White phosphorus is composed of tetrahedral molecules of P4 in which every P atom is connected to three other P atoms. In the Lewis structure of P4, there are 1. 6 bonding pairs and 4 lone pairs of electrons. 2. 6 bonding pairs and no lone pairs of electrons. 3. 5 bonding pairs and 4 lone pairs of electrons. 4. 3 bonding pairs and 4 lone pairs of electrons. 5. 6 bonding pairs and 2 lone pairs of electrons.

Answer:

Correct option -D

Explanation:

Mass defect:

The difference between the sum of the masses of individual nucleons that form an atomic nucleus and the mass of the nucleus.

Hence, i- is correct

Nuclear fission:

It is the process of splitting a nucleus into two nuclei with smaler masses.

Hence, ii- is correct

Nuclear fission involved bombarding with nuclei [tex]^{235}_{92}U[/tex] with [tex]^{4}_{2}He[/tex]

Nuclear fission of [tex]^{235}_{92}U[/tex] with neutrons forms two smaller mass nuclei.

[tex]^{1}_{0}n+^{235}_{92}U \rightarrow ^{91}_{36}Kr+^{142}_{56}B+3^{1}_{0}n+Energy[/tex]

Hence, iii- is incorrect

Therefore, option -D is correct

Statements i) and ii) are correct. Mass defect refers to the difference in mass between that of a nucleus and the sum of the masses of its component nucleons. Nuclear fission refers to the splitting of a heavier nucleus into two nuclei with smaller mass numbers.

The correct statement(s) from the ones provided are: i) Mass defect is the difference in mass between that of a nucleus and the sum of the masses of its component nucleons. This is because the mass of a nucleus is not just the sum of the individual masses of the protons and neutrons. Some mass is converted into energy to provide the binding force that holds the nucleus together, so the mass of the nucleus has a slight deficit (or defect) compared to the combined masses of its components.

ii) Nuclear fission involves the splitting of a heavier nucleus into two nuclei with smaller mass numbers. This process usually does not occur naturally, but is induced by bombardment with neutrons. The first reported nuclear fission occurred in 1939 when uranium-235 atoms were bombarded with slow-moving neutrons that split the uranium nuclei into smaller fragments.

Thus, the correct option is D. i) and ii) only.

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Answer:

27.7 g

Explanation:

To solve this problem we use the definition of molarity:

From the problem we're given M = 2.5 M and volume = 0.100 L. We use the above formula and calculate the required moles of calcium chloride (CaCl₂):

Finally we use the molecular weight of calcium chloride to calculate the required mass:

Rounding to the nearest tenth of a gram the answer is 27.7 g.

To make a 2.5 M stock solution of calcium chloride, you would need to weigh out approximately 276.2 grams of calcium chloride.

To make a 2.5 M stock solution of calcium chloride, you need to calculate the number of grams of calcium chloride required. The molecular weight of calcium chloride is 110.98 g/mol. The formula to calculate the number of grams is:

Grams = Moles x Molecular Weight

Since you want to make 100 mL of a 2.5 M solution, you first need to convert mL to moles using the formula:

Moles = Volume (L) x Molarity (M)

Finally, you can substitute the calculated moles into the first formula to find the grams, and round to the nearest tenth of a gram.

Using these calculations, you would need to weigh out approximately 276.2 grams of calcium chloride to make the 2.5 M stock solution.

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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:

c) spontaneous

Explanation:

According the equation of Gibb's free energy -

∆G = ∆H -T∆S

∆G = is the change in gibb's free energy

∆H = is the change in enthalpy

T = temperature

∆S = is the change in entropy .

And , the sign of the  ΔG , determines whether the reaction is Spontaneous or non Spontaneous or at equilibrium ,

i.e. ,

if

From the question ,

The value for ΔG is negative ,

Hence ,

ΔG < 0 , the reaction is Spontaneous

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:

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)

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:

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.

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).

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

Answer: 0.94 V

Explanation:

For the given chemical reaction :

[tex]Zn(s)+Cu^{2+}(aq)\rightarrow Cu(s)+Zn^{2+}[/tex]

Using Nernst equation :

[tex]E_{cell}=E^o_{cell}-\frac{2.303RT}{nF}\log \frac{[Zn^{2+}]}{[Cu^{2+}]}[/tex]

where,

F = Faraday constant = 96500 C

R = gas constant = 8.314 J/mol.K

T = room temperature = [tex]298K[/tex]

n = number of electrons in oxidation-reduction reaction = 2

[tex]E^o_{cell}[/tex] = standard electrode potential of the cell = +1.10 V

[tex]E_{cell}[/tex] = emf of the cell = ?

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

[tex]E_{cell}=+1.10-\frac{2.303\times (8.314)\times (298)}{2\times 96500}\log \frac{2.5}{1.0\times 10^{-5}}[/tex]

[tex]E_{cell}=+1.10-0.16V=0.94V[/tex]

The cell potential for this reaction is 0.94 V

The cell potential for the given reaction depends on the concentration of the reactants. We can use the Nernst equation to calculate the cell potential when the concentration of [Cu2+]=1.0⋅10−5M and [Zn2+]=2.5M.

The standard cell potential (E°cell) for the given reaction is +1.10V. To find the cell potential for this reaction when the concentrations of [Cu2+]=1.0⋅10−5M and [Zn2+] = 2.5M, we need to use the Nernst equation:

Ecell = E°cell - (0.0592/n) x log(Q)

Here, n is the number of electrons involved in the reaction, and Q is the reaction quotient, which is equal to [Zn2+]/[Cu2+]. In this case, n = 2. Plugging in the values, we get:

Ecell = 1.10V - (0.0592/2) x log((2.5M)/ (1.0⋅10−5M))

Solving this equation will give us the cell potential for the given concentrations.

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Answer:

119 kCal per serving.

Explanation:

The heat energy necessary to elevates water's temperature from 23.4°C to 37.9°C can be calculated by the equation below:

Q = mcΔT

Q: heat energy

m: mass in g

c: specific heat capacity in cal/g°C

ΔT = temperature variation in °C

m is the mass of water, considering the density of water to be 1g/mL, 100 mL of water weights  100g. Therefore:

Q = 100 g x 1.00 cal/g°C x (37.9 - 23.4)°C

Q = 1450 cal

1450 cal ____ 0.341 g peanuts

x             ____  28 g peanuts

x = 119061.58 cal

This means that the cal from fat per serving of peanuts is at least 119 kCal.

The temperature change of the water indicated 1.45 Calories for 0.341 grams of peanut. Therefore, a serving size of 28.0 grams contains option 2) 119 Calories from fat.

To determine the Calories from fat per serving, we must first calculate the energy released by the peanut during combustion and then scale it to a full serving size.

Thus, the Calories from fat per serving is option 2) 119 Cal/serving.

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:

1. Strong electrolytes: HCl and NaOH.

Weak electrolytes: CH₃COOH and NH₃.

Nonelectrolytes: (NH₂)₂CO (urea) and CH₃OH (methanol).

2. C) [tex]C) Sr(C₂H₃O₂)₂ (aq) + Na₂SO₄ (aq) → SrSO₄ (s) + 2NaC₂H₃O₂ (aq)[/tex]

3. The spectator ions are Na⁺ and C₂H₃O₂⁻.

4. Written in explanation section.

Explanation:

A characteristic of strong electrolytes is that the solute is assumed to be 100 percent dissociated  into ions in solution, therefore they are great conductors of electricity. Eg: HCl and NaOH.

On the other hand, weak electrolytes are not completely dissociated in solution, therefore they are poor conductors of electricity. Eg: CH₃COOH and NH₃.

(By dissociation we mean the breaking up of the compound into cations and anions).

     A nonelectrolyte does not conduct electricity when dissolved in water. Eg: (NH₂)₂CO (urea) and CH₃OH (methanol).

     2. The products formed on each reaction are (always remember to balance the equations):

A) [tex]2 LiOH (aq) + Na₂S (aq) → Li₂S (aq) + 2NaOH (aq)[/tex]

B) [tex](NH₄)₂SO4 (aq) +2 LiCl (aq) → 2NH₄Cl (aq) + Li₂SO₄ (aq)[/tex]

C) [tex]C) Sr(C₂H₃O₂)₂ (aq) + Na₂SO₄ (aq) → SrSO₄ (s) + 2NaC₂H₃O₂ (aq)[/tex]

D) [tex]D) KNO₃ (aq) + NaOH (aq) → NaNO₃ (aq) + KOH (aq) [/tex]

The reaction C will produce SrSO₄, a white color precipitate.

    3. When ionic compounds dissolve in water, they break apart into their component cations and anions. To be more realistic, the equations should show the dissociation of dissolved ionic compounds into ions. This is called a ionic equation which shows dissolved species as free ions. To see whether a precipitate might form from this solution, we first combine the cation and anion from different compounds, and refer to the solubility rules. The spectator ions are ions that are not involved in the overall reaction.

Therefore, for the equation chosen above:

[tex]Sr²⁺ + 2C₂H₃O₂⁻ + 2Na⁺ + SO₄²⁻ → SrSO₄ + 2Na⁺ + 2C₂H₃O₂⁻[/tex]

Because spectator ions appear on both sides of an equation, they can be eliminated from the ionic equation.

[tex]Sr²⁺ + SO₄²⁻ → SrSO₄[/tex]

Finally, we end up with the net ionic equation, which shows only the species that actually take part in the reaction.

In this reaction, the spectator ions are Na⁺ and C₂H₃O₂⁻.

     4.  Molecular equations:

A) [tex]2 LiOH (aq) + Na₂S (aq) → Li₂S (aq) + 2NaOH (aq)[/tex]

B) [tex](NH₄)₂SO4 (aq) +2 LiCl (aq) → 2NH₄Cl (aq) + Li₂SO₄ (aq)[/tex]

C) [tex]C) Sr(C₂H₃O₂)₂ (aq) + Na₂SO₄ (aq) → SrSO₄ (s) + 2NaC₂H₃O₂ (aq)[/tex]

D) [tex]D) KNO₃ (aq) + NaOH (aq) → NaNO₃ (aq) + KOH (aq) [/tex]

         Complete ionic equations:

A) [tex]2Li⁺ + OH⁻ + 2Na⁺ + S²⁻ → 2Li⁺ + S²⁻ + 2Na⁺ +OH⁻[/tex]

B) [tex]2NH₄⁺ + SO₄²⁻ + 2Li⁺ + 2Cl⁻ → 2Li⁺ +S²⁻ +2NH₄⁺ +2Cl⁻ + 2Li⁺ + SO₄²⁻[/tex]

C) [tex]Sr²⁺ + 2C₂H₃O₂⁻ + 2Na⁺ + SO₄²⁻ → SrSO₄ + 2Na⁺ + 2C₂H₃O₂⁻[/tex]

D) [tex]K⁺ + NO₃⁻ + Na⁺ +OH⁻ → Na⁺ + NO₃⁻ + K⁺ + OH⁻[/tex]

        If all products are aqueous, a net ionic equation cannot be written because all ions are canceled out as spectator ions. Therefore, no precipitation reaction occurs. The only net equation can be written for reaction C):

C) [tex]Sr²⁺ + SO₄²⁻ → SrSO₄[/tex]

Answer:

F⁻(aq) + H⁺(aq) ⇄ HF(aq)

Explanation:

When aqueous solutions of potassium fluoride and hydrochloric acid are mixed, an aqueous solution of potassium chloride and hydrofluoric acid results. The corresponding molecular equation is:

KF(aq) + HCl(aq) ⇄ KCl(aq) + HF(aq)

The full ionic equation includes all the ions and the molecular species. HF is a weak acid so it exists mainly in the molecular form.

K⁺(aq) + F⁻(aq) + H⁺(aq) + Cl⁻(aq) ⇄ K⁺(aq) + Cl⁻(aq) + HF(aq)

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

F⁻(aq) + H⁺(aq) ⇄ HF(aq)

Answer:

The molecular formula for the organic compound is C₈H₈O₂

Explanation:

70.58 % C , 5.9 % H , and 23.50 % O

This is a percent by mass, so:

in 100 g of compound I have 70.58 g of C, 5.9 g of H and 23.5 g of O.

We need to apply the formula for freezing point depression, the colligative property.

ΔT = Kf . m

Where ΔT means T°F (pure solvent) − T°F (solution)

Kf, the cryoscopic constant and m is molality (mol of solute in 1kg of solvent); Kf for cyclohexane is 20.8 °C/m

6.59°C - 3.29°C = 20.8 °C/m . m

3.3°C = 20.8 °C/m . m

3.3°C /20.8 m/°C = molal

0.158 m/kg

In 1kg of solvent I have 0.158 moles, but I have 718.1 g of cyclohexane

1000 g _____ 0.158 moles of solute

718.1 g _____ 0.114 moles of solute

To find out the molar mass of my compound I have to divide

mass/mol → 15.5 g /0.114 m = 136.04 g/m

Now I have to work with the percent and think this rules of three:

100 g of compound has ____ 70.58 g of C __5.9 g of H __23.5 g of O

136.04 g of compound has ___  96.02 g of C __8.02 g of H __ 32 g of O

Molar mass of C = 12 g/m

Molar mass of H = 1 g/m

Molar mass of O = 16 g/m

g / molar mass = moles

C: 96 g / 12 g/m = 8

H: 8 g / 1 g/m =8

O: 32 g/ 16 g/m = 2

The molecular formula for the organic compound is C₈H₈O₂

To find the molecular formula of the organic compound, calculate the empirical formula first, which is CH₃O. Then calculate the molar mass of the compound, which is 29g/mol. Divide the molar mass of the compound by the molar mass of the empirical formula to find the molecular formula ratio. The molecular formula is CH₃O.

To determine the molecular formula of the organic compound, we need to calculate the empirical formula first.

The percentages of carbon, hydrogen, and oxygen in the compound represent the mass ratios of these elements. Assume we have 100g of the compound, which means we have 70.58g of carbon, 5.9g of hydrogen, and 23.50g of oxygen. Now we convert these masses to moles using the molar masses: 12g/mol for carbon, 1g/mol for hydrogen, and 16g/mol for oxygen.

Next, we divide the moles of each element by the smallest number of moles to get the simplest mole ratio. The empirical formula of the compound is CH3O.

To find the molecular formula, we need to know the molar mass of the compound. This can be calculated by adding the atomic masses of the elements in the empirical formula: 12g/mol for carbon, 1g/mol for hydrogen, 16g/mol for oxygen. The sum is 29g/mol.

Since the molar mass of the compound is 29g/mol and 15.5g of the compound is dissolved in 718.1g of cyclohexane, we can calculate the moles of the compound: 15.5g / 29g/mol = 0.534moles.

The molecular formula is found by multiplying the empirical formula by an integer value to match the molar mass. To do this, divide the molar mass of the compound by the molar mass of the empirical formula: 29g/mol / 29g/mol = 1.

Therefore, the molecular formula of the organic compound is CH3O.

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Answer:

Explanation:

Use gas equation to calculate mol of H2O gas:  

P = pressure = 389/760 = 0.512atm  

V = volume = 4L  

n=???

R = 0.082057

T = 121+273 = 394K  

0.512 * 4 = n*0.082057*394

n = (0.512*4) / (0.082057*394)  

n = 2.048 / 32.33

n = 0.063 mol H2O  

Molar5 mass H2O = 18g/mol  

Mass of water = 0.063*18

Mass of water valour = 1.14g

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:

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:

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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Answer:

a) 40 %

b) [tex]4.04~g~CO_2[/tex]

c) [tex]5.53x10^2^3~molecules~of~O_2[/tex]

Explanation:

For a) we will have to calculate the molar mass of [tex]C_6H_1_2O_6[/tex], so the first step is to find the atomic mass of each atom and multiply by the amount of atoms in the molecule.

C => 12*(6) = 72

H => 1*(12) = 12

O => 6*(16) = 96

Molar mass = 180 g/mol

Then we can calculate the percentage by mass:

[tex]Percentage~=~\frac{72}{180}*100=40[/tex]

For b) we have to start with the reaction of glucose:

[tex]C_6H_1_2O_6~+~6O_2~->~6CO_2~+~6H_2O[/tex]

Then we have to convert the grams of glucose to moles, the moles of glucose to moles of carbon dioxide and finally the moles of carbon dioxide to grams. To do this we have to take into account the following conversion ratios:

-) 180 g of glucose = 1 mol glucose

-) 1 mol glucose = 6 mol carbon dioxide

-) 1 mol carbon dioxide = 44 g carbon dioxide

[tex]16.55~g~C_6H_1_2O_6\frac{1~mol~C_6H_1_2O_6}{180~g~C_6H_1_2O_6}\frac{6~mol~CO_2}{1~mol~C_6H_1_2O_6}\frac{44~g~CO_2}{1~mol~CO_2}=4.04~g~CO_2[/tex]

For C, we have to start with the conversion from grams of glucose to moles, the moles of glucose to moles of oxygen and finally the moles of oxygen to molecules. To do this we have to take into account the following conversion ratios:

-) 180 g of glucose = 1 mol glucose

-) 1 mol glucose = 6 mol oxygen

-) 1 mol oxygen = 6.023x10^23 molecules of O2

[tex]16.55~g~C_6H_1_2O_6\frac{1~mol~C_6H_1_2O_6}{180~g~C_6H_1_2O_6}\frac{6~mol~O_2}{1~mol~C_6H_1_2O_6}\frac{6.023x10^2^3~molecules~O_2}{1~mol~O_2}=~5.53x10^2^3~molecules~of~O_2[/tex]

This detailed answer explains the mass percent of carbon in glucose, the mass of CO2 produced by its combustion, and the number of oxygen molecules needed for complete combustion.

a) Mass Percent of Carbon in Glucose: Glucose has a molar mass of 180.16 g/mol. The molar mass of the carbon in one mole of glucose is 72.06 g (6 carbons in a molecule of glucose). Therefore, the mass percent of carbon in glucose is 40%.

b) Mass of CO2 Produced: The combustion of 16.55 g of glucose will produce 44.01 g of CO2.

c) Oxygen Molecules Needed: For the complete combustion of 16.55 g of glucose, 23 molecules of O2 are required.

Answer:

The standard potential for this galvanic cell is 1.25 V

Explanation:

Step 1

Oxidation: Fe(s) → Fe^2+ (aq) + 2e-     E° = 0.447 V

Reduction: Ag+(aq) + e- → Ag(s)          E° = 0.7996 V

The electron number for the oxidation is 2; the electron number for the reduction is 1.

By multiplying the reduction reaction by 2, we make the electron number the same between the two half-equations

2 Ag+(aq) + 2e- → 2Ag(s)          E° = 0.7996 V

Step 2: The overall balanced equation

Fe(s) + 2Ag+(aq) → Fe^2+(aq) + 2ag(s)   E° = 0.447 V + 0.7996 V = 1.25V

The standard potential for this galvanic cell is 1.25 V

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:

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:

[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

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

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:

a) 2-chloro-2,3-dimethylbutane

b) 1-chloro-2,3-dimethylbutane

Explanation:

The monochlorination is a reaction in which we have to add only 1 Cl to the molecule. In this case, we will have to add a Cl to a primary carbon (a) and to a tertiary carbon (b).

In the monochlorination of the primary carbon, we can choose any methyl carbon. For the monochlorination of the terciary carbon we have to choose an CH carbon.

(See the figure)

Final answer:

When 2,3-dimethylbutane undergoes monochlorination under radical substitution conditions, multiple constitutional isomers can be formed due to the presence of primary and tertiary carbon atoms.

Explanation:

When 2,3-dimethylbutane undergoes monochlorination under radical substitution conditions, multiple constitutional isomers can be formed. The monochlorination reaction involves the substitution of a hydrogen atom on the carbon chain with a chlorine atom. Since 2,3-dimethylbutane has two different types of carbon atoms (primary and tertiary), the resulting constitutional isomers will also differ.

(a) Primary alkyl halide: The primary carbon atom is bonded to one other carbon atom and one hydrogen atom. So, a constitutional isomer could be formed by substituting the hydrogen atom with a chlorine atom.

(b) Tertiary alkyl halide: The tertiary carbon atom is bonded to three other carbon atoms. So, a constitutional isomer could be formed by substituting one of the carbon atoms with a chlorine atom.

Answer:

0,07448M of phosphate buffer

Explanation:

sodium monohydrogenphosphate (Na₂HP) and sodium dihydrogenphosphate (NaH₂P) react with HCl thus:

Na₂HP + HCl ⇄ NaH₂P + NaCl (1)

NaH₂P + HCl ⇄ H₃P + NaCl (2)

The first endpoint is due the reaction (1), When all phosphate buffer is as NaH₂P form, begins the second reaction. That means that the second endpoint is due the total concentration of phosphate that is obtained thus:

0,01862L of HCl×[tex]\frac{0,1000mol}{L}[/tex]= 1,862x10⁻³moles of HCl ≡ moles of phosphate buffer.

The concentration is:

[tex]\frac{1,862x10^{-3}moles}{0,02500L}[/tex] = 0,07448M of phosphate buffer

I hope it helps!