Calculate the empirical formula for each of the following substances. (Express answer as a chemical formula) 1) 2.90 g of Ag and 0.125 g of N 2) 2.22 g of Na and 1.55 g of S 3) 2.11 g of Na, 0.0900 g of H, 2.94 g of S, and 5.86 g of O 4) 1.84 g of K, 0.657 g of N, and 2.25 g of O

Answers

Answer 1

Answer:

1) Ag3N

2)Na2S

3)NaHSO4

4) KNO3

Explanation:

We divide each mass by the element's relative atomic mass

1) 2.90/108-Ag, 0.125/14-N

0.027-Ag, 0.0089-N

Divide by the lowest ratio

0.027/0.0089-Ag, 0.0089/0.0089 N

3-Ag, 1-N

Empirical formula- Ag3N

2)2.22/23-Na, 1.55/32-S

0.097-Na, 0.048-S

Divide by the lowest ratio

0.097/0.048-Na, 0.048/0.048-S

2-Na, 1-S

Empirical formula- Na2S

3) 2.11/23-Na, 0.0900/1-H, 2.94/32-S,5.86/16-O

0.09-Na, 0.09-H, 0.09-S,0.366-O

Divide by the lowest ratio

0.09/0.09-Na, 0.09/0.09-H, 0.09/0.09-S, 0.366/0.09-O

1-Na, 1-H, 1-S, 4-O

Empirical formula- NaHSO4

4)1.84/39, 0.657/14-N, 2.25/16-O

0.047-K, 0.047-N, 0.14-O

Divide through by the lowest ratio

0.047/0.047-K, 0.047/0.047-N, 0.14/0.047-O

1-K, 1-N, O-3

Empirical formula- KNO3


Related Questions

An unknown element, X, reacts with oxygen to form two oxides with the general formula X_2O_3 and X_2O_5. The pure element is strongly paramagnetic, and the oxides are amphoteric. Which of the following elements could be represented by X?a) Scb) Bc) Nd) Sbe) S

Answers

Answer:

The unknown element is Sb

Explanation:

The first thing we must note is that the unknown element must be a member of group 15 in the periodic table. This is clear from the fact that the two oxides formed are X2O3 and X2O5. This implies that the unknown element X must have a valency of 3 or 5. This corresponds to our knowledge that the outermost electron configuration of group 15 elements is ns2np3. Hence, group fifteen elements can have a valency of 3 or 5.

The electronic configuration of antimony is; [Kr]4d10 5s2 5p3. This implies that the atom is paramagnetic since there are three unpaired 5p electrons. The oxides of antimony are known to be amphoteric. An ampohoteric oxide reacts with both acid and base, hence the answer.

Starting with 0.250L of a buffer solution containing 0.250 M benzoic acid (C 6H 5COOH) and 0.20 M sodium benzoate (C 6H 5COONa), what will the pH of the solution be after the addition of 25.0 mL of 0.100M HCl? (K a (C 6H 5COOH) = 6.5 x 10 -5)

Answers

Answer:

pH = 4.05

Explanation:

The pH of the benzoic buffer can be determined using H-H equation:

pH = pKa + log [A⁻] / [HA]

Where pKa is -logKa = 4.187

pH = 4.187 + log [Sodium Benzoate] / [Benzoic Acid]

Where [] can be understood as moles of each specie.

Thus, to find pH of the buffer we need to calculate moles of benzoic acid and sodium benzoate.

Initial moles:

Initial moles of benzoic acid and sodium benzoate are:

Acid: 250mL = 0.250L ₓ (0.250 moles / L) = 0.0625 moles of benzoic acid

Benzoate : 250mL = 0.20L ₓ (0.250 moles / L) = 0.050 moles of sodium benzoate

Moles after reaction:

Now, 0.0250L×(0.100mol/L) = 0.0025 moles of HCl are added to the buffer reacting with sodium benzoate, C₆H₅COONa, producing more benzoic acid, as follows:

HCl + C₆H₅COONa → C₆H₅COOH + NaCl

That means after reaction moles of both species are:

Benzoic acid: 0.0625 mol + 0.0025mol (Moles produced) = 0.065 moles

Sodium Benzoate: 0.050mol - 0.0025mol (Moles that react) = 0.0475 moles

Replacing in H-H equation:

pH = 4.187 + log [0.0475] / [0.065]

pH = 4.05

tank contains helium gas at 490 mm Hg, nitrogen gas at 0.75 atm and neon at 520 torr. What is the total pressure in atm? 2.1 atm 0.55 atm 1.5 atm 5.1 atm 51 atm

Answers

Answer:

2.1 atm

Explanation:

Before we get the total pressure, we have to ensure all the gases have the same pressure unit.

