Hydrazine (N2H4) is used as a rocket fuel. The combustion products are nitrogen (N2) and water (H2O). Use the following information about bond energies to estimate the change in enthalpy associated with the combustion of 1 mol of hydrazine

Bond Bond Energy (kJ/mol)

N–N 163

N=N 418

N N 941 <- triple bond

N–H 388

O–O 146

O=O 495

O–H 463

a. −583 kJ/mol d. −358 kJ/mol b. +2,450 kJ/mol e. −1,970 kJ/mol c. −1,078 kJ/mol

Answer:

For Kp = 1,4; [tex]P_{[A] = 0,22[/tex], [tex]P_{[B] = 0,56atm[/tex]

For Kp = 2,0x10⁻⁴; [tex]P_{[A] = 0,495[/tex], [tex]P_{[B] = 0,01atm[/tex]

For Kp = 2,0x10⁵; [tex]P_{[A] = 5x10^{-6}[/tex], [tex]P_{[B] = 0,99999atm[/tex]

Explanation:

For the reaction:

A(g) ⇄ 2B(g)

kp is: [tex]kp = \frac{P_{[B]}^2}{P_{[A]}}[/tex]

If initial pressure of B is 1,0atm and initial pressure of A is 0,0atm the equilibrium pressures are:

[tex]P_{[A] = 0,0atm + X[/tex]

[tex]P_{[B] = 1,0atm - 2X[/tex]

Replacing for Kp= 1,4:

[tex]1,4 = \frac{(1-2X)^2}{X}[/tex]

1,4X = 4X² - 4X + 1

0 = 4X² - 5,4X + 1

Solving for X:

X = 0,22 -Right answer-

X = 1,13

Replacing:

[tex]P_{[A] = 0,22[/tex]

[tex]P_{[B] = 1,0atm - 0,44atm = 0,56atm[/tex]

For Kp= 2,0x10⁻⁴:

[tex]2,0x10^{-4} = \frac{(1-2X)^2}{X}[/tex]

2,0x10^{-4}X = 4X² - 4X + 1

0 = 4X² - 4,0002X + 1

Solving for X:

X = 0,495atm

Replacing:

[tex]P_{[A] = 0,495atm[/tex]

[tex]P_{[B] = 1,0atm - 0,99atm = 0,01atm[/tex]

For Kp= 2,0x10⁵:

[tex]2,0x10^5 = \frac{(1-2X)^2}{X}[/tex]

2,0x10^5X = 4X² - 4X + 1

0 = 4X² - 2,00004x10^5X + 1

Solving for X:

X = 5x10⁻⁶ -Right answer-

Replacing:

[tex]P_{[A] = 5x10^{-6}[/tex]

[tex]P_{[B] = 1,0atm - 0,00001atm = 0,99999atm[/tex]

I hope it helps!

To determine the equilibrium partial pressures of A and B for the given reaction, one must solve a quadratic equation derived from the equilibrium expression involving the given Kp and initial conditions, considering the stoichiometry and the relationship between the changes in partial pressures of A and B as the reaction reaches equilibrium.

When considering the equilibrium partial pressures of A and B for the reaction A(g) = 2B(g), we need to calculate how the initial conditions and the equilibrium constant (Kp) affect the final partial pressures at equilibrium. Starting with an initial partial pressure of B as 1.0 atm and A as 0.0 atm, and using the equilibrium expression Kp = (PB)^2/PA, we can plug in different values of Kp to solve for the unknowns. Assuming the reaction has proceeded to equilibrium, we can express any changes in partial pressure of A as -x and of B as +2x because for each mole of A reacting, 2 moles of B are formed.

For example, with Kp = 1.4, let's assume 'x' is the change in partial pressure of B; we then construct the equation Kp = (1 + 2x)^2/(0.0 + x). Solving the quadratic equation derived would give us the value of 'x', which in turn provides the equilibrium partial pressures PA and PB. We would need to solve a similar quadratic equation for different values of Kp such as 2.0 * 10^-4 and 2.0 * 10^5 following the same approach. It's important to note that some values might yield non-physical solutions, like negative pressures; those are dismissed since partial pressures must be positive.

Answer:

ΔE is -4250 J

Explanation:

Step1: Data given

3.30 kJ of heat is released

There is 950 J of work done

Step 2: Calculate ΔE

ΔE = q + w  

 ⇒ with q = heat of energy released = 3.3 kJ = 3300 J ( negative because the heat is released)

⇒ work done = 950 J ( also negative )

ΔE = -3300J-950J= -4250 J

ΔE is -4250 J

Answer: [tex]\Delta H=-16.5 kcal[/tex]

Explanation:

The balanced chemical reaction is,

[tex]C_6H_{12}O_6(s)\longrightarrow 2C_2H_5OH(l)+2CO_2(g)[/tex]

The expression for enthalpy change is,

[tex]\Delta H=\sum [n\times \Delta H_f(product)]-\sum [n\times \Delta H_f(reactant)][/tex]

[tex]\Delta H=[(n_{C_2H_5OH}\times \Delta H_{C_2H_5OH})+(n_{CO_2}\times \Delta H_{CO_2})]-[(n_{C_6H_{12}O_6}\times \Delta H_{C_6H_{12}O_6})][/tex]

where,

n = number of moles

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

[tex]\Delta H=[(2\times -66.4)+(2\times -93.9)]-[(1\times -304.5)][/tex]

[tex]\Delta H=-16.5kcal[/tex]

Therefore, the enthalpy change for this reaction is -16.5 kcal

The Lewis symbol for Ca is Ca²⁺, while F is F⁻. The Ca atom loses two electrons, and the F atom gains one electron. The most likely product is CaF₂.

(a) The Lewis symbol for Ca is Ca2+, while the Lewis symbol for F is F-. The action that occurs between Ca and F atoms is a redox reaction, where one atom loses electrons (oxidation) and the other gains electrons (reduction).
(b) The chemical formula of the most likely product is CaF2 since one Ca atom can bond with two F atoms.
(c) In the reaction, two electrons are transferred, with Ca losing two electrons and each F atom gaining one electron.
(d) Ca atom loses electrons in the reaction.

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Answer : The amount of heat required is, 639.3 KJ

Solution :

The conversions involved in this process are :

[tex](1):Cr(s)(1760^oC)\rightarrow Cr(s)(1860^oC)\\\\(2):Cr(s)(1860^oC)\rightarrow Cr(l)(1860^oC)\\\\(3):Cr(l)(1860^oC)\rightarrow Cr(l)(2060^oC)[/tex]

Now we have to calculate the enthalpy change.

[tex]\Delta H=[m\times c_{p,s}\times (T_{final}-T_{initial})]+n\times \Delta H_{fusion}+[m\times c_{p,l}\times (T_{final}-T_{initial})][/tex]

where,

[tex]\Delta H[/tex] = enthalpy change or heat required = ?

m = mass of Cr = 126.3 g

[tex]c_{p,s}[/tex] = specific heat of solid Cr = [tex]44.8J/g^oC[/tex]

[tex]c_{p,l}[/tex] = specific heat of liquid Cr = [tex]0.94J/g^oC[/tex]

n = number of moles of Cr = [tex]\frac{\text{Mass of Cr}}{\text{Molar mass of Cr}}=\frac{126.3g}{52g/mole}=2.428mole[/tex]

[tex]\Delta H_{fusion}[/tex] = enthalpy change for fusion = 20.5 KJ/mole = 20500 J/mole

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

[tex]\Delta H=[126.3g\times 44.8J/g^oC\times (1860-(1760))^oC]+2.428mole\times 20500J/mole+[126.3g\times 0.94J/g^oC\times (2060-1860)^oC][/tex]

[tex]\Delta H=639342.4J=639.3KJ[/tex]     (1 KJ = 1000 J)

Therefore, the amount of heat required is, 639.3 KJ

To determine the amounts of heat for each heating step, we need to calculate the heat required to raise the temperature from solid Cr at 1760°C to its melting point at 1860°C, the heat of fusion to convert the solid Cr at its melting point to liquid Cr, and the heat required to raise the temperature from liquid Cr at 1860°C to the final temperature of 2060°C.

To determine the amounts of heat for each heating step, we need to calculate the heat required to raise the temperature from solid Cr at 1760°C to its melting point at 1860°C, the heat of fusion to convert the solid Cr at its melting point to liquid Cr, and the heat required to raise the temperature from liquid Cr at 1860°C to the final temperature of 2060°C.

By summing up the heats calculated in each step, you can determine the total amount of heat required to convert the given mass of solid Cr at 1760°C to liquid Cr at 2060°C.

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Answer:

Therewill be produced 170.6 grams NH3, there will remain 25 moles of N2, this is 700 grams

Explanation:

Step 1: Data given

Number of moles hydrogen = 30 moles

Number of moles nitrogen = 30 moles

Yield = 50 %

Molar mass of N2 = 28 g/mol

Molar mass of H2 = 2.02 g/mol

Molar mass of NH3 = 17.03 g/mol

Step 2: The balanced equation

N2 + 3H2 → 2NH3

Step 3: Calculate limiting reactant

For 1 mol of N2, we need 3 moles of H2 to produce 2 moles of NH3

Hydrogen is the limiting reactant.

The 30 moles will be completely be consumed.

N2 is in excess. There will react 30/3 =10 moles

There will remain 30 -10 = 20 moles (this in the case of a 100% yield)

In a 50 % yield, there will remain 20 + 0,5*10 = 25 moles. there will react 5 moles.

Step 4: Calculate moles of NH3

There will be produced, 30/ (3/2) = 20 moles of NH3 (In case of 100% yield)

For a 50% yield there will be produced, 10 moles of NH3

Step 5: Calculate the mass of NH3

Mass of NH3 = mol NH3 * Molar mass NH3

Mass of NH3 = 20 moles * 17.03

Mass of NH3 = 340.6 grams = theoretical yield ( 100% yield)

Step 6: Calculate actual mass

50% yield = actual mass / theoretical mass

actual mass = 0.5 * 340.6

actual mass = 170.3 grams

Step 7: The mass of nitrogen remaining

There remain 20 moles of nitrogen + 50% of 10 moles = 25 moles remain

Mass of nitrogen = 25 moles * 28 g/mol

Mass of nitrogen = 700 grams

Answer:

25 moles

Explanation:

1 mole of nitrogen reacts with 3 moles of hydrogen to give 2 moles of ammonia. 30 moles of hydrogen will react with 10 moles of nitrogen to give 20 moles of ammonia. As the actual yield is 50%, ammonia formed is 10 moles, the amount of nitrogen reacted is 5 moles, and the amount of hydrogen reacted is 15 moles. The mass of the remaining hydrogen is 15 moles and of the remaining nitrogen is 25 moles.

Answer:

7 valence electrons

Explanation:

Astatine has the atomic number 85. Thus, its electron configuration is:

[Xe] 4f¹⁴ 5d¹⁰ 6s² 6p⁵

As we can see, in the last level (6) it has 2 + 5 = 7 electrons, that is, astatine has 7 electrons in its valence shell. In the Lewis dot structure (attached) we write 3 pairs of electrons and 1 unpaired electron around the symbol of At.

428.7 kJ of heat are absorbed when 25.00 g of NH₃(g) reacts in the presence of excess O₂(g) to produce NO(g) and H₂O(l).

Let's consider the following thermochemical equation.

4 NH₃(g) + 5 O₂(g) → 4 NO(g) + 6 H₂O(l)  ΔH° = 1168 kJ

The standard enthalpy of the reaction is positive, which means that the reaction is endothermic, that is, heat is absorbed.

We will convert 25.00 g of NH₃ to moles using its molar mass (17.03 g/mol).

[tex]25.00 g \times \frac{1mol}{17.03g} = 1.468 mol[/tex]

1168 kJ of heat are absorbed when 4 moles of NH₃ react. The heat absorbed when 1.468 moles of NH₃ react is:

[tex]1.468 mol \times \frac{1168kJ}{4mol} = 428.7 kJ[/tex]

428.7 kJ of heat are absorbed when 25.00 g of NH₃(g) reacts in the presence of excess O₂(g) to produce NO(g) and H₂O(l).

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Chemical reactions are defined as the reaction in which two or more reactants combine to form single or more products. The heat can be absorbed or evolved in the reaction. The heat absorbed is 428.7kJ when ammonia reacts with excess oxygen.

Given that,

The chemical equation between ammonia and oxygen is:

Now, the enthalpy of the reaction is positive, such that the reaction is endothermic. In endothermic reaction heat is absorbed.

Now, converting the mass of ammonia into moles, we get:

1168 kJ of heat is absorbed when 4 moles of ammonia react with oxygen. The heat absorbed in 1.468 moles, will be:

Therefore, when 25 grams of ammonia reacts in the presence of oxygen 428.7 kJ will be absorbed. Hence, it is an endothermic reaction.

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Answer:

The final volume will be 24.7 cm³

Explanation:

Step 1: Data given:

Initial temperature = 180 °C

initial volume = 13 cm³ = 13 mL

The mixture is heated to a fina,l temperature of 587 °C

Pressure and amount = constant

Step 2: Calculate final volume

V1/T1 = V2/T2

with V1 = the initial volume V1 = 13 mL = 13*10^-3

with T1 = the initial temperature = 180 °C = 453 Kelvin

with V2 = the final volume = TO BE DETERMINED

with T2 = the final temperature = 587 °C = 860 Kelvin

V2 = (V1*T2)/T1

V2 = (13 mL *860 Kelvin) /453 Kelvin

V2 = 24.68 mL = 24.7 cm³

The final volume will be 24.7 cm³

Answer:

C.) At room temperature and pressure, because intermolecular interactions are minimized and the particles are relatively far apart.

Explanation:

For gas to behave as an ideal gas there are 2 basic assumptions:

In which instance is a gas most likely to behave as an ideal gas?

A.) At low temperatures, because the molecules are always far apart. FALSE. At low temperatures, molecules are closer and IMF are more appreciable.

B.) When the molecules are highly polar, because IMF are more likely. FALSE. When IMF are stronger the gas does not behave as an ideal gas.

C.) At room temperature and pressure, because intermolecular interactions are minimized and the particles are relatively far apart. TRUE.

D.) At high pressures, because the distance between molecules is likely to be small in relation to the size of the molecules. FALSE. At high pressures, the distance between molecules is small and IMF are strong.

Answer:

The answers are:

Purines:

C. contain four ring nitrogen atoms.

E. contain two heterocyclic rings.

Pyrimidines:

C. contain only two ring nitrogen atoms.

E. contain one heterocyclic ring.

Explanation:

Purines and Pyrimidines are nitrogenous bases which are the building blocks of nucleic acids (DNA and RNA).

Purines are composed by two fused heterocyclic rings, one of them is a 6-ring and the other is a 5-ring. Each ring contains two nitrogen atoms which form part of the ring. Thus, the nitrogen positions in purines are: 1', 3', 7' and 9'. Depending on the functional groups bonded to the two-ring structure, a purine base can be Guanidine (G) or Adenine (A).

The structure of Pyrimidines is a single heterocycle ring wich contains two nitrogen atoms in positions 1' and 3'. Depending of the functional groups, they can be: Cytosine (C), Thymidine (T) and Uracil (U, which is found in RNA).

Answer : The enthalpy of combustion per mole of [tex]C_6H_{12}O_6[/tex] is -2815.8 kJ/mol

Explanation :

Enthalpy change : It is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles. It is represented as [tex]\Delta H^o[/tex]

The equation used to calculate enthalpy change is of a reaction is:  

[tex]\Delta H^o_{rxn}=\sum [n\times \Delta H^o_f(product)]-\sum [n\times \Delta H^o_f(reactant)][/tex]

The equilibrium reaction follows:

[tex]C_6H_{12}O_6(s)+6O_2(g)\rightleftharpoons 6CO_2(g)+6H_2O(l)[/tex]

The equation for the enthalpy change of the above reaction is:

[tex]\Delta H^o_{rxn}=[(n_{(CO_2)}\times \Delta H^o_f_{(CO_2)})+(n_{(H_2O)}\times \Delta H^o_f_{(H_2O)})]-[(n_{(C_6H_{12}O_6)}\times \Delta H^o_f_{(C_6H_{12}O_6)})+(n_{(O_2)}\times \Delta H^o_f_{(O_2)})][/tex]

We are given:

[tex]\Delta H^o_f_{(C_6H_{12}O_6(s))}=-1260kJ/mol\\\Delta H^o_f_{(O_2(g))}=0kJ/mol\\\Delta H^o_f_{(CO_2(g))}=-393.5kJ/mol\\\Delta H^o_f_{(H_2O(l))}=-285.8kJ/mol[/tex]

Putting values in above equation, we get:

[tex]\Delta H^o_{rxn}=[(6\times -393.5)+(6\times -285.8)]-[(1\times -1260)+(6\times 0)]=-2815.8kJ/mol[/tex]

Therefore, the enthalpy of combustion per mole of [tex]C_6H_{12}O_6[/tex] is -2815.8 kJ/mol

Answer:

muahhh

Explanation:

The answer is a. "The melting point of the fatty acid decreases as the number of double bonds increases" because when there are double bonded carbons it cancels the polarity of the carboxyl in the opposite direction, increasing that effect as double bonded carbons add to the molecule, giving the net dipole moment to the molecule. As the net dipole moment of the molecule decrease, the melting point of the fatty acid also decreases.

In addition, when the fatty acid is major than 8 carbons and have a double bond that prevents the fatty acid from crystal lattice formation, so the melting point will be lower also for this reason in this case.

Final answer:

The melting point of a fatty acid decreases as the number of double bonds increases because double bonds create kinks in the fatty acid chains, preventing tight packing and weakening intermolecular forces. The correct answer is option a.

Explanation:

For fatty acids with the same number of carbon atoms, the melting point changes as the number of double bonds in the fatty acid changes. The correct option is (a) - the melting point of the fatty acid decreases as the number of double bonds increases. This decrease in melting point occurs because unsaturated fatty acids, which have one or more double bonds, tend to have kinks or bends at the location of these bonds.

These kinks inhibit the fatty acids from packing closely together, resulting in weaker intermolecular Van der Waals forces (specifically London dispersion forces) and thus a lower melting point. In contrast, saturated fatty acids without double bonds have straight chains and can pack tightly together, leading to stronger intermolecular forces and higher melting points.

A good illustration of this concept is the substantial difference in melting points between palmitoleic acid, which contains a cis-double bond and melts over 60℃ lower than its saturated counterpart, palmitic acid. When considering the physical properties of fatty acids, it's important to note that more saturated fatty acids are typically more solid at room temperature due to their higher melting points, while unsaturated fatty acids are usually liquid.

the concentration of [tex]\(Sn^{2+}\)[/tex] is approximately [tex]\(3.3 \times 10^{-2} \, \text{M}\)[/tex].

To solve this problem, we can use the Nernst equation, which relates the cell potential [tex](\(E_{\text{cell}}\))[/tex] to the standard cell potential [tex](\(E^{\circ}_{\text{cell}}\))[/tex] and the concentrations of the species involved in the redox reaction:

[tex]\[ E_{\text{cell}} = E^{\circ}_{\text{cell}} - \frac{0.0592}{n} \log \left( \frac{[\text{cathode product}]^m}{[\text{anode product}]^n} \right) \][/tex]

Given the standard reduction potentials, we can determine that the number of electrons involved in the half-reactions is [tex]\(n = 2\)[/tex]. The given cell potential is [tex]\(E_{\text{cell}} = 0.660 \, \text{V}\)[/tex]. We need to find the concentration of [tex]\(Sn^{2+}\)[/tex].

First, let's write the Nernst equation for the given cell:

[tex]\[ 0.660 \, \text{V} = (E^{\circ}_{\text{cell}}) - \frac{0.0592}{2} \log \left( \frac{[\text{Sn}^{2+}]}{[\text{Zn}^{2+}]} \right) \][/tex]

We're given the standard reduction potentials, which are [tex]\(E^{\circ}_{\text{cell}} = -0.76 \, \text{V}\)[/tex] for the zinc half-reaction and [tex]\(E^{\circ}_{\text{cell}} = -0.136 \, \text{V}\)[/tex] for the tin half-reaction.

Plugging in the values and solving for [tex]\([\text{Sn}^{2+}]\)[/tex]:

[tex]\[ 0.660 \, \text{V} = (-0.76 \, \text{V}) - \frac{0.0592}{2} \log \left( \frac{[\text{Sn}^{2+}]}{2.5 \times 10^{-3}} \right) \][/tex]

[tex]\[ 0.660 \, \text{V} = (-0.76 \, \text{V}) - 0.0296 \log \left( \frac{[\text{Sn}^{2+}]}{2.5 \times 10^{-3}} \right) \][/tex]

[tex]\[ 0.660 + 0.76 = 0.0296 \log \left( \frac{[\text{Sn}^{2+}]}{2.5 \times 10^{-3}} \right) \][/tex]

[tex]\[ 0.0296 \log \left( \frac{[\text{Sn}^{2+}]}{2.5 \times 10^{-3}} \right) = 0.76 + 0.660 \][/tex]

[tex]\[ \log \left( \frac{[\text{