SiH4 Draw the molecule by placing atoms on the grid and connecting them with bonds. Do not identify the charge on each of these species. Include all lone pairs of electrons. To change the symbol of an atom, double-click on the atom and enter the letter of the new atom. +- CHONSPFBrClIXMore

There are three conditions for simplifying the Michaelis-Menten equation: when [S] is significantly less than Km, when [S] equals Km, and when [S] is much larger than Km. These are associated with less than 10% and half of the active sites being occupied by the substrate, and with doubling [S] either almost doubling the rate or having little effect.

The Michaelis-Menten equation expresses the relationship between the initial enzymatic reaction velocity, V₀, and the substrate concentration, [S]. It allows for a simplification of the relationship between [S] and rate, given certain conditions:

Understanding these conditions can significantly simplify the usage of the Michaelis-Menten equation in exploring enzyme kinetics.

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Answer:

There will be 15.05 Joules of heat released.

Explanation:

Step 1: Data given

Mass of mercury = 5.00 grams

The specific heat of mercury = 0.14 J/g*K = 0.14 J/g°C

Raise of temperature from 15.0°C to 36.5 °C

Step 2: Calculate heat released

q = m*c*ΔT

⇒ with q = The heat released in Joules

⇒ with m = the mass of mercury = 5.00 grams

⇒ with c = The specific heat of mercury = 0.14 J/g°C

⇒ with ΔT = The change of temperature = T2 - T1 = 36.5°C - 15.0°C = 21.5 °C

q = 5.00g * 0.14 J/g°C * 21.5 °C = 15.05 J

There will be 15.05 Joules of heat released.

The quantity of heat needed to raise 5g of mercury  from 15.0°C to 36.5°C is  15.05 Joules

Given Data

We know that the expression for the amount of heat is given as

Q =. mcΔT

Substituting our given data into the expression we have

Q = 5*0.14*(36.5 - 15)

Q = 0.7*21.5

Q = 15.05 Joules

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Answer:

C

Explanation:

If 2 moles of potassium chlorate produced 2 moles of MgO, then 2*6/2 moles of MgO are produced when 6 moles of potassium perchlorate is used.

Answer:

The heat energy transfer required to raise the temperature of 13.0 kg liquid ammonia from -50.0 °C to 0.0 °C is 36,896.44 kJ.

Explanation:

The process involved in this problem are :

[tex](1):NH_3(l)(-50^oC)\rightarrow NH_3(l)(-33.4^oC)\\\\(2):NH_3(l)(-33.4^oC)\rightarrow NH_3(g)(-33.4^oC)[/tex]

[tex](3):NH_3(g)(-33.4^oC)\rightarrow NH_3(g)(-0.0^oC)[/tex]

Now we have to calculate the amount of heat released or absorbed in both processes.

For process 1 :

[tex]Q_1=m\times c_{1}\times (T_{final}-T_{initial})[/tex]

where,

[tex]Q_1[/tex] = amount of heat absorbed = ?

m = mass of ammonia = 13000 g

[tex]c_1[/tex] = specific heat of liquid ammonia  = [tex]4.7J/g^oC[/tex]

[tex]T_1[/tex] = initial temperature = [tex]-50.0^oC[/tex]

[tex]T_2[/tex] = final temperature = [tex]-33.4^oC[/tex]

Now put all the given values in [tex]Q_3[/tex], we get:

[tex]Q_1=13000 g\times 4.7 J/g^oC\times ((-33.4)-(-50.0))^oC[/tex]

[tex]Q_1=1,014,260 J=1.1014.260 kJ[/tex]

For process 2 :

[tex]Q_2=n\times \Delta H_{fusion}[/tex]

where,

[tex]Q_2[/tex] = amount of heat absorbed = ?

m = mass of solid ammonia = 13.0 Kg  = 13000 g

n = Moles of ammonia = [tex]\frac{13000 g}{17 g/mol}=764.71 mol[/tex]

[tex]\Delta H_{vap}[/tex] = enthalpy change for vaporization=23.5 kJ/mol

Now put all the given values in [tex]Q_1[/tex], we get:

[tex]Q_2=764.71 mol\times 23.5 kJ/mol=17,970.6 kJ[/tex]

For process 3 :

[tex]Q_3=m\times c_{p,l}\times (T_{final}-T_{initial})[/tex]

where,

[tex]Q_3[/tex] = amount of heat absorbed = ?

m = mass of ammonia = 13000 g

[tex]c_2[/tex] = specific heat of gaseous ammonia  = [tex]2.2J/g^oC[/tex]

[tex]T_1[/tex] = initial temperature = [tex]-33.4^oC[/tex]

[tex]T_2[/tex] = final temperature = [tex]0.0^oC[/tex]

Now put all the given values in [tex]Q_3[/tex], we get:

[tex]Q_3=13000 g\times 2.2J/g^oC\times (0.0-(-33.4))^oC[/tex]

[tex]Q_3=955,240 J=955.240 kJ[/tex]

The heat energy transfer required to raise the temperature of 13.0 kg liquid ammonia from -50.0 °C to 0.0 °C  = Q

[tex]Q=Q_1+Q_2+Q_3=17,970.6 kJ+17,970.6 kJ+955.240 kJ[/tex]

Q = 36,896.44 kJ

Answer:

Ionic crystal

Explanation:

An ionic crystal has a high melting point. In an ionic crystal, the ions are tightly held in electrostatic attraction by their oppositely charged neighbors forming a rigid three dimensional lattice. However, when this solid melts, the rigid crystal structure collapses and the individual ions become free and mobile. Hence the melt conducts electricity.

Answer:

Their particles are arranged randomly

Explanation:

The random arrangement of molecules cause the translational motion of the liquid and Brownian motion of the gas

Answer: principal quantum number (n), azimuthal quantum number (l) and magnetic quantum number (m)

Explanation:

According to quantum mechanics, there are three sets of quantum numbers which accurately describes the electrons position; principal quantum number (n), azimuthal quantum number (l) and magnetic quantum number (m). These three quantum numbers are obtained from Schrödinger wave equation.

Answer:

CH3Cl + Cl2 -> CH2ClCH2Cl + HCl

Explanation:

Chlorination is an addition reaction that involves addition of Chlorine to a chemical compound. Methane undergo addition reaction with chlorine to produce chloromethane, which undergo further further chlorination to produce dichloromethane.

For the chlorination of methane, one of the hydrogen atoms of methane is replaced by a chlorine atom of the chlorine gas (molecule)

CH4         +          Cl2          ->           CH3Cl        +          HCl

methane         chlorine               chloromethane     Hydrochloric acid

Using their structure:

        H                                                                  H

         I                                                                    I

  H - C - H            +       Cl - Cl        ->            H - C - Cl               +         HCl

         I                                                                   I

        H                                                                  H

     methane                                         chloromethane

In further chlorination of chloromethane, one of the hydrogen atom of chloromethane is replaced by the chlorine atom in the chlorine gas (molecule)

Chlorination of Chloromethane to dichloromethane is represented below:

       CH3Cl         +          Cl2          ->        CH2ClCH2Cl        +          HCl

Chloromethane         Chlorine               dichloromethane     Hydrochloric acid

Using their structure:

        Cl                                                                 Cl

         I                                                                    I

  H - C - H            +       Cl - Cl        ->            H - C - Cl               +         HCl

         I                                                                   I

        H                                                                  H

Chloromethane                 chlorine                        dichloromethane

Organic compound    Inorganic compound

From the chemical equation above, an organic compound (Chloromethane) reacts with an inorganic compound (Chlorine) .

Answer:

The molar mass of the solid is 511.3 g/mol

Explanation:

Step 1: Data given

Mass of the sample = 0.588 grams

Mass of benzene = 11.5 grams

The solution freezes at 5.02 °C

The freezing point of benzene is 5.51 °C

The freezing point depression constant, kf = 4.90 °C/m

Step 2:  Determine the temperature change

Δt = 5.51 - 5.02 = 0.49°C

Step 3: Determine number of moles

Δt = i*Kf*m

⇒ with i = the number of dissolved particles the solute produces = 1

⇒ with Kf = the molal freezing point depression constant Kf = 4.90 °C*Kg/mol

⇒ with m =  the molality of the solute

0.49 °C = (1) (4.90 °C kg/mol) (x / 0.01150 kg)

x = 0.00115 mol

Step 4: Calculate molar mass

0.588 grams / 0.00115 mol = 511.3 g/mol

The molar mass of the solid is 511.3 g/mol

Answer:

Micelle Formation

Explanation:

The soap molecule has a nonpolar and a polar cationic end when it is dissolved in water. When it dissolves in water, the sodium stearate congregates to form small spheres (called micelles) with the on the inside and the on the surface. The anionic end can attract and interacts with the polar water molecules, while the interact with the nonpolar grease. This allows the soapy water to remove the grease by trapping the grease inside the micelle.

Because greased is electrically neutral and water is polarity, water alone will not simply remove oil off dishes or hands. The grease, on the other hand, dissolves when soap is added to water.

This traps the greasy within the micelle, allowing the soapy liquid to remove it.

So,

In case 1: Hydro-carbon  

In case 2: Anionic  

In case 3: Hydro-carbon  

In case 4: Anionic  

In case 5: Anionic  

In case 6: Hydro-carbon

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Answer:

-1

Explanation:

The relation between Kp and Kc is given below:

[tex]K_p= K_c\times (RT)^{\Delta n}[/tex]

Where,  

Kp is the pressure equilibrium constant

Kc is the molar equilibrium constant

R is gas constant

T is the temperature in Kelvins

Δn = (No. of moles of gaseous products)-(No. of moles of gaseous reactants)

For the first equilibrium reaction:

[tex]2SO_2_{(g)}+O_2_{(g)}\rightleftharpoons2SO_3_{(g)} [/tex]

Δn = (2)-(2+1) = -1  

Thus, Kp is:

[tex]K_p=  K_c\times (RT)^{-1}[/tex]

Answer:

Doping with galium or indium will yield a p-type semiconductor while doping with arsenic, antimony or phosphorus will yield an n-type semiconductor.

Explanation:

Doping refers to improving the conductivity of a semiconductor by addition of impurities. A trivalent impurity leads to p-type semiconductor while a pentavalent impurity leads to an n-type semiconductor.

Germanium, being a group 4A semiconductor, has four valence electrons. When a group 3A element (with fewer valence electrons) is used as a dopant in germanium, it creates "holes" in the crystal lattice, resulting in p-type semiconductors.

When a group 5A element (with more valence electrons) is used as a dopant in germanium, it introduces extra electrons into the crystal lattice, creating an excess of negative charge carriers and resulting in n-type semiconductors.

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Final answer:

The initial rate of formation of P2 in the dimerization reaction of a protein with a concentration of 1.6 × 10^-4 M is approximately 1.59 × 10^-10 M/s.

Explanation:

To calculate the initial rate of formation of P2 in the dimerization reaction of a protein, we can use the rate equation:

rate = k[P]^2

where k is the rate constant and [P] is the concentration of the protein.

Plugging in the given values:

rate = (6.2 × 10^-3 M^-1s^-1)(1.6 × 10^-4 M)^2

Simplifying:

rate = 6.2 × 10^-3 M^-1s^-1 * (1.6 × 10^-4 M)^2

rate = 6.2 × 10^-3 M^-1s^-1 * 2.56 × 10^-8 M^2

rate ≈ 1.59 × 10^-10 M/s

Therefore, the initial rate of formation of P2 is approximately 1.59 × 10^-10 M/s.

To find the mass of carbon tetrachloride needed, multiply the volumes by the density. The student needs to weigh out 135 g of carbon tetrachloride, rounded to the correct number of significant digits.

To solve this problem, we need to multiply the volume of carbon tetrachloride needed by the student (85.0 mL) by its density (1.59 g/cm^3). However, note that the volume is given in milliliters (mL) while the density is given in grams per cubic centimeter (g/cm^3). You need to know that 1 mL is equivalent to 1 cm^3. So, we don’t have to convert the units. Hence, the calculation becomes:

Mass = Density x Volume = 1.59 g/cm^3 x 85.0 mL = 135.15 g.

Since the given values only have three significant digits, the answer should also be rounded to three significant digits so the student needs to weigh out 135 g of carbon tetrachloride.

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Answer:

The system is at equilibrium

Explanation:

Let's consider the following reversible reaction.

2 SO₂(g) + O₂(g) ⇄ 2 SO₃(g)

To determine whether a reaction is at equilibrium or not, we have to calculate the reaction quotient (Q).

[tex]Q=\frac{[SO_{3}]^{2}}{[SO_{2}]^{2}.[O_{2}]} =\frac{(9.00)^{2} }{(6.00)^{2}\times 0.45} =5.0[/tex]

Since Q = K (equilibrium constant), regardless of the significant figures, the system is at equilibrium.

The correct answer is at equilibrium.

The initial concentrations given correspond to the equilibrium concentrations for the reaction at 1300 K, and no shift in the reaction direction is necessary to achieve equilibrium because [tex]\( Q = K \).[/tex]

The equilibrium position of the reaction under the given conditions, we need to compare the initial concentrations with the equilibrium concentrations and the equilibrium constant [tex]\( K \).[/tex]

Given.

- Equilibrium constant K = 5.00

- Initial concentrations.

- [tex]\([SO_2]_{\text{initial}} = 6.00 \)[/tex] M.

- [tex]\([O_2]_{\text{initial}} = 0.45 \)[/tex]  M.

- [tex]\([SO_3]_{\text{initial}} = 9.00 \)[/tex] M.

The reaction is. [tex]\( 2 SO_2(g) + O_2(g) \leftrightarrow 2 SO_3(g) \)[/tex]

1.Calculate Reaction Quotient Q

[tex]\[ Q = \frac{[\text{SO}_3]^2}{[\text{SO}_2]^2 [\text{O}_2]} \][/tex]

Substitute the initial concentrations into the expression for [tex]\( Q \).[/tex]

[tex]\[ Q = \frac{(9.00)^2}{(6.00)^2 \times 0.45} \][/tex]

[tex]\[ Q = \frac{81.00}{16.20} \][/tex]

[tex]\[ Q \approx 5.00 \][/tex]

2. Compare Q with K.

- Q = 5.00

- K = 5.00

Since Q the reaction quotient is equal to K the equilibrium constant the system is already at equilibrium.

Answer:

Solid KOH will corrode the skin but a solution of KOH in triethyleneglycol will not.

The temperature is monitored with an external thermometer also because the sand looses some heat to energy exchange with the surrounding.

Explanation:

KOH is a deliquescent solid which is very corrosive upon contact with skin. Its solution in an organic liquid is not corrosive.

Secondly, when exposed to the surrounding, heat is lost according to the laws of thermodynamics.

Answer:

Sample: 1.67

Blank: 0.070

Explanation:

The absorbance of a solution (A) is explained by the Beer-Lambert law.

A = ε . l . c

where,

ε is the absorptivity of the species

l is the optical path length

c is the molar concentration of the species

As we can see, the absorbance is directly proportional to the path length, that is, the length of the cuvette. If l is increased 5 times (1.00 cm to 5.00 cm), the absorbance will also be increased 5 times.

The absorbance of the sample will be 5 × 0.333 = 1.67

The absorbance of the blank will be 5 × 0.014 = 0.070

According to Beer's law, if the cuvette size increases fivefold from 1.00 cm to 5.00 cm, the absorbance will also increase fivefold. Therefore, the absorbance of the sample would become 1.665, and the reagent blank would become 0.070.

The absorbance of a solution is related to the path length (in this case, cuvette size) according to Beer's law, which states that A = εlc, where A is absorbance, ε is molar absorptivity, l is path length, and c is concentration. Thus, absorbance is directly proportional to the path length.

If the path length increases from 1.00 cm to 5.00 cm, the absorbance also increases by the same factor. Therefore, the absorbance of the sample and the reagent blank in a 5.00 cm cuvette would be:

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Answer:

high C:N; N supply; energy

Explanation:

Nitrogen supply is required for microbial growth to synthesize nutrients such as amino acids and proteins. For a high C:N ratio, the amount of nitrogen supply is considerably small compared with the amount of carbon. As a result, a low amount of nutrients is released during decomposition.

In the given question, the organic matter with a __high C:N___ ratio results in a low net release of nutrients during decomposition. This is because microbial growth is more limited by __N supply____ than it is by __energy_____.

Answer:

For a: The empirical formula for the given compound is [tex]CH[/tex]

For b: The empirical and molecular formula for the given organic compound are [tex]C_{10}H_{20}O[/tex]

Explanation:

The chemical equation for the combustion of hydrocarbon follows:

[tex]C_xH_y+O_2\rightarrow CO_2+H_2O[/tex]

where, 'x', and 'y' are the subscripts of Carbon and hydrogen respectively.

We are given:

Conversion factor used:  1 g = 1000 mg

Mass of [tex]CO_2=5.86mg=5.86\times 10^{-3}g[/tex]

Mass of [tex]H_2O=1.37mg=1.37\times 10^{-3}g[/tex]

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44g of carbon dioxide, 12 g of carbon is contained.

So, in [tex]5.86\times 10^{-3}g[/tex]  of carbon dioxide, [tex]\frac{12}{44}\times 5.86\times 10^{-3}=1.60\times 10^{-3}g[/tex] of carbon will be contained.

For calculating the mass of hydrogen:

In 18g of water, 2 g of hydrogen is contained.

So, in [tex]1.37\times 10^{-3}g[/tex] of water, [tex]\frac{2}{18}\times 1.37\times 10^{-3}=0.152\times 10^{-3}g[/tex] of hydrogen will be contained.

To formulate the empirical formula, we need to follow some steps:

Moles of Carbon = [tex]\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{1.60\times 10^{-3}g}{12g/mole}=0.133\times 10^{-3}moles[/tex]

Moles of Hydrogen = [tex]\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.152\times 10^{-3}g}{1g/mole}=0.152\times 10^{-3}moles[/tex]

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is [tex]0.133\times 10^{-3}[/tex] moles.

For Carbon = [tex]\frac{0.133\times 10^{-3}}{0.133\times 10^{-3}}=1[/tex]

For Hydrogen = [tex]\frac{0.152\times 10^{-3}}{0.133\times 10^{-3}}=1.14\approx 1[/tex]

The ratio of C : H = 1 : 1

Hence, the empirical formula for the given compound is [tex]CH[/tex]

The chemical equation for the combustion of menthol follows:

[tex]C_xH_yO_z+O_2\rightarrow CO_2+H_2O[/tex]

where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.

We are given:

Mass of [tex]CO_2[/tex]  = 0.2829 g

Mass of [tex]H_2O[/tex] = 0.1159 g

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44g of carbon dioxide, 12 g of carbon is contained.

So, in 0.2829  g of carbon dioxide, [tex]\frac{12}{44}\times 0.2829=0.077g[/tex] of carbon will be contained.

For calculating the mass of hydrogen:

In 18g of water, 2 g of hydrogen is contained.

So, in 0.1159 g of water, [tex]\frac{2}{18}\times 0.1159=0.013g[/tex] of hydrogen will be contained.

Mass of oxygen in the compound = (0.1005) - (0.077 + 0.013) = 0.105 g

To formulate the empirical formula, we need to follow some steps:

Moles of Carbon = [tex]\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{0.077g}{12g/mole}=0.0064moles[/tex]

Moles of Hydrogen = [tex]\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.013g}{1g/mole}=0.013moles[/tex]

Moles of Oxygen = [tex]\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{0.0105g}{16g/mole}=0.00065moles[/tex]

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.00065 moles.

For Carbon = [tex]\frac{0.0064}{0.00065}=9.84\approx 10[/tex]

For Hydrogen = [tex]\frac{0.013}{0.00065}=20[/tex]

For Oxygen = [tex]\frac{0.00065}{0.00065}=1[/tex]

The ratio of C : H : O = 10 : 20 : 1

The empirical formula for the given compound is [tex]C_{10}H_{20}O[/tex]

For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.

The equation used to calculate the valency is:

[tex]n=\frac{\text{Molecular mass}}{\text{Empirical mass}}[/tex]

We are given:

Mass of molecular formula = 156 g/mol

Mass of empirical formula = 156 g/mol

Putting values in above equation, we get:

[tex]n=\frac{156g/mol}{156g/mol}=1[/tex]

Multiplying this valency by the subscript of every element of empirical formula, we get:

[tex]C_{(1\times 10)}H_{(1\times 20)}O_{(1\times 1)}=C_{10}H_{20}O[/tex]

Hence, the empirical and molecular formula for the given organic compound are [tex]C_{10}H_{20}O[/tex]

Final answer:

The empirical formula of toluene, derived from combustion yielding 5.86 mg of CO2 and 1.37 mg of H2O, is C7H8. For menthol, from 0.1005 g combusted to give 0.2829 g of CO2 and 0.1159 g of H2O and a molar mass of 156 g/mol, its empirical and molecular formulas are both C10H20O.

Explanation:

Empirical and Molecular Formulas of Organic Compounds

Combustion analysis is commonly used to determine the empirical formulas of organic compounds. The process involves burning a sample of the compound to produce CO2 and H2O, which can then be analyzed to determine the amounts of carbon and hydrogen in the original compound.

The empirical formula of toluene, given that combustion yields 5.86 mg of CO2 and 1.37 mg of H2O with only carbon and hydrogen present, can be found by:

Converting the mass of CO2 to moles of carbon.

Converting the mass of H2O to moles of hydrogen.

Determining the simplest whole number ratio of carbon to hydrogen atoms to find the empirical formula.

Following these steps, we can determine that the empirical formula of toluene is C7H8.

For menthol, with a 0.1005-g sample yielding 0.2829 g of CO2 and 0.1159 g of H2O, and a known molar mass of 156 g/mol, we:

Convert the masses of CO2 and H2O to moles of carbon and hydrogen, respectively.

Assume the rest of the mass is due to oxygen and calculate its moles.

Determine the empirical formula.

Calculate the molecular formula using the empirical formula and the given molar mass.

By doing so, we find that menthol's empirical formula is C10H20O and its molecular formula is also C10H20O.

Answer:

k = 0,0026 s⁻¹

Explanation:

To calculate the rate constant it is necessary to find out the order of reaction. The R² nearest 1 will be the order of reaction.

For zeroth order the integrated rate law is:

[A] = [A]₀ -kt

The graph of [A] vs t gives a correlation coefficient R² of 0,9944.

The first order is:

ln [A] = ln [A]₀ -kt

The graph of ln [A] vs t gives a R² of 1

The second order is:

1/[A] = 1/[A]₀ -kt

The graph of 1/[A] vs t gives a R² of 0,9942

As R² = 1 for first order, the descomposition of azomethane follows this kinetics order. The lineal correlation is:

y = b - mx

y = 1,5077 - 0,0026x

ln [A] = ln [A]₀ -kt

That means:

-k = - 0,0026 s⁻¹

k = 0,0026 s⁻¹

I hope it helps!

Answer:

q = 38,5 kJ

Explanation:

In its melting point, at 0°C, water is liquid. The boiling point of water is 100°C. It is possible to estimate the heat you required to raise the temperature of water from 0°C to 100°C using:

q = C×m×ΔT

Where C is specific heat of water (4,184J/g°C), m is mass of water (92,0g) and ΔT is change in temperature (100°C-0°C = 100°C)

Replacing:

q = 4,184J/g°C×92,0g×100°C

q = 38493 J, in kilojoules:

q = 38,5 kJ

I hope it helps!

The heat energy required to raise the temperature of 92.0g of water from its melting point to its boiling point is calculated using the formula q = mcΔT. Using given values and converting to kilojoules, the answer is 38.51kJ.

To find the heat required to raise the temperature of water from its melting point to its boiling point, we can make use of the formula for calculating heat (q):

q = mcΔT

Where:

In your case, m = 92g, c = 4.184 J/(g??C), and ΔT = 100°C (boiling point of water) - 0°C (melting point of water) = 100°C.

Substitute these values into the formula:

q = (92g)(4.184 J/g°C)(100°C)

Calculate to get q = 38506.88 J, in kilojoules convert by dividing by 1000 to get 38.51 kJ.

Hence, the heat required to raise the temperature of 92.0g of water from its melting point to its boiling point is 38.51 kJ.

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Answer:

HCN

Explanation:

Acids are the species which furnish hydrogen ions in the solution or is capable of forming bonds with electron pair species as they are electron deficient species.

When an acid donates a proton, it changes into a base which is known as its conjugate base.

Bases are the species which furnish hydroxide ions in the solution or is capable of forming bonds with electron deficient species as they are electron rich species. When a base accepts a proton, it changes into a acid which is known as its conjugate acid.

The acid and the base which is only differ by absence or presence of the proton are known as acid conjugate base pair.

Also, the strongest acid leads to the weakest conjugate base and vice versa.

Thus, Out of HI, HCN, [tex]HNO_3[/tex], [tex]HClO_4[/tex] and HCl , the weakest acid is:- HCN

Thus, HCN corresponds to the strongest conjugate base.

Final answer:

Lithium aluminum hydride reduces carboxylic acids to primary alcohols via an aldehyde intermediate, although the aldehyde is not isolated and is reduced further to a primary alcohol in one step. So the correct option is C.

Explanation:

Lithium aluminum hydride reduces carboxylic acids to primary alcohols by first forming an aldehyde as the intermediate. The reduction process does not stop at the aldehyde stage but proceeds to reduce the aldehyde further to a primary alcohol. When a carboxylic acid is reduced, the process typically bypasses the isolation of the intermediate aldehyde due to the strong reducing power of lithium aluminum hydride (LiAlH4), which is capable of fully reducing the carboxylic acid to the alcohol in one step.

Answer:

b. The final state of the substance is a gas.

d. The sample is initially a liquid. One or more phase changes will occur.

Explanation:

Methane has the following properties:

*"Normal" refers to normal pressure (1 atm).

According to this, we can affirm:

A sample of methane at a pressure of 1.00 atm and a temperature of 93.1 K is heated at constant pressure to a temperature of 158 K. Which of the following are true? Choose all that apply.

a. The liquid initially present will solidify.  FALSE. The liquid will vaporize.

b. The final state of the substance is a gas.  TRUE.

c. The sample is initially a solid.  FALSE. The sample is initially a liquid.

d. The sample is initially a liquid. One or more phase changes will occur. TRUE.

The substance would start as a gas, then transition to a solid without becoming a liquid due to temperature conditions.

The substance would begin as a gas and as the pressure increases, it would compress and eventually solidify without liquefying as the temperature is below the triple point temperature.

Describe the phase changes from -80°C to 500°C at 2 atm.

The sample is initially a gas, condenses to a solid as the pressure increases, and then melts when the pressure is increased further to give a liquid.

Answer:

94.56KJ

Explanation:

1.7 grams of CH4 contains 1.7/16 moles of CH4.

If 1 mole of CH4 give 890KJ

1.7/16 moles of CH4 gives 1.7/16*890 = 94.56KJ

Answer:

Orbital mixing or sigma pi crossover

Explanation:

Looking at the diagram below, the ordering of molecular orbitals differ between carbon and oxygen molecules. The reason for this is that, with increase in atomic number, the energies of sigma 2p bonding molecular orbital and pi 2p molecular orbital come close together but are energetically far apart in the lighter elements. Following this ordering of orbitals, the two degenerate pi- 2p antibonding orbitals are singly filled accounting for the paramagnetism of oxygen while the two degenerate pi-2p bonding orbitals in carbon molecule is doubly occupied hence the molecule is diamagnetic.

O₂ is paramagnetic with a bond order of 2 due to two unpaired electrons in antibonding molecular orbitals. C₂ is diamagnetic and also has a bond order of 2, but with all electrons paired due to two additional electrons filling the antibonding orbitals when compared to B₂.

The student is asking about the paramagnetic nature of O₂ and the diamagnetic nature of C₂ despite both having a bond order of 2. According to molecular orbital theory, the electronic configuration of O₂ includes two unpaired electrons in antibonding
(*12py, *12pz) molecular orbitals. These unpaired electrons cause O₂ to be paramagnetic. Calculating the bond order, we find that there are eight electrons in bonding orbitals and four electrons in antibonding orbitals, resulting in a bond order of 2 (8 - 4 = 2), which confirms the double covalent bond in O₂. In contrast, the C₂ molecule, with two additional electrons compared to B₂, fills up its antibonding orbitals completely, leading to all electrons being paired and thus making C₂ diamagnetic.

Explanation:

As per Brønsted-Lowry concept of acids and bases, chemical species which donate proton are called Brønsted-Lowry acids.

The chemical species which accept proton are called Brønsted-Lowry base.

(a) [tex]HNO_3 + H_2O \rightarrow H_3O^+ + NO_3^-[/tex]

[tex]HNO_3[/tex] is Bronsted lowry acid and [tex]NO_3^-[/tex] is its conjugate base.

[tex]H_2O[/tex] is Bronsted lowry base and [tex]H_3O^+[/tex] is its conjugate acid.

(b)

[tex]CN^- + H_2O \rightarrow HCN + OH^-[/tex]

[tex]CN^-[/tex] is Bronsted lowry base and HCN is its conjugate acid.

[tex]H_2O[/tex] is Bronsted lowry acid and [tex]OH^-[/tex] is its conjugate base.

(c)

[tex]H_2SO_4 + Cl^- \rightarrow HCl + HSO_4^-[/tex]

[tex]H_2SO_4[/tex] is Bronsted lowry acid and [tex]HSO_4^-[/tex] is its conjugate base.

Cl^- is Bronsted lowry base and HCl is its conjugate acid.

(d)

[tex]HSO_4^-+OH^- \rightarrow SO_4^{2-}+H_2O[/tex]

[tex]HSO_4^-[/tex] is Bronsted lowry acid and [tex]SO_4^{2-}[/tex] is its conjugate base.

OH^- is Bronsted lowry base and [tex]H_2O[/tex] is its conjugate acid.

(e)

[tex]O_{2-}+H_2O \rightarrow 2OH^-[/tex]

[tex]O_{2-}[/tex] is Bronsted lowry base and OH- is its conjugate acid.

[tex]H_2O[/tex] is Bronsted lowry acid and OH- is its conjugate base.

The identification and labelling of the Brønsted-Lowry acid and base, and their respective conjugate are done below.

Brønsted-Lowry acids are chemical species that donate proton (H+) while the chemical species which accept proton (H+) are called Brønsted-Lowry base

Based on this question, the conjugate acid and base of the equations given are as follows:

For equation a:

For equation b:

For equation c:

For equation d:

For equation e:

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Answer: the heat-sensitive glassware that were given are : Volumetric and Graduated cylinder.

Explanation:glass material that reacts to ambient temperatures radiated off of other surfaces like hands or water is known as heat sensitive glassware. They are not meant to be heated and could shatter if exposed to a heat source. Examples from the video includes Volumetric and Graduated cylinder. Hope this helps. Thanks.

The glassware in the video that are shown to be heat sensitive are volumetric and graduated cylinders.

The exposure of heat to the system has been resulted in the atoms of the system to excite. The material that radiate hands or water with the exposure of the heat are termed as heat-sensitive materials.

The heat sensitive  materials are not heated at the high temperatures, as they may shatter. This has been the result, as there has been an increase in the volume of the glassware with the explosion of heat.

The heat results in the glass molecules to raise the temperature. The excited electrons emit the radiations, that has been the representation of the material to be heat-sensitive.

The glassware in the video that are shown to be heat sensitive are volumetric and graduated cylinders.

Learn more about heat-sensitive glassware, here:

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Answer:

A

Explanation:

Iron has the ground state electronic configuration [Ar]3d64s2

Fe2+ has the electronic configuration [Ar]3d6.

In an octahedral crystal field, there are two sets of degenerate orbitals; the lower lying three t2g orbitals, and the higher level two degenerate eg orbitals. Strong field ligands cause high octahedral crystal field splitting, there by separating the two sets of degenerate orbitals by a tremendous amount of energy. This energy is much greater than the pairing energy required to pair the six electrons in three degenerate orbitals. Since CN- is a strong field ligand, it leads to pairing of six electrons in three degenerate orbitals

The Fe(CN)64- ion is diamagnetic due to the significant energy difference caused by strong field ligands like CN-, resulting in pairing of electrons in the 3 low energy d orbitals before filling the higher energy ones.

The answer is A. There are 3 low lying d orbitals, which will be filled with 6 electrons before filling the 2, assumed to be degenerate, higher energy orbitals. When a metal ion is coordinated to ligands, such as in Fe(CN)64-, the degeneracy of the 3d orbitals is broken (they have different energies), due to the electrostatic interactions between the ligands and the orbitals. In the case of strong-field ligands like CN-, the energy difference is significant enough to cause pairing of electrons in the 3 low energy d orbitals before the 2 high energy orbitals are populated, resulting in a diamagnetic species.

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The electron configuration shows the arrangement of electrons in atoms.

The electron configuration gives a description of the location of the electron in an atom.

Note that;

n = Principal quantum number

l = orbital quantum number

m = magnetic quantum number

s = spin quantum number

n = 4, l = 3, m = 3, -2, -1, 0, 1, 2, 3, s= ±1/2

n = 3, l = 0, m = 0, s = ±1/2

n = 4, l = 0, m = 0, s = ±1/2

n = 4, l = 0, m = 0, s = ±1/2

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Final answer:

The quantum numbers for specific electrons in the 4f6, 3s1, 28Ni, and 22Ti configurations have been provided according to their respective principal, azimuthal, magnetic, and spin quantum numbers.

Explanation:

The four quantum numbers are principal (n), azimuthal (l), magnetic (ml), and spin (ms). Here are their specified values for the given electrons:

Final answer:

The quantum numbers for specific electrons in the 4f6, 3s1, 28Ni, and 22Ti configurations have been provided according to their respective principal, azimuthal, magnetic, and spin quantum numbers.

Explanation:

The four quantum numbers are principal (n), azimuthal (l), magnetic (ml), and spin (ms). Here are their specified values for the given electrons: