques listed below. Include in your answer the specific structural information you get from each method.
 IR
 UV-VIS
 NMR
 Mass Spec

Answers

delete please .....................................

Increasing which factor will cause the gravitational force between two objects to decrease?
weights of the objects
distance between the objects
acceleration of the objects
masses of the objects

Answers

Answer:

B

Explanation:

Increasing distance between the objects factor will cause the gravitational force between two objects to decrease. Therefore, option B is correct.

What causes gravitational force to decrease?

The gravitational force grows in proportion to the size of the masses . The gravitational force weakens rapidly as the distance between masses grows. Unless at least one of the objects has a lot of mass, detecting gravitational force is extremely difficult.

Gravity is affected by object size and distance between objects. Mass is a unit of measurement for the amount of matter in an object.

The force of gravity is proportional to the masses of the two objects and inversely proportional to the square of the distance between them. This means that the force of gravity increases with mass but decreases as the distance between objects increases.

Thus, option B is correct.

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Compound has a molar mass of and the following composition: elementmass % carbon47.09% hydrogen6.59% chlorine46.33% Write the molecular formula of .

Answers

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]

Candle wax melts low temperature, it is not conductive to electricity, it is insoluble in water and partially soluble in solvents nonpolar, like gasoline. Than type of links are present in the candle wax?

A. Electrostatics.
B. Apolar.
C. lónicos.
D. Hydrogen bridges.

Answers

electrostatic and ionic are definitely not the answer because they have high melting point

hydrogen bonds are too weak and not permanent.

so the answer is apolar as it is soluble in polar solvents (water)

Answer:

B. Nonpolar

Explanation:

The low melting point tells you the compound is not ionic, metallic, or a network solid.

It is almost certainly a molecular solid.

It does not conduct electricity, so it is not metallic (which we have already ruled out).

It is insoluble in polar solvents (water) and soluble in nonpolar solvents (gasoline).

Since like dissolves like, the molecule is nonpolar.

The type of links must be nonpolar.

Chlorine monoxide and dichlorine dioxide are involved in the catalytic destruction of stratospheric ozone. They are related by the equation:

2ClO(g) ⇌ Cl2O2(g) for which Kc is 4.96×10^11 at 273 K.

For an equilibrium mixture in which [Cl2O2] is 6.00 x 10^-6M, what is [ClO]?

Answers

Answer:

[ClO] = 3.48×10¯⁹ M.

Explanation:

The following data were obtained from the question:

Equilibrium constant (Kc) = 4.96×10¹¹

Concentration of Cl2O2, [Cl2O2] = 6x10¯⁶ M.

Concentration of ClO, [ClO] =.?

The equation for the reaction is given below:

2ClO(g) ⇌ Cl2O2(g)

The equilibrium constant for a reaction is simply defined as the ratio of the concentration of product raised to their coefficient to the concentration of the reactant raised to their coefficient.

The equilibrium constant, Kc for the reaction is given by:

Kc = [Cl2O2] / [ClO]²

Thus, we can calculate the concentration of ClO, [ClO] as follow:

Kc = [Cl2O2] / [ClO]²

4.96×10¹¹ = 6x10¯⁶ / [ClO]²

Cross multiply

4.96×10¹¹ × [ClO]² = 6x10¯⁶

Divide both side by 4.96×10¹¹

[ClO]² = 6x10¯⁶ / 4.96×10¹¹

[ClO]² = 1.21×10¯¹⁷

Take the square root of both side

[ClO] = √ (1.21×10¯¹⁷)

[ClO] = 3.48×10¯⁹ M

Therefore, the concentration of ClO, [ClO] is 3.48×10¯⁹ M.

Identify a homogeneous catalyst. Identify a homogeneous catalyst. Pd in H2 gas N2 and H2 catalyzed by Fe SO2 over vanadium (V) oxide Pt with methane H2SO4 with concentrated HCl

Answers

Answer:

H2SO4 with concentrated HCl

Consider the following reaction at 298.15 K: Co(s)+Fe2+(aq,1.47 M)⟶Co2+(aq,0.33 M)+Fe(s) If the standard reduction potential for cobalt(II) is −0.28 V and the standard reduction potential for iron(II) is −0.447 V, what is the cell potential in volts for this cell? Report your answer with two significant figures.

Answers

Answer:

The correct answer is 0.186 V

Explanation:

The two hemirreactions are:

Reduction: Fe²⁺ + 2 e- → Fe(s)  

Oxidation : Co(s)  → Co²⁺ + 2 e-

Thus, we calculate the standard cell potential (Eº) from the difference between the reduction potentials of cobalt and iron, respectively,  as follows:

Eº = Eº(Fe²⁺/Fe(s)) - Eº(Co²⁺/Co(s)) = -0.28 V - (-0.447 V) = 0.167 V

Then, we use the Nernst equation to calculate the cell potential (E) at 298.15 K:

E= Eº - (0.0592 V/n) x log Q

Where:

n: number of electrons that are transferred in the reaction. In this case, n= 2.

