Which compound is most likely to form intermolecular hydrogen bonds?
A.) C4H10
B.) NaH
C.) C2H4OH
D.) C2H5SH

Answer:

The chemical term in the equation for the precipitate of AgCl(s) is n=3.54*10^-3

Explanation:

the quantity of AgCl(s) in moles is:

n = 0.508g / 143.32 g/mol = 3.54*10^-3 mol

to verify it the mass of AgNO3 involved in the reaction should be

n AgNO3 required = n = 3.54*10^-3 mol

the mass of n involved should be higher than n AgNO3

n existing = V*N = 0.523 mol/L * 35*10^-3 L = 18.305*10^-3 mol

The chemical term of the reaction

AgNO3(aq) +  NaCl(aq) →  AgCl(s) + NaNO3(aq)

Is when appears red brick in the solution indicating the formation of the precipitate

Answer: concentration in Molarity it's actually Molarity

Molarity=concentration/molar mass

Concentration=50g/0.3dm³=166.67g/dm³

Molar mass=180g/mol

Molarity=0.926mol/dm³

Answer:

Answer is =0. 925 mol L−1

Explanation:

C₆H₁₂O₆ is formula of Glucose  

For Molarity, we must know the following things  

• the number of moles of solute present in solution

• the total volume of the solution

we know the mass of one mole of Glucose = 180.156 g/mol

Number of moles = Given Mass of substance / Mass of one mole

No of moles = 50g / 180.156 g/mol

                   = 0.277 moles  

Now we know that  molarity is expressed per liter of solution. Since you dissolve 0.277 moles of potassium chloride in 300. mL of solution, you can say that 1.0 L will contain

For 300 ml of solution, no of moles are = 0.277 moles

For 1 ml of solution, no of moles are = 0.277/300 moles

For 1(1000) ml of solution, no of moles are=  0. 277/300 x 1000

                                                                 = 0.925 moles/ L

Answer is =0. 925 mol L−1

Answer:

The binding energy per nucleon = 1.368*10^-12  (option D)

Explanation:

Step 1: Data given

The mass of a proton is 1.00728 amu

The mass of a neutron is 1.00867 amu

The mass of a cobalt-60 nucleus is59.9338 amu

Step 2: Calculate binding energy

The mass defect = the difference between the mass of a nucleus and the total mass of its constituent particles.

Cobalt60 has 27 protons and 33 neutrons.

The mass of 27 protons = 27*1.00728 u = 27.19656 u

The mass of 33 neutrons = 33*1.00867 u = 33.28611 u

Total mass of protons + neutrons = 27.19656 u + 33.28611 u = 60.48267 u

Mass of a cobalt60 nucleus = 59.9338 amu

Mass defect = Δm = 0.54887 u

ΔE =c²*Δm

ΔE = (3.00 *10^8 m/s)² *(0.54887 amu))*(1.00 g/ 6.02 *10^23 amu)*(1kg/1000g)

Step 3: Calculate binding energy per nucleon

ΔE = 8.21 * 10^-11 J

8.21* 10^-11 J / 59.9338 = 1.368 *10^-12

The binding energy per nucleon = 1.368*10^-12  (option D)

To calculate the binding energy per nucleon of a Co nucleus, we calculate the mass defect, use Einstein's equation to calculate the binding energy, and then divide by the number of nucleons in the nucleus. The binding energy per nucleon can be converted to MeV using a conversion factor.

To calculate the binding energy per nucleon (in J) of a Co nucleus, we first need to determine the total binding energy. The mass defect for a Co nucleus is the difference between the total mass of the nucleus and the sum of the masses of the individual protons and neutrons. The mass defect can be calculated by subtracting the actual mass of the Co nucleus (59.9338 amu) from the sum of the individual masses of protons (1.00728 amu) and neutrons (1.00867 amu). Once we have the mass defect, we can use Einstein's equation E=mc² to calculate the binding energy. Finally, to find the binding energy per nucleon, we divide the total binding energy by the number of nucleons in the nucleus.

Using the given values, we have:

Mass defect (Δm) = (59.9338 amu) - (27 * 1.00728 amu + 33 * 1.00867 amu)

Binding energy (E) = Δm * (c²) = Δm * (3 × 10^8 m/s)²

Binding energy per nucleon = E / number of nucleons in Co nucleus

Substituting the values and performing the calculations, we get the binding energy per nucleon in joules. We can then convert the binding energy per nucleon from joules to MeV by using the conversion factor, 1 MeV = 1.602 × 10^-13 J.

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The value of ΔH is expected to be negative in processes where heat is released, which includes the condensation of steam to liquid water (II) and the freezing of water (III), both exothermic processes. Therefore, the correct answer is option D.) II and III only.

The question asks for which processes the value of ΔH (enthalpy change) is expected to be negative, indicating an exothermic reaction where heat is released. We can analyze each process one by one:

Therefore, the processes where ΔH is expected to be negative (exothermic) are options II and III, where steam condenses and water freezes.

Answer

The answer and procedures of the exercise are attached in the following archives.

Explanation  

You will find the procedures, formulas or necessary explanations in the archive attached below. If you have any question ask and I will aclare your doubts kindly.  

Answer:

why a solution is a home generous mixture?