Nitrogen gas = 0.75 atm

Helium = 490mmHg

To convert mmHg to atm;

760 mmHg = 1 atm

490 = x

x = 460 / 760 = 0.645 atm

Neon = 520 torr

Converting torr to atm;

760 torr = 1 atm

520 torr = x

x = 520 / 760 = 0.684 atm

The total pressure is then given as;

0.75 + 0.684 + 0.645 = 2.1 atm

Which of the following best describes a salt bridge? a) A pathway composed of salt water that ions pass through. b) A pathway between the cathode and anode in which ions are reduced. c) A pathway by which counterions can flow between the half-cells with the solutions in the half-cell completely mixing.

Answers

Answer: A

Explanation:

"Calculate the pH during the titration of 20.00 mL of 0.1000 M HF(aq) with 0.2000 M NaOH(aq) after 9.4 mL of the base have been added. Ka of hydrofluoric acid

Answers

Answer:

The answer is " 10.39"

Explanation:

Calculating acid moles:

[tex]= 0.02000 \ L \times 0.1000 \ M \\\\= 0.002000[/tex]

Calculating NaOH moles:

[tex]= 0.02012 \ L \times 0.1000 \ M \\\\= 0.002012[/tex]

calculating excess in OH-  Moles:

[tex]= 0.002012 - 0.002000\\\\=0.000012[/tex]

calculating total volume:

[tex]= 20.00 + 20.12\\\\ = 40.12 mL \\\\= 0.04012 L[/tex]

[tex][OH-]= \frac{0.000012} { 0.0472}[/tex]

           [tex]=0.00025 M[/tex]

[tex]pOH = - \log 0.00025[/tex]

        = 3.6

[tex]pH = 14 - pOH[/tex]

      = 10.39

Calculate the free energy of formation of NaBr(s) given the following information: NaBr(s) → Na(s) + 1/2Br2(l), ∆G° = 349 kJ/mol

Answers

The given question is incomplete, the complete question is:

Calculate the free energy of formation of NaBr(s) given the following information: NaBr(s) → Na(s) + 1/2Br2(l), ΔG° = 349 kJ/mol

A) –309 kJ/mol

B) –329 kJ/mol

C) None of the above

D) –349 kJ/mol

E) –369 kJ/mol

Answer:

The correct answer is option D, that is, -349 kJ/mol.

Explanation:

Based on the given information, the reaction is:  

NaBr (s) ⇔ Na (s) + 1/2 Br₂ (l), the ΔG° of the reaction given is 349 kJ per mole. In the given question, it is clearly mentioned that there is a need to determine the free energy of the formation of NaBr. Thus, there is a need to keep Na (s) and Br₂ (l) at the reactant side and NaBr (s) at the product side.  

Therefore, there is a need to reverse the reaction and change the sign on ΔG.  

Now the reaction will become,  

Na (s) + 1/2 Br₂ (l) ⇔ NaBr (s), and the ΔG° will now become -349 kJ per mole. Hence, -349 kJ per mole is the free energy of the formation of NaBr (s).  

Identify the elements that correspond to the following generalized electron configuration: (noble gas]ns2(n − 2) f6
Express your answer as the element symbol. If there is more than one answer, separate them by a comma.

Answers

Answer:

Samarium:

Electron configuration:

Samarium

Explanation:

Samarium is a chemical element that belongs to the lanthanoid series. The lanthanoids are the chemical elements that follow lanthanum. They are all known to possess 4f orbitals. The 4f electrons are found in the antepenultimate shell of the elements of the lanthanoid series and they do not take part in chemical bonding. They are neither removed in bonding nor do they take part in crystal field stabilization of lanthanoid complexes.