Q: ratio between the concentrations of products over reactants, calculated as follows:

[tex]Q = \frac{ [Co^{2+} ]}{[Fe^{2+} ]} = \frac{0.33 M}{1.47 M} = 0.2244[/tex]

Finally, we introduce Eº= 0.167 V, n= 2, Q=0.2244, to obtain E:

E= 0.167 V - (0.0592 V/2) x log (0.2244) = 0.186 V

A sample of ammonia gas was allowed to come to equilibrium at 400 K. 2NH3(g) <----> N2(g) 3H2(g) At equilibrium, it was found that the concentration of H2 was 0.0484 M, the concentration of N2 was 0.0161 M, and the concentration of NH3 was 0.295 M. What was the initial concentration of ammonia

Answers

Answer:

0.327 M

Explanation:

Step 1: Write the balanced equation

2 NH₃(g) ⇄ N₂(g) + 3H₂(g)

Step 2: Make an ICE chart

        2 NH₃(g) ⇄ N₂(g) + 3 H₂(g)

I              x             0            0

C          -2y            +y         +3y

E         x-2y             y           3y

Step 3: Find the value of y

The concentration of N₂ at equilibrium is 0.0161 M. Then,

y = 0.0161

Step 4: Find the value of x

The concentration of NH₃ at equilibrium is 0.295 M. Then,

x-2y = 0.295

x-2(0.0161) = 0.295

x = 0.327

From the graph of Density vs. Concentration, created in Graph 1, what was the relationship between the concentration of the sugar solution and the density of the sugar solution?

Answers

The graph is not given in the question, so, the required graph is attached below:

Answer:

According to the graph, the relationship between the density of the sugar solution and the concentration of the sugar solution is directly proportional to each other as they both are increasing exponentially.

The graph shows that, the density of sugar solution will increase with the increase in concentration of sugar in the solution.

105/22 • (1.251 - 0.620)=

Answers

Answer:

105/22*(1.251-0.620)

105/22*0.631

4.772*0.631

3.011132

Hope it helps

Answer:

3.0

Explanation:

First, complete the operations inside the parenthesis according to the normal rules for significant figures. Because there are subsequent calculations, keep at least one extra significant figure when possible: (4.7727) × (0.631).

The final product will be rounded to two significant figures because it can’t be more precise than the least precise number in the problem, 22. The final product is 3.0.

If an individual proton has mass 1.007825 amu, and an individual neutron has mass 1.008665 amu, what's the calculated mass of a neptunium-236 nucleus? options: A) 237.92482 amu B) 236.99873 amu C) 237.96682 amu D) 237.04817 amu

Answers

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.

g The electronic structure of which ONE of the following species cannot be adequately described by a single Lewis formula? (In other words, the electronic structure of which one can only be described by drawing two or more resonance structures?) A) C2H4 B) SO3 2– C) SO3 D) C3H8 E) HCN

Answers

Answer:

C) SO3

Explanation:

Lewis formula shows the bonding between atoms of a molecule and expresses the lone pair present in the atoms.

SO3 or Sulfur trioxide cannot be adequately described by a single Lewis formula because it has majorly 3 resonance structures because Sulfur does not follow the octet rule and can expand electrons in its outer shell.

Hence, the correct answer is C) SO3

What is the value of the equilibrium constant, K, for a reaction for which ∆G° is equal to –5.20 kJ at 50°C?

Answers

Answer:

6.93

Explanation:

Step 1: Given data

Standard Gibbs free energy (∆G°): -5.20 kJTemperature (T): 50°CEquilibrium constant (K): ?

Step 2: Convert the temperature to the Kelvin scale

We will use the following expression.

K = °C + 273.15

K = 50°C + 273.15

K = 323 K

Step 3: Calculate K

We will use the following expression.

∆G° = -R × T × ln K

-5.20 × 10³ J = -(8.314 J/mol.K) × 323 K × ln K

K = 6.93

If 11.2 g of naphthalene, C10H8, is dissolved in 107.8 g of chloroform, CHCl3, what is the molality of the solution

Answers

Answer:

CHC12

Explanation:

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

How many molecules are there in 3.5 moles of carbon dioxide? A. 63.21 x 10^23 B. 21.07 x 10^23 C. 42.14 x 10^23 D. 6.02 x 10^23

Answers

Answer:

B. 21.07 x 10²³ molecules

Explanation:

Avogadro's Number: 6.022 x 10²³

Step 1: Set up equation

[tex]3.5 mols CO_2(\frac{6.022(10^{23}) moleculesCO_2}{1 mol CO_2})[/tex]

Step 2: Multiply and cancel out units

3.5(6.022 x 10²³) = 21.07 x 10²³ molecules CO₂

Step 3: Convert to proper scientific notation

≈ 2.11 x 10²³ molecules CO₂

The complex ion Fe(CN)63- is paramagnetic with one unpaired electron. The complex ion Fe(SCN)63- has five unpaired electrons. Where does SCN- lie in the spectrochemical series with respect to CN-?

Answers

Answer:

SCN- is a weak field ligand while CN- is a strong field ligand

Explanation:

The spectrochemical series is an arrangement of ligands according to their magnitude of crystal field splitting. Ligands that cause only a small degree of crystal field splitting are called weak field ligands while ligands that cause large crystal field splitting are called strong field ligands.

Strong field ligands often lead to the formation of low spin complexes with the least number of unpaired electrons while high spin complexes are formed by weak field ligands.

CN- is a strong field ligand as it lies towards the right hand side of the spectrochemical series.

SCN- is a weak field ligand hence it forms a high spin complex having the maximum number of unpaired electrons for Fe^3+, hence the answer.

SCN⁻ lies in the lower (weak field) region of the spectrochemical series while CN⁻ lies in the higher (stronger field) region.

CN⁻ is a strong field ligand with a large splitting constant, and it is high up in the spectrochemical series.

Conversely, SCN⁻ is a weak field ligand with a low splitting constant, and it is lower in the spectrochemical series.

Hence, SCN⁻ lies in the lower (weak field) region of the spectrochemical series while CN⁻ lies in the higher (stronger field) region.

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How many grams of sodium chloride are required to make 2.00 L of a solution with a concentration of 0.100 M?

Answers

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

We discussed the different types of intermolecular forces in this lesson, which can affect the boiling point of a substance.
1. Which of these has the highest boiling point?
A) Ar
B) Kr
C) Xe
D) Ne
2. Which substance has the highest boiling point?
A) CH4
B) He
C) HF
D) Cl2

Answers

Answer:

1, C, Xe 2, B,He

Explanation:

1, cause as u go down a group the boiling point increases.

2, boiling point of single element is greater than a compound

According to  periodic trends in periodic table boiling point increases down the  group and hence Xe has highest boiling point and more amount of heat is required to boil an element hence He has highest boiling point.

What is periodic table?

Periodic table is a tabular arrangement of elements in the form of a table. In the periodic table, elements are arranged according to the modern periodic law which states that the properties of elements are a periodic function of their atomic numbers.

It is called as periodic because properties repeat after regular intervals of atomic numbers . It is a tabular arrangement consisting of seven horizontal rows called periods and eighteen vertical columns called groups.

Elements present in the same group have same number of valence electrons and hence have similar properties while elements present in the same period show gradual variation in properties due to addition of one electron for each successive element in a period.

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Given that π = n M R T, rearrange the equation to solve for V

Answers

Answer:

V= n/M

Explanation:

From;

π = nRT/V = MRT

Where;

n= number of moles

R= gas constant

T= absolute temperature

M= molar mass

V= volume of the solution

π= osmotic pressure

Thus;

nRT/V = MRT

nRT = VMRT

V= nRT/MRT

V= n/M

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

Answers

Answer:

8.68

Explanation:

pOH = 8.68

all you need is contained in the sheet

Answer:

Approximately [tex]8.68[/tex].

Explanation:

The [tex]\rm pOH[/tex] of a solution can be found from the hydroxide ion concentration [tex]\rm \left[OH^{-}\right][/tex] with the following equation:

[tex]\displaystyle \rm pOH = -\log_{10} \rm \left[OH^{-}\right][/tex].

On the other hand, the ion-product constant of water, [tex]K_{\text{w}}[/tex], relates the hydroxide ion concentration [tex]\rm \left[OH^{-}\right][/tex] of a solution to its hydronium ion concentration [tex]\rm \left[{H_3O}^{+}\right][/tex]:

[tex]K_\text{w} = \rm \left[{H_3O}^{+}\right] \cdot \rm \left[OH^{-}\right][/tex].

At [tex]25 \; ^\circ \rm C[/tex], [tex]K_{\text{w}} \approx 1.0 \times 10^{-14}[/tex]. For this particular [tex]25 \; ^\circ \rm C[/tex] solution, [tex]\rm \left[{H_3O}^{+}\right] = 4.8 \times 10^{-6}\; \rm mol \cdot L^{-1}[/tex].

Hence the [tex]\rm \left[OH^{-}\right][/tex] of this solution:

[tex]\begin{aligned}\left[\mathrm{OH}^{-}\right] &= \frac{K_\text{w}}{\rm \left[{H_3O}^{+}\right]} \\ &= \frac{1.0 \times 10^{-14}}{4.8 \times 10^{-6}}\; \rm mol\cdot L^{-1} \approx 2.08333 \times 10^{-9}\; \rm mol\cdot L^{-1}\end{aligned}[/tex].

Therefore, the [tex]\rm pOH[/tex] of this solution would be:

[tex]\begin{aligned}\rm pOH &= -\log_{10} \rm \left[OH^{-}\right] \\ &\approx -\log_{10} \left(4.8 \times 10^{-6}\right) \approx 8.68\end{aligned}[/tex].

Note that by convention, the number of decimal places in [tex]\rm pOH[/tex] should be the same as the number of significant figures in [tex]\rm \left[OH^{-}\right][/tex].

For example, because the [tex]\rm \left[{H_3O}^{+}\right][/tex] from the question has two significant figures, the [tex]\rm \left[OH^{-}\right][/tex] here also has two significant figures. As a result, the [tex]\rm pOH[/tex] in the result should have two decimal places.

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