Answer:

The concentration of the H₃PO₄ solution is 0,245M

Explanation:

The first titration is:

KHP + NaOH → KP⁻ + Na⁺ + H₂O

0,495g of KHP are:

0,495g×[tex]\frac{1mol}{204,22g}[/tex]= 2,42x10⁻³ moles of KHP

As 1 mole of KHP reacts with 1 mole of NaOH, moles of NaOH are 2,42x10⁻³ moles.

As volume required was 28,56mL, the concentration of the NaOH solution is:

2,42x10⁻³ moles / 0,02856L = 0,0849M

The titration of the phosporic acid with NaOH occurs as follows:

H₃PO₄ + NaOH → H₂PO₄⁻ + Na⁺ + H₂O

If were required 29,88mL of NaOH, the moles of NaOH spent were:

0,0849M×0,02988L = 2,54x10⁻³ moles of NaOH that are the same than H₃PO₄ moles.

As the volume of the solution of H₃PO₄ was 10,33mL, the concentration of the H₃PO₄ solution is:

2,54x10⁻³ moles of H₃PO₄ / 0,01033L = 0,245M

I hope it helps!

Final answer:

The concentration of the phosphoric acid is calculated using the molarity of a standardized NaOH solution obtained from a titration with KHP. The determined NaOH molarity is then applied to find the moles of phosphoric acid reacted in its own titration, which leads to the final concentration of the acid.

Explanation:

To determine the concentration of the phosphoric acid (H3PO4), we first need to calculate the molarity of the sodium hydroxide (NaOH) solution using the titration with potassium hydrogen phthalate (KHP). Since 0.495 grams of KHP was titrated with 28.56 mL of NaOH, we can calculate the moles of KHP used:

Now, using the concentration of NaOH, we can find the concentration of phosphoric acid. Given that 29.88 mL of the NaOH solution is required to titrate the 10.33 mL of H3PO4, we have:

0.2577 moles of ions

Explanation:

The dissociation of Rb₂SO₄ in water:

Rb₂SO₄ (s) + H₂O (l) → 2 Rb²⁺ (aq) + SO₄²⁻ (aq)

where:

(s) - solid

(l) - liquid

(aq) - aqueous

(aq) - aqueous

Knowing the dissociation of Rb₂SO₄ in water, we devise the following reasoning:

if         1 mol of Rb₂SO₄ dissociate in 2 moles of Rb²⁺ and 1 mole of SO₄²⁻

then   0.0859 moles of Rb₂SO₄ dissociate in X moles of Rb²⁺ and Y moles of SO₄²⁻

X = (0.0859 × 2) / 1 = 0.1718 moles of Rb²⁺

Y = (0.0859 × 1) / 1 = 0.0859 moles of SO₄²⁻

total moles of ions = moles of Rb²⁺ ions + moles of SO₄²⁻ ions

total moles of ions = 0.1718 + 0.0859 = 0.2577 moles

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Total moles of ions released when 0.0859 mol of Rb2SO4 dissolves completely in water are 0.2577 mol.

In the International System of Units, Mole is the base unit of the amount of any substance.

The dissociation of [tex]Rb_2SO_4[/tex] in water:

[tex]\rm Rb_2SO_4 (s) + H_2O (l) = 2 Rb^2^+ (aq) + SO_4^\;^2^- (aq)[/tex]

If 1 mol of [tex]Rb_2SO_4[/tex] dissociating into 2 moles of Rb²⁺ and 1 mole of SO₄²⁻

Then, 0.0859 mol of [tex]Rb_2SO_4[/tex] will dissociate into?

[tex]X = \dfrac{(0.0859 \times 2) }{1 } = 0.1718 moles\; of\; Rb^2^+[/tex]

[tex]Y= \dfrac{(0.0859 \times 2) }{1 } = 0.0859\; moles\; of\; So_4\;^2^-[/tex]

Total moles of ions  = moles of Rb²⁺ ions + moles of SO₄²⁻ ions

Total moles of ions = 0.1718 + 0.0859 = 0.2577 moles

Thus, the moles of ions are 0.2577 mol.

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An exothermic chemical reaction occurs when a system becomes warmer and the potential energy of the chemical substances increases.

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Final answer:

The correct answer is A, wherein during an exothermic reaction, the system warms up due to a decrease in potential energy of the substances involved.

Explanation:

The correct answer to this question is option A: a system becomes warmer, and the chemical substances undergo a decrease in potential energy.  During an exothermic chemical reaction, energy is released into the surroundings in the form of heat, causing the temperature of the surrounding system to increase. This energy release is due to the reactants possessing higher potential energy than the products; as the reaction occurs, the potential energy stored within the chemical bonds of the reactants is converted into kinetic energy (heat), significantly reducing the potential energy within the resulting products. This principle is foundational in understanding energy changes during chemical reactions, which is crucial in fields ranging from chemical engineering to environmental science.

Answer:

A.) Rate=k[A]

Explanation:

The rate law has the following general form:

Rate = k . [A]ᵃ . [B]ᵇ

where,

k is the rate constant

[A] and [B] are the molar concentrations of the reactants

a and b are the reaction orders with respect to A and B

a + b is the overall order of reaction

The rate expression for a first order reaction could be

A.) Rate=k[A]          YES. The reaction order is 1.

B.) Rate=k[A]²[B]     NO. The reaction order is 2 + 1 = 3.

C.) Rate=k[A][B]      NO. The reaction order is 1 + 1 = 2.

