In the simulation, open the Micro mode, then select the solutions indicated below from the dropdown list above the beaker. The beaker will fill up to the 0.50 L mark with the solution. Arrange the solutions in increasing order of basicity.

drain cleaner, hand soap, blood, milk, orange juice, soda pop

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:

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:

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.

Final answer:

The molality of the sodium hydroxide (NaOH) solution is 0.54 mol/kg.

Explanation:

The molality of a solution is defined as the number of moles of solute per kilogram of solvent. In this case, the solute is sodium hydroxide (NaOH) and the solvent is water.

To calculate the molality, we need to first convert the given mass of NaOH to moles using its molar mass, which is 40.0 g/mol. Then, we need to convert the mass of the solution to kilograms using the density of the solution, which is 1.15 g/mL.

Using the given information:

Mass of NaOH = 24.8 g/L

Density of solution = 1.15 g/mL

Molar mass of NaOH = 40.0 g/mol

The molality can be calculated as follows:

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:

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:

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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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

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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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:

HCl solution - H30+ and Cl ions. pH 3

NaCl - Na+ and Cl-. pH 7

Explanation:

a) HCl solution - the hydrogen ion combines with water molecule to form the hydronium molecule which is responsible for acidity. The chloride ion is also found in solution.

pH = -log [H+] = -log(10^-3) = 3

b) NaCl 10-3 g. The solid dissocates in water forming the Na+ and Cl- ions. None of these ions affect pH

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.

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:

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:

Answer:

The answer would be A. the number of electrons in the outermost electron shell.

Explanation:

These are called valence electrons which are transferred, shared, and rearranged by creating covalent bonds producing new substances.

Hope this helped! :)

The type of chemical reaction an atom chooses is determined by the number of the outermost electrons in the outermost shell of an atom.

About the importance of electrons the below points should be noted;

Therefore the answer is the number of electrons in the outermost electron shell.

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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:

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:

MG is less polar  

Explanation:

The structures of crystal violet (CV) and malachite green (MG) are shown in Figures 1 and 2, respectively.  

The obvious difference is that CV has an extra dimethylamino group (polar).

Alumina is a polar adsorbent, so it retains the more polar substances more strongly and they are eluted last.

MG is less polar than CV, so it is retained less strongly and is eluted first.

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.

Answer:

[H₃O⁺] = 2.63×10⁻¹⁰ M

As pH = 9.57, the solution is basic

Explanation:

We must know this knowledge:

[OH⁻] . [H₃O⁺] = 1×10⁻¹⁴

3.8×10⁻⁵ . [H₃O⁺] = 1×10⁻¹⁴

[H₃O⁺] = 1×10⁻¹⁴ / 3.8×10⁻⁵ → 2.63×10⁻¹⁰ M

Let's determine the pH to state if the solution is acidic or basic

pH < 7 → acidic ; pH > 7 → basi

pH = - log [H₃O⁺]

pH = - log 2.63×10⁻¹⁰ → 9.57

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

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

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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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.

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

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

Answer: The bonds are intermediate between double and single bonds

Explanation:

A closer look at the diagram below shows that the bonds in sulphur IV oxide are intermediate between double and single bonds. Hence they do not have the exact bond angle of single bonds. This is why the bond angle is not exactly 120°. There are two resonance structures in the diagram that clearly show this point.

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 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:

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: