50 mL of 0.1 M acetic acid is mixed with 50 mL of 0.1 M sodium acetate (the conjugate base). The Ka of acetic acid is approximately 1. 74 X 10 -5. What is the pH of the resulting solution?

Answer:

Fronts

Explanation:

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

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Answer:

C.

Explanation:

Since the mass number is the number of protons and neutrons added together, the answer is 35. Since the questions are respectively electron quantity and mass number, the only answer choice with 35 as the second choice is C, so that is the correct answer.

HERE IS THE COMPLETE QUESTION

what is the calculated value of the cell potential at 298 k for an electrochemical cell with the following reaction, when the Cl2 pressure is 1.31 atm, the cl- concentration is at  1.28M, and the Ni2+ concentration is 1.24M

[tex]Cl_{2(g)} + Ni _{(s)} \to 2Cl^-_{aq} + Ni^{2+}_{aq}[/tex]

Answer:

[tex]\mathbf{E_{cell} = +1.4962 \ V}[/tex]

Explanation:

From the given question;

We can see that Nickel is oxidized and Chlorine is reduced. We also known that Oxidation occurs at the anode while reduction occurs at the cathode.

SO the Oxidation and the reduction reaction can be expressed as shown below.

Oxidation :      [tex]Ni_{s} \to Ni ^{2+}_{aq} + 2e^- \ \ \ \ E^0 = -0.26[/tex]

Reduction :      [tex]Cl_{2(g)} + 2e^- \to 2Cl^- _{(aq)} \ \ \ \ E^0} = 1.36[/tex]

                        [tex]Cl_{2(g)} + Ni _{(s)} \to 2Cl^-_{aq} + Ni^{2+}_{aq}[/tex]

[tex]E^0_{cell} = E^0_{cathode}-E^0_{anode}[/tex]

[tex]E^0_{cell} = E^0_{Cl_2/Cl^-}-E^0_{Ni^{2+}/Ni}[/tex]

[tex]E^0_{cell} = +1.36 - (-0.26)[/tex]

[tex]E^0_{cell} = +1.62 \ V[/tex]

By applying Nernst Equation; we have :

[tex]E_{cell} = E^0_{cell}- (\dfrac{0.0591}{n}) \ log Q[/tex]

where

n= 2 moles

[tex]Q= \dfrac{[Cl^-]^2[Ni^{2+}]}{P_{Cl_2}}[/tex]

[tex]E_{cell} = +1.62 \ V}- (\dfrac{0.0591}{2}) \ log \dfrac{[Cl^-]^2[Ni^{2+}]}{P_{Cl_2}}[/tex]

[tex]E_{cell} = +1.62 \ V}- (\dfrac{0.0591}{2}) \ log \dfrac{[1.28]^2[1.24]}{1.31*10^{-4}}[/tex]

[tex]E_{cell} = +1.62 \ V}- (0.02955) \ log \dfrac{[1.6384][1.24]}{1.31*10^{-4}}[/tex]

[tex]E_{cell} = +1.62 \ V}- (0.02955) \ log \dfrac{[1.6384*1.24]}{1.31}*10^{4}[/tex]

[tex]E_{cell} = +1.62 \ V}- (0.02955) \ log \dfrac{[2.031616]}{1.31}*10^{4}[/tex]

[tex]E_{cell} = +1.62 \ V}- (0.02955) * \ log (1.55085*10^{4})[/tex]

[tex]E_{cell} = +1.62 \ V}- (0.02955) * \ log (1.55085)+ log \ 10^{4}[/tex]

[tex]E_{cell} = +1.62 \ V}- (0.02955) * \ 0.1905 + 4[/tex]

[tex]E_{cell} = +1.62 \ V}- (0.02955) * \ 4.1905[/tex]

[tex]E_{cell} = +1.62 \ V}- 0.1238[/tex]

[tex]\mathbf{E_{cell} = +1.4962 \ V}[/tex]

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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Answer:

8.59 g

2.25 g

Explanation:

According to the given situation the calculation of grams of PbO and grams of NaCL is shown below:-

Moles of Pb(OH)CL is

[tex]= \frac{Mass}{Molar\ mass}[/tex]

[tex]= \frac{10.0 g}{259.65g / mol}[/tex]

= 0.0385 mol

Mass of PbO needed is

[tex]= 0.385mol Pb(OH) Cl\times \frac{1 mol PbO}{1molpb (OH) cl} \times \frac{223.2g PbO}{1mol PbO}[/tex]

After solving the above equation we will get

= 8.59 g

Mass of NaCL needed is

[tex]= \frac{1mol\ NaCl}{1molPb\ (OH)Cl} \times \frac{58.45NaCl}{1mol NaCl}[/tex]

After solving the above equation we will get

= 2.25 g

Therefore we have applied the above formula.