The electronic configuration of samarium is; 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10 5p6 6s2 4f6 while the condensed, short hand electronic configuration is [Xe] 4f6 6s2. This corresponds to (noble gas]ns2(n − 2) f6 as required by the question, hence the answer provided above.

Which statement best describes covalent bonding?

Answers

Answer:

Option C. Electrons are shared between two atoms

Explanation:

Covalent bonding is a type of bonding which exist between two non metals.

In this bonding, electrons are shared between the two atoms involved in order to attain a stable octet configuration.

This can be seen when hydrogen atom combine with chlorine atom to form hydrogen chloride as shown below:

H + Cl —> HCl

Hydrogen has 1 electron in it's outmost shell and it requires 1 electron to attain a stable configuration.

Chlorine has 7 electrons in it's outmost shell and requires 1 electron to attain a stable configuration.

During bonding, both hydrogen and chlorine will contribute 1 electron each to form bond, thereby attaining a stable configuration. The bond formed in this case is called covalent bond as both atoms involved shared electron to attain a stable configuration.

C. Electrons are shared between two atoms.

What is Covalent Bonding?

A covalent bond is fashioned among non-metals which have comparable electronegativities. Neither atom is "strong" sufficient to draw electrons from the other.It is formed when pairs of electrons are shared by atoms.Atoms will covalently bond with different atoms that allows you to benefit extra stability, that is gained through forming a complete electron shell. By sharing their outer most (valence) electrons, atoms can replenish their outer electron shell and gain stability.

For example:

In H₂ molecule; there is a covalent bond formation between two hydrogen atoms as the electron from each hydrogen atom is shared leading to the formation of hydrogen molecule.

Learn more:

https://brainly.com/question/3447218

Which of the following would have a fixed shape and volume? (3 points) nitrogen gas solid wood liquid water neon gas

Answers

Answer:

Solid Wood

Explanation:

Wood is like a solid block, whereas gases flow freely and liquids spread to fill the shape of their container.

Please let me know if I misunderstood the question, by the way.

What do we call temperature changes caused by changes in air pressure?

Answers

Answer:

Fronts

Explanation:

For example, there are hot and cold fronts which cause the air to become warmer or cooler in a specific region!

Hope this helps! Please mark as brainiest!

The reaction: A + 3 B → D + F was studied and the following mechanism was determined. A + B C (fast) C + B → D + E (slow) E + B → F (very fast) The species, C, is properly described as

Answers

Answer:

Intermediate.

Explanation:

Hello,

In this case, we can rewrite the steps as:

[tex]A + B \rightarrow C\ \ (fast)\\\\C + B \rightarrow D + E\ \ (slow)\\\\E + B \rightarrow F \ \ (very fast)[/tex]

Thus, we can notice that in the fast step, C is present as a product but after that is consumed in the slow step, for that reason, and by cause of its formation-consumption behavior, it is properly described as an intermediate as it is not neither a starting-up substance (reactant in the first step) nor a final substance (product in the final step).

Best regards.

Balance the following equations: (c) H2(g)+I2(s)⟶HI(s)H2(g)+I2(s)⟶HI(s)

Answers

Answer: [tex]H_2(g)+I_2(g)\rightarrow 2HI(s)[/tex]

Explanation:

According to the law of conservation of mass, mass can neither be created nor be destroyed. Thus the mass of products has to be equal to the mass of reactants. The number of atoms of each element has to be same on reactant and product side. Thus chemical equations are balanced.

Thus in the reactants, there are 2 atoms of hydrogen and 2 atoms of iodine .Thus there has to be 2 atoms of hydrogen and 2 atoms of iodine in the product as well. Thus a coefficient of 2 is placed in front of HI.

The balanced chemical reaction is:

[tex]H_2(g)+I_2(g)\rightarrow 2HI(s)[/tex]

a ______ consumer is a heterotroph that directly eats an autotroph. A primary B. Quaterany C tertiary D secondary

Answers

Primary consumer just remember the “primary” (first) consumer so the one that eats a heterotroph

Question 11: How does the energy of a photon emitted when the electron moves from the 3rd orbital to the 2nd orbital compare to the energy of a photon absorbed when the electron moves from the 2nd orbital to the 3rd orbital?

