KBr(aq) AgNO3(aq) Classify each chemical reaction: reaction type of reaction (check all that apply) combination precipitation single replacement combustion double replacement acid-base decomposition combination precipitation single replacement combustion double replacement acid-base decomposition combination precipitation single replacement combustion double replacement acid-base decomposition combination precipitation single replacement combustion double replacement acid-base decomposition

Answer: ΔG =23.169kJ/mol

Explanation:

Solution

To Calculate Gibbs free energy ΔG for the reaction above we use the equation ΔG=ΔH−TΔS.

Where

ΔH= 38.468 kJ/mol = 38468 J/mol

∆S = +51.4 J mol−1 K−1).

T = 25◦C =298k

ΔG= 38468J/mol−298k(51.4 J mol−1 K−1).

ΔG = 38468 J/mol - 15317.2J/mol

ΔG = 23168.8J/mol

ΔG =23.169kJ/mol

Ozone, according to the VSEPR theory, has a bent or 'V' shaped geometry due to the repulsion of electron pairs. This is because it has one lone pair and two bonding domains.

      O

    /    \

   O    O

The structure of ozone, or O₃, can be drawn according to the VSEPR theory. The central atom is one oxygen atom while the other two oxygen atoms are attached to the central one. Then, there is one lone pair on the central atom, creating a 'bent' or 'V' shape in its geometry.

The VSEPR (Valence Shell Electron Pair Repulsion) theory suggests that electron pairs will repel each other as much as possible, resulting in specific molecular geometries. Ozone is a molecular geometry example of a molecule with 3 total sites of electrons, 2 bonding domains, and one non-bonding domain. This leads to a 'bent' or 'V' shape because the non-bonding pair pushes the two bonding domains closer together.

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Answer:

Al 72.61%

Mg 27.39%

Explanation:

To obtain the mass percentages, we need to place the individual masses over the total mass and multiply by 100%.

If we observe clearly, we can see that the parameters given are the moles. We need to convert the moles to mass.

To do this ,we need to multiply the moles by the atomic masses. The atomic mass of aluminum is 27 while that of magnesium is 24.

Now, the mass of aluminum is thus = 27 * 0.0898 = 2.4246g

The mass of magnesium is 0.0381 * 24 = 0.9144g

We can now calculate the mass percentage.

The total mass is 0.9144 + 2.4246 = 3.339g

% mass of Al = 2.4246/3.339 * 100 = 72.61%

% mass of Mg = 0.9144/3.39 * 100 = 27.39%

Answer:

Mass % Al = 72.3 %

Mass % Mg = 27.7 %

Explanation:

Step 1: Data given

Number of moles Al = 0.0898 moles

Number of moles Mg = 0.0381 moles

Molar mass Al = 26.98 g/mol

Molar mass Mg = 24.3 g/mol

Step 2: Calculate mass Al

Mass Al = moles Al * molar mass Al

Mass Al = 0.0898 moles * 26.98 g/mol

Mass Al = 2.42 grams

Step 3: Calculate mass Mg

Mass Mg = 0.0381 moles * 24.3 g/mol

Mass Mg = 0.926 grams

Step 4: Calculate total mass

Total mass = mass Al + mass Mg

Total mass = 2.42 grams + 0.926 grams

Total mass = 3.346 grams

Step 5: Calculate mass %

Mass % Al = (mass Al/ total mass) * 100%

Mass % Al = (2.42 grams / 3.346 grams ) *100%

Mass % Al = 72.3 %

Mass % Mg = (mass Mg/ total mass)*100%

Mass % Mg = (0.926 / 3.346) *100 %

Mass % Mg = 27.7 %

Answer:

Ethical Behavior

Explanation:

It's the ethical behavior how companies work or do business that have the positive impact on the community. They not only think about making money but also about the welfare of the society. They are concerned about the products they made and it's impact on the environment. Ethical behaviour is based on the human perception of right and wrong. That kind of behaviour whichis  not required by the law, but is done for the betterment of the society is ethical behaviour.

Silver Mining's voluntary decision to install environmentally friendly devices, despite no legal requirement, exemplifies business ethical behavior.

In this instance, Silver Mining's decision to install scrubbers on the smokestacks of their new facility, even though not legally required, is a clear example of business ethical behavior. This action demonstrates the company prioritizing environmental protection over potential costs. While the act also aligns with legal behavior, it is not driven by legal necessity. It's a voluntary measure taken for the greater good, thus making it an ethical business decision.

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Answer : The concentration of a solution with an absorbance of 0.420 is, 0.162 M

Explanation :

Using Beer-Lambert's law :

[tex]A=\epsilon \times C\times l[/tex]

As per question, at constant path-length there is a direct relation between absorbance and concentration.

[tex]\frac{A_1}{A_2}=\frac{C_1}{C_2}[/tex]

where,

A = absorbance of solution

C = concentration of solution

l = path length

[tex]A_1[/tex] = initial absorbance = 0.350

[tex]A_2[/tex] = final absorbance = 0.420

[tex]C_1[/tex] = initial concentration = 0.135 M

[tex]C_2[/tex] = final concentration = ?

Now put all the given value in the above relation, we get:

[tex]\frac{0.350}{0.420}=\frac{0.135}{C_2}[/tex]

[tex]C_2=0.162M[/tex]

Thus, the concentration of a solution with an absorbance of 0.420 is, 0.162 M

Explanation:

Molarity of solution = 1.00 M = 1.00 mol/L

In 1 L of solution 1.00 moles of calcium chloride is present.

Mass of solute or calcium chloride = m

[tex]m = 1 mol\times 111 g/mol = 111 g[/tex]

Mass of solution = M

Volume of solution = V = 1L = 1000 mL

Density of solution , d= 1.07 g/mL

[tex]M=d\times V=1.07 g/mL\times 1000 mL=1,070 g[/tex]

1) The value of %(m/M):

[tex]\frac{m}{M}\times 100=\frac{111 g}{1,070 g}\times 100=10.37\%[/tex]

2) The value of %(m/V):

[tex]\frac{m}{V}\times 100=\frac{111 g}{1000 L}\times 100=11.1\%[/tex]

[tex]Molality = \frac{\text{Moles of compound }}{\text{mass of solvent in kg}}[/tex]

[tex]Normality=\frac{\text{Moles of compound }}{n\times \text{volume of solution in L}}[/tex]

n = Equivalent mass

n = [tex]\frac{\text{molar mass of ion}}{\text{charge on an ion}}[/tex]

3) Normality of calcium ions:

Moles of calcium ion = 1 mol (1 [tex]CaCl_2[/tex] mole has 1 mole of calcium ion)

[tex]n=\frac{40 g/mol}{2}=20 [/tex]

[tex]=\frac{1 mol}{20 g/mol\times 1L}=0.050 N[/tex]

4) Normality of chlorine ions:

Moles of chlorine ion = 2 mol (1 [tex]CaCl_2[/tex] mole has 2 mole of chlorine ion)

[tex]n=\frac{35.5 g/mol}{1}=35.5[/tex]

[tex]=\frac{2 mol}{35.5 g/mol\times 1L}=0.056 N[/tex]

Moles of calcium chloride = 1.00 mol

Mass of solvent =  Mass of solution - mass of solute

= 1,070 g - 111 g = 959  g = 0.959 kg ( 1 g =0.001 kg)

5) Molality of the solution :

[tex]\frac{1 mol}{0.959 kg}=1.043 mol/kg[/tex]

Moles of calcium chloride = [tex]n_1=1mol[/tex]

Mass of solvent = 959 g

Moles of water = [tex]n_2=\frac{959 g}{18 g/mol}=53.28 mol[/tex]

Mass of solvent = 959 g

6) Mole fraction of calcium chloride =

[tex]\chi_1=\frac{n_1}{n_1+n_2}=\frac{1mol}{1 mol+53.28 mol}=0.01842[/tex]

7) Mole fraction of water =

[tex]\chi_2=\frac{n_2}{n_1+n_2}=\frac{53.28 mol}{1mol+53.28 mol}=0.9816[/tex]

8) Mass of solution = m'

Volume of the solution= v = 100 mL

Density of solution = d = 1.07 g/mL

[tex]m'=d\times v=1.07 g/ml\times 100 g= 107 g[/tex]

Mass of 100 mL of this solution 107 grams of solution.

9) Volume of solution = V = 100 mL

Mass of solution = M'' = 107 g

Mass of solute = m

The value of %(m/V) of solution = 11.1%

[tex]11.1\%=\frac{m}{100 mL}\times 100[/tex]

m = 11.1 g

Mass of solvent = M''- m = 107 g -11.1 g = 95.9 g

95.9 grams of water was present in 100 mL of given solution.

Answer:

Options A and B are correct.

Sodium salts and Carbon based fuels satisfy the criteria.

Explanation:

Sodium salts and Carbon Based fuels produce bright yellow emissions that can mask other emission colors.

Magnesium Salts produce no emissions, although, Magnesium metal burns brightly white while Aluminium salts produce white emissions.

Hope this Helps!!!

Options A and B are correct.

Sodium salts and Carbon based fuels satisfy the criteria.