Answer:

(a) The empirical formula of the compound is

m(CxHy) + m(O2) = m(CO) + m(CO2) + m(H2O).

(b) The grams of O2 that were used in the reaction is 1.146 g

(c) The amount of O2 that would have been required for complete combustion is 1.401 g.

Explanation:

a. m(CxHy) + m(O2) = m(CO) + m(CO2) + m(H2O)

(b) Using law of conservation of mass from above

m(O2) = m(CO) + m(CO2) + m(H2O) - m(CxHy)

m(O2) = 0.446 + 0.700 + 0.430 - 0.430

m(O2) = 1.146 g

The grams of O2 that were used in the reaction is 1.146 g

(c) for complete combustion, we need to oxidized CO to CO2

Then, 2CO +O2 = 2CO2

m(add)(O2) = M(O2)*¢(O2)/2 = M(O2) * {(m(CO))/(2M(CO))}

m(add)(O2) = 32 * {(0.446)/(2*28)} = 0.255 g

Note; Molar mass of O2 = 32, CO = 28

m(total)(O2) = m(O2) + m(add)(O2)

m(total)(O2) = 1.146 + 0.255 = 1.401 g

The amount of that grams would have been required for complete combustion is 1.401 g.

Note (add) and (total) were used subscript to "m"

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

Answer:

intermolecular dipole-dipole hydrogen bonds

Explanation:

Water is a polar molecule. Recall that the central atom in water is oxygen. The molecule is bent, hence it has an overall dipole moment directed towards the oxygen atom. Since it has a permanent dipole moment, we expect that it will show dipole-dipole interactions in the liquid state.

Similarly, water contains hydrogen and oxygen. Recall that hydrogen bonds are formed when hydrogen is covalently bonded to highly electronegative elements. Hence, water in the liquid state exhibits strong hydrogen bonding. The unique type of dipole-dipole interaction in liquid water is actually hydrogen bonding, hence the answer.

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,

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.

Answer:

THE NEW FREEZING POINT IS -4.196 °C

Explanation:

ΔTf = 1 Kf m

molarity of MgCl2:

Molar mass = (24 + 35.5 *2) g/mol

molar mass = 95 g/mol

7.15 g of MgCl2 in 100 g of water

7.15 g = 100 g

(7.15 * 100 / 1000) = 1000 g or 1 L or 1 dm3

= 0.715 g /dm3

Molarity in mol/dm3 = molarity in g/dm3 / molar mass

= 0.715 g /dm3 / 95 g/mol

m = 0.00752 mol/ dm3

So therefore:

ΔTf = i Kf m

1 = 3 (1 Mg and 2 Cl)

Kf = 1.86 °C/m

M = 0.752 moles

So we have:

ΔTf = 3 * 1.86 * 0.752

ΔTf = 4.196 °C

The new freezing point therefore will be 0 °C - 4.196 °C which is equals to - 4.196 °C

Answer:

8.33mL or .0083L

Explanation:

Use m1 * V1 = m2 * V2

6.00M(x) = 0.100M(500mL)

solve for x

x= (.1 * 500) / 6

x=8.333 mL

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.

Answer:

Explanation:

Phosphoric acid H₃PO₄ is produced from reaction of water and tetraphosphorus decoxide  P₄O₁₀ as follows

P₄O₁₀ + 6 H₂O = 4 H₃PO₄

In this reaction 6 molecules of H₂O and one molecule of  phosphorus compound P₄O₁₀ is needed to produce phosphoric acid , ie the conversion factor of water to acid is 6 / 1

This ratio is called mole ratio of water to H₃PO₄.

So the required ratio is 6 : 1 .

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.  

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

Answer:

61.5%.

Explanation:

We usually define percentage yield in chemistry as follows;

Percentage yield= actual yield/theoretical yield ×100

In this case we have;

Actual yield of the guaifenesin 400mg

Theoretical yield of the guaifenesin 650 mg

Hence;

Percentage yield of the guaifenesin =

400/650 ×100 = 61.5%

Hence the theoretical yield of the guaifenesin is 61.5%.

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

Answer:

[tex]\large \boxed{1.64\times 10^{-5}\text{ mol/L }}[/tex]

Explanation:

Ag₂CO₃(s) ⇌2Ag⁺(aq) + CO₃²⁻(aq); Ksp = 8.10 × 10⁻¹²

                           2x      0.007 50 + x

[tex]K_{sp} =\text{[Ag$^{+}$]$^{2}$[CO$_{3}^{2-}$]} = (2x)^{2}\times 0.00750 = 8.10 \times 10^{-12}\\0.0300x^{2} = 8.10 \times 10^{-12}\\x^{2} = 2.70 \times 10^{-10}\\x = \sqrt{2.70 \times 10^{-10}} = \mathbf{1.64\times 10^{5}} \textbf{ mol/L}\\\text{The maximum concentration of Ag$^{+}$ is $\large \boxed{\mathbf{1.64\times 10^{-5}}\textbf{ mol/L }}$}[/tex]

The amount of the sample in space is referred to as concentration, in this the Maximum concentration of [tex]Ag^+[/tex] is [tex](3.28 \times 10^{-5}\ M)[/tex].

        Concentration of [tex]Na_2CO_3 = 0.00750 M[/tex]

           [tex](CO_3)^{2-} = Na_2CO_3 = 0.00750\ M\\\\Ksp \ \ Ag_2CO_3 =( Ag^{+})^2 (CO_3)^{2-}\\\\8.10 \times 10^{-12} = (Ag^+)^2 \times (0.00750\ M)\\\\(Ag^+)^2 = \frac{(8.10 \times 10^{-12})}{ (0.00750\ M)}\\\\(Ag^+)^2 = 1.08 \times 10^{-9}\ M^2\\\\(Ag^+) = (1.08 \times 10^{-9}\ M^2)^{\frac{1}{2}}\\\\\[(Ag^+)\] = (3.28 \times 10^{-5}\ M)\\\\[/tex]

So, the Maximum concentration of [tex]Ag^+[/tex] is [tex](3.28 \times 10^{-5}\ M)[/tex].

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Answer:

A catalyst cannot change an exothermic reaction into an endothermic reaction or vice versa.

Explanation:

Catalyst is basically a substance that enables a chemical reaction to occur at a faster rate as compared to the reaction without catalysis. It lowers the activation energy and temperature for a chemical reaction and a catalyst itself does not goes through any permanent chemical change. This means it does not get used in the process.

Exothermic and endothermic are the chemical reaction. Exothermic reactions absorb energy. This energy is absorbed in the form of heat. When the energy is released in the form of heat then this reaction is called endothermic. So one absorbs the heat and the other  releases it.

As we know that the catalyst does not undergo change at the end of the reaction so the energy or heat whether is absorbed or emitted or you can say whether the reaction is exothermic or endothermic, the total energy stays unchanged during the reaction. So with and without a catalyst, if both have same reactants and products and the difference in enthalpy between products and reactants will be the same.

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

Answer:

Uranium-238 undergoes alpha decay to form Thorium-234 as daughter product.

Explanation:

Alpha decay is indicative of loss of the equivalents of a helium particle emission. The reaction equation for this reaction is shown below:

[tex]_{92} ^{238} U_{}[/tex]→ [tex]_{90} ^{234} Th_{} + _{2} ^{4} He_{}[/tex]

I hope this explanation is clear and explanatory.

Answer:

A = 0.7871g

B = 0.8417g

C = 0.3192g

D = 0.3897g

Explanation:

Hello,

The number of grams of sodium (Na) in 2g of NaCl is

Molar mass of NaCl = 58.44g/mol

Molar mass of Na = 23g

23g of Na = 58.44g of NaCl

x g of Na = 2g of NaCl

x = (2 × 23) / 58.44

x = 0.7871g

Therefore, 0.7871g of Na is present in 2g of NaCl

2.

2g of Na₃PO₄

Molar mass of Na₃PO₄ = 163.94g/mol

Molar mass of Na = 23g

(3 × 23)g of Na = 163.94 g of Na₃PO₄

x g of Na = 2g of Na₃PO₄

x = (2 × 69) / 163.94

x = 0.8417g

0.8417g of Na is present in 2g of Na₃PO₄

3.

2g of NaC₇H₅O₂

Molar mass of NaC₇H₅O₂ = 144.103g/mol

Molar mass of Na = 23g

23g of Na = 144.103g of NaC₇H₅O₂

x g of Na = 2g of NaC₇H₅O₂

x = (2 × 23) / 144.103

x = 0.3192g

0.3192g of Na is present in 2g of NaC₇H₅O₂

4. 2g of Na₂C₆H₆O₇

Molar mass of Na₂C₆H₆O₇ = 236.08g/mol

Molar mass of Na = 23g

(2 × 23)g of Na = 236.08g of Na₂C₆H₆O₇

x g = 2g of Na₂C₆H₆O₇

x = (46 × 2) / 236.08

x = 0.3897g

0.3897g is present in 2g of Na₂C₆H₆O₇

Answer:

B - Making an Observation

Explanation:

Making an observation best describes the act of using senses or tools to gather information. Therefore, option B is correct.

The five senses—sight, taste, touch, hearing, and smell—gather data about our surroundings that the brain interprets. Based on prior experience (and subsequent learning), as well as by combining the data from each sensor, we make sense of this information.

Information gleaned from your five senses is referred to as an observation. These are smell, taste, touch, hearing, and sight. When you see a bird or hear it sing, you notice it.

The term observation, which is also used to sense five aspects of the world including vision, taste, touch, smell, and hearing, is used to describe utilizing the senses to examine the world, employing tools to take measurements, and looking at prior research findings.

Thus, option B is correct.

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Answer:

Thin-layer chromatography

Explanation:

Here in this question, we are concerned with giving the best technique that could be use for monitoring the progress of the reaction that involves the oxidation of alcohols.

The most useful technique here for monitoring the progress of the reaction would be the thin layer chromatography.

Firstly, we should know the reason why we are using a chromatographic technique. This is intuitive. The reason is that the oxidizing agent being used is a colored substance. So basically, the progress of the oxidation reaction would solely rest on the fact that we have a decrease in color saturation of the said oxidizing agent.

This is best done using the thin layer chromatography because we have a mobile phase that moves quickly over the stationary phase and asides this, it moves evenly.

This makes the fact that color changes can be quickly and easily identified and we can tell if the oxidation reaction has gone to completion or not

Answer:

even I have the same dought

Answer: -1,030 kJ

Explanation:

To calculate the number of moles we use the equation:

[tex]\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar mass}}[/tex]     .....(1)

Putting values in equation 1, we get:

[tex]\text{Moles of methanol}=\frac{45.3g}{32g/mol}=1.42mol[/tex]

The balanced chemical reaction is:

[tex]CH_3OH(l)+3O_2(g)\rightarrow 2CO_2(g)+4H_2O(g)[/tex]  [tex]\Delta H=-726kJ[/tex]

Given :

Energy released when 1 mole of [tex]CH_3OH[/tex] combusts = 726k J

Thus Energy released when 1.42 moles of [tex]CH_3OH[/tex] combusts =  [tex]\frac{726kJ}{1}\times 1.42=1030J[/tex]

Thus 1030 kJ of heat is released and as [tex]\Delta H[/tex] is negative for exothermic reaction, [tex]\Delta H=-1030kJ[/tex]

Answer: single replacement reaction

Explanation:

A single replacement reaction is one in which a more reactive element displaces a less reactive element from its salt solution. Thus one element should be different from another element.

A general single displacement reaction can be represented as :

[tex]X+YZ\rightarrow XZ+Y[/tex]

The reaction [tex]2AgNO_3(aq)+Zn(s)\rightarrow 2Ag(s)+Zn(NO_3)_2(aq)[/tex]

When zinc metal is added to aqueous silver nitrate, zinc being more reactive than silver displaces silver atom from its salt solution and lead to formation of zinc nitrate and silver metal.

Answer:A medication break can ease side effects. A lack of appetite, weight loss, sleep troubles, headaches, and stomach pain are common side effects of ADHD medication.

Explanation: It may boost your child’s growth. Some ADHD medications can slow a child’s growth in height, especially during the first 2 years of taking it. While height delays are temporary and kids typically catch up later, going off medication may lead to fewer growth delays.

It won’t hurt your child. Taking a child off ADHD medication may cause their ADHD symptoms to reappear. But it won’t make them sick or cause other side effects.

Answer:

To obtain the grams of fat that the ground round has, knowing that it weighs 1.33 pounds we must pass this value to grams. Since 1 pound equals 453.59 grams, 1.33 pounds equals 603.27 (453.59 x 1.33).

Now, to obtain 29 percent of 603.27, we must make the following calculation: 603.27 / 100 x 29, which gives a total of 174.94 grams.

In this way, your reasoning is correct and it is probably a mistake in the book.