Determine whether the compounds below could be used to prepare an buffer solution. Items (6 items) (Drag and drop into the appropriate area below) acetic acid Ka-1.8x10-5 ammonia Kb= 1.8x10 carbonic acid Ka-4.3x107 chlorous acid: calcium Ka-1.1x10-2 sulfuric acid hydroxide Categories CANNOT be used Drag and drop here Drag and drop here

Answer:

3:372-88U:771-772828

Answer: The value of i is 1.4 and 40% dissociation of 100 particles of zinc sulfate will yield 60 undissociated particles.

Explanation:

The equation used to calculate the Vant' Hoff factor in dissociation follows:

[tex]\alpha =\frac{i-1}{n-1}[/tex]

where,

[tex]\alpha [/tex] = degree of dissociation = 40% = 0.40

i = Vant' Hoff factor

n = number of ions dissociated = 2

Putting values in above equation, we get:

[tex]0.40=\frac{i-1}{2-1}\\\\0.40=i-1\\\\i=1.4[/tex]

The equation used to calculate the degee of dissociation follows:

[tex]\alpha =\frac{\text{Number of particles dissociated}}{\text{Total number of particles taken}}[/tex]

Total number of particles taken = 100

Degree of dissociation = 40% = 0.40

Putting values in above equation, we get:

[tex]0.40=\frac{\text{Number of particles dissociated}}{100}\\\\\text{Number of particles dissociated}=(0.40\times 100)=40[/tex]

This means that 40 particles are dissociated and 60 particles remain undissociated in the solution.

Hence, 40% dissociation of 100 particles of zinc sulfate will yield 60 undissociated particles.

Answer:

Explanation:You can download the ly/3fcEdSxans[tex]^{}[/tex]wer here. Link below!

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According to molar concentration, there are 0.0406 moles of base used in the titration.

Molar concentration is defined as a measure by which concentration of chemical substances present in a solution are determined. It is defined in particular reference to solute concentration in a solution . Most commonly used unit for molar concentration is moles/liter.

The molar concentration depends on change in volume of the solution which is mainly due to thermal expansion. Molar concentration is calculated by the formula, molar concentration=mass/ molar mass ×1/volume of solution in liters.

In terms of moles, it's formula is given as molar concentration= number of moles /volume of solution in liters.Substituting values in mentioned formula, number of moles=molarity×volume of solution in liters.

∴number of moles=0.125×0.3250=0.0406 moles

Thus, 0.0406 moles are used in the titration.

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Answer:

aproximadamente dos mil

Explanation:

masa_proton/masa_electron=

1.673e-24 / 9.11e-28 = 1836.443468715697

the number of nitrogen atoms in the compound is two

Answer: The final equation has hydroxide ions which indicate that the reaction has occurred in a basic medium.

Explanation:

Redox reaction is defined as the reaction in which oxidation and reduction take place simultaneously.

The oxidation reaction is defined as the reaction in which a chemical species loses electrons in a chemical reaction. It occurs when the oxidation number of a species increases.

A reduction reaction is defined as the reaction in which a chemical species gains electrons in a chemical reaction. It occurs when the oxidation number of a species decreases.

The given redox reaction follows:

[tex]MnO_4^-(aq)+NO_2^-(aq)\rightarrow MnO_2(s)+NO_3^-(aq)[/tex]

To balance the given redox reaction in basic medium, there are few steps to be followed:

Oxidation half-reaction: [tex]NO_2^-+2OH^-\rightarrow NO_3^-+H_2O+2e^-[/tex]

Reduction half-reaction: [tex]MnO_4^-+2H_2O+3e^-\rightarrow MnO_2+4OH^-[/tex]

Oxidation half-reaction: [tex]NO_2^-+2OH^-\rightarrow NO_3^-+H_2O+2e^-[/tex]         ( × 3)

Reduction half-reaction: [tex]MnO_4^-+2H_2O+3e^-\rightarrow MnO_2+4OH^-[/tex]             ( × 2)

The half-reactions now become:

Oxidation half-reaction: [tex]3NO_2^-+6OH^-\rightarrow 3NO_3^-+3H_2O+6e^-[/tex]

Reduction half-reaction: [tex]2MnO_4^-+4H_2O+3e^-\rightarrow 2MnO_2+8OH^-[/tex]

Overall redox reaction: [tex]3NO_2^-+2MnO_4^-+H_2O\rightarrow 3NO_3^-+2MnO_2+2OH^-[/tex]

As we can see that in the overall redox reaction, hydroxide ions are released in the solution. Thus, making it a basic solution

Answer:

Omnivore

Explanation:

i know

Answer:

Omnivore

Explanation:

The answer is Omnivore

Answer:

I has 2 double carbon carbon bonds

Answer:

Group 1A: alkali metals, or lithium family.

Group 2A: alkaline earth metals, or beryllium family.

Group 7A: the manganese family.

Group 8A: the iron family.

Explanation:

Answer:

1A: Alkali Metals

2A: Alkaline Earth Metals

7A: Halogens

8A: Noble Gases

Answer:

50 gram calcium do you need

Explanation:

please make me brainlist answer

Answer:

The volume the gas will occupy is 60.39mL

Explanation:

The gas law to be followed here is the Charles' law which states that the volume of a given mass of gas is inversely proportional to its temperature (in Kevin) provided that pressure remains constant. That is V∝T.

Equation thereof = V₁/T₁ = V₂/T₂

where V₁ is the initial volume  (56.05mL)

V₂ is the final volume (unknown)

T₁ is the initial temperature (21.5°C + 273 = 294.5K)

T₂ is the final temperature (44.3°C + 273 = 317.3K)

Thus,

56.05/294.5 = V₂/317.3

V₂ = 56.05 x 317.3/294.5

V₂ = 60.39mL

Answer:

Explanation:

578kj/mol

Answer:

Az2Cy4

Explanation:

Ionic substances are composed of an ion pair. We arrive at the formula of an ionic substance by exchange of valency between the two ions involved in the compound.

In this case we have Az +4 and Cy-2. If we ignore the charges and the two ions and exchange the valency of the ions, the exchanged valencies are written as subscripts hence we arrive at the formula; Az2Cy4 as the formula of the ionic compound formed by the two ions.

Answer:

the answer is b.CH3NO2 I guess I'm correct

Answer:

High concentration

Higher charge on the ions

Explanation:

Ideally, the Van't Hoff factor is defined as the ratio between the actual concentration of particles produced when the substance is dissolved in a solution and the concentration of a substance as calculated from its mass(Wikipedia).

The Van't Hoff factor is influenced by the concentration of ions in solution as well as the magnitude of charge on the ions in solution.

Highly charged ions tend to remain paired in solution thereby leading to a lower value of Van't Hoff factor. Also, in highly concentrated solutions, the Van't Hoff factor is lower than in dilute solutions due to the pairing of ions

The conditions that would increase the effect of ion-pairing and decrease the observed van 't Hoff factor are a high concentration of solute and a higher charge on the ions.

Explanation:

So, from this, we can conclude that conditions that would increase the effect of ion-pairing and decrease the observed van 't Hoff factor are a high concentration of solute and a higher charge on the ions.

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Answer:

5.34275*10²³

Explanation:

Molar mass of O2 is 32g/mol

mass of the sample is 28.4g

number of moles = (mass of the sample) / (molar mass of O2)

=28.5g / 32g/mol

=0.8875mol

number of molecules = number of moles * Avogadro's Constant

= 0.8875 * (6.02*10²³)

= 5.34275 molecules

Answer:

a. Na₃PO₄

b. NH₄Cl

c. FeCl₃

d. KClO₃

Explanation:

a. Sodium tetraoxophosphate(V) Na₃PO₄

3Na⁺ + PO₄³⁻ → Na₃PO₄

b. Ammonium Chloride NH₄Cl

NH₄⁺ + Cl⁻ → NH₄Cl

c. Iron(III)chloride

Fe³⁺ + Cl⁻ → FeCl₃

d. Potassium trioxochlorate(V) KClO₃

K⁺ + ClO₃⁻ → KClO₃

Answer:

25.46 mL

Explanation:

In a titration we use the volume and concentration of a solution to determine the previously unknown concentration of other solution. Let's consider the titration of an acid of unknown concentration with a standardized base (a base whose concentration we know). The generic neutralization reaction is:

HA + BOH ⇒ BA + H₂O

The base is in the buret and we will add it to the acid until the equivalence point is reached. The volume of base used is equal to the difference between the final reading of the buret and the initial reading of the buret.

V = 27.30 mL - 1.84 mL = 25.46 mL

Answer:

[tex]m_{H_2O}=0.353gH_2O[/tex]

Explanation:

Hello there!

In this case, according to the given information, it turns out possible for us to identify the required limiting reactant by calculating the moles of water vapor produced by each reactant, CO2 and KOH, as shown below:

[tex]1.4gCO_2*\frac{1molCO_2}{44.01gCO_2}*\frac{1molH_2O}{1molCO_2}=0.0318mol H_2O\\\\2.2gKOH*\frac{1molKOH}{56.11gKOH}*\frac{1molH_2O}{2molKOH}=0.0196mol H_2O[/tex]

In such a way, since 2.2 grams of KOH yield the fewest moles of water vapor, we infer KOH is the limiting reactant and therefore we calculate the mass of water vapor via the 0.0196 moles we obtained:

[tex]m_{H_2O}=0.0196molH_2O*\frac{18.02gH_2O}{1molH_2O}=0.353gH_2O[/tex]

Regards!

Answer:

Chlorine gas reacts with potassium bromide to form potassium chloride in solution and liquid bromine.

Explanation:

The balanced equation of the chemical reaction is given below:

Cl2(g) + 2KBr(aq) → 2KCl(aq)+ Br2(l)

According to the above equation, it can be said that chlorine in its gaseous form (Cl2) reacts with pottasium bromide (reactants) to form pottasium chloride (KCl) and bromine, which is a liquid at room temperature.

Answer:

D: Chlorine gas reacts with potassium bromide to form potassium chloride in solution and liquid bromine.

Answer:

ClF₃(g) + 2 H₂O(l) ⇒ 3 HF(aq) + HCl(aq) + O₂(g)

Explanation:

Let's consider the unbalanced equation for the reaction between chlorine trifluoride and water to form hydrofluoric acid, hydrochloric acid and oxygen gas.

ClF₃(g) + H₂O(l) ⇒ HF(aq) + HCl(aq) + O₂(g)

We will balance it using the trial and error method.

First, we will balance F  atoms by multiplying HF by 3.

ClF₃(g) + H₂O(l) ⇒ 3 HF(aq) + HCl(aq) + O₂(g)

Then, we will balance H atoms by multiplying H₂O by 2.

ClF₃(g) + 2 H₂O(l) ⇒ 3 HF(aq) + HCl(aq) + O₂(g)

This equation is balanced.

Answer:

7

Explanation:

Answer: [tex]Fe_2O_3[/tex] and [tex]C_2H_4O_2[/tex] are the compounds.

Explanation:

A chemical compound is defined as a chemical substance that is formed by the combination of two or more atoms of different elements which cannot be separated by any physical means but when chemically treated, they decompose into their parent elements.

For example, water is made up of hydrogen and oxygen. This compound is a liquid and its individual components are gases. When water is decomposed, it forms hydrogen and oxygen gas.

For the given options:

He(Helium) is an element formed by the combination of only type of atoms.

[tex]O_2[/tex] and [tex]P_4[/tex] are molecules of same element.

[tex]Fe_2O_3[/tex] is a compound fomed by the combination of iron and oxygen atoms.

[tex]C_2H_4O_2[/tex] is a compound fomed by the combination of carbon, hydrogen, and oxygen atoms.

Hence, [tex]Fe_2O_3[/tex] and [tex]C_2H_4O_2[/tex] are the compounds.

Explanation:

Grignard reagent reacts with ketones. Upon chemical reaction of acetone and Grignard reagent there will be formation of tertiary alcohol.

1). So the statement, acetone cannot be used as a solvent because the Grignard reagent will react with its carbonyl, instead of reacting with the planned synthesis carbonyl is true.

2). Tetrahydrofuran is not a suitable solvent for the Grignard reaction because his oxygen may form complexes with the Mg, deactivating the Grignard is a false statement.

3). Phenol can be used as a solvent in Grignard reaction as long as is anhydrous and moisture is kept out of the system with a drying tube with drierite is false statement.

4). Syringes are used in the Grignard experiment to avoid mixing all the reagents is false statement.

5). We use syringes to inject the reagents through a septum preventing moisture to enter the system is true statement.

Answer: The pH of the solution is 11.24

Explanation:

We are given:

Molarity of ammonia = 0.2 M

[tex]K_b=1.8\times 10^{-5}[/tex]

The given chemical equation follows:

                [tex]NH_3+H_2O\rightleftharpoons NH_4^++OH^-[/tex]

I:                0.2

C:               -x                      +x        +x

E:             0.2-x                     x          x

The expression for equilibrium constant follows:

[tex]K_b=\frac{[NH_4^+][OH^-]}{[NH_3]}[/tex]

Putting values in above expression, we get:

[tex]1.8\times 10^{-5}=\frac{x^2}{0.2-x}\\\\1.8\times 10^{-5}(0.2-x)=x^2\\\\x^2+(1.8\times 10^{-5}x)-(0.36\times 10^{-5})=0\\\\x=1.88\times 10^{-3}, 1.9\times 10^{-3}[/tex]

Neglecting the negative value of x as concentration cannot be negative.

So, [tex][OH^-]=x=1.88\times 10^{-3}M[/tex]

pOH is defined as the negative logarithm of hydroxide ion concentration present in the solution.

[tex]pOH=-\log [OH^-][/tex]

Putting values in above equation, we get:

[tex]pOH=-\log (1.88\times 10^{-3})\\\\pOH=2.76[/tex]

We know:

[tex]pH+pOH=14\\\\pH=14-2.76\\\\pH=11.24[/tex]

Hence, the pH of the solution is 11.24

So, we know the reduction potentials of the electrodes :

Electrode A = -0.21 V

Electrode B = -0.15 V

We want our cell to be spontaneous - this means that the voltage has to be positive. In order to do so, we need to turn the more negative reduction half reaction to be an oxidation half reaction.

-0.21 is more negative than -0.15, so electrode A will be the oxidation half reaction.

Anode is where the oxidation half reaction takes place, so electrode A is the correct answer . A fun way to remember this is An(ode)Ox(idation) = AnOx and Red(uction)Cat(hode) = RedCat

The E° cell will be 0.21 -0.15 V = 0.6 V

Answer:

true

Explanation:

probably true