The following information should be considered:

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Answer:

18 Carats because of 75% Purity...

Explanation:

Red Gold Jewelry always contains a mixing of 25% copper by mass to make it durable and strong. That according to the simple parallel ratio rule, gives us a purity level of gold to be exactly 18 Carats. In other words the Red Gold is 75% pure...

The question pertains to the calculation of gold purity in carats. Without an exact gold percentage in the red gold jewelry, a precise purity or carat value cannot be identified.

The question relates to the understanding of carats and the purity of gold used in jewellery. Generally, 24 carats is defined as 100% pure gold. Hence, the number of carats indicates the proportion of gold in an alloy.

The question asks about a piece of red gold jewellery but lacks the necessary percentage to determine its carat value. If we had an exact percentage of gold in the item, we could calculate the carat value by multiplying the given percentage by 24.

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Explanation:

Ionic equation

NaCl(aq) --> Na+(aq) + Cl-(aq)

Na2SO4(aq) --> 2Na+(aq) + SO4^2-(aq)

In NaCl solution, 1 mole of Na+ is dissociated in 1 liter of solution while in Na2SO4, 2 moles of Na+ is dissociated in 1 liter of solution.

Molecular weight of NA2SO4 = (23*2) + 32 + (16*4)

= 142 g/mol

Molecular weight of NaCl = 23 + 35.5

= 58.5 g/mol

Masses

% Mass of NA+ in Na2SO4 = mass of Na+/total mass of Na2SO4 * 100

= 46/142 * 100

= 32.4%

% Mass of NA+ in NaCl = mass of Na+/total mass of NaCl * 100

= 23/58.5 * 100

= 39.3%

Therefore, the % mass of Na+ in NaCl and Na2SO4 are different so it cannot be used.

You cannot substitute Na2SO4 directly for NaCl based on mass since they have different molar masses. The same mass of Na2SO4 will provide more Na+ ions than NaCl, leading to a change in the Na+ ion concentration.

No, you cannot substitute the same number of grams of Na2SO4 for the NaCl in a solution. This is because NaCl and Na2SO4 have different molar masses and therefore different numbers of moles per gram. The concentration of a solution is determined by the number of moles of solute per unit volume of solvent, not the mass. Hence, using the same mass of a different compound would alter the concentration of Na+ ions in the solution.

For instance, if one mole of NaCl gives us one mole of Na+, one mole of Na2SO4 will provide two moles of Na+. In other words, the same mass of Na2SO4 contains more Na+ ions than the same mass of NaCl. So using the same mass of Na2SO4 in place of NaCl will result in a solution with a higher Na+ ion concentration.

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Answer:

Primary alcohol.

Explanation:

Jones reagent is mixture of chromium trioxide (CrO3) and sulfuric acid (H2SO4) dissolved in a mixture of acetone and water. Alternatively, potassium dichromate (K2CrO7) can be used in place of chromium trioxide because of its carcinogenic nature.

This oxidation reaction is an organic reaction for the oxidation of primary alcohols to aldehydes then carboxylic acid and secondary alcohols to ketones.

Lucas reagent is a solution of anhydrous zinc chloride (ZnCl) in concentrated hydrochloric acid. The reaction involves substitution reaction in which the chloride replaces a hydroxyl group.

A positive test is indicated by a change in appearance of the solution, from clear and colourless to fog-like, which shows the formation of a chloroalkane. Accurate results for this test are observed in tertiary alcohols, as they form alkyl halides the fastest due to the stability of their intermediate tertiary carbocation.

Therefore, an alcohol in which there was a positive Jones test and a negative Lucas test indicates the presence of primary alcohol.

This is because:

A primary alcohol would test positive to Jones test but in Lucas test, the substitution reaction is the slowest as compared to the secondary and tertiary alcohols.

1° alcohols < 2° alcohols < 3° alcohols

So a primary alcohol will give a negative result to Lucas reagent.

Answer:

              1,1,1-trichloroethane being a saturated halogenated alkane will undergo substitution reaction via free radical mechanism. The mechanism is divided in three steps,

Step 1: Initiation:

                          In this step the reaction is started by treating the chlorine gas either with UV light or by sunlight. This results in the formation of free radical.

Step 2: Propagation:

                          In this step the radical formed will react with the hydrogen atom resulting in formation of HCl and generating free radical of corresponding alkane. Hence, the radical will agian react with Cl2 molecule generating another Chlorine radical and corresponding halogenated compound i.e. 1,1,1,2-tetrachloroethane.

Step 3: Termination:

                           This is the last step. In this step the reaction is stopped/terminated. The free radicals react with each other forming a single bonds and stopping the formation of further radicals.