Answers

Answer:

Explanation:

The energy of a photon emitted when the electron moves from the 3rd orbital to the 2nd orbital is exactly same as the energy of a photon absorbed when the electron moves from the 2nd orbital to the 3rd orbital

An aqueous solution of nitric acid is standardized by titration with a 0.110 M solution of calcium hydroxide. If 21.1 mL of base are required to neutralize 23.8 mL of the acid, what is the molarity of the nitric acid solution? ? M Nitric acid

Answers

Answer:

[tex]\large \boxed{\text{0.195 mol/L}}[/tex]

Explanation:

(a) Balanced equation

2HNO₃ + Ca(OH)₂ ⟶ Ca(NO₃)₂ + 2H₂O  

(b) Moles of Ca(OH)₂

[tex]\text{Moles of Ca(OH)}_{2} = \text{21.1 mL Ca(OH)}_{2} \times \dfrac{\text{0.110 mmol Ca(OH)}_{2}}{\text{1 mL Ca(OH)}_{2}}\\= \text{2.321 mmol Ca(OH)}_{2}[/tex]

(c) Moles of HNO₃

The molar ratio is 2 mol HNO₃:1 mol Ca(OH)₂

[tex]\text{Moles of HNO}_{3} = \text{2.321 mmol Ca(OH)}_{2} \times\dfrac{\text{2 mmol HNO}_{3}}{\text{1 mmol Ca(OH)}_{2}}= \text{4.642 mmol HNO}_{3}[/tex]

(d) Molar concentration of HNO₃

[tex]c = \dfrac{\text{moles of solute}}{\text{litres of solution}}\\\\c = \dfrac{n}{V}\\\\c= \dfrac{\text{4.642 mmol}}{\text{23.8 mL}} = \text{0.195 mol$\cdot$L$^{-1}$}\\\\\text{The molar concentration of the Ca(OH)$_{2}$ is $\large \boxed{\textbf{0.195 mol/L}}$}[/tex]

Calculate the pH for the following 1.0M weak acid solutions:a. HCOOH Ka = 1.8 x 10-4 [

Answers

Answer: pH=2.38

Explanation:

To calculate the pH, let's first write out the equation. Then, we will make an ICE chart. The I in ICE is initial quantity. In this case, it is the initial concentration. The C in ICE is change in each quantity. The E is equilibrium.

            HCOOH ⇄ H⁺ + HCOO⁻

I               1.0M          0          0

C              -x            +x        +x

E            1.0-x            x          x

For the steps below, refer to the ICE chart above.

1. Since we were given the initial of HCOOH, we can fill this into the chart.

2. Since we were not given the initial for H⁺ and HCOO⁻, we will put 0 in their place.

3. For the change, we need to add concentration to the products to make the reaction reach equilibrium. We would add on the products and subtract from the reactants to equalize the reaction. Since we don't know how much the change in, we can use variable x.

4. We were given the Kₐ of the solution. We know [tex]K_{a} =\frac{product}{reactant}=\frac{[H^+][HCOO^-]}{[HCOOH]}[/tex].

5. The problem states that the Kₐ=1.8×10⁻⁴. All we have to so is to plug it in and to solve for x.

[tex]1.8*10^-^4 =\frac{x^2}{0.1-x}[/tex]

6. Once we plug this into the quadratic equation, we get x=0.00415.

7. The equilibrium concentration of [H⁺]=0.00415. pH is -log(H⁺).

-log(0.00415)=2.38

Our pH for the weak acid solution is 2.38.

The heat capacity of liquid water is 4.18 J/g·°C and the heat of vaporization is 40.7 kJ/mol. How many kilojoules of heat must be provided to convert 1.00 g of liquid water at 67°C into 1.00 g of steam at 100°C?

Answers

Answer:

The correct answer would be - 2.4KJ or, 2400J

Explanation:

Given:

heat capacity of liquid water - 4.18 J/g·°C

heat of vaporization - 40.7 kJ/mol

Mass of water = 1g

Moles of water = mass/molar mass

= 1g/18.016g

= 0.055 moles

Then,

Total heat required = q1(to raise the temperature to 100) + q2(change from the liquid phase to gas/steam)

= m *s*dt + moles * heat of vaporization

= (1g * 4.18 j/gc * (100-67)) + 0.055* 40.7 KJ

= 137.94J + 2.26KJ

=0.138KJ + 2.26KJ

=2.4KJ or, 2400J

Thus, the correct answer would be - 2.4KJ or, 2400J

When a strip of magnesium metal is placed in an aqueous solution of copper(II) nitrate, elemental copper coats the surface of the magnesium strip and aqueous magnesium nitrate forms.
1. As a reactant, what is the charge on copper?
2. As a product, what is the charge on copper?

