Be sure to answer all parts. Coal gasification is a multistep process to convert coal into cleaner-burning fuels. In one step, a coal sample reacts with superheated steam: C(s) + H2O(g) → CO(g) + H2(g) ΔH o rxn = 129.7 kJ (a) Combine the reaction above with the following two to write an overall reaction for the production of methane: CO(g) + H2O(g) → CO2(g) + H2(g) ΔH o rxn = −41 kJ CO(g) + 3H2(g) → CH4(g) + H2O(g) ΔH o rxn = −206 kJ In the overall reaction, include the physical states of each product and reactant. (b) Calculate ΔH o rxn for this overall change. 12.03 kJ (c) Using the value in (b) and calculating ΔH o rxn for the combustion of methane, find the total heat for gasifying 6.28 kg of coal and burning the methane formed. Assume water forms as a gas and the molar mass of coal is 12.00 g/mol.

Explanation:

Magnesium oxide

Answer:

109.09°C

Explanation:

Given that:

the capacity of the cooling car system = 5.6 gal

volume of solute = volume of the water; since a 50/50 blend of engine coolant and water (by volume) is used.

∴ [tex]\frac{5.60}{2}gallons = 2.80 gallons[/tex]

Afterwards, the mass of the solute and the mass of the water can be determined as shown below:

mass of solute = [tex](M__1}) = Density*Volume[/tex]

                          [tex]= 1.1g/mL *2.80*\frac{3785.41mL}{1gallon}[/tex]

                         [tex]= 11659.06grams[/tex]

On the other hand; the mass of water = [tex](M__2})= Density*Volume[/tex]

                         [tex]= 0.998g/mL *2.80*\frac{3785.41mL}{1gallon}[/tex]

                        [tex]= 10577.95 grams[/tex]

Molarity = [tex]\frac{massof solute*1000}{molarmassof solute*massofwater}[/tex]

              =  [tex]\frac{11659.06*1000}{62.07*10577.95}[/tex]

              = 17.757 m

              ≅ 17.76 m

∴  the boiling point of the solution is calculated using the  boiling‑point elevation constant for water and the Molarity.

[tex]\Delta T_{boiling} = k_{boiling}M[/tex]

where,

[tex]k_{boiling}[/tex] = 0.512 °C/m

[tex]\Delta T_{boiling}[/tex] =  100°C + 17.56 × 0.512

              = 109.09 °C

Answer:

43.47 g

Explanation:

The boiling point elevation is described as:

Where ΔT is the difference in boiling points: 120.6-118.4 = 2.2 °C

K is the boiling point elevation constant, K= 2.40 °C·kg·mol⁻¹

and m is the molality of the solution (molality = mol solute/kg solvent).

So first we calculate the molality of the solution:

Now we calculate the moles of benzamide (C₇H₇NO, MW=315g/mol), using the given mass of the liquid X.

Finally we convert moles of C₇H₇NO into grams, using its molecular weight:

Final answer:

To calculate the mass of C7H7NO dissolved in the solution, use the formula for boiling point elevation in a solution.

Explanation:

The mass of C7H7NO that was dissolved in the solution can be calculated using the formula:

ΔTb = i * K * m

Where ΔTb is the boiling point elevation, i is the van't Hoff factor (number of particles the solute dissociates into), K is the boiling point elevation constant, and m is the molality of the solution.

Substitute the given values into the equation to find the mass of C7H7NO dissolved.

Answer:

The rate of the reaction increased by a factor of 1012.32

Explanation:

Applying Arrhenius equation

ln(k₂/k₁) = Ea/R(1/T₁ - 1/T₂)

where;

k₂/k₁ is the ratio of the rates which is the factor

Ea is the activation energy = 274 kJ/mol.

T₁ is the initial temperature = 231⁰C = 504 k

T₂ is the final temperature = 293⁰C = 566 k

R is gas constant = 8.314 J/Kmol

Substituting this values into the equation above;

ln(k₂/k₁) = 274000/8.314(1/504 - 1/566)

ln(k₂/k₁) = 32956.4589 (0.00198-0.00177)

ln(k₂/k₁)  = 6.92

k₂/k₁ = exp(6.92)

k₂/k₁ = 1012.32

The rate of the reaction increased by 1012.32

Answer:

By a factor of 2.25.

Explanation:

Using the Arrhenius equations for the given conditions:

k1 = A*(exp^(-Ea/(RT1))

k2 = A*(exp^(-Ea/(RT2))

T1 = 231°C

= 231 + 273.15 K

= 574.15 K

T2 = 293°C

= 293 + 273.15 K

= 566.15 K

Ea = 274 kJ mol^-1

R= 0.008314 kJ/mol.K

Now divide the second by the first:

k2/k1 = exp^(-Ea/R * (1/T2 - (1/T1))

= 0.444

2.25k2 = k1

1. The fraction of the atomic volume occupied by the nucleus is approximately 0.0000064 or 6.4 x 10^-6.

2. The density of a proton is approximately 5.77 x 10^20 g/cm^3.

Part 1: Fraction of atomic volume occupied by nucleus

**1. Calculate volumes of atom and nucleus:**

- Convert Å to m:

   - Atom diameter: 2.50 Å * 10^-10 m/Å = 2.50 x 10^-10 m

   - Nucleus diameter: 9.00 x 10^-5 Å * 10^-10 m/Å = 9.00 x 10^-15 m

- Calculate radii:

   - Atom radius: 2.50 x 10^-10 m / 2 = 1.25 x 10^-10 m

   - Nucleus radius: 9.00 x 10^-15 m / 2 = 4.50 x 10^-15 m

- Calculate volumes of spheres using the formula (4/3)πr³:

   - Atom volume: (4/3)π * (1.25 x 10^-10 m)³ ≈ 8.18 x 10^-31 m³

   - Nucleus volume: (4/3)π * (4.50 x 10^-15 m)³ ≈ 52.36 x 10^-44 m³

**2. Calculate fraction of volume occupied by nucleus:**

- Divide nucleus volume by atom volume:

   - Fraction = 52.36 x 10^-44 m³ / 8.18 x 10^-31 m³ ≈ 0.0000064

Therefore, the fraction of the atomic volume occupied by the nucleus is approximately 0.0000064 or 6.4 x 10^-6.

Part 2: Density of a proton

**1. Convert mass of proton to kg:**

- 1 amu = 1.66057 x 10^-27 kg

- Proton mass: 1.0073 amu * 1.66057 x 10^-27 kg/amu ≈ 1.6726 x 10^-27 kg

**2. Calculate volume of a proton from its diameter:**

- Proton radius: 1.72 x 10^-15 m / 2 = 8.60 x 10^-16 m

- Proton volume: (4/3)π * (8.60 x 10^-16 m)³ ≈ 2.90 x 10^-45 m³

**3. Calculate density:**

- Divide proton mass by its volume:

   - Density = 1.6726 x 10^-27 kg / 2.90 x 10^-45 m³ ≈ 5.77 x 10^17 kg/m³

**4. Convert density to g/cm^3:**

- 1 kg/m³ = 1000 g/cm³

- Density ≈ 5.77 x 10^17 kg/m³ * 1000 g/cm³ ≈ 5.77 x 10^20 g/cm³

Therefore, the density of a proton is approximately 5.77 x 10^20 g/cm^3.

Answer:

[tex]n=3.0\times 10^{13}[/tex]

Explanation:

Charge on 1 electron = [tex]-1.6\times 10^{-19}\ C[/tex]

The expression for charge is:-

[tex]Charge=n\times q_e[/tex]

Given that:- Charge = [tex]-4.8\times 10^{-6}\ C[/tex]

[tex]-4.8\times 10^{-6}=n\times (-1.6\times 10^{-19})[/tex]

[tex]n=\frac{4.8\times 10^{-6}}{1.6\times 10^{-19}}=3.0\times 10^{13}[/tex]

Total number of electrons, n = [tex]3.0\times 10^{13}[/tex]

The total number of electrons the charge represent is 3.0 × 10¹³

From the question, we are to calculate the total number of electrons

Using the formula,

Q = ne

Where Q is the total charge

n is the number of electrons

and e is the charge of an electron

From the given information,

q = - 4.8 × 10⁻⁶ C

(NOTE: The negative sign indicates the type of charge)

e = 1.6 × 10⁻¹⁹ C

Then,

4.8 × 10⁻⁶ = n × 1.6 × 10⁻¹⁹

n = [tex]\frac{4.8 \times 10^{-6} }{1.6 \times 10^{-19} }[/tex]

n = 3.0 × 10¹³

Hence, the total number of electrons the charge represent is 3.0 × 10¹³

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Answer:

0.218 M of Pb(NO3)2

Explanation:

Equation of the reaction

Pb(NO3)2(aq) + 2NaCl(aq) --> PbCl2(s) + 2NaNO3(aq)

1 mole of Pb(NO3)2 reacts to precipitate 1 mole of PbCl2

Molar mass of PbCl2 = 207 + (35.5*2)

= 278 g/mol

Number of moles of PbCl2 precipitated = mass/molar mass

= 12.11/278

= 0.04356 mol

Since 0.04356 moles of PbCl2 was precipitated, therefore by stoichiometry; 0.04356 moles of Pb(NO3)2 reacted.

Molarity is defined as the number of moles of solute in 1 liter of solution.

Molarity = number of moles/volumes

= 0.04356/0.2

= 0.218 M

Answer:

The answers indicate that wavelength is inversely proportional to the energy of light (photon)

Explanation:

Energy of photon E = hc/λ

where;

h is Planck's constant = 6.626 X 10⁻³⁴js

c is the speed of light (photon) = 3 X 10⁸ m/s

λ is the wavelength of the photon

⇒For ultraviolet ray, with wavelength λ = 1 x 10⁻⁸ m

E = (6.626 X 10⁻³⁴ X 3 X 10⁸)/ (1 x 10⁻⁸)

E = 19.878 10⁻¹⁸ J

⇒For Visible light, with wavelength λ = 5 x 10⁻⁷ m

E = (6.626 X 10⁻³⁴ X 3 X 10⁸)/ (5 x 10⁻⁷)

E = 3.9756 X 10⁻¹⁹ J

⇒For Infrared, with wavelength λ = 1 x 10⁴ m

E = (6.626 X 10⁻³⁴ X 3 X 10⁸)/ (1 x 10⁴)

E = 19.878 X 10⁻³⁰ J

From the result above, ultraviolet ray has the shortest wavelength, but it has the highest energy among other lights.

Also infrared has the highest wavelength but the least energy among other lights.

Hence, wavelength is inversely proportional to the energy of light (photon).

The energies of one photon of ultraviolet (λ = 1 x 10⁻⁸ m), visible (λ = 5 x 10⁻⁷ m), and infrared (λ = 1 x 10⁴ m) light are [tex]1.988 \times 10^{-17} \, \text{J} \),[/tex]  [tex]3.98 \times 10^{-19} \, \text{J} \),[/tex] [tex]1.99 \times 10^{-30} \, \text{J} \)[/tex] respectively. The answers indicate that as the wavelength of light decreases, the energy of the photons increases.

To calculate the energy of a photon, we use the formula:

[tex]\[ E = \frac{hc}{\lambda} \][/tex]

where:

- E is the energy of the photon

- h  is Planck's constant [tex](\( 6.626 \times 10^{-34} \, \text{J} \cdot \text{s} \))[/tex]

- c is the speed of light in a vacuum [tex](\( 3.00 \times 10^8 \, \text{m/s} \))[/tex]

- [tex]\( \lambda \)[/tex] is the wavelength of the photon

Let's calculate the energy for each type of light:

Ultraviolet Light (λ = 1 x 10⁻⁸ m)

[tex]\[ E_\text{UV} = \frac{6.626 \times 10^{-34} \, \text{J} \cdot \text{s} \times 3.00 \times 10^8 \, \text{m/s}}{1 \times 10^{-8} \, \text{m}} \][/tex]

Visible Light (λ = 5 x 10⁻⁷ m)

[tex]\[ E_\text{Visible} = \frac{6.626 \times 10^{-34} \, \text{J} \cdot \text{s} \times 3.00 \times 10^8 \, \text{m/s}}{5 \times 10^{-7} \, \text{m}} \][/tex]

Infrared Light (λ = 1 x 10⁴ m)

[tex]\[ E_\text{IR} = \frac{6.626 \times 10^{-34} \, \text{J} \cdot \text{s} \times 3.00 \times 10^8 \, \text{m/s}}{1 \times 10^4 \, \text{m}} \][/tex]

Let's compute these values:

1. Ultraviolet Light:

[tex]\[ E_\text{UV} = \frac{6.626 \times 10^{-34} \times 3.00 \times 10^8}{1 \times 10^{-8}} \][/tex]

[tex]\[ E_\text{UV} = \frac{19.878 \times 10^{-26}}{1 \times 10^{-8}} \][/tex]

[tex]\[ E_\text{UV} = 19.878 \times 10^{-18} \][/tex]

[tex]\[ E_\text{UV} = 1.988 \times 10^{-17} \, \text{J} \][/tex]

2. Visible Light:

[tex]\[ E_\text{Visible} = \frac{6.626 \times 10^{-34} \times 3.00 \times 10^8}{5 \times 10^{-7}} \]\[ E_\text{Visible} = \frac{19.878 \times 10^{-26}}{5 \times 10^{-7}} \]\[ E_\text{Visible} = 3.9756 \times 10^{-19} \]\[ E_\text{Visible} = 3.98 \times 10^{-19} \, \text{J} \][/tex]

3. Infrared Light:

[tex]\[ E_\text{IR} = \frac{6.626 \times 10^{-34} \times 3.00 \times 10^8}{1 \times 10^4} \]\[ E_\text{IR} = \frac{19.878 \times 10^{-26}}{1 \times 10^4} \]\[ E_\text{IR} = 1.9878 \times 10^{-30} \]\[ E_\text{IR} = 1.99 \times 10^{-30} \, \text{J} \][/tex]

Summary of Energies

- Ultraviolet Light (λ = 1 x 10⁻⁸ m): [tex]\( E_\text{UV} = 1.988 \times 10^{-17} \, \text{J} \)[/tex]

- Visible Light (λ = 5 x 10⁻⁷ m): [tex]\( E_\text{Visible} = 3.98 \times 10^{-19} \, \text{J} \)[/tex]

- Infrared Light (λ = 1 x 10⁴ m): [tex]\( E_\text{IR} = 1.99 \times 10^{-30} \, \text{J} \)[/tex]

Relationship between Wavelength and Energy

Specifically:

- Ultraviolet light has the shortest wavelength and the highest energy.

- Visible light has a moderate wavelength and moderate energy.

- Infrared light has the longest wavelength and the lowest energy.

This inverse relationship between wavelength and photon energy is consistent with the equation [tex]\( E = \frac{hc}{\lambda} \)[/tex]. As [tex]\( \lambda \) (wavelength) decreases, \( E \)[/tex] (energy) increases, and vice versa.

Glycine, an amino acid, changes its structure and net charge in response to pH due to two dissociable protons. At pH 2 it's fully protonated with a net charge of +1. At pH 7, its carboxyl group loses a proton leaving the net charge to 0. By pH 10, both the carboxyl and amino group have lost protons, so the net charge is -1.

The amino acid glycine (H2N–CH2–COOH) has two dissociable protons, one associated with its carboxyl group (pK 2.35) and one with its amino group (pK 9.78). At very low pH values, protons are abundant, meaning glycine is in its fully protonated form with a net charge of +1. Thus, at pH 2, the structure is H3N+–CH2–COOH and the net charge is +1.

At pH 7, the proton on the carboxyl group is lost, leaving the carboxylate form (COO-) and the proceed ammonium group (NH3+). Our structure therefore becomes H3N+–CH2–COO- with a net charge of zero.

Finally, at pH 10, glycine loses both of its protons. The structure now is H2N–CH2–COO- and the net charge is -1.

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The 2p sublevel has quantum numbers n=2, l=1, and possible ml values of -1, 0, +1, with a maximum of 6 electrons. The 5f sublevel has quantum numbers n=5, l=3, and possible ml values ranging from -3 to +3, holding up to 14 electrons.

The n, l, and possible ml values for 2p and 5f sublevels are derived from the quantum numbers that define the properties of electrons in atoms. For the 2p sublevel, n is 2 (the principal quantum number indicating the second shell), l is 1 (the angular momentum quantum number corresponding to a p sublevel), and the possible ml values range from -1 to +1 (which are -1, 0, and 1), making for three possible orientations.

For the 5f sublevel, n is 5 (indicating the fifth shell), l is 3 (f sublevel), and the possible ml values range from -3 to +3 (which are -3, -2, -1, 0, 1, 2, 3), resulting in a total of seven possible orientations. Using the formula maximum number of electrons that can be in a subshell = 2(2l + 1), we can calculate that the 2p sublevel can hold a maximum of 6 electrons and the 5f sublevel can hold up to 14 electrons.

Explanation:

Formula according to the radius ratio rule is as follows.

             [tex]\frac{r_{+}}{r_{-}} = \frac{73}{184}[/tex]

                          = 0.397

According to the radius ratio rule, as the calculated value is 0.397 and it lies in between 0.225 to 0.414. Therefore, it means that the type of void is tetrahedral.

Thus, we can conclude that the given compound is most likely to adopt closest-packed array with lithium ions occupying tetrahedral holes.

Answer:

closest-packed array with lithium ions occupying tetrahedral holes

Explanation:

Given the small size of lithium ions and that they are present in twice the amount as selenide ions, they must occupy tetrahedral holes in a closest-packed array.

The alkali metal cation in the unknown metal hydroxide that has been reacted with hydrochloric acid is Potassium (K+). The identity of the cation was determined by performing a stoichiometric calculation based on the balanced chemical reaction and the molar masses of the alkali metal hydroxide.

This is a stoichiometry problem that requires understanding of acid-base titration reactions. In a neutralization reaction, an acid reacts with a base to form water plus a salt. Here, the hydrochloric acid (HCl) is reacting with the alkali metal hydroxide (MOH) to form water (H2O) and a salt (MCl). The balanced chemical reaction is HCl + MOH -> H2O + MCl.

The mole ratio between HCl and MOH in this reaction is 1:1. This means that for each mole of HCl used, one mole of MOH will react. From the volume (17.0 mL) and molarity (2.5 M) of HCl used, we can calculate the number of moles of HCl, which will also be the number of moles of MOH because of the 1:1 mole ratio.

mol HCl = Molarity x Volume = 2.50 mol/L x 17.0 x 10-3 L = 0.0425 mol. Hence, the number of moles of MOH = 0.0425 mol.

The molar mass of alkali metal hydroxide (MOH) is its mass divided by the number of moles. Hence, Molar mass = mass (g) / moles = 4.36 g / 0.0425 mol = 102.59 g/mol. This molar mass is closest to the molar mass of Potassium Hydroxide (KOH), which is 39.10 (K) + 15.9994 (O) + 1.00784 (H) = 56.11 g/mol. So, the alkali metal cation in the unknown metal hydroxide is K+ (Potassium).

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The provided unknown sample contains Rubidium (Rb⁺) as its alkali metal cation. This conclusion is based on titration calculations, yielding a molar mass closest to Rb.

To identify the alkali metal cation (Li⁺, Na⁺, K⁺, Rb⁺, Cs⁺) from the given sample, we need to perform a titration calculation. Here are the steps:

First, calculate the moles of HCl used in the titration:

The balanced equation for the reaction is:

MOH + HCl → MCl + H2O

Since the reaction is 1:1, moles of MOH = moles of HCl = 0.0425 mol

Next, calculate the molar mass of the alkali metal hydroxide (MOH):

Given MOH is an alkali metal hydroxide, its molar mass is the sum of the molar masses of M, O, and H, which is represented as:

M + 17.01 (since the molar mass of OH is 17.01 g/mol)

Therefore, M = 102.59 - 17.01 ≈ 85.58 g/mol

Match this result with the molar masses of the alkali metals:

The metal with a molar mass closest to 85.58 g/mol is Rb (Rubidium).

Therefore, the alkali metal cation is Rb⁺ (Rubidium).

Answer:

195.52 g/mol

Explanation:

Answer:

195.52 g/mol

Explanation:

1.) ΔT = Kb * molality

2.) (100.22-100) = 0.512 * molality

3.) 0.22 = 0.512 * molality

4.) 0.22/0.512 =( 0.512 * molality)/ 0.512

5.) 0.4297 = molality

6.) molality = mol of solute/ kg of solvent

7.) 0.4297 = mol of solute / (0.750- 0.058125)

8.) 0.4297 = mol of solute / 0.69175

9.) (0.4297 = mol of solute / 0.69175) * 0.69175

10.) mol of solute = 0.29729

11.) mol = mass / mw therefore mw = mass/ mol

12.) mw = 58.125g / 0.29729mol

13.) mw = 195.125 g/mol

Please be mindful I did not consistently write conversions or write out units to show cancellations.

To determine the molecular weight of the unknown compound, one would need to calculate the boiling point elevation, then the molality of the solution, and finally rearrange the formula to solve for the molar mass based on the given masses of solute and solvent.

The student's question pertains to the calculation of the molecular weight of an unknown compound using the boiling point elevation method in Chemistry. To find the molecular weight of the unknown compound, the boiling point elevation (ΔTb) first needs to be determined, using the observed boiling point and the normal boiling point of water (100°C). Then, using the boiling point elevation constant (Kb for water), and the mass of the solute and solvent, the molality (m) of the solution can be calculated. The molar mass (M) of the unknown compound is then found by rearranging the formula ΔTb = i ⋅ Kb ⋅ m, where i is the van't Hoff factor (which is 1 for a nonelectrolyte compound) and solving for M.

Since we know the mass of the solute and solvent and the boiling point elevation, the molality can be calculated with m = (moles of solute) / (kilograms of solvent). The moles of solute is molar mass dependent, so when we rearrange to solve for molar mass, we get M = (mass of solute in grams) / (molality ⋅ kg of solvent).

To solve the problem, the steps are as follows:

Answer:

(a) 6g. Shell 6, n = 6. Subshell g, l = 4. Number of orbitals in sublevel = 9

(b) 4s. Shell 4, n = 4. Subshell s, l = 0. Number of orbitals in sublevel = 1

(c) 3d. Shell 3, n = 3. Subshell d, l = 2. Number of orbitals in sublevel = 5

Explanation:

The rules for electron quantum numbers are:

1. Shell number, 1 ≤ n, n = 1, 2, 3...

2. Subshell number, 0 ≤ l ≤ n − 1, orbital s - 0, p - 1, d - 2, f - 3

3. Orbital energy shift, -l ≤ ml ≤ l

4. Spin, either -1/2 or +1/2

So,

(a) 6g. Shell 6, n = 6. Subshell g, l = 4. Number of orbitals in sublevel = 2l+1 = 9

(b) 4s. Shell 4, n = 4. Subshell s, l = 0. Number of orbitals in sublevel = 2l+1 = 1

(c) 3d. Shell 3, n = 3. Subshell d, l = 2. Number of orbitals in sublevel = 2l+1 = 5

(a) 6g Shell 6, n = 6. Subshell g, l = 4. Number of orbitals in sublevel = 9

(b) 4s Shell 4, n = 4. Subshell s, l = 0. Number of orbitals in sublevel = 1

(c) 3d Shell 3, n = 3. Subshell d, l = 2. Number of orbitals in sublevel = 5

There are a total of four quantum numbers: the principal quantum number (n), the orbital angular momentum quantum number (l), the magnetic quantum number (ml), and the electron spin quantum number (ms).

The rules for electron quantum numbers are:

1. Shell number, 1 ≤ n, n = 1, 2, 3...

2. Subshell number, 0 ≤ l ≤ n − 1, orbital s - 0, p - 1, d - 2, f - 3

3. Orbital energy shift, -l ≤ ml ≤ l

4. Spin, either -1/2 or +1/2

(a) 6g. Shell 6, n = 6. Subshell g, l = 4. Number of orbitals in sublevel = 2l+1 = 9

(b) 4s. Shell 4, n = 4. Subshell s, l = 0. Number of orbitals in sublevel = 2l+1 = 1

(c) 3d. Shell 3, n = 3. Subshell d, l = 2. Number of orbitals in sublevel = 2l+1 = 5

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Answer:

Since ΔEN > 0, the bond is covalent polar and the molecule is polar (dipole). Since ΔEN > 0, the bond is covalent polar and the molecule is polar (dipole). HI and ClF interact through a dipole-dipole force

Explanation:

Final answer:

Dipole-dipole forces act between a hydrogen chloride molecule and a hydrogen iodide molecule due to the partially positive hydrogen atom and partially negative chlorine atom in a hydrogen chloride molecule.

Explanation:

Dipole-dipole forces, which are attractive forces between polar molecules, act between a hydrogen chloride molecule and a hydrogen iodide molecule. These forces occur because a hydrogen chloride molecule has a partially positive hydrogen atom and a partially negative chlorine atom. In a collection of many hydrogen chloride molecules, the oppositely charged regions of neighboring molecules will align themselves near each other.

Answer:

1)- 0.11 m Fe(NO₃)₃  ⇒ A- Lowest freezing point

2)- 0.18 m NaOH ⇒ D- Highest freezing point

3)- 0.21 m FeSO₄ ⇒ B- Second lowest freezing point

4)- 0.38 m Glucose ⇒ C- Third lowest freezing point

Explanation:

Freezing point depression is given by the following equation:

ΔTf= Kf x m x i

As it is a depression point (final temperature is lower than initial temperature), as higher is ΔTf, lower is the freezing point. Kf is the cryoscopic constant. For water, Kf= 1.853 K·kg/mol. At high m (molality of solute) and i (Van't Hoff factor, dissociated species), the freezing point will be low.

Kf is the same for all solution, so we can simply calculate m x i and order the solutions from high m x i (lowest freezing point) to low m x i (highest freezing point):

      m x i = 0.11 x 4 = 0.44

      A) Lowest freezing point

    m x i = 0.18 x 2= 0.36

    D) Highest freezing point

    m x i = 0.21 x 1= 0.42

    B) Second lowest freezing point

    m x i = 0.38 x 1 = 0.38

    C) Third lowest freezing point

A fall in the hotness and coldness at which the matter freezes is called freezing point depression.

It can be calculated using:

[tex]\Delta \text{T}_{\text{f}} &= \text{K}_{\text{f}} \times \text{m} \times \text{ i}[/tex]

For all the solution [tex]\text{k}_{\text{f}}[/tex] value is constant hence, m × i can be calculated to know the order of lowest and highest freezing points.

The correct matches are:

1) 0.11 m Fe(NO₃)₃ ⇒ Option A. Lowest freezing point

2) 0.18 m NaOH ⇒ Option D. Highest freezing point

3) 0.21 m FeSO₄ ⇒ Option B. Second lowest freezing point

4) 0.38 m Glucose ⇒ Option C. Third lowest freezing point

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Answer:

The answer the question, what is the pressure of the gas, in pascals is 101195.73 Pa

Explanation:

Firstly we list out the known variables

The known variables are

mass of piston = 1.1 kg, diameter of piston, D = 10.0 cm = 0.1 m

mass of weight = 2.0 kg atmospheric pressure = 1.0 atm

In this question the quantity required is the presssure of the gas in the cylinder after placing a weght on the piston

To solve this, we note that Pressure = Force per unit area

= Force/area, hence

We compute the area of the piston thus

Area = (πD²)÷4 = 0.0079 m²

While the sum of the mass of the piston and the added weight = 1.1 kg + 2.0 kg = 3.1 kg

The weight of the added mass and piston that is their force on the gas = W = Mass × gravity = 3.1 kg × 9.81 m/s² = 30.411 N

Therefore the pressure  = 30.411N/(0.0079 m²) = 3870.73 Pascals

The pressure of the gas = pressure due to the piston and the added weight + pressure due to the atmosphere

thus pressure of the gas = 3870.73 Pa + 1.0 atm =3870.73 Pa + 101325 Pa =101195.73 Pa

The pressure of the gas, in pascals is 101195.73 Pa

The rate changes by a factor of 1/16 when the concentration of H+ is decreased by a factor of 4 in the given rate law.

The reaction with the rate law Rate = k{BrO3-}{Br-}{H+}2 indicates that the rate of the reaction is directly proportional to the concentration of H+ raised to the second power. Thus, if the concentration of H+ is decreased by a factor of 4, the rate of the reaction would decrease by a factor of 42 or 16. Therefore, the rate changes by a factor of 1/16 when the concentration of H+ decreases by a factor of 4.

The rate law for the given reaction is Rate = k{BrO3-}{Br-}{H+}^2. If the concentration of H+ is decreased by a factor of 4, it means the new concentration of H+ would be 1/4 of the original concentration. Since the rate law is quadratic with respect to H+, the rate would change by a factor of (1/4)^2, which is 1/16. Therefore, the rate would decrease by a factor of 1/16 or 0.0625.

The rate of the reaction changes by a factor of [tex]\(\frac{1}{16}\)[/tex] (or decreases to [tex]\(\frac{1}{16}\)[/tex] of its original rate) when the concentration of[tex]\( \text{H}^+ \)[/tex] is decreased by a factor of 4.

Given the reaction with the rate law:

[tex]\[ \text{Rate} = k[\text{BrO}_3^-][\text{Br}^-][\text{H}^+]^2 \][/tex]

We need to determine how the rate changes if the concentration of[tex]\( \text{H}^+ \)[/tex] is decreased by a factor of 4.

Step-by-Step Explanation:

1. Initial Rate Law Expression:

[tex]\[ \text{Rate}_1 = k[\text{BrO}_3^-][\text{Br}^-][\text{H}^+]^2 \][/tex]

2. Change in [tex]\( \text{H}^+ \)[/tex] Concentration:

  Let's denote the initial concentration of[tex]\( \text{H}^+ \)[/tex]as [tex]\([ \text{H}^+ ]_1\)[/tex]. If the concentration of[tex]\( \text{H}^+ \)[/tex] is decreased by a factor of 4, the new concentration will be:

[tex]\[ [ \text{H}^+ ]_2 = \frac{[ \text{H}^+ ]_1}{4} \][/tex]

3. New Rate Law Expression:

  Substitute the new concentration of [tex]\( \text{H}^+ \)[/tex] into the rate law:

[tex]\[ \text{Rate}_2 = k[\text{BrO}_3^-][\text{Br}^-]\left( \frac{[ \text{H}^+ ]_1}{4} \right)^2 \][/tex]

4. Simplify the Expression:

[tex]\[ \text{Rate}_2 = k[\text{BrO}_3^-][\text{Br}^-]\left( \frac{[ \text{H}^+ ]_1^2}{4^2} \right) \][/tex]

[tex]\[ \text{Rate}_2 = k[\text{BrO}_3^-][\text{Br}^-]\left( \frac{[ \text{H}^+ ]_1^2}{16} \right) \][/tex]

5. Relate [tex]\( \text{Rate}_2 \) to \( \text{Rate}_1 \)[/tex] :

  From the initial rate law, we know that:

 [tex]\[ \text{Rate}_1 = k[\text{BrO}_3^-][\text{Br}^-][\text{H}^+]_1^2 \][/tex]

  So, we can write:

[tex]\[ \text{Rate}_2 = \frac{\text{Rate}_1}{16} \][/tex]

6. Factor by Which the Rate Changes:

  The rate decreases by a factor of 16.

Answer: 4376.6g/mol

Explanation:Please see attachment for explanation

The molar mass of the protein is 4376.6 g/mol.

The molar mass of a chemical compound is defined as the mass of a sample divided by the amount of substance in that sample, measured in moles.

Solution:

Given, the mass of protein is 3.3×10-2 mg

Mass converted into gram = 3.3×10-2 mg = M 0.000033 g

Therefore, xg of the protein dissolve in 1 L

xg pf protein =[tex]\bold{\dfrac{0.000033}{0.00025} = 0.132\; g/L}[/tex]

Given, the osmotic pressure of 0.56 torr.

[tex]\bold{\pi = 0.56\;torr =\dfrac{0.56}{760} = 0.00737\; atm p}[/tex]

Thus, the volume of the solution is 0.007 L

Given, Temperature = 25 °C

Converting into kelvin = 25+273 = 298 k.

R = 0.082 atm.

Now, by the formula  [tex]\bold{M_2= \dfrac{W_2RT}{\pi V}}[/tex]

[tex]\bold{M_2= \dfrac{0.132\times0.082\times298}{0.00737\times0.007} =4376.6 \;g/mol}[/tex]

Thus, the molar mass of protein is 4376.6 g/mol.

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Explanation:

Pseudo-noble gas configuration which can also be called pseudo inert configuration is when elements have fully filled d-orbital, along with s- and p- orbitals. Examples are ions of elements, this is because they lose their valence electrons to be fully filled in order to have a stable octet.

Example of Group 3A(13) ion is Aluminium ion; Al3+. Al3+ takes up the configuration of Neon when it loses its 3 valence electrons.

Electronic configuration of Al and Al3+

Al = [Ne] 3s2 3p1

Al3+ = 1s2 2s2 2p6

A pseudo-noble gas configuration occurs when an ion or atom has an electron configuration that resembles that of a noble gas. An example of an ion from Group 3A(13) that has a pseudo-noble gas configuration is aluminium (Al3+).

A pseudo-noble gas configuration refers to the electron configuration of an ion or atom that resembles that of a noble gas. This occurs when an ion gains or loses electrons to acquire the same electron configuration as a noble gas. An example of an ion from Group 3A(13) that has a pseudo-noble gas configuration is aluminium (Al3+).

Aluminum has an electron configuration of 1s2 2s2 2p6 3s2 3p1 in its ground state. When it loses its three valence electrons, its configuration becomes 1s2 2s2 2p6, which is the same as that of the noble gas, neon (Ne). Therefore, aluminium has a pseudo-noble gas configuration.

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The equilibrium molarity of SO4 2- in a 0.049 M aqueous solution of H2SO4 will be approximately 0.049 M after the initial dissociation, as sulfuric acid is a strong acid and dissociates completely in its first step, and the second dissociation is generally weaker.

The question involves the dissociation of sulfuric acid (H2SO4) in water, which occurs in two steps. The first dissociation is strong, with the equation H2SO4 (aq) → 2H+ (aq) + SO42- (aq). Because sulfuric acid is a strong acid, the initial dissociation is essentially complete, and the equilibrium molarity of SO42- will be equal to the initial molarity of H2SO4, barring any further reactions.

Given a 0.049 M solution of H2SO4, after the first dissociation, we have 0.049 M of SO42-. The second dissociation of HSO4- to form SO42- is weak. However, without the acid dissociation constant (Ka) value or any other provided equilibrium concentrations, one cannot calculate the additional contribution of SO42- from the second dissociation. Therefore the most straightforward answer, assuming the second dissociation's contribution is negligible compared to the first, is that the equilibrium molarity of SO42- due to the first dissociation is approximately 0.049 M.

Explanation:

When we move across a period from left to right then there will occur an increase in electronegativity and also there will occur an increase in non-metallic character of the elements.

As both arsenic (As) and antimony (Sb) are group 15 elements. And, on moving down the group increases metallic character hence, Sb will be more metallic than As. Therefore, As is less metallic in nature than Sb.  

Silicon (Si) is a group 14 element and phosphorus (P) is a group 15 element. And, both of them lie in period 3 and since, non-metallic character increases on moving from left to right along a period. Therefore, phosphorus (P) is less metallic than silicon (Si).    

Sodium (Na) is a group 1A element (also known as alkali metal) and beryllium (Be) is a group 2A element (also known as alkaline earth metal). Hence, Be is less metallic than Na because on moving from left to right there occurs an increase in non-metallic character.

Answer:

Explanation:

a) The pH can be define as the measurement of the hydrogen or hydroxide ion concentration in a solution. The pH for blood is 7.4. In severe acidosis condition the pH of blood is 7.35 or lower.

b) Severe acidosis is a condition which is caused due to the overproduction of acid and building up of acid in the blood. This occurs due to the excessive loss of bicarbonate from the blood or buildup of carbon dioxide  in the blood. This leads to poor functioning of the lungs and depressed breathing.

The severe acidosis is critical condition because because it can adversely affect the cell membranes, muscle contraction, and function of the kidneys and neural activity of the body.  

Final answer:

pH is a scale used to measure how acidic or alkaline a substance is, with normal blood pH being between 7.35 and 7.45. Severe acidosis, often due to diabetic ketoacidosis, disrupts normal bodily functions, impairs oxygen transport, and can lead to life-threatening symptoms like dehydration, lethargy, and coma if untreated.

Explanation:

pH is a measure of the acidity or alkalinity of a solution, where a pH below 7 is acidic, and a pH above 7 is alkaline. The normal pH of blood is tightly regulated between 7.35 and 7.45. Severe acidosis is a condition where the blood pH falls significantly below this range.

Severe acidosis is problematic because it can disrupt biological processes. Specifically, it affects the ability of hemoglobin to transport oxygen and may lead to symptoms like labored breathing, dehydration, lethargy, and loss of appetite. The condition can be life-threatening, as it may lead to a coma if not promptly treated.

Diabetic ketoacidosis is a common cause of acidosis in diabetic individuals, occurring when excessive ketone bodies in the blood lower the pH well below normal levels, often leading to a value around 6.9, thus disrupting the acid-base balance and necessitating medical intervention.

Answer : The  mass of [tex]KC_2H_5CO_2[/tex] is, 24.5 grams

Explanation : Given,

[tex]K_a=1.3\times 10^{-5}[/tex]

pH = 5.02

Concentration of [tex]HC_2H_5CO_2[/tex] = 1.30 M

Volume of solution = 125 mL = 0.125 L

First we have to calculate the value of [tex]pK_a[/tex].

The expression used for the calculation of [tex]pK_a[/tex] is,

[tex]pK_a=-\log (K_a)[/tex]

Now put the value of [tex]K_a[/tex] in this expression, we get:

[tex]pK_a=-\log (1.3\times 10^{-5})[/tex]

[tex]pK_a=5-\log (1.3)[/tex]

[tex]pK_a=4.89[/tex]

Now we have to calculate the concentration of [tex]KC_2H_5CO_2[/tex]

Using Henderson Hesselbach equation :

[tex]pH=pK_a+\log \frac{[Salt]}{[Acid]}[/tex]

[tex]pH=pK_a+\log \frac{[KC_2H_5CO_2]}{[HC_2H_5CO_2]}[/tex]

Now put all the given values in this expression, we get:

[tex]5.02=4.89+\log (\frac{[KC_2H_5CO_2]}{1.30})[/tex]

[tex][KC_2H_5CO_2]=1.75M[/tex]

Now we have to calculate the moles of [tex]KC_2H_5CO_2[/tex]

[tex]\text{Moles of }KC_2H_5CO_2=1.75M\times 0.125L=0.219mol[/tex]

Now we have to calculate the mass of [tex]KC_2H_5CO_2[/tex]

[tex]\text{Mass of }KC_2H_5CO_2=\text{Moles of }KC_2H_5CO_2\times \text{Molar mass of }KC_2H_5CO_2[/tex]

Molar mass of [tex]KC_2H_5CO_2[/tex] = 112 g/mol

[tex]\text{Mass of }KC_2H_5CO_2=0.219mol\times 112g/mol=24.5g[/tex]

Thus, the mass of [tex]KC_2H_5CO_2[/tex] is, 24.5 grams

To find the mass of KC₂H₃CO₂ needed to create a buffer at pH 5.02, the Henderson-Hasselbalch equation is used to calculate the ratio of conjugate base to acid, then the molar mass is used to convert moles to mass, the concentration of A⁻ is approximately 1.755 M.

The question asks for the mass of potassium propanoate (KC₂H₃CO₂) needed to create a buffer with a specific pH from a solution of propanoic acid (KC₂H₃CO₂). To solve this, we apply the Henderson-Hasselbalch equation:

[tex]\[ pH = pKa + \log \left( \frac{[A^-]}{[HA]} \right) \][/tex]

Given variables:

Rearranging the equation to solve for A⁻:

A⁻ = [tex][HA] \times 10^{(pH - pKa)}[/tex]

A⁻ = 1.755M

After calculating A⁻, it's then converted from molarity to moles given the volume of the solution. Finally, the mass of KC₂H₃CO₂ is 1.755M by multiplying the number of moles of A⁻ by its molar mass.

Answer:

The quantity of sodium fluoride is 17,676,14 kilograms is needed per year for a city of 50,000 people.

Explanation:

Population of the city = 50,000

Volume of water consumed by an individual in day= 115.0 gallons

Volume of water consumed by 50,000 people in day: V

V = 115.0 gallons × 50,000 = 5,750,000 gallons

1 gallon = 3.79 L

[tex]V=5,750,000 gallons=5,750,000\times 3.79 L = 21,792,500 L[/tex]

[tex]V=21,792,500 L=21,792,500,000 mL[/tex]

( 1L = 1000 mL)

Mass of water = m

Density of water = d = 1.0 g/mL

[tex]m=d\times V=1.0 g/mL\times 21,792,500,000 mL=21,792,500,000 g[/tex]

Concentration of Fluorine in water = 1 ppm = 1 gram of F /1 million grams of water

Then mass of fluorine present in 21,792,500,000 g of water:

[tex]\frac{1}{10^6}\times 21,792,500,000 g=21,792.5 g[/tex]

Mass of sodium fluoride with 21,792.5 g of F = M

Percentage of fluorine in NaF = 45.0 %

[tex]45\%=\frac{21,792.5 g}{M}\times 100[/tex]

M = 48,427.78 g = 48.427 kg ( 1g = 0.001 kg)

48.427 kg of NaF should be added to water in day for city population.

Amount of NaF needed per year for city with 50,000 population :

(1 year = 365 days)

[tex]48.427 kg\times 365=17,676.14 kg[/tex]

The quantity of sodium fluoride is 17,676,14 kilograms is needed per year for a city of 50,000 people.

Requiring a fluorine concentration of 1 ppm in a city's water supply for 50,000 people, each consuming 115 gallons of water per day, would necessitate 17,677 kg of sodium fluoride per annum.

First, we'd need to calculate the total amount of water consumed by the entire city in a year. This would be 50,000 people x 115 gallons/person/day x 365 days/year = 2,098,750,000 gallons/year. After converting this to liters (1 gallon = 3.79 L), the total water consumption is 7,954,625,000 L/year or 7,954,625,000,000 g/year (since 1L of water = 1g).

For a fluorine concentration of 1 ppm (1 g of fluorine/1,000,000 g of water), the yearly requirement of fluorine is 7,954,625 g of fluorine. Sodium fluoride is 45.0 percent fluorine by mass, so to get this amount of fluorine, we will need 7,954,625 g / 0.45 = 17,676,944 g or 17,677 kg of sodium fluoride per year.

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The unit cell described is called the simple cubic unit cell, which contains one atom total due to each corner atom being shared by eight unit cells.

The name of the unit cell of the crystal lattice you've described is the simple cubic unit cell, also known as primitive cubic unit cell. In a simple cubic lattice, the unit cell that repeats itself in all directions to form the entire lattice is a cube with atoms at its corners. These atoms effectively 'touch' each other, and each corner atom is shared among eight unit cells; hence, each unit cell contains one-eighth of an atom at each of its eight corners, totaling one atom per unit cell.

Answer:

0.144 M

Explanation:

First, we will calculate the mass/volume percent (% m/v) using the following expression.

% m/v = % m/m × density

% m/v = 2.50 % × 1.04 g/cm³

% m/v = 2.60 %

This means that there are 2.60 grams of solute per 100 cm³ (100 mL) of solution. The molarity of glucose is:

M = mass of glucose / molar mass of glucose × liters of solution

M = 2.60 g / 180.16 g/mol × 0.100 L

M = 0.144 M

Answer:

E) The rate of the reaction is directly proportional to the concentration of the reactant.

Explanation:

Give the characteristic of a first order reaction having only one reactant.

A) The rate of the reaction is not proportional to the concentration of the reactant.

B) The rate of the reaction is proportional to the square of the concentration of the reactant.

C) The rate of the reaction is proportional to the square root of the concentration of the reactant.

D) The rate of the reaction is proportional to the natural logarithm of the concentration of the reactant.

E) The rate of the reaction is directly proportional to the concentration of the reactant.

Answer:

The rate of the reaction is directly proportional to the concentration of the reactant.

Explanation:

Answer:

233 g

Explanation:

Let's consider the following reaction.

CaCO₃ → CaO + CO₂

We can establish the following relations:

The mass of carbon dioxide produced from 530 g of calcium carbonate is:

[tex]530gCaCO_{3}\frac{1molCaCO_{3}}{100.09gCaCO_{3}} ,\frac{1molCO_{2}}{1molCaCO_{3}} .\frac{44.01gCO_{2}}{1molCO_{2}} =233gCO_{2}[/tex]

233 gram of carbon dioxide gas is produced from the decomposition of calcium carbonate.

[tex]CaCO_3\rightarrow CaO + CO_2[/tex]

From the above decomposition, we can say that the molar ratio of calcium carbonate to carbon dioxide is 1:1.

The molar mass of calcium carbonate is 100.09 g/mol.

The molar ratio of calcium carbonate to carbon dioxide is 1:1.

The molar mass of carbon dioxide is 44.01 g/mol

The mass of carbon dioxide produced from 530 g of calcium carbonate is:

[tex]{530\ gCaCO_3}\times\frac{1molCaCO_3}{100gCaCO_3} \times\frac{1molCO_2}{1molCaCO_3} \times\frac{44.01\ gCO_2}{1molCO_2} =233\ gCO_2[/tex]

So, the 233 g [tex]CO_2[/tex] will be produced.

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Answer:

Molality for the solution is 2.22 m

Explanation:

Molality is a sort of concentration. Indicated the moles of solute in 1kg of solvent. → mol/kg

Let's determine the moles of solute (mass / molar mass)

17.85 g / 92 g/mol = 0.194 moles

Let's convert the mass of solvent (g) to kg

87.4 g . 1kg / 1000 g = 0.0874 kg

Mol/kg → Molality

0.194 mol / 0.0874 kg = 2.22 m