The mechanism is shown below,

Answer:

9.57

Explanation:

Given that:

[tex]pK_{b}=-\log\ K_{b}=-\log(8.0\times 10^{-5})=4.1[/tex]

Considering the Henderson- Hasselbalch equation for the calculation of the pOH of the basic buffer solution as:

[tex]pOH=pK_b+log\frac{[conjugate\ acid]}{[base]}[/tex]

So,  

[tex]pOH=4.1+\log\frac{0.540}{0.250}=4.43[/tex]

pH + pOH = 14  

So, pH = 14 - 4.43 = 9.57

Answer:

35.26 rad

Explanation:

Let's assume the cube in the figure below. If it's in the first octant, then origin (0, 0, 0) is one of the vertices and it's also the common vertex of the diagonals (OB and OE).

The point B is at the y-axis, so since the length is 8, it is (8, 0, 8), and the point E is (8, 8, 8). The vectors of the diagonals are the subtraction of the coordinates of the two points, so OB = <8, 0, 8> and OE = <8, 8, 8>. The angle between two vectors in the tridimensional space is:

θ = cos⁻¹[(OB · OE)/(|OB|·|OE|)]

The module (| |) of a vector <x, y, z> is √(x² + y² + z²)

θ = cos⁻¹[(<8, 0, 8> · <8, 8, 8>)/(√(8² + 0² + 8²) · √(8² + 8² + 8²))]

θ = cos⁻¹[(8*8 + 8*0 + 8*8)/(√128 ·√192)]

θ = cos⁻¹[128/156.77]

θ = cos⁻¹[0.8165]

θ = 35.26 rad

Answer: The concentration of salt (potassium phosphate), phosphoric acid and KOH in the solution is 0.0342 M, 0.0533 M and 0 M respectively.

Explanation:

To calculate the number of moles for given molarity, we use the equation:

[tex]\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}[/tex]     .....(1)

For KOH:

Initial molarity of KOH solution = 0.205 M

Volume of solution = 25.0 mL = 0.025 L   (Conversion factor:   1 L = 1000 mL)

Putting values in equation 1, we get:

[tex]0.205M=\frac{\text{Moles of KOH}}{0.025L}\\\\\text{Moles of KOH}=(0.205mol/L\times 0.025L)=5.123\times 10^{-3}mol[/tex]

For phosphoric acid:

Initial molarity of phosphoric acid solution = 0.175 M

Volume of solution = 25.0 mL = 0.025 L

Putting values in equation 1, we get:

[tex]0.175M=\frac{\text{Moles of }H_3PO_4}{0.025L}\\\\\text{Moles of }H_3PO_4=(0.175mol/L\times 0.025L)=4.375\times 10^{-3}mol[/tex]

The chemical equation for the reaction of KOH and phosphoric acid follows:

[tex]3KOH+H_3PO_4\rightarrow K_3PO_4+3H_2O[/tex]

By Stoichiometry of the reaction:

3 moles of KOH reacts with 1 mole of phosphoric acid

So, [tex]5.123\times 10^{-3}[/tex] moles of KOH will react with = [tex]\frac{1}{3}\times 5.123\times 10^{-3}=1.708\times 10^{-3}mol[/tex] of phosphoric acid

As, given amount of phosphoric acid is more than the required amount. So, it is considered as an excess reagent.

Thus, KOH is considered as a limiting reagent because it limits the formation of product.

Excess moles of phosphoric acid = [tex](4.375-1.708)\times 10^{-3}=2.667\times 10^{-3}mol[/tex]

By Stoichiometry of the reaction:

3 moles of KOH produces 1 mole of potassium phosphate

So, [tex]5.123\times 10^{-3}[/tex] moles of KOH will produce = [tex]\frac{1}{3}\times 5.123\times 10^{-3}=1.708\times 10^{-3}moles[/tex] of potassium phosphate

Moles of potassium phosphate = [tex]1.708\times 10^{-3}moles[/tex]

Volume of solution = [25.0 + 25.0] = 50.0 mL = 0.050 L

Putting values in equation 1, we get:

[tex]\text{Molarity of potassium phosphate}=\frac{1.708\times 10^{-3}}{0.050}=0.0342M[/tex]

Moles of excess phosphoric acid = [tex]2.667\times 10^{-3}moles[/tex]

Volume of solution = [25.0 + 25.0] = 50.0 mL = 0.050 L

Putting values in equation 1, we get:

[tex]\text{Molarity of phosphoric acid}=\frac{2.667\times 10^{-3}}{0.050}=0.0533M[/tex]

Moles of KOH remained = 0 moles

Volume of solution = [25.0 + 25.0] = 50.0 mL = 0.050 L

Putting values in equation 1, we get:

[tex]\text{Molarity of KOH}=\frac{0}{0.050}=0M[/tex]

Hence, the concentration of salt (potassium phosphate), phosphoric acid and KOH in the solution is 0.0342 M, 0.0533 M and 0 M respectively.

A triglyceride formed from glycerol and myristic acid involves esterifying three myristic acid molecules with a glycerol backbone through a dehydration reaction, releasing three water molecules and forming a simple triglyceride.

The triglyceride formed from the esterification of glycerol and three molecules of myristic acid is constructed by attaching each of the three fatty acid molecules to the glycerol backbone through a dehydration reaction. In this process, each fatty acid's carboxyl group (COOH) reacts with one of the hydroxyl groups (OH) on the glycerol molecule, resulting in the formation of an ester bond and the release of water.

The chemical structure of glycerol (H₂C-OH | HC-OH | H₂C-OH) bonded with three myristic acid molecules (which have the formula CH₃(CH₂)₌₁₂COOH) will show three ester linkages replacing the hydroxyl groups of glycerol with the alkyl chains of myristic acid.

Each dehydration synthesis reaction between the carboxyl group of a fatty acid and a hydroxyl group of glycerol results in the release of one molecule of water, leading to a total of three water molecules released when forming the triacylglycerol. If all three OH groups on the glycerol molecule are esterified with myristic acid, the molecule formed is a simple triglyceride because it contains only one type of fatty acid.

Answer:

2 moles of Methane gas will occupy 44.8 L

Explanation:

If one mole of any gas occupies 22.4 L under certain conditions of temperature and pressure, and those conditions are assumed in this question, then we comfortably solve this problem as follows;

1 mole of Methane gas ---------------> 22.4 L

2 moles of Methane gas -------------->?

Cross and multiply, 2 moles of Methane gas = 2 X 22.4 L = 44.8L

Therefore, 2 moles of Methane gas will occupy 44.8 L, if the conditions of temperature and pressure are maintained.

OPTION A IS THE RIGHT SOLUTION.

Answer:

The answer is A

Explanation:

Answer:

6.25×10¹⁹ e⁻

Explanation:

Let's apply a rule of three:

1 e⁻ has 1.60×10⁻¹⁹ C

There, we can think:

1.60×10⁻¹⁹ C of charge are made by 1 e⁻

If we want to produce 1 C, we would need ( 1 . 1) / 1.60×10⁻¹⁹

= 6.25×10¹⁹ e⁻

Answer:

The reaction requires anti-Markovnikov product without sigmatropic rearrangement.

Explanation:

The reaction is known to begin with the concerted syn addition of B and H across the double bond, with the boron adding to the less substituted carbon atom.

The second step of the reaction involves hydrogen peroxide and a base such as NaOH are added, NaOH deprotonates the hydrogen peroxide.

The resulting NaOOH then attacks the boron and sets up the key migration step, where the carbon-boron bond migrates to the oxygen bound to boron, breaking the weak oxygen-oxygen bond . Then the -OH expelled then returns to form a bond on the boron resulting in a deprotonated alcohol (alkoxide). The alkoxide is then protonated by water or some other comparably acidic species.

Hydroboration is a syn addition that gives an anti-Markovnikov product without sigmatropic rearrangement.

The reasons why hydroboration–oxidation reagents were chosen to effect the transformation will be:

It should be noted that hydroboration–oxidation is vital for the production of alcohol. It's needed for the transformation rather than using other reagents because hydroboration-oxidation reagents yield the anti-Markovnikov product of addition.

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Answer:

the range of d metal radii is not very great.

Explanation:

The difference in metallic radii are not great hence the metallic ions are almost similar in size across the series. As a result of this, they can easily take up positions in the lattice of other transition metals leading to the formation of transition metal alloys. This explains the wide range of transition metal alloys used for various purposes in industry.

We can calculate the vapor pressure of liquid cobalt at a given temperature by using the Clausius-Clapeyron equation and the given vapor pressure at another temperature. This involves substituting known values into the equation and solving for the desired vapor pressure.

The question is asking for the calculated vapor pressure of liquid cobalt at a certain temperature based on its known vapor pressure at another temperature. This involves using the Clausius-Clapeyron equation, which describes the relationship between the vapor pressure of a substance and its temperature. Let's denote the initial conditions (i.e., 400 mm Hg at 3.03x10^3 K) as P1 and T1, and the conditions we want to find (i.e., vapor pressure at 3.07x10^3K) as P2 and T2.

First, convert the molar heat of vaporization from kJ/mol to J/mol by multiplying by 1000, which gives 450000 J/mol. Next, the Clausius-Clapeyron equation can be rearranged to solve for P2:

P2 = P1 * exp [ -ΔHvap (1/T2 - 1/T1) / R ]

where ΔHvap is the molar heat of vaporization, R is the ideal gas constant (8.314 J/mol·K). Substituting all known values into this equation will give the vapor pressure of liquid Co at the desired temperature.

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The vapor pressure of liquid cobalt at 3.07x10^3 K is approximately 3748.64 mm Hg.

The vapor pressure of liquid cobalt at a temperature of 3.07x10^3 K can be determined using the Clausius-Clapeyron equation, which relates the vapor pressure of a substance to its temperature. The equation is given by:

[tex]\[ \ln\left(\frac{P_2}{P_1}\right) = -\frac{\Delta H_{\text{vap}}}{R}\left(\frac{1}{T_2} - \frac{1}{T_1}\right) \][/tex]

First, we need to convert [tex]\( \Delta H_{\text{vap}} \)[/tex] from kJ/mol to J/mol to match the units of [tex]\( R \)[/tex]:

[tex]\[ \Delta H_{\text{vap}} = 450 \text{ kJ/mol} \times 1000 \text{ J/kJ} = 450,000 \text{ J/mol} \][/tex]

Now we can plug the values into the Clausius-Clapeyron equation:

[tex]\[ \ln\left(\frac{P_2}{400 \text{ mm Hg}}\right) = -\frac{450,000 \text{ J/mol}}{8.314 \text{ J/(mol·K)}}\left(\frac{1}{3.07x10^3 \text{ K}} - \frac{1}{3.03x10^3 \text{ K}}\right) \][/tex]

Solving for [tex]\( P_2 \):[/tex]

[tex]\[ \ln\left(\frac{P_2}{400}\right) = -\frac{450,000}{8.314}\left(\frac{1}{3.07x10^3} - \frac{1}{3.03x10^3}\right) \] \[ \ln\left(\frac{P_2}{400}\right) = -\frac{450,000}{8.314}\left(\frac{3.03x10^3 - 3.07x10^3}{(3.07x10^3)(3.03x10^3)}\right) \] \[ \ln\left(\frac{P_2}{400}\right) = -\frac{450,000}{8.314}\left(\frac{-40}{3.07x10^3x3.03x10^3}\right) \] \[ \ln\left(\frac{P_2}{400}\right) = -\frac{450,000}{8.314}\left(\frac{-40}{9.3051x10^6}\right) \][/tex]

[tex]\[ P_2 \approx 3748.64 \text{ mm Hg} \][/tex]

Therefore, the vapor pressure of liquid cobalt at 3.07x10^3 K is approximately 3748.64 mm Hg.

The answer & explanation for this question is given in the attachment below.

Answer:

0.00077 qt

Explanation:

Density -

Density of a substance is given by the mass of the substance divided by the volume of the substance .

Hence , d = m / V

V = volume

m = mass ,

d = density ,

From the question ,

The mass mercury = 100 g

Density of mercury = 13.6 g/cm³ .

Hence , by using the above formula ,and putting the corresponding values , the volume of mercury is calculated as -

d = m / V

13.6 g/cm³ = 100 g  / V

V = 7.35 cm³

1 cm³ = 0.001 L

V = 7.35 * 0.001 L = 0.0073 L

Since ,

1 L = 1.06 qt

V = 0.0073* 1.06 qt = 0.0077 qt

Answer:

The concentration is [-1 + sqrt(1+0.11t)]/0.1542 M

Explanation:

Let the concentration of CH3CHO after selected reaction times be y

Rate = Ky^2 = change in concentration of CH3CHO/time

K = 0.0771 M^-1 s^-1

Change in concentration of CH3CHO = 0.358 - y

0.0771y^2 = 0.358-y/t

0.0771ty^2 = 0.358 - y

0.0771ty^2 + y - 0.358 = 0

The value of y must be positive and is obtained in terms of t using the quadratic formula

y = [-1 + sqrt(1^2 -4(0.0771t)(-0.358)]/2(0.0771) = [-1 + sqrt(1+0.11t)]/0.1542 M

Final answer:

The question involves calculating the instantaneous rate of a second order decomposition reaction of acetaldehyde using the given rate constant and concentration values.

Explanation:

The question deals with a second order reaction describing the decomposition of acetaldehyde (CH3CHO) into methane (CH4) and carbon monoxide (CO). A second order reaction rate is dependent on the square of the concentration of one reactant or the product of two reactants concentrations. The rate constant provided (0.0771 M-1 s-1 or 4.71 × 10-8 L mol-1 s-1) is used alongside the concentration of acetaldehyde to determine the instantaneous rate of reaction or, in some scenarios, to deduce the remaining concentration of acetaldehyde at a given time.

Let the concentration of CH3CHO after selected reaction times be y

Rate = Ky^2 = change in concentration of CH3CHO/time

K = 0.0771 M^-1 s^-1

Change in concentration of CH3CHO = 0.358 - y

0.0771y^2 = 0.358-y/t

0.0771ty^2 = 0.358 - y

0.0771ty^2 + y - 0.358 = 0

The value of y must be positive and is obtained in terms of t using the quadratic formula

y = [-1 + sqrt(1^2 -4(0.0771t)(-0.358)]/2(0.0771) = [-1 + sqrt(1+0.11t)]/0.1542 M

Chloroform-Chloroform and Acetone-Acetone interactions are weaker than chloroform-acetone interactions.

Explanation:

Raoult's law states, In a solution, vapor pressure is equal to the sum of the vapor pressures of individual component and this implies when it is multiplied by the mole fraction of that component in the solution. Raoult's Law is expressed by the formula:

                         P solution = Χ solvent [tex]\times[/tex] P0 solvent.

Raoult's law assumes that the components in the mixture are ideal, it assumes that the Chloroform-Chloroform interactions, Chloroform-Acetone, and Acetone-Acetone are the same. A negative deviation is shown by the measurement of observed vapor pressure which is less than that of calculated. Hence the interaction of Chloroform-Chloroform is weaker than the Chloroform-Acetone interactions.  

A solution of chloroform and acetone exhibits a negative deviation from Raoult's law, which means that the interactions between chloroform and acetone are stronger than those of the individual components

A solution of chloroform (CHCl3) and acetone((CH3)2CO) exhibits a negative deviation from Raoult's law. This means the solution has a lower vapor pressure than predicted by Raoult's Law due to the strength of the new intermolecular interactions being stronger than what was present in the pure components. The correct answer is X. chloroform-chloroform and acetone-acetone interactions are weaker than chloroform-acetone interactions. This is because in a solution exhibiting negative deviation, the interactions between the different molecules in the mixture are stronger than those of the pure components. This stronger interaction between chloroform and acetone molecules reduces the overall vapor pressure, leading to a negative deviation from Raoult's Law.

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Answer:

Balanced molecular reaction

Na2S2O5 + H2O ----> 2NaHSO3

Net ionic reaction

(S2O5)^(2-) + H2O ----> 2((HSO3)^(1-))

Explanation:

The sodium bisulfite compound is produced in this manner because Sodium metabisulfite has better preservative properties. It is more stable compared to sodium bisulfite and readily dissolves in water to give the required sodium bisulfite.

Sodium bisulfite sold in the market contains sodium metabisulfite as it is the more stable one of the pair.

And of all the ways of preparing the compound, this is the cheapest and easiest one.

Answer: check explanation.

Explanation:

The balanced molecular equation for the reaction of sodium metabisulfite (Na2S2O5) with water to produce sodium bisulfite (NaHSO3) is given below;

Na2S2O5 + H2O --------> 2NaHSO3.

Therefore, the net ionic equation for this reaction is given below;

S2O5^2- + H2O ------> 2HSO3^-1.

S2O5^2- + H+ ------------> 2HSO3^-1

=====> Why is sodium bisulfite prepared using this method?

sodium bisulfite is prepared using this method because of the following reasons;

(1). It is the most easy way of synthesisizing/producing sodium bisulfite (NaHSO3).

(2). In order to produce sodium bisulfite (NaHSO3) through this method, it does not require high cost,that is to say that it is financial friendly.

(3). Because of high preservative properties of sodium metabisulfite (Na2S2O5).

(4). It reacts with water with ease to produce sodium bisulfite (NaHSO3).

Answer:

Option D. 4.02 kJ

Explanation:

A simple calorimetry problem

Q = m . C . ΔT

ΔT = Final T° - Initial T°

C = Specific heat capacity

m = mass

Let's replace the data

Q = 125 g . 2.42 J/g∘C . (34.8°C -21.5 °C)

Q= 4023.25 J

We must convert the answer to kJ

4023.25 J . 1kJ /1000 =4.02kJ

Answer:

The trend in metallic character as a function of position in the periodic table is that the metallic character increases as you go down a group. Since the ionization energy decreases going down a group (or increases going up a group), the increased ability for metals lower in a group to lose electrons makes them more reactive.

This is not the same for the atomic size, as you go down a column of the periodic table, the atomic radii increase. This is because the valence electron shell is getting a larger and there is a larger principal quantum number, so the valence shell lies physically farther away from the nucleus.

Similarly, it is also different for the ionization energy trend, as you go down the periodic table, it becomes easier to remove an electron from an atom (i.e., IE decreases) because the valence electron is farther away from the nucleus.

Final answer:

The metallic character in the periodic table decreases across a period and increases down a group. It trends similarly to atomic size but oppositely to ionization energy.

Explanation:

Metallic character: The metallic trend follows the trend of the atomic radius. It increases within a group of the periodic table from the top to the bottom and decreases within a period from left to right. Metallic character relates to the ease of losing an electron in a chemical reaction and is opposite to the trend of ionization energy.

Atomic size shows a trend that parallels the metallic character. Atomic size increases down a group because of the increase in electron shells, which makes the valence electrons less tightly held. Conversely, atomic size decreases from left to right within a period due to an increasing effective nuclear charge, which draws electrons closer to the nucleus, reducing the size of the atom.

In summary, the metallic character and atomic size increase from right to left in a period and from top to bottom in a group, while ionization energy generally shows the opposite trend. Hence, the trend in metallic character is similar to the trend in atomic size but opposite to the trend in ionization energy.

Answer:

emits (radiates) , energy difference

Explanation:

According to the Bohr theory, when an electron jumps from higher orbital to the lower orbital, it radiates energy which is equal to the energy difference between the orbitals.

Mathematically, it can be shown as:-

The expression for Bohr energy is shown below as:-

[tex]E_n=-2.179\times 10^{-18}\times \frac{1}{n^2}\ Joules[/tex]

For transitions:

[tex]Energy\ Difference,\ \Delta E= E_f-E_i =-2.179\times 10^{-18}(\frac{1}{n_f^2}-\frac{1}{n_i^2})\ J=2.179\times 10^{-18}(\frac{1}{n_i^2} - \dfrac{1}{n_f^2})\ J[/tex]

[tex]\Delta E=2.179\times 10^{-18}(\frac{1}{n_i^2} - \dfrac{1}{n_f^2})\ J[/tex]

Also, [tex]\Delta E=\frac {h\times c}{\lambda}[/tex]

Where,  

h is Plank's constant having value [tex]6.626\times 10^{-34}\ Js[/tex]

c is the speed of light having value [tex]3\times 10^8\ m/s[/tex]

According to the Bohr model of the atom, when an electron transitions from a higher-energy orbit to a lower-energy orbit, it emits electromagnetic energy equal to the energy difference between the two orbits.

The Bohr model of the atom states that when an electron transitions from a higher-energy orbit to a lower-energy orbit, it emits electromagnetic energy with an energy equal to the difference between the two orbits. This energy is released in the form of a photon.

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Answer:

6.6 is the [tex]pK_a[/tex] of the weak acid.

Explanation:

To calculate the pH of acidic buffer, we use the equation given by Henderson Hasselbalch:

[tex]pH=pK_a+\log(\frac{[salt]}{[acid]})[/tex]

We are given:

[tex]pK_a[/tex] = negative logarithm of acid dissociation constant =?

The ratio of conjugate base to acid is = [tex]\frac{[salt]}{acid}=4[/tex]

pH = 7.2

Putting values in above equation, we get:

[tex]7.2=pK_a+\log(4)[/tex]

[tex]pK_a=7.2-\log(4)=6.598\approx 6.6[/tex]

6.6 is the [tex]pK_a[/tex] of the weak acid.

Answer:

The percent yield of a reaction is 48.05%.

Explanation:

[tex]WO_3+3H_2\rightarrow W+3H_2O[/tex]

Volume of water obtained from the reaction , V= 5.76 mL

Mass of water = m = Experimental yield of water

Density of water = d = 1.00 g/mL

[tex]M=d\times V = 1.00 g/mL\times 5.76 mL=5.76 g[/tex]

Theoretical yield of water : T

Moles of tungsten(VI) oxide = [tex]\frac{51.5 g}{232 g/mol}=0.2220 mol[/tex]

According to recation 1 mole of tungsten(VI) oxide gives 3 moles of water, then 0.2220 moles of tungsten(VI) oxide will give:

[tex]\frac{3}{1}\times 0.2220 mol=0.6660 mol[/tex]

Mass of 0.6660 moles of water:

0.666 mol × 18 g/mol = 11.988 g

Theoretical yield of water : T = 11.988 g

To calculate the percentage yield of reaction , we use the equation:

[tex]\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100[/tex]

[tex]=\frac{m}{T}\times 100=\frac{5.76 g}{11.988 g}\times 100=48.05\%[/tex]

The percent yield of a reaction is 48.05%.

Answer:The percent yield of a reaction is 48.05%.

Explanation: