A 8.00 L tank at 26.9 C is filled with 5.53 g of dinitrogen difluoride gas and 17.3 g of sulfur hexafluoride gas. You can assume both gases behave as ideal gases under these conditions. Calculate the mole fraction and partial pressure of each gas, and the total pressure in the tank

Answer:

34.7mL

Explanation:

First we have to convert our grams of Zinc to moles of zinc so we can relate that number to our chemical equation.

So: 6.25g Zn x (1 mol / 65.39 g) = 0.0956 mol Zn

All that was done above was multiplying the grams of zinc by the reciprocal of zincs molar mass so our units would cancel and leave us with moles of zinc.

So now we need to go to HCl!

To do that we multiply by the molar coefficients in the chemical equation:

[tex]\frac{0.0956g Zn}{1 } (\frac{2 mol HCl}{1molZn})[/tex]

This leaves us with 2(0.0956) = 0.1912 mol HCl

Now we use the relationship M= moles / volume , to calculate our volume

Rearranging we get that V = moles / M

Now we plug in: V = 0.1912 mol HCl / 5.50 M HCl

V= 0.0347 L

To change this to milliliters we multiply by 1000 so:

34.7 mL

Explanation:

It'd be better to use cyclohexane.  The possible explanation is that the freezing temperature will change by 20.1 degrees for each mole of substance added to 1 kg of cyclohexane, although the same amount added to naphthalene will change its freezing point just by 6.94 degrees.

It is so much easier to identify a larger change more adequately than a smaller one.  You would actually not have a 1 molal solution in operation, so the variations in freezing points would be even smaller than the ones already described.

Final answer:

Cyclohexane, with its higher freezing point depression constant, would give a more accurate determination of molar mass than naphthalene because it results in more significant temperature changes, making precise measurements easier.

Explanation:

The determination of the molar mass of a substance using freezing point depression will be more accurate with a solvent that has a higher freezing point depression constant (Kƒ). In this case, cyclohexane, with a Kƒ of 20.1°C/m, would give a more accurate determination of molar mass compared with naphthalene, which has a Kƒ of 6.94°C/m, assuming that the substance is equally soluble in both. This is because a higher Kƒ value indicates a greater change in freezing point per mole of solute, making the temperature change easier to measure precisely.

Answer: The value of equilibrium constant for reverse reaction is [tex](\frac{1}{95})^2[/tex]

Explanation:

The given chemical equation follows:

[tex]N_2O_5(g)\rightarrow 2NO_2(g)+\frac{1}{2}O_2(g)[/tex]

The equilibrium constant for the above equation is 95.

We need to calculate the equilibrium constant for the reverse equation of above chemical equation, which is:

[tex]O_2(g)+4NO_2(g)\rightarrow 2N_2O_5(g)[/tex]

The equilibrium constant for the reverse reaction will be the reciprocal of the initial reaction.

If the equation is multiplied by a factor of '2', the equilibrium constant of the reverse reaction will be the square of the equilibrium constant  of initial reaction.

The value of equilibrium constant for reverse reaction is:

[tex]K_{eq}'=(\frac{1}{95})^2[/tex]

Hence, the value of equilibrium constant for reverse reaction is [tex](\frac{1}{95})^2[/tex]

The equilibrium constant for the reaction O2(g) + 4 NO2(g) --> 2 N2O5(g) at 25ºC is (1/95)^2. This is due to Le Chatelier's principle stating that K for the reverse reaction is the reciprocal of the original one and it's squared when the reaction is doubled.

The second reaction in your question is the reverse reaction of the first one, but also multiplied by two. According to Le Chatelier's principle, the equilibrium constant (K) of the reverse reaction is the reciprocal of the original one. So the K of the reverse of 1: N2O5(g) --> 2 NO2(g) + ½ O2(g) would be 1/95. However, the actual reaction to the question is twice that.

When you double a reaction, you square its equilibrium constant. Thus, the K for the reaction: O2(g) + 4 NO2(g) --> 2 N2O5(g) would be (1/95)^2.

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Final answer:

The concentration of the chemist's mercury(II) iodide solution is 0.019 μmol/L after rounding to two significant digits.

Explanation:

The concentration of the chemist's mercury(II) iodide solution can be calculated using the formula for molarity, which is the number of moles of solute divided by the volume of the solution in liters.

In this case, we are given 0.0067μmol of HgI2 and a volume of 0.350 L:

Molarity (μM) = 0.0067μmol / 0.350 L = 0.01914μmol/L

After rounding to two significant digits, the concentration of the solution is 0.019 μM.

Answer:

D) Fe3+ + 3e− → Fe (s) Eo = −0.036V

Explanation:

In electrolysis we have to add  electrical energy for redox reactions non-spontaneous because ΔºG is positive.

These  4 reduction reactions  are all non-spontaneous because their reduction potentials are negative   (ΔGº = - nFεº , ΔGº will be positive). So the ion most easily reduced is the least negative, in this case  Fe3+ + 3e− → Fe (s) Eo = −0.036V.

Answer:

The formula of Estrone = [tex]C_{18}H_{22}O_2[/tex]

Explanation:

Mass of water obtained = 1.388 g

Molar mass of water = 18 g/mol

Moles of [tex]H_2O[/tex] = 1.388 g /18 g/mol = 0.07711 moles

2 moles of hydrogen atoms are present in 1 mole of water. So,

Moles of H = 2 x 0.07711 = 0.1542 moles

Molar mass of H atom = 1.008 g/mol

Mass of H in molecule = 0.1542 x 1.008 = 0.1555 g

Mass of carbon dioxide obtained = 5.543 g

Molar mass of carbon dioxide = 44.01 g/mol

Moles of [tex]CO_2[/tex] = 5.543 g  /44.01 g/mol = 0.126 moles

1 mole of carbon atoms are present in 1 mole of carbon dioxide. So,

Moles of C = 0.126 moles

Molar mass of C atom = 12.0107 g/mol

Mass of C in molecule = 0.126 x 12.0107 = 1.5133 g

Given that the Estrone only contains hydrogen, oxygen and carbon. So,

Mass of O in the sample = Total mass - Mass of C  - Mass of H

Mass of the sample = 1.893 g

Mass of O in sample =  1.893 - 1.5133 - 0.1555 = 0.2242 g  

Molar mass of O = 15.999 g/mol

Moles of O  = 0.2242  / 15.999  = 0.01401 moles

Taking the simplest ratio for H, O and C as:

0.1542 : 0.01401 : 0.126

= 11 :1 : 9

The empirical formula is = [tex]C_9H_{11}O[/tex]

Molecular formulas is the actual number of atoms of each element in the compound while empirical formulas is the simplest or reduced ratio of the elements in the compound.

Thus,  

Molecular mass = n × Empirical mass

Where, n is any positive number from 1, 2, 3...

Mass from the Empirical formula = 9×12 + 11×1 + 16= 135 g/mol

Given, Molar mass = 270.36 g/mol

So,  

Molecular mass = n × Empirical mass

270.36 = n × 135

⇒ n = 2

The formula of Estrone = [tex]C_{18}H_{22}O_2[/tex]

Final answer:

To find the molecular formula of estrone, calculate the empirical formula first by finding the moles of carbon and hydrogen. Then, determine the mole ratio between carbon and hydrogen and round to the nearest whole number to get the empirical formula. Finally, divide the molar mass of estrone by the molar mass of the empirical formula to find the ratio, and round to the nearest whole number to get the molecular formula.

Explanation:

To find the molecular formula for estrone, we need to determine the empirical formula first. The empirical formula represents the simplest whole-number ratio of the elements in a compound. We can use the information from the combustion analysis to calculate the empirical formula.

First, we need to determine the moles of carbon and hydrogen in the estrone sample.

Moles of carbon = mass of CO₂ / molar mass of CO₂

Moles of carbon = 5.543 g / 44.01 g/mol = 0.1259 mol

Moles of hydrogen = mass of H₂O / molar mass of H₂O

Moles of hydrogen = 1.388 g / 18.015 g/mol = 0.0770 mol

Next, we need to find the mole ratio between carbon and hydrogen by dividing the moles of each element by the smallest mole value.

Dividing both moles of carbon and hydrogen by 0.0770 mol, we get the ratio 1.6367 : 1.

Finally, we round the ratio to the nearest whole number to obtain the empirical formula. In this case, the empirical formula is C2H2O.

To find the molecular formula, we need to know the molar mass of estrone. The molar mass of estrone is given as 270.36 g/mol. Dividing the molar mass of estrone by the molar mass of the empirical formula (42.08 g/mol), we get a ratio of approximately 6.42.

Rounding the ratio to the nearest whole number, we find that the molecular formula for estrone is C6H6O3.

Answer:

B. Strong attractive forces hold the gas molecules together.

Explanation:

Kinetic molecular theory postulates:-

Hence, The correct option which is not a postulate of kinetic molecular theory is:- B. Strong attractive forces hold the gas molecules together.

Answer:

ΔG°rxn = -72.9 kJ

Explanation:

Let's consider the following reaction.

HCN(g) + 2 H₂(g) → CH₃NH₂(g)

We can calculate the standard Gibbs free energy of the reaction (ΔG°rxn) using the following expression:

ΔG°rxn = ΔH° - T.ΔS°

where,

ΔH° is the standard enthalpy of the reaction

T is the absolute temperature

ΔS° is the standard entropy of the reaction

ΔG°rxn = -158.0 KJ - 387 K × (-219.9 × 10⁻³ J/K)

ΔG°rxn = -72.9 kJ

The estimated ΔG°rxn for the given reaction HCN(g) + 2 H2(g) → CH3NH2(g) at 387K, calculated using the equation ΔG = ΔH - TΔS and the given values ΔH° = −158.0 kJ and ΔS° = −219.9 J/K, is -72.86 kJ.

To estimate ΔG ° rxn for the given reaction HCN(g) + 2 H2(g) → CH3NH2(g) at 387K, we need to use the equation ΔG = ΔH - TΔS. Given that ΔH° = −158.0 kJ and ΔS° = −219.9 J/K (remember to convert kJ to J, so ΔH is -158000 J), we substitute the values into the equation: ΔG = (-158000 J) - (387K * -219.9 J/K) =  -158 kJ + 85.14 kJ = -72.86 kJ. So, the estimated ΔG°rxn for the reaction under these conditions is -72.86 kJ.

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Answer:

A) CaCO₃

C) No precipitate

Explanation:

To answer these questions we need to consider the solubility rules.

Identify the precipitate or lack thereof for the following:

CaCl₂(aq) + K₂CO₃(aq) ⇒ CaCO₃(s) + 2 KCl(aq)

FeCl₂(aq) + (NH₄)₂SO₄(aq) ⇒ FeSO₄(aq) + 2 NH₄Cl(aq)

Answer: The mole fraction of hydrogen gas at 20°C is 0.975  

Explanation:

We are given:

Vapor pressure of water vapor at 20°C = 17.5 torr

Total pressure at 20°C = 700.0 torr

Vapor pressure of hydrogen gas at 20°C = (700.0 - 17.5) torr = 682.5 torr

To calculate the mole fraction of hydrogen gas at 20°C, we use the equation given by Raoult's law, which is:

[tex]p_{H_2}=p_T\times \chi_{H_2}[/tex]

where,

[tex]p_{H_2}[/tex] = partial pressure of hydrogen gas = 682.5 torr

[tex]p_T[/tex] = total pressure = 700.0 torr

[tex]\chi_{H_2}[/tex] = mole fraction of hydrogen gas = ?

Putting values in above equation, we get:

[tex]682.5torr=700.0torr\times \chi_{H_2}\\\\\chi_{H_2}=\frac{682.5}{700.0}=0.975[/tex]

Hence, the mole fraction of hydrogen gas at 20°C is 0.975

Answer: 0.975

Explanation:

First, calculate the partial pressure of hydrogen gas. The partial pressure of hydrogen gas, PH2, is the difference between the total pressure and the partial pressure of water vapor.

PH2=700.0 torr−17.5torr=682.5torr

Now, use the following equation to calculate the mole fraction of hydrogen gas.

PH2 = XH2 × Ptotal

XH2 = PH2 / Ptotal

= 682.5 torr / 700.0 torr = 0.975

Answer:

(a) rate =  4.82 x 10⁻³s⁻¹  [N2O5]

(b) rate =   1.16 x 10⁻⁴  M/s

(c) rate =   2.32 x 10⁻⁴ M/s

(d) rate =   5.80 x 10⁻⁵ M/s

Explanation:  

We are told the rate law is first order in N₂O₅, and its rate constant is 4.82 x 10⁻³s⁻¹ . This means the rate is proportional to the molar concentration   of   N₂O₅, so

(a) rate = k [N2O5] = 4.82 x 10⁻³s⁻¹ x [N2O5]

(b) rate = 4.82×10⁻³s⁻¹  x 0.0240 M =  1.16 x 10⁻⁴ M/s

(c) Since the reaction is first order if the concentration of  N₂O₅ is double the rate will double too:  2 x   1.16 x 10⁻⁴ M/s = 2.32 x 10⁻⁴ M/s

(d) Again since the reaction is halved to 0.0120 M, the rate will be halved to

1.16 x 10⁻⁴ M/s / 2 =  5.80 x 10⁻⁵ M/s

Answer:

a) r = 4.82x10⁻³*[N2O5]

b) 1.16x10⁻⁴ M/s

c) The rate is doubled too (2.32x10⁻⁴ M/s)

d) The rate is halved too (5.78x10⁻⁴ M/s)

Explanation:

a) The rate law of a generic reaction (A + B → C + D) can be expressed by:

r = k*[A]ᵃ*[B]ᵇ

Where k is the rate constant, [X] is the concentration of the compound X, and a and b are the coefficients of the reaction (which can be different from the ones of the chemical equation).

In this case, there is only one reactant, and the reaction is first order, which means that a = 1. So, the rate law is:

r = k*[N2O5]

r = 4.82x10⁻³*[N2O5]

b) Substituing the value of the concentration in the rate law:

r = 4.82x10⁻³*0.0240

r = 1.16x10⁻⁴ M/s

c) When [N2O5] = 0.0480 M,

r = 4.82x10⁻³*0.0480

r = 2.32x10⁻⁴ M/s

So, the rate is doubled too.

d) When [N2O5] = 0.0120 M,

r = 4.82x10⁻³*0.0120

r = 5.78x10⁻⁴ M/s

So, the rate is halved too.

Answer:

[tex]\Delta G^{\circ}=-15902 J/mol[/tex]

Explanation:

In this problem we only have information of the equilibrium, so we need to find a expression of the free energy in function of the constant of equilireium (Keq):

[tex]\Delta G^{\circ}=-R*T*ln(K_{eq})[/tex]

Being Keq:

[tex]K_{eq}=\frac{[fructose][Pi]}{[Fructose-1-P]}[/tex]

Initial conditions:

[tex][Fructose-1-P]=0.2M[/tex]

[tex][Fructose]=0M[/tex]

[tex][Pi]=0M[/tex]

Equilibrium conditions:

[tex][Fructose-1-P]=6.52*10^{-5}M[/tex]

[tex][Fructose]=0.2M-6.52*10^{-5}M[/tex]

[tex][Pi]=0.2M-6.52*10^{-5}M[/tex]

[tex]K_{eq}=\frac{(0.2M-6.52*10^{-5}M)*(0.2M-6.52*10^{-5}M)}{6.52*10^{-5}M}[/tex]

[tex]K_{eq}=613.1[/tex]

Free-energy for T=298K (standard):

[tex]\Delta G^{\circ}=-8.314\frac{J}{mol*K}*298K*ln(613.1)[/tex]

[tex]\Delta G^{\circ}=-15902 J/mol[/tex]

Answer: 0.069 mmol/L

Explanation:

Molarity of a solution is defined as the number of moles of solute dissolved per Liter of the solution.

[tex]Molarity=\frac{n\times 1000}{V_s}[/tex]

where,

n= moles of solute  = [tex]3.46\mu mol[/tex] =  [tex]0.001\times 3.46=3.46\times 10^{-3}mmol[/tex]

1micro (µ) = 0.001 milli (m)

[tex]V_s[/tex] = volume of solution in ml= 50 ml

[tex]Molarity=\frac{3.46\times 10^{-3}\times 1000}{50ml}=0.069mmol/L[/tex]

Thus concentration in mmol/L of the chemist's potassium dichromate solution is 0.069.

Final answer:

To calculate the concentration of the potassium dichromate solution in mmol/L, convert 3.46μmol to mmol, then divide by the volume in liters, resulting in a concentration of 0.069 mmol/L.

Explanation:

The question asks to calculate the concentration of a potassium dichromate solution in millimoles per liter (mmol/L) given that 3.46μmol of potassium dichromate is dissolved in a 50 mL volumetric flask.

First, convert micromoles (μmol) to millimoles (mmol). Recall that 1μmol = 0.001 mmol:

3.46μmol = 3.46 x 0.001 = 0.00346 mmol

Next, we need to find the concentration in mmol/L. To do this, we use the formula for concentration:

Concentration (mmol/L) = (amount of solute in mmol) / (volume of solution in L)

Since the volume of the solution is 50 mL, we convert this to liters (L) because concentration is expressed in mmol/L. Recall that 1000 mL = 1 L.

50 mL = 0.05 L

Now, we can calculate the concentration:

After rounding to two significant digits, the concentration is 0.069 mmol/L.

Answer:

It’s either 34.2 OR 34.3 either would be correct!

Explanation:

The correct answer is (E) [tex]34.17\text{ mL}[/tex].

To solve this problem, we will use stoichiometry to relate the moles of [tex]AgNO_3[/tex] needed to the moles of Ag2SO4 formed. Here is the step-by-step solution:

1. First, we need to write down the balanced chemical equation:

[tex]\[ 2\text{AgNO}_3(aq) + \text{H}_2\text{SO}_4(aq) \rightarrow \text{Ag}_2\text{SO}_4(s) + 2\text{H}_2\text{O}(l) \][/tex]

2. From the equation, we can see that [tex]2[/tex] moles of [tex]AgNO_3[/tex] produce 1 mole of [tex]Ag_2SO_4[/tex] This gives us the mole ratio:

[tex]\[ \frac{2\text{ moles AgNO}_3}{1\text{ mole Ag}_2\text{SO}_4} \][/tex]

3. Calculate the molar mass of [tex]Ag_2SO_4[/tex] using the atomic masses of Ag [tex](107.87 g/mol)[/tex] and [tex]S (32.07 g/mol)[/tex]

[tex]\[ \text{Molar mass of Ag}_2\text{SO}_4 = 2 \times 107.87\text{ g/mol (Ag)} + 32.07\text{ g/mol (S)} + 4 \times 16.00\text{ g/mol (O)} \][/tex]

[tex]\[ \text{Molar mass of Ag}_2\text{SO}_4 = 2 \times 107.87 + 32.07 + 4 \times 16.00 \][/tex]

[tex]\[ \text{Molar mass of Ag}_2\text{SO}_4 = 215.74 + 32.07 + 64.00 \][/tex]

[tex]\[ \text{Molar mass of Ag}_2\text{SO}_4 = 311.81\text{ g/mol} \][/tex]

4. Use the given mass of [tex]Ag_2SO_4[/tex] to find the moles of [tex]Ag_2SO_4[/tex] produced:

[tex]\[ \text{Moles of Ag}_2\text{SO}_4 = \frac{\text{mass of Ag}_2\text{SO}_4}{\text{molar mass of Ag}_2\text{SO}_4} \][/tex]

[tex]\[ \text{Moles of Ag}_2\text{SO}_4 = \frac{0.657\text{ g}}{311.81\text{ g/mol}} \][/tex]

[tex]\[ \text{Moles of Ag}_2\text{SO}_4 \approx 0.002097\text{ moles} \][/tex]

5. Using the mole ratio from step 2, calculate the moles of AgNO3 needed:

[tex]\[ \text{Moles of AgNO}_3 = 2 \times \text{Moles of Ag}_2\text{SO}_4 \][/tex]

[tex]\[ \text{Moles of AgNO}_3 = 2 \times 0.002097 \][/tex]

[tex]\[ \text{Moles of AgNO}_3 = 0.004194\text{ moles} \][/tex]

6. Now, we can use the molarity of AgNO3 to find the volume needed:

[tex]\[ \text{Molarity (M)} = \frac{\text{moles of solute (mol)}}{\text{volume of solution (L)}} \][/tex]

[tex]\[ \text{Volume of AgNO}_3 = \frac{\text{moles of AgNO}_3}{\text{Molarity of AgNO}_3} \][/tex]

[tex]\[ \text{Volume of AgNO}_3 = \frac{0.004194\text{ moles}}{0.123\text{ M}} \][/tex]

[tex]\[ \text{Volume of AgNO}_3 \approx 0.03417\text{ L} \][/tex]

7. Convert the volume from liters to milliliters (since 1 L = 1000 mL):

[tex]\[ \text{Volume of AgNO}_3 = 0.03417\text{ L} \times 1000\text{ mL/L} \][/tex]

[tex]\[ \text{Volume of AgNO}_3 = 34.17\text{ mL} \][/tex]

Answer

23.52 atm pressure is exerted by a column of methanol

Explanation:

formula for pressure  

Pressure = force / area  

And we know that Force F= mass x acceleration (ma)

And mass = volume x density

And volume = cross sectional area x height?

so that...

P = F/A

Putting value of F

P = (mg)/A

Putting value for mass

P = (ρ V) g / A

Putting value for volume

P = ρ (A h) g / A

Canceling A  

P = ρgh

Now add the value  

now watch the units...

P = (0.787g/cm³) x 9.80 m/s² x 305m...  units= g x m² / (cm³ s²)  

and if you recall for pressure unit Pressure ...

1 Pa = 1 N / m² = (1 kg m / s²) / m²

so know

[g x m² / (cm³ s²)] x (1 kg / 1000g) x (100 cm / m)³  

converts the units to (1 kg m² / s²) / m

= (1 kg m / s²) / m²

Addition of all values  

P = (0.787g/cm³) x 9.80 m/s² x 305m x (1 kg / 1000g) x (100 cm / m)³ x (1 atm / 101325 Pa)

Pressure= 23.52 atm

Final answer:

The expression for the equilibrium constant (Kc) for the reaction 2SO₃ --> O₂ + 2SO₂ + 198 kJ/mol is C.) [O₂][SO₂]^2/[SO₃]^2, according to the law of equilibrium.

Explanation:

The question involves understanding how to write the expression for the equilibrium constant (Kc) for a given chemical reaction. Given the reaction 2SO₃ --> O₂ + 2SO₂ + 198 kJ/mol, the correct expression for Kc reflects the concentration of products over reactants, raised to the power of their stoichiometric coefficients in the balanced equation.

The correct expression for Kc is therefore C. [O₂][SO₂]^2/[SO₃]^2. This follows from the equilibrium law, which states that Kc is calculated by taking the concentration of the products, [O₂] and [SO₂] squared (because the coefficient of SO₂ is 2), over the concentration of the reactants, [SO₃] squared (because the coefficient of SO₃ is also 2).

The entropy change for the surroundings in the given reaction can be calculated using the formula ΔS° surroundings = -ΔH° reaction / T.

The entropy change for the surroundings in the given reaction can be calculated using the formula: ΔS° surroundings = -ΔH° reaction / T. In this case, the enthalpy change of the reaction is -802 kJ mol⁻¹ and the temperature is 298 K. Plugging in these values, we can calculate the entropy change for the surroundings.

ΔS° surroundings = -(-802 kJ mol⁻¹) / 298 K

ΔS° surroundings = 2.69 kJ K⁻¹ mol⁻¹

Therefore, the entropy change for the surroundings is 2.69 kJ K⁻¹ mol⁻¹.

Answer:

The percentage composition of each element in H2O2 is 5.88% H and 94.12% O (Option D).

Explanation:

Step 1: Data given

Molar mass of H = 1.0 g/mol

Molar mass of O = 16 g/mol

Molar mass of H2O2 = 2*1.0 + 2*16 = 34.0 g/mol

Step 2: Calculate % hydrogen

% Hydrogen = ((2*1.0) / 34.0) * 100 %

% hydrogen = 5.88 %

Step 3: Calculate % oxygen

% Oxygen = ((2*16)/34)

% oxygen = 94.12 %

We can control this by the following equation

100 % - 5.88 % = 94.12 %

The percentage composition of each element in H2O2 is 5.88% H and 94.12% O (Option D).

The percentage composition of hydrogen peroxide (H₂O₂) is D) 5.88% H and 94.12% O.

The calculation can be represented as follows:

1) Molar masses:

2) Molar mass of [tex]H_2O_2:[/tex]

3) Percentage of hydrogen:

4) Percentage of oxygen:

Therefore, the percentage composition of [tex]$\ce{H_2O_2}$[/tex] is approximately 5.88% H and 94.12% O.

Comparing with the given options, we can see that the closest match is D) 5.88% H and 94.12% O

Answer:

A. tetrahedral

Explanation:

With the above information in mind, the correct answer would then have to be either a) or b), however the complex  [Zn(NH₃)₂Cl₂]²⁺ has 4 ligands. This means that the correct answer is a), because the trigonal bipyramidal geometry has 5 ligands.

Answer:

The correct answer is The rate of decomposition of N2O4 is equal to the rate of formation of NO2.

Explanation:

According to theory of reaction kinetics we all know that at equilibrium the amount of reactant is equal to the amount of product.

   In other word it can be stated that the rate of the decomposition of N2O4 is equal to the rate of the formation of NO2.

Final answer:

At equilibrium, the partial pressure of N2O4 is equal to the partial pressure of NO2 in the flask.

Explanation:

Once equilibrium has been established, the partial pressure of N2O4(g) is equal to the partial pressure of NO2(g) in the flask. This is because the equilibrium constant (Kp) expression for the decomposition of N2O4 is (PNO2)^2 / PN2O4, which means the partial pressures of the two gases are directly related. Therefore, answer choice B, 'The partial pressure of N2O4(g) is equal to the partial pressure of NO2(g),' best describes the system within the flask at equilibrium.

Answer : The number of free electrons per cubic meter for silver are [tex]7.62\times 10^{28}m^{-3}[/tex]

Explanation :

To calculate the number of free electrons per cubic meter for silver by using the following equation:

[tex]n=1.3\times N_{Ag}[/tex]

[tex]n=1.3\times \frac{\rho_{Ag}\times N_A}{A_{Ag}}[/tex]

where,

[tex]\rho_{Ag}[/tex] = density of Ag = [tex]10.5g/cm^3[/tex]

[tex]N_A[/tex] = Avogadro's number = [tex]6.022\times 10^{23}atoms/mol[/tex]

[tex]A_{Ag}[/tex] = 107.87 g/mol

Now put all the given values in the above formula, we get:

[tex]n=1.3\times \frac{\rho_{Ag}\times N_A}{A_{Ag}}[/tex]

[tex]n=1.3\times \frac{(10.5g/cm^3)\times (6.022\times 10^{23}atoms/mol)}{107.87g/mol}[/tex]

[tex]n=7.62\times 10^{22}cm^{-3}[/tex]

[tex]n=7.62\times 10^{28}m^{-3}[/tex]

conversion used : [tex]1cm^{-3}=10^6m^{-3}[/tex]

Therefore, the number of free electrons per cubic meter for silver are [tex]7.62\times 10^{28}m^{-3}[/tex]

Final answer:

To calculate the number of free electrons per cubic meter for silver, we can use the density and molar mass of silver to determine the number of silver atoms per cubic meter. Since there is one free electron per silver atom, we can then calculate the number of free electrons per cubic meter.

Explanation:

To calculate the number of free electrons per cubic meter for silver, we first need to calculate the number of silver atoms per cubic meter. We can do this by using the density of silver, which is given as 10.5 g/cm³. Converting this to kg/m³ gives us a density of 10500 kg/m³. Next, we need to find the molar mass of silver, which is 107.87 g/mol. Using Avogadro's number (6.02 × 10^23 atoms/mol), we can calculate the number of silver atoms per cubic meter.

n = (10500 kg/m³) / (107.87 g/mol * (1 kg / 1000 g) * (6.02 × 10^23 atoms/mol))

n ≈ 9.48 x 10^28 atoms/m³

Since there is one free electron per silver atom, we can conclude that there are approximately 9.48 x 10^28 free electrons per cubic meter of silver.

Answer:

B. Hund's rule

Explanation:

Hund's rule -

According to Hund's rule ,

As, the electrons are negatively charged and hence , like poles repel , so they repel each other in order to stabilizes themselves and hence ,  the electron firstly occupies a vacant orbital , before actually pairing up , so as to reduce repulsion .

The rule of maximum multiplicity states that ,

The term with the lowest energy is the one with the highest value of the spin multiplicity .

hence , the statement given in the question is about Hund's rule .

Answer:

its actually A

Explanation:

Answer:

[tex][H_{3}O^{+}]=x M = 2.5\times 10^{-5}M[/tex] and pH = 4.6

Explanation:

Construct an ICE table to calculate changes in concentration at equilibrium.

[tex]C_{6}H_{5}COOH+H_{2}O\rightleftharpoons C_{6}H_{5}COO^{-}+H_{3}O^{+}[/tex]

I(M): 0.17                                    0.42                0

C(M): -x                                        +x                  +x

E(M): 0.17-x                                0.42+x             x

So, [tex]\frac{[C_{6}H_{5}COO^{-}][H_{3}O^{+}]}{[C_{6}H_{5}COOH]}=K_{a}(C_{6}H_{5}COOH)[/tex]

or, [tex]\frac{(0.42+x)x}{(0.17-x)}=6.3\times 10^{-5}[/tex]

or, [tex]x^{2}+0.4201x-(1.071\times 10^{-5})=0[/tex]

So, [tex]x=\frac{-0.4201+\sqrt{(0.4201)^{2}+(4\times 1\times 1.071\times 10^{-5})}}{(2\times 1)}M[/tex]

([tex]ax^{2}+bx+c=0\Rightarrow x=\frac{-b\pm \sqrt{b^{2}-4ac}}{2a};x< 0.17M[/tex])

So, [tex]x=2.5\times 10^{-5}[/tex]M

Hence [tex][H_{3}O^{+}]=x M = 2.5\times 10^{-5}M[/tex]

[tex]pH=-log[H_{3}O^{+}]=-logx=-log(2.5\times 10^{-5})=4.6[/tex]

The pH is 4.44, and the [H3O+] is 3.63 × 10^-5 M.

The question asks about the calculation of the hydronium ion concentration ([H3O+]) and the pH of a buffer solution consisting of benzoic acid (C6H5COOH) and sodium benzoate (C6H5COONa).

Write the equilibrium expression for benzoic acid dissociation:
C6H5COOH ↔ C6H5COO- + H3O+.

Use the Henderson-Hasselbalch equation:
pH = pKa + log ([A-]/[HA]),
where [A-] = concentration of benzoate ion and [HA] = concentration of benzoic acid.

Insert the values into the Henderson-Hasselbalch equation:
pH = 4.20 + log (0.42/0.17).
Calculate the pH.

Determine [H3O+] from pH:
[H3O+] = 10-pH.

Using the provided concentrations and the pKa of benzoic acid (4.20), the pH and [H3O+] can be calculated with high precision. For this specific buffer:

The pH is calculated to be 4.44.

The [H3O+] can then be found as 3.63 × 10-5 M.

Answer:

41.3 minutes

Explanation:

Since the reaction is a first order reaction, therefore, half life is independent of the initial concentration, or in this case, pressure.

[tex]t_{1/2}= \frac{0.693}{K}[/tex]

So, fraction of original pressure = [tex]\frac{1}{2}^2[/tex]

n here is number of half life

therefore, [tex]\frac{1}{8}= \frac{1}{2}^3[/tex]

⇒ n= 3

it took 124 minutes to drop pressure to 1/8 of original value, half life = 124/3= 41.3 minutes.

Answer : The half-life of this reaction at this temperature is, 41.5 seconds.

Explanation :

First we have to calculate the rate constant.

Expression for rate law for first order kinetics is given by:

[tex]k=\frac{2.303}{t}\log\frac{a}{a-x}[/tex]

where,

k = rate constant  = ?

t = time passed by the sample  = 124 s

a = let initial amount of the reactant  = X

a - x = amount left after decay process = [tex]\frac{1}{8}\times (X)=\frac{X}{8}[/tex]

Now put all the given values in above equation, we get

[tex]k=\frac{2.303}{124s}\log\frac{X}{(\frac{X}{8})}}[/tex]

[tex]k=0.0167s^{-1}[/tex]

Now we have to calculate the half-life.

[tex]k=\frac{0.693}{t_{1/2}}[/tex]

[tex]t_{1/2}=\frac{0.693}{0.0167s^{-1}}[/tex]

[tex]t_{1/2}=41.5s[/tex]

Therefore, the half-life of this reaction at this temperature is, 41.5 seconds.

Answer:

The enthalpy of reaction for the reaction of chlorine with ozone is -162.5 kJ.

Explanation:

[tex]ClO ( g ) + O_3 ( g )\rightarrow Cl ( g ) + 2 O_2 ( g ),\Delta H^o_{1,rxn} =-122.8 kJ [/tex]..[1]

[tex]2 O_3 ( g )\rightarrow 3O_2 ( g ),\Delta H^o_{2,rxn} = -285.3 kJ [/tex]..[2]

[tex]O_3(g) + Cl(g)\rightarrow ClO (g)+O_2(g),\Delta H^o_{3,rxn}=?[/tex]..[3]

The enthalpy of reaction for the reaction of chlorine with ozone can be calculated by using Hess's law:

[2] - [1] = [3]

[tex]\Delta H^o_{3,rxn}=\Delta H^o_{2,rxn}-\Delta H^o_{1,rxn}[/tex]

[tex]=-285.3 kJ-(-122.8 kJ)=162.5 kJ[/tex]

The enthalpy of reaction for the reaction of chlorine with ozone is -162.5 kJ.

Answer:

(b) Trigonal planar

Explanation:

The molecular geometry is the one that stabilizes better the bonds and the free electron pairs. If the molecule is nonpolar (overall dipole moment zero), so, there's no free electron pairs at the central atom. So, the molecule has the central atom A surrounded by three atoms of B, which is the trigonal planar geometry.

In an AB3 molecule, where A and B differ in electronegativity, for the molecule to have an overall dipole moment of zero, the molecular shape must be calculated in a way that the polar bonds sum up to zero. The correct molecular shape in this case is Trigonal Planar. In this geometry, all three B atoms are symmetrically distributed in a plane around the atom A (120° apart), which leads to cancellation of dipole moments.

In order for a molecule with disparity in electronegativity (polar bonds) to have an overall dipole moment of zero, the molecular structure needs to be in such a way that the dipole vectors cancel each other out. This typically happens when the molecule is symmetric. In the case of an AB3 molecule where A and B differ in electronegativity, the geometrical structures under consideration that could meet this condition are: Trigonal pyramidal, Trigonal planar, T-shaped, and Tetrahedral.

Upon reviewing these, the correct answer is (b) Trigonal planar. This is because this geometry allows for all three B atoms to be distributed in a plane around the A atom symmetrically (120° apart), allowing the dipole moments from the polar bonds to cancel each other out, which results in an overall zero dipole moment. The other geometries do not allow such cancellation, hence they can't be the correct answer.

To understand better, trying to draw the shapes out or use molecular 3D models can help visualise the planar shape and how the dipole vectors cancel each other out.

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Answer:

The correct answer is: 'Nitrogen is relatively inert'.

Explanation:

As, atomic number of nitrogen is 7 and its electronic distribution is 2, 5. Hence, there are 5 valence electrons present in a nitrogen atom. And, to attain stability it will gain three electrons from a donor atom. Hence, it will make a triple bond.

[tex][N]=1s^22s^22p^3[/tex]

Therefore, nitrogen has 5 valence electrons and makes 3 bonds in neutral compounds.Thus nitrogen will combine with another nitrogen atom to completes its octet to form [tex]N_2[/tex] gas molecule. Due to the presence of this triple bond, the gas molecule is almost inert as its bond dissociation energy is very high.

Hence, the correct option is:- nitrogen is relatively inert.

Answer:

[tex]X_{H_2}=\0.3584[/tex]

[tex]X_{CO_2}=\0.6416[/tex]

Explanation:

According to the Dalton's law of partial pressure, the total pressure of the gaseous mixture is equal to the sum of the pressure of the individual gases.

Partial pressure of carbon dioxide = 401 mmHg

Partial pressure of hydrogen gas = 224 mmHg

Total pressure,P = sum of the partial pressure  of the gases = 401 + 224 mmHg = 625 mmHg

Also, the partial pressure of the gas is equal to the product of the mole fraction and total pressure.

So,

[tex]P_{CO_2}=X_{CO_2}\times P[/tex]

[tex]X_{CO_2}=\frac{P_{CO_2}}{P}[/tex]

[tex]X_{CO_2}=\frac{401\ mmHg}{625\ mmHg}[/tex]

[tex]X_{CO_2}=\0.6416[/tex]

Similarly,

[tex]X_{H_2}=\frac{P_{H_2}}{P}[/tex]

[tex]X_{H_2}=\frac{224\ mmHg}{625\ mmHg}[/tex]

[tex]X_{H_2}=\0.3584[/tex]

Answer:

Complete ionic equation:

2H²⁺(aq) + SO₄²⁻(aq) + CaCO₃(s)   →  H₂O(l) + CO₂(g) + Ca²⁺(aq) + SO₄²⁻(aq)

Net ionic equation:

2H²⁺(aq) + CaCO₃(s)   →  H₂O(l) + CO₂(g) + Ca²⁺(aq)

Explanation:

Chemical equation:

H₂SO₄(aq) + CaCO₃(s)   →  H₂O(l) + CO₂(g) + CaSO₄(aq)

Balanced chemical equation:

H₂SO₄(aq) + CaCO₃(s)   →  H₂O(l) + CO₂(g) + CaSO₄(aq)

Complete ionic equation:

2H²⁺(aq) + SO₄²⁻(aq) + CaCO₃(s)   →  H₂O(l) + CO₂(g) + Ca²⁺(aq) + SO₄²⁻(aq)

Net ionic equation:

2H²⁺(aq) + CaCO₃(s)   →  H₂O(l) + CO₂(g) + Ca²⁺(aq)

Answer:

Complete ionic equation: 2 H⁺(aq) + SO₄²⁻(aq) + CaCO₃(s) → H₂O(I) + CO₂(g) + Ca²⁺(aq) + SO₄²⁻(aq)

Net ionic equation: 2 H⁺(aq) + CaCO₃(s) → H₂O(I) + CO₂(g) + Ca²⁺(aq)

Explanation:

The molecular equation includes all the species in the molecular form.

H₂SO₄(aq) + CaCO₃(s) → H₂O(I) + CO₂(g) + CaSO₄(aq)

The complete ionic equation includes all the ions and the molecular species.

2 H⁺(aq) + SO₄²⁻(aq) + CaCO₃(s) → H₂O(I) + CO₂(g) + Ca²⁺(aq) + SO₄²⁻(aq)

The net ionic equation includes only the ions that participate in the reaction (not spectator ions) and the molecular species.

2 H⁺(aq) + CaCO₃(s) → H₂O(I) + CO₂(g) + Ca²⁺(aq)

Answer:

The volume occupied is 25.7 L

Explanation:

Let's replace all the data in the formula

P . V = N . k . T

N = nNA

1 gm.mole . 6.02x10²³

k = Boltzmann's contant

T = T° in K

1 atm =  1.013 × 10⁵ N/m2

1.013 × 10⁵ N/m2 . Volume = 6.02x10²³  . 1.38065 × 10⁻²³ N · m/K . 313.7K

Volume = (6.02x10²³ . 1.38065 × 10⁻²³ N · m/K . 313.7K) / 1.013 × 10⁵ m2/N

Volume = 2607.32 N.m / 1.013 × 10⁵ m2/N = 0.0257 m³

1 dm³ = 1 L

1m³  = 1000 dm³

25.7 L

The volume of 1 gm·mole of gas at atmospheric pressure and a temperature of 313.7 K is approximately 25.7 Liters, as calculated using the Ideal Gas Law PV=nRT, with n=1, P=1 atm, T=313.7K, and R=0.08206 L·atm/mol·K.

To calculate the volume occupied by 1 gm·mole of the gas at atmospheric pressure and a temperature of 313.7 K, you'll want to use the Ideal Gas Law (PV=nRT), where P is the pressure of a gas, V is the volume it occupies, n is the number of moles of the gas, T is the temperature, and R is the universal gas constant. We'll use the given values: n=1 gm·mole, P=1 atm, T=313.7K, and also utilize the known universal gas constant for these units R=0.08206 L·atm/mol·K.

The ideal gas equation PV=nRT can be rearranged to solve for the volume of the gas: V=nRT/P. Substituting the values, V=(1 gm·mole * 0.08206 L·atm/mol·K * 313.7K) / 1 atm. By performing the calculation, we find that the volume V is approximately 25.7 Liters.

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