Answers

Answer:

1. 2+ ([tex]Cu^{2+}[/tex]).

2. 0 ([tex]Cu^0[/tex]).

Explanation:

Hello,

In this case, the described chemical reaction is a redox reaction in fact, since the oxidation states of both magnesium and copper change as shown due to the displacement:

[tex]Mg(s)+Cu(NO_3)_2(aq)\rightarrow Mg(NO_3)_2(aq)+Cu(s)[/tex]

Therefore:

1. Since copper is the cation in the copper (II) nitrate, the (II) means that its charge is 2+ ([tex]Cu^{2+}[/tex]).

2. Since copper is alone, it means no electrons are being neither shared not given, its charge is 0 ([tex]Cu^0[/tex]).

Best regards.

How many moles of ammonia are in 0.175 L of a 6.50 M aqueous ammonia solution?

Answers

Answer:

Concentration (C) = number of moles (n) / volume (v)

Therefore

number of moles (n) = concentration × volume

Concentration = 6.50M

Volume =0.175 L = 0.175dm³

n = 6.50 × 0.175

n = 1.138 moles

Moles of ammonia in the solution is

1.138 moles.

Hope this helps

If the pKaof HCHO2is 3.74 and the pH of an HCHO2/NaCHO2solution is 3.11, which of the following is true?
A. [HCHO2] < [NaCHO2]
B. [HCHO2] = [NaCHO2]
C. [HCHO2] << [NaCHO2]
D. [HCHO2] > [NaCHO2]
E. It is not possible to make a buffer of this pH from HCHO2 and NaCHO2

Answers

Answer:

D. [HCHO₂] > [NaCHO₂]

Explanation:

Formic acid, HCHO₂, is a weak acid that, in presence of its conjugate base, NaCHO₂ (CHO₂⁻), produce a buffer following H-H equation:

pH = pKa + log [CHO₂⁻] / [HCHO₂]

As pKa of the acid is 3.74 and pH of the solution is 3.11:

3.11 = 3.74 + log [CHO₂⁻] / [HCHO₂]

-0.63 = log [CHO₂⁻] / [HCHO₂]

0.2344 = [CHO₂⁻] / [HCHO₂]

A ratio [CHO₂⁻] / [HCHO₂] < 1, means:

[HCHO₂] > [CHO₂⁻]

Magnesium oxide can be produced by heating magnesium metal in the presence of oxygen. When 10.1 g of Mg reacts with 10.5 g of O2, 11.9 g of MgO are collected.
a) Write the balanced equation for this reaction, including all states of matter.
b) What is the limiting reactant in this reaction?
c) What is the theoretical yield for this reaction?
d) What is the percent yield for this reaction?

Answers

Answer:

A. 2Mg(s) + O2(g) —> 2MgO(s)

B. Mg is the limiting reactant.

C. Theoretical yield of MgO is 16.83g.

D. The percentage yield is 70.7%

Explanation:

A. The balanced equation for the reaction. This is given below:

2Mg(s) + O2(g) —> 2MgO(s)

B. Determination of the limiting reactant.

First, we shall determine the mass of Mg and O2 that reacted and the mass of MgO produced from the balanced equation. This is illustrated below:

Molar mass of Mg = 24g/mol

Mass of Mg from the balanced equation = 2 x 24 = 48g.

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

Mass of O2 from the balanced equation = 1 x 32 = 32g

Molar mass of MgO = 24 + 16 = 40g/mol

Mass of MgO from the balanced equation = 2 x 40 = 80g

Summary:

From the balanced equation above,

48g of Mg reacted with 32g of O2 to produce 80g of MgO.

Now, we can obtain the limiting reactant as follow:

From the balanced equation above,

48g of Mg reacted with 32g of O2.

Therefore, 10.1g of Mg will react with = (10.1 x 32)/48 = 6.73g of O2.

From the calculations made above, we can see that only 6.73g out of 10.5g of O2 given is needed to react completely with 10.1g of Mg.

Therefore, Mg is the limiting reactant and O2 is the excess reactant.

C. Determination of the theoretical yield of MgO.

The limiting reactant is used in this case as it will produce the maximum yield of the reaction since all of I is used up in the reaction.

The theoretical yield can be obtain as illustrated below:

From the balanced equation above,

48g of Mg reacted to produce 80g of MgO.

Therefore, 10.1g of Mg will react to produce = (10.1 x 80)/48 = 16.83g of MgO.

Therefore, the theoretical yield of MgO is 16.83g.

D. Determination of the percentage yield.

This is illustrated below:

Actual yield of MgO = 11.9g

Theoretical yield of MgO = 16.83g

Percentage yield =..?

Percentage yield = Actual yield /Theoretical yield x 100

Percentage yield = 11.9/16.83 x 100

Percentage yield = 70.7%

Add distilled water to the beaker until the volume
totals 15 mL.
Record the amount of oil that dissolved.

Answers

Answer:

i guess oil never dissolve in water. As like dissolve like. water is polar so it dissolves only polar substances

Explanation:

Answer:

None

Explanation:

Answer on Edge 2022

Devise a detailed experimental procedure to purify ~ 20 grams of benzoic acid that is contaminated with sodium chloride. Justification of the steps (including solubility calculations) that are included in the procedure. In other words, explain why the steps are being included.

Answers

Answer:

Based on the difference in solubility one can perform the process of purification of the benzoic acid contaminated with sodium chloride. The benzoic acid does not get soluble in cold water, while the sodium chloride is soluble in cold water.  

Thus, for separation, the supplementation of cold water can be done into the mixture in the experiment of purifying benzoic acid from sodium chloride. In the process, the mixture is placed on the ice bath and is stirred well, in the end, the solution is filtered. The filtrate contains sodium chloride and on the filter paper pure benzoic acid is collected.  

Without doing any calculations, match the following thermodynamic properties with their appropriate numerical sign for the following endothermic reaction. 2N2(g) + O2(g)2N2O(g) Clear All > 0 Hrxn < 0 Srxn = 0 Grxn > 0 low T, < 0 high T Suniverse < 0 low T, > 0 high T

Answers

Answer:

∆H > 0

∆Srxn <0

∆G >0

∆Suniverse <0

Explanation:

We are informed that the reaction is endothermic. An endothermic reaction is one in which energy is absorbed hence ∆H is positive at all temperatures.

Similarly, absorption of energy leads to a decrease in entropy of the reaction system. Hence the change in entropy of the reaction ∆Sreaction is negative at all temperatures.

The change in free energy for the reaction is positive at all temperatures since ∆S reaction is negative then from ∆G= ∆H - T∆S, we see that given the positive value of ∆H, ∆G must always return a positive value at all temperatures.

Since entropy of the surrounding= - ∆H/T, given that ∆H is positive, ∆S surrounding will be negative at all temperatures. This is so because an endothermic reaction causes the surrounding to cool down.

If 2 moles of helium undergo a temperature increase of 100 K at constant pressure, how much energy has been transferred to the helium as heat

Answers

Answer:

[tex]Q=4154J[/tex]

Explanation:

Hello,

In this case, the involved heat in this heating process is considered to be computed via:

[tex]Q=nCp\Delta T[/tex]

Whereas we assume a constant molar specific heat of helium which is 20.77 J/(mol*K), thus, the transferred energy in the form of heat turns out:

[tex]Q=2mol*20.77\frac{J}{mol*K} *100K\\\\Q=4154J[/tex]

Regards.

A 5.22 × 10−3−mol sample of HY is dissolved in enough H2O to form 0.088 L of solution. If the pH of the solution is 2.37, what is the Ka of HY?

Answers

Answer:

3.07 × 10⁻⁴

Explanation:

Step 1: Calculate the concentration of H⁺

We will use the definition of pH.

[tex]pH = -log [H^{+} ]\\\[ [H^{+} ] = antilog -pH = antilog -2.37 = 4.27 \times 10^{-3} M[/tex]

Step 2: Calculate the concentration of HY

5.22 × 10⁻³ mol of HY are dissolved in 0.088 L. The concentration of the acid (Ca) is:

[tex]Ca = \frac{5.22 \times 10^{-3} mol }{0.088L} = 0.0593M[/tex]

Step 3: Calculate the acid dissociation constant (Ka)

We will use the following expression.

[tex]Ka = \frac{[H^{+}]^{2} }{Ca} = \frac{(4.27 \times 10^{-3} )^{2} }{0.0593} = 3.07 \times 10^{-4}[/tex]

The number of water molecules per ionic molecule in a hydrate is called the ___________. A. electronegativity B. anhydrous number C. hydration number D. polarity

Answers

Answer:

C. hydration number

Explanation:

When we dissolve an ionic compound (a charged species) the charges can interact with the water molecule. In the case of cations (positive charges) the negative dipole of water (generated in the oxygen) will interact with the positive charge at the same time the anions (negative charges) the positive dipole of water (generated in the hydrogen).

The amount of water molecules that can interact with a single ion (cation or anion) is called hydration number. In the example, we have a hydration number of "4" for the sodium cation.

I hope it helps!

Calculate the pH of mixing 24 mL of 1M acetic acid with 76 mL of 1M sodium acetate. For the purpose of this calculation, assume the Ka of acetic acid is 1.8 X 10-5. You must include units to obtain full credit. You must show all your work to obtain any credit.

Answers

Answer:

pH = 5.24

Explanation:

Mixture of acetic acid with acetate ion is a buffer (Mixture of a weak acid with its conjugate base). The pH of a buffer can be determined using Henderson-Hasselbalch equation:

pH = pKa + log₁₀ [A⁻] / [HA]

Where pKa is -log Ka = 4.74; [A⁻] is the concentration of conjugate base (Acetate ion) and [HA] is molar concentration of the weak acid.

Concentration of the acetic acid in the 100mL≡0.1L (76mL + 24mL) solution is:

[HA] = 0.024L ₓ (1mol / L) / 0.1L = 0.24M

[A⁻] = 0.076L ₓ (1mol / L) / 0.1L = 0.76M

Replacing in H-H equation:

pH = 4.74 + log₁₀ [0.76M] / [0.24M]

pH = 5.24

What is the name of Mn(CO3)2

Answers

Answer:

Mn is manganese and CO₃ is carbonate. Since the charge for CO₃ is -2 and the subscript is 2, the charge of Mn must be +4 so the answer is manganese (IV) carbonate.

Manganese (IV) carbonate is the name of Mn(CO[tex]_3[/tex])[tex]_2[/tex]. The only names used to identify salts are those of the cation or the anion.

The chemical formula of the anion (such as chloride or acetate) comes first in the name of a salt, which is followed with the identity of the cation (such as sodium or ammonium). They are created when acids and bases react, and they are always composed of either metal cations or cations made of ammonium. Manganese is Mn, and carbonate is CO[tex]_3[/tex]. The solution equals manganese (IV) carbonate since the charge for CO[tex]_3[/tex] is -2 but the subscript is 2, meaning that the charge of Mn has to be +4.

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Dissolving NaOH(s) in water is exothermic. Two calorimetry experiments are set up. Experiment 1: 2 g of NaOH are dissolved in 100 mL of water Experiment 2: 4 g of NaOH are dissolved in 200 mL of water Which of the following statements is true?a. both temperature changes will be the sameb. the second temeprature change will be approximately twice the firstc. the second temperature change will be approximately four times the firstd. the second temperature change will be approximately one-half of the firste. the second temperature change will be approximately one-fourth the first

Answers

Answer:

a. both temperature changes will be the same

Explanation:

When sodium hydroxide (NaOH) is dissolved in water, a determined amount is released to the solution following the equation:

Q = m×C×ΔT

Where Q is the heat released, m is the mass of the solution, C is the specific heat and ΔH is change in temperature.

Specific heat of both solutions is the same (Because the solutions are in fact the same). Specific heat = C.

m is mass of solutions: 102g for experiment 1 and 204g for experiment 2.

And Q is the heat released: If 2g release X heat, 4g release 2X.

Thus, ΔT in the experiments is:

Experiment 1:

X / 102C = ΔT

Experiment 2:

2X / 204C = ΔT

X / 102C = ΔT

That means,

a. both temperature changes will be the same

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