D.) Rate=k[A]²[B]²  NO. The reaction order is 2 + 2 = 4

Answer:

The anode half reaction is : [tex]Fe(s)\rightarrow Fe^{2+}(aq.)+2e^{-}[/tex]

Explanation:

In electrochemical cell, oxidation occurs in anode and reduction occurs in cathode.

In oxidation, electrons are being released by a species. In reduction, electrons are being consumed by a species.

We can split the given cell reaction into two half-cell reaction such as-

Oxidation (anode): [tex]Fe(s)\rightarrow Fe^{2+}(aq.)+2e^{-}[/tex]

Reduction (cathode): [tex]Sn^{4+}(aq.)+2e^{-}\rightarrow Sn^{2+}(aq.)[/tex]

------------------------------------------------------------------------------------------------------------

overall: [tex]Fe(s)+Sn^{4+}(aq.)\rightarrow Fe^{2+}(aq.)+Sn^{2+}(aq.)[/tex]

So the anode half reaction is : [tex]Fe(s)\rightarrow Fe^{2+}(aq.)+2e^{-}[/tex]

In the electrochemical cell with redox reaction, Sn4+ (aq) + Fe (s) → Sn2+ (aq) + Fe2+ (aq), the anode half-reaction is Fe (s) → Fe2+ (aq) + 2e-, as Fe is oxidized from 0 to +2.

In the given chemical reaction, Sn4+ (aq) + Fe (s) → Sn2+ (aq) + Fe2+ (aq), we see that tin (Sn) and iron (Fe) change their oxidation states. Tin is being reduced from an oxidation state of +4 to +2, and iron is being oxidized from 0 to +2.

Type of reaction happening at anode is always an oxidation. Oxidation is defined as a reaction where a substance loses electrons. Since Fe goes from Fe to Fe2+, it loses 2 electrons in the process. Therefore, the anode half-reaction for this electrochemical cell is: Fe (s) → Fe2+ (aq) + 2e-

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First, calculate the total heat absorbed by the solution, then divide this value by the number of moles of Zn reacted. The resulting ΔH_rxn is -162.839 kJ/mol.

The question asks us to calculate the heat of reaction, or ΔH_rxn, per mole of Zinc (Zn) in the reaction.

First, we calculate the total heat absorbed by the solution using the formula q=mcΔT, where q is the heat absorbed, m is the mass of the solution, c is the specific heat capacity, and ΔT is the change in temperature.

The mass of the solution can be calculated by multiplying the volume of solution by its density, hence m = V . d = 50.0 mL x 1.02 g/mL = 51.0 g. The change in temperature is 23.7°C - 22.5°C = 1.2°C. Therefore, the heat absorbed by the solution, q, can be calculated as q = (51.0 g) * (4.18 J/g°C) * (1.2°C) = 256.536 J.

Then, to find the heat per mole, ΔH_rxn, we divide this value by the number of moles of Zn reacted. The molar mass of Zn is 65.38 g/mol, so 0.103 g is 0.103 g / 65.38 g/mol = 0.001575 mol of Zn. Therefore, ΔH_rxn = (256.536 J) / (0.001575 mol) = -162,839 J/mol = -162.839 kJ/mol. We get a negative value because the reaction is exothermic.

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ΔHrxn per mole of Zn is approximately -162.27 kJ/mol.

To find the enthalpy change (ΔHrxn) per mole of Zn, we need to follow several steps:

Therefore, ΔHrxn per mole of Zn is approximately -162.27 kJ/mol.

Answer:

a. 3 electrons

b. 13 neutrons

c. 12 mass number

d. 10 electrons

e. 10 electrons

Explanation:

A. For an element , the atomic number of a given element is equal to the number of protons , and

For a neutral atom , the number of electrons equals the number of protons ,

Hence ,

The atomic number of Li = 3 ,

Hence ,

Number of electron for Li = 3 .

B.

Number of neutrons = mass number - atomic number ,

For Mg-25

For Magnesium , atomic number = 12 ,

Number of neutrons = 25 - 12 = 13 neutrons

C. Mass number of an element is equal to the sum of number of protons and number of neutrons ,

Hence , mass number = 5 + 7 = 12 .

D. For a negatively charged ion , add the negative charge value with the number of electrons of the neutral atom ,

For Oxygen , atomic mass = 8 ,

hence , number of electrons = 8

For O²⁻ = 8 + 2 = 10 electrons

E. For a positively charged ion , subtract the positive charge value with the number of electrons of the neutral atom ,

For Magnesium , atomic number = 12 ,

hence , number of electrons = 12 ,

for ,  Mg²⁺ , number of electrons = 12 - 2 = 10 electrons ,

A neutral atom of Lithium has 3 electrons. Mg-25 has 13 neutrons. An atom with 5 protons and 7 neutrons has a mass number of 12. O2- has 10 electrons and Mg2+ has 10 electrons.

A. A neutral atom of lithium contains 3 electrons, equal to its atomic number. B. An atom of Mg-25 has 12 protons (since magnesium's atomic number is 12), and as Mg-25 indicates a mass number of 25, subtracting the atomic number from the mass number gives us 13 neutrons. C. The mass number of an atom is the total number of protons and neutrons in its nucleus. So for an atom with 5 protons and 7 neutrons, the mass number is 12. D. O2- is an ion of oxygen that has gained 2 electrons. Oxygen has 8 electrons in its neutral state so O2- has 10 electrons. E. Mg2+ means that magnesium has lost 2 electrons, so Mg2+ has 10 electrons (since in its neutral state it has 12).

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Answer:

m H2O = 56 g

Explanation:

∴ The heat ceded (-) by the Aluminum part is equal to the heat received (+) by the water:

⇒ - (mCΔT)Al = (mCΔT)H2O

∴ m Al = 25.0 g

∴ Mw Al = 26.981 g/mol

⇒ n Al = (25.0g)×(mol/26.981gAl) = 0.927 mol Al

⇒ Q Al = - (0.927 mol)(24.03 J/mol°C)(26.8 - 86.4)°C

⇒ Q Al = 1327.64 J

∴ mH2O = Q Al / ( C×ΔT) = 1327.64 J / (4.18 J/g.°C)(26.8 - 21.1)°C

⇒ mH2O = 55.722 g ≅ 56 g

To calculate the mass of water in the calorimeter, use the conservation of energy principle and the specific heat capacity formula. The mass of water in the calorimeter is 61 g.

To calculate the mass of water in the calorimeter, you can use the conservation of energy principle. The heat lost by the aluminum equals the heat gained by the water. The heat gained by water can be calculated using the specific heat capacity formula: Q = mcΔT.

From the given information, we know the initial temperature of the water (21.1°C), the final temperature of the water (26.8°C), and its specific heat capacity (4.18 J/g°C). Considering the amount of heat lost by the aluminum and the heat gained by the water, you can calculate the mass of water in grams to be 61 g.

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Answer: The standard cell potential of the cell is -0.71 V

Explanation:

The half reactions follows:

Oxidation half reaction:  [tex]Mg\rightarrow Mg^{2+}+2e^-;E^o_{Mg^{2+}/Mg}=-2.37V[/tex]  ( × 3)

Reduction half reaction:  [tex]Al^{3+}(aq.)+3e^-\rightarrow Al(s);E^o_{Al^{3+}/Al}=-1.66V[/tex]  ( × 2)

The balanced cell reaction follows:

[tex]2Al^{3+}(aq.)+3Mg(s)\rightarrow 2Al(s)+3Mg^{2+}(aq.)[/tex]

To calculate the [tex]E^o_{cell}[/tex] of the reaction, we use the equation:

[tex]E^o_{cell}=E^o_{cathode}-E^o_{anode}[/tex]

Substance getting oxidized always act as anode and the one getting reduced always act as cathode.

Putting values in above equation, we get:

[tex]E^o_{cell}=-2.37-(-1.66)=-0.71V[/tex]

Hence, the standard cell potential of the cell is -0.71 V

Answer:

a) 727.5 kJ

Explanation:

Step 1: Data given

Mass of the piece of copper = 6.22 kg

Initial temperature of the copper = 20.5 °C

Final temperature of the copper = 324.3 °C

Specific heat of copper = 0.385 J/g°C

Step 2:

Q = m*c*ΔT

⇒ with Q = heat transfer (in J)

⇒ with m = the mass of the object (in grams) = 6220 grams

⇒ with c = the specific heat capacity = 0.385 J/g°C

⇒ with ΔT = T2 -T1 = 324.3 -  20.5 = 303.8

Q = 6220 grams * 0.385 J/g°C * 303.8 °C

Q = 727509.9 J = 727.5 kJ

b) This heat capacity is the heat capacity given for a copper at a temperature of 25°C

The heat absorbed by the copper metal is 727.5 kJ.

To calculate the heat absorbed by the copper metal, we can use the formula Q = mcΔT, where Q is the heat absorbed, m is the mass of the copper, c is the specific heat of copper, and ΔT is the change in temperature. Plugging in the given values, we get Q = 6.22 kg * 0.385 J/g°C * (324.3°C - 20.5°C).

Simplifying the equation, we get Q = 6.22 kg * 0.385 J/g°C * 303.8°C. Converting grams to kilograms, and simplifying further, we get Q = 727.5 kJ.

The expected heat capacity at the classical limit is not provided in the information given, so we cannot determine how close the calculated heat capacity is to the expected value.

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To calculate the pKa for the acid-base indicator using the given absorbance and molar absorptivity values, apply Beer's Law to determine the concentrations of HIn and In-, then use the Henderson-Hasselbalch equation.

To calculate the pKa for the acid-base indicator HIn, we must use the absorbance data provided and apply Beer's Law, which relates the absorbance to the concentrations of the protonated form (HIn) and deprotonated form (In-) of the indicator.

We are given that at pH 6.12, the absorbance (A) is 0.818, the molar absorptivity (ε) of HIn is 2929 M-1 cm-1 and that of In- is 20060 M-1 cm-1, and the total concentration of the indicator (C) is 0.000127 M.

From Beer's Law we know that:

A = εHInb[HIn] + εnb[In-]

where b is the path length of the cuvette used, which is 1 cm. The concentration of HIn and In- sum up to the total concentration of the indicator:

[HIn] + [In-] = C

We must also consider the Henderson-Hasselbalch equation, which relates the pKa to the pH and the ratio of deprotonated to protonated forms:

pH = pKa + log([In-]/[HIn])

Using the absorbance at pH 6.12 and the total indicator concentration, we can calculate the fractions of HIn and In- and then use the Henderson-Hasselbalch equation to solve for the pKa.

Answer : The correct option is, (C) spontaneous only at low temperatures.

Explanation :

According to Gibb's equation:

[tex]\Delta G=\Delta H-T\Delta S[/tex]

[tex]\Delta G[/tex] = Gibbs free energy  

[tex]\Delta H[/tex] = enthalpy change

[tex]\Delta S[/tex] = entropy change  

T = temperature in Kelvin

As we know that:

[tex]\Delta G[/tex]= +ve, reaction is non spontaneous

[tex]\Delta G[/tex]= -ve, reaction is spontaneous

[tex]\Delta G[/tex]= 0, reaction is in equilibrium

The given chemical reaction is:

[tex]2S(s)+3O_2(g)\rightarrow 2SO_3(g)[/tex]

As we are given that, the given reaction is exothermic that means the enthalpy change is negative.

In this reaction, the randomness of reactant molecules are more and as we move towards the formation of product the randomness become less that means the degree of disorderedness decreases. So, the entropy will also decreases that means the change in entropy is negative.

Now we have to determine the spontaneity of this reaction when ΔH is negative and ΔS is negative.

As, [tex]\Delta G=\Delta H-T\Delta S[/tex]

[tex]\Delta G=(-ve)-T(-ve)[/tex]

[tex]\Delta G=(+ve)[/tex]   (at high temperature) (non-spontaneous)

[tex]\Delta G=(-ve)[/tex]   (at low temperature) (spontaneous)

Thus, the reaction is spontaneous only at low temperatures.

Answer:

A) [tex]\frac{1mol(SCN)_{2}}{2molNaSCN}[/tex]

B) 0.025 mol (SCN)₂

C) 2 mol (SCN)₂

D) [tex]\frac{1molMnSO_{4}}{2molH_{2}SO_{4}}[/tex]

E) 3.5504 mol H₂SO₄

Explanation:

2NaSCN +2H₂SO₄ + MnO₂ → (SCN)₂ + 2H₂O + MnSO₄ + Na₂SO₄

A) A conversion factor could be [tex]\frac{1mol(SCN)_{2}}{2molNaSCN}[/tex] , as it has the units that we want to convert to in the numerator, and the units that we want to convert from in the denominator.

B) 0.05 mol NaSCN *  [tex]\frac{1mol(SCN)_{2}}{2molNaSCN}[/tex] = 0.025 mol (SCN)₂

C) With 4 moles of NaSCN and 3 moles of MnSO₄, the reactant is NaSCN so we use that value to calculate the moles of product formed:

4 mol NaSCN * [tex]\frac{1mol(SCN)_{2}}{2molNaSCN}[/tex] = 2 mol (SCN)₂

D) [tex]\frac{1molMnSO_{4}}{2molH_{2}SO_{4}}[/tex]

E) 1.7752 mol MnSO₄ * [tex]\frac{2molH_{2}SO_{4}}{1molMnSO_{4}}[/tex] = 3.5504 mol H₂SO₄

Answer: High P and Low T

Where P is the Pressure and T is the Temperature.

Explanation:

1. High Pressure : The above chemical equation has the reactant and product sides.

The product side has total moles of 3 and reactant side has total moles of 4.

To obtain maximum yield in the production of CH4 , pressure must be high because it will favour the side the less number of moles(volume) which is the product side.

2.Low Temperature : The enthaply change indicated as negative shows it is an exothermic reaction. And for an exothermic reaction, the temperature must be lower so as to favor the forward reaction and to make up for the heat/energy loss.

Explanation:

The given data is as follows.

        Space craft fuel rate = 353 L/min

As 1 liter equals 1000 ml and 1 min equals 60 seconds.

So,     [tex]353 \times \frac{1000 ml}{60 sec}[/tex]

           = 5883.33 ml/sec

It is also given that density of the fuel is 0.7 g/ml and standard enthalpy of combustion of fuel is -57.9 kJ/g.

Fuel rate per second is 5883.33 ml.

             [tex]5883.33 ml \times 0.7 g/ml[/tex]

               = 4118.33 g

Hence, calculate the maximum power as follows.

          Power = Fuel consumption rate × (-enthalpy of combustion)

                      = 4118.33 g/s \times 57.9 kJ/g

                      = 238451.36 kJ/s

or,                  = 238451.36 kW

Thus, we can conclude that maximum power produced by given spacecraft is 238451.36 kW.

Answer:

[tex]HOCl[/tex], [tex]K_a=2.0\times 10^{-8}[/tex]

Explanation:

pH is defined as the negative logarithm of the concentration of hydrogen ions.

Thus,  

pH = - log [H⁺]

Strong acids dissociate completely and thus, 0.5 M of a solution of a strong acid yields a pH of 1.0 .

The expression of the pH of the calculation of weak acid is:-

[tex]pH=-log(\sqrt{k_a\times C})[/tex]

Where, C is the concentration = 0.5 M

Given, pH = 4.0

So, for [tex]HOCl[/tex], [tex]K_a=2.0\times 10^{-8}[/tex]

[tex]pH=-log(\sqrt{2.0\times 10^{-8}\times 0.5})[/tex]

pH = 4.0

Hence, the acid is HOCl.

Final answer:

The acid with a 0.50 M solution resulting in a pH of 4.0 is likely hydrofluoric acid (HF), due to its intermediate acid strength indicated by its Ka value of 6.8 x 10^-4.

Explanation:

The question involves determining the identity of an unknown acid based on its molarity and pH balance. A 0.50 M solution of the acid has a pH of 4.0. The pH is a measure of the hydrogen ion concentration in a solution, and the pKa value is the negative logarithm of the acid dissociation constant (Ka), which provides insight into the strength of an acid. Strong acids fully dissociate in water, resulting in a lower pH for a given concentration, while weak acids do not completely dissociate, leading to higher pH values.

Given the pH of 4.0 for a 0.5 M solution, we are looking for an acid with a pKa close to 4, since this would indicate a weak acid that only partially dissociates. Among the given options, HF with a Ka value of 6.8 x 10-4 would likely be the correct acid since it has an adequate strength to result in a pH of 4 for a 0.5 M solution. On the other hand, HBr is a strong acid and would yield a much lower pH for the same concentration, while HOCl and C6H5OH have much lower Ka values, meaning they are weaker acids and would result in a higher pH than 4.

Answer:

4 shared pairs

Explanation:

According to the VESPR, a molecule with a central atom C linked to 4 Cl would form a tetrahedral, whose angles are 109.5°. When more lone pairs are added, the repulsion is greater due to the unshared electrons leading to a distorted tetrahedral. The more lone pairs, the lower the bond angle.

Answer:

70.8 mmHg

Explanation:

To find partial pressure of carbon dioxide we first find the mole fraction of carbon dioxide

n (carbon dioxide) = 5/44.01 g/mol = 0.11361 mol

n (Helium) = 3.75 g/4 g/mol = 0.9375 mol

The partial pressure of carbon dioxide will be the mole fraction of carbon dioxide multiplied by the total pressure

Partial pressure = 0.11361g/(0.11361g + 0.9375 g)* 655 mmHg

                          = 70.8 mmHg

Answer:

The calcium chloride is an electrolyte salt, so the frezzing point of solution must be higher than ethylene glycol.

Explanation:

This is the colligative property for this question: Frezzing point depression

ΔTf = Kf · molal · i

ΔTf = T° fussion solvent pure - T° fussion solution

As both solutions are the same in molality and the solvent is water, the formula stands the same but the Calcium Chloride is a salt which is dissociated in water like this:

CaCl₂ → Ca²⁺  + 2Cl⁻

We have 3 moles of ions, so this value modiffy the formula with the Van't Hoff Factor (number of ion particles per individual molecule of solute).

Ethyleneglycol  is a non-electrolytic organic compound (It is often used 1 as the i).

Let's see the formula in both:

0° - T° fussion solution = Kf · molal · 3 → CaCl2

0° - T° fussion solution = Kf · molal · 1 → C2H6O2

Answer :

The rate of the reaction at t=0 is, [tex]9.0\times 10^{-13}M.s^{-1}[/tex]

The overall order of reaction is, second order reaction.

Explanation :

Rate law : It is defined as the expression which expresses the rate of the reaction in terms of molar concentration of the reactants with each term raised to the power their stoichiometric coefficient of that reactant in the balanced chemical equation.

The general reaction is:

[tex]A+B\rightarrow C+D[/tex]

The general rate law expression for the reaction is:

[tex]\text{Rate}=k[A]^a[B]^b[/tex]

where,

a = order with respect to A

b = order with respect to B

R = rate  law

k = rate constant

[tex][A][/tex] and [tex][B][/tex] = concentration of A and B reactant

Now we have to determine the rate law for the given reaction.

The given balanced equations is:

[tex]O_3(g)+NO(g)\rightarrow O_2(g)+NO_2(g)[/tex]

In this reaction, [tex]O_3[/tex] and [tex]NO[/tex] are the reactants.

The rate law expression for the reaction is:

[tex]\text{Rate}=k[O_2][NO][/tex]    ..........(1)

Given :

Rate constant = [tex]k=3.0\times 10^{-5}M^{-1}s^{-1}[/tex]

Concentration of [tex]O_3[/tex] = [tex]5.0\times 10^{-4}M[/tex]

Concentration of [tex]NO[/tex] = [tex]6.0\times 10^{-5}M[/tex]

Now put all the given values in equation 1, we get:

[tex]\text{Rate}=(3.0\times 10^{-5}M^{-1}s^{-1})\times (5.0\times 10^{-4}M)\times (6.0\times 10^{-5}M)[/tex]

[tex]\text{Rate}=9.0\times 10^{-13}M.s^{-1}[/tex]

Thus, the rate of the reaction at t=0 is, [tex]9.0\times 10^{-13}M.s^{-1}[/tex]

Now we have to determine the overall order of reaction.

From the given rate law expression we conclude that,

The order of reaction with respect to [tex]O_3[/tex] is, first order reaction.

The order of reaction with respect to [tex]NO[/tex] is, first order reaction.

Thus, overall order of reaction will be:

Overall order of reaction = 1 + 1 = 2

Thus, the overall order of reaction is second order reaction.

The calculated rate of the reaction at t=0 is 9.0×10−1 M×s−1, and the overall order of the reaction is second order, resulting from being first order with respect to each reactant.

The rate of the reaction O3(g) + NO(g) → O2(g) + NO2(g) at t=0 can be calculated using the given rate law and concentrations. The rate law is rate = k[O3][NO], where k is the rate constant, and the brackets denote the concentration of the reactants O3 and NO respectively. To calculate the reaction rate at t=0, we plug in the given values: rate = (3.0×106 M−1×s−1)(5.0×10−4 M)(6.0×10−5 M), which gives a rate of 9.0×10−1 M×s−1.

The overall order of the reaction can be determined by looking at the exponents of the reactant concentrations in the rate law. Since the rate law is rate = k[O3]1[NO]1, it is first order with respect to both O3 and NO. This means that the overall order of the reaction is the sum of the individual orders, which is 1 + 1 = 2, making the reaction second order overall.

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Answer:

The molar solubility of distilled water is [tex]$1.25 \times 10-3$[/tex] (Option B)

Explanation:

Given:

[tex]Pbl_2 -7.9 \times 10^-^9.[/tex]

Expression for solubility constant is,

[tex]$K_{s p}=\left[P b^{2+}\right]\left[I^{-}\right]^{2}$[/tex]

The given equation is,

[tex]$K_{s p}=7.9 \times 10^{-9}$[/tex]

Equation is to calculated as,  

[tex]$7.9 \times 10^{-9}=(s) \times(2 s)^{2}$[/tex]

[tex]$7.9 \times 10^{-9}=4 s^{3}$[/tex]

[tex]$s=1.25 \times 10^{-3} \mathrm{~mol} / L$[/tex]

The solubility product in distilled water is [tex]$s=1.25 \times 10^{-3} \mathrm{~mol} / L$[/tex].

Learn more about bonds refer:

The correct answer is C.)[tex]5.0x10^-^4.[/tex]

To determine the molar solubility of PbI2 in distilled water, we need to use the solubility product constant (Ksp) for PbI2, which is given as 7.9x10^-9.

 The chemical equation for the dissociation of PbI2 in water is:

[tex]\[ PbI_2(s) \rightleftharpoons Pb^{2+}(aq) + 2I^-(aq) \][/tex]

 The solubility product expression for this equilibrium is:

[tex]\[ Ksp = [Pb^{2+}][I^-]^2 \][/tex]

Let's denote the molar solubility of PbI2 as s. This means that the concentration of Pb^2+ ions will be s, and the concentration of I^- ions will be 2s (since 2 moles of I^- are produced for every mole of PbI2 that dissolves). We can now express the Ksp equation in terms of s:

[tex]\[ Ksp = s \cdot (2s)^2 \][/tex]

[tex]\[ Ksp = s \cdot 4s^2 \][/tex]

[tex]\[ Ksp = 4s^3 \][/tex]

 Now we can solve for s using the given Ksp value:

[tex]\[ 7.9 \times 10^{-9} = 4s^3 \][/tex]

[tex]\[ s^3 = \frac{7.9 \times 10^{-9}}{4} \][/tex]

[tex]\[ s^3 = 1.975 \times 10^{-9} \][/tex]

[tex]\[ s = \sqrt[3]{1.975 \times 10^{-9}} \][/tex]

[tex]\[ s \approx 5.82 \times 10^{-4} \][/tex]

 Since we are looking for the molar solubility, we can round this value to the appropriate number of significant figures, which is two significant figures based on the Ksp value given [tex](7.9x10^-9[/tex] has two significant figures). Therefore, the molar solubility of PbI2 in distilled water is approximately:

[tex]\[ s \approx 5.8 \times 10^{-4} \][/tex]

 The closest answer choice to this calculated value is C.) 5.0x10^-4.

Answer:

molarity of acid =0.0132 M

Explanation:

We are considering that the unknown acid is monoprotic. Let the acid is HA.

The reaction between NaOH and acid will be:

[tex]NaOH+HA--->NaA+H_{2}O[/tex]

Thus one mole of acid will react with one mole of base.

The moles of base reacted = molarity of NaOH X volume of NaOH

The volume of NaOH used = Final burette reading - Initial reading

Volume of NaOH used = 22.50-0.55= 21.95 mL

Moles of NaOH = 0.1517X21.95=3.33 mmole

The moles of acid reacted = 3.33 mmole

The molarity of acid will be = [tex]\frac{mmole}{volumne(mL)}=\frac{0.33}{25}=0.0132M[/tex]

The problem is about acid-base titration used in Chemistry. A 1:1 reaction between NaOH and the unknown acid is assumed. The molarity of the unknown acid is calculated as 0.1332 M.

This problem pertains to acid-base titration, a common method in Chemistry for determining the concentration of an unknown acid or base. The initial and final buret readings indicate the volume of NaOH used in the reaction: 22.50 mL - 0.55 mL = 21.95 mL, which is 0.02195 L. Using the molarity of NaOH (0.1517 M), we can find the moles of NaOH used: 0.1517 M * 0.02195 L = 0.00333 mol. Since the reaction between NaOH and the unknown acid is assumed to be 1:1, the moles of the unknown acid should also be 0.00333. Therefore, to obtain the molarity of the unknown acid, divide the moles of the acid by the volume of the acid in liters (25.0 mL = 0.025 L): 0.00333 mol / 0.025 L = 0.1332 M. Therefore, the molarity of the unknown acid is 0.1332 M.

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Answer:

ΔU = -538kJ/mol

Explanation:

Step 1: Data given

Mass of compound X = 0.485 grams

Mass of water = 3000 grams

Temperature rise = 0.285 °C

Heat capacity of the calorimeter = 3.81 kJ/°C

Specific heat of water = 4.184 J/g°C

MW of X = 56.0 g/mol

Step 2: Calculate q

∆U = ΔH - PΔV

Since the bomb calorimeter has a constant volume  ΔV = 0

⇒ ΔU = ΔH

qrxn = - (q(water) + q(bomb))

q(bomb) = 3810 J/°C * 0.285 = 1085.85 J

q(water) = 3000g * 4.184 J/g°C * 0.285°C = 3577.32 J

qrxn = q(water) + q(bomb)

qrxn = 4663.17 J = 4.66 kJ (Since this is an exothermic, the heat is released. (q is positive, ΔH is negative).

Step 3: Calculate moles of compound

Moles = mass / molar mass

Moles = 0.485 grams / 56.0 g/mol

Moles = 0.00866 moles

Step 4: Calculate ΔU

= 4663.17 J /0.00866 moles = 538472 j/mol = 538.5 kJ/mol

Since the reaction is exothermic, ΔU is negative

ΔU = -538kJ/mol

Answer:

D.) Chlorine has greater electronegativity and a larger first ionization energy

Explanation:

first we have to consider where sodium and chlorine are in the atomic table. Both are in the third row and both are nearly opposite ends of third row. sodium have high tendency to lose an electron while chlorine have higher tendency to gain an electron. so because of this fact chlorine is more electronegative than sodium.

As we move left to right in the periodic table, there is increase in the number of protons in atom and therefore there is increase in attraction between the nucleus and electrons and shrinking down the atom which is general trend when we move from left to right in the periodic table. so chlorine will be smaller than sodium because of the shrinking of the atom and smaller atoms have higher ionization energy because their electrons are close to the nucleus. They require more energy to remove. so this will make  chlorine have both a smaller atomic radius and larger ionization energy. so D is the right answer.

Final answer:

The correct statement regarding sodium and chlorine is that chlorine has a larger atomic radius and greater electronegativity.

Explanation:

The correct statement regarding sodium and chlorine is Option C: Chlorine has a larger atomic radius and greater electronegativity.

Chlorine, with 17 electrons, has a strong attraction for electrons and tends to gain an electron to form a chloride ion (Cl-). Sodium, with 11 electrons, has a lower ionization potential and tends to lose an electron to form a sodium ion (Na+). Therefore, chlorine has a greater electronegativity and a larger atomic radius compared to sodium.

Additionally, when sodium and chlorine react, sodium donates an electron to chlorine, forming an ionic bond and creating the compound sodium chloride (NaCl).

Answer: The specific heat of iridium is 0.130 J/g°C

Explanation:

When iridium is dipped in water, the amount of heat released by iridium will be equal to the amount of heat absorbed by water.

[tex]Heat_{\text{absorbed}}=Heat_{\text{released}}[/tex]

The equation used to calculate heat released or absorbed follows:

[tex]Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})[/tex]

[tex]m_1\times c_1\times (T_{final}-T_1)=-[m_2\times c_2\times (T_{final}-T_2)][/tex]      ......(1)

where,

q = heat absorbed or released

[tex]m_1[/tex] = mass of iridium = 23.9 g

[tex]m_2[/tex] = mass of water = 20.0 g

[tex]T_{final}[/tex] = final temperature = 22.6°C

[tex]T_1[/tex] = initial temperature of iridium = 89.7°C

[tex]T_2[/tex] = initial temperature of water = 20.1°C

[tex]c_1[/tex] = specific heat of iridium = ?

[tex]c_2[/tex] = specific heat of water = 4.18 J/g°C

Putting values in equation 1, we get:

[tex]23.9\times c_1\times (22.6-89.7)=-[20\times 4.18\times (22.6-20.1)][/tex]

[tex]c_1=0.130J/g^oC[/tex]

Hence, the specific heat of iridium is 0.130 J/g°C

Taking into account the definition of calorimetry, the specific heat of iridium is 0.13 [tex]\frac{J}{gC}[/tex].

Calorimetry is the measurement and calculation of the amounts of heat exchanged by a body or a system.

Sensible heat is defined as the amount of heat that a body absorbs or releases without any changes in its physical state (phase change).

So, the equation that allows to calculate heat exchanges is:

Q = c× m× ΔT

where:

In this case, you know:

Replacing in the expression to calculate heat exchanges:

For iridium: Qiridium= c × 23.9 g× (22.6 C - 89.7 C)

For water: Qwater= 4.186 [tex]\frac{J}{gC}[/tex]× 20 g× (22.6 C - 20.1 C)

If two isolated bodies or systems exchange energy in the form of heat, the quantity received by one of them is equal to the quantity transferred by the other body. That is, the total energy exchanged remains constant, it is conserved.

Then, the heat that the gold gives up will be equal to the heat that the water receives. Therefore:

- Qiridium = + Qwater

- c × 23.9 g× (22.6 C - 89.7 C)= 4.18 [tex]\frac{J}{gC}[/tex]× 20 g× (22.6 C - 20.1 C)

Solving:

c × 23.9 g× ( 89.7 C - 22.6 C)= 4.18 [tex]\frac{J}{gC}[/tex]× 20 g× (22.6 C - 20.1 C)

c × 23.9 g× 67.1 C= 4.18 [tex]\frac{J}{gC}[/tex]× 20 g× 2.5 C

c × 1603.69 g×C= 209 J

[tex]c=\frac{209 J}{1603.69 gC}[/tex]

c= 0.13 [tex]\frac{J}{gC}[/tex]

Finally, the specific heat of iridium is 0.13 [tex]\frac{J}{gC}[/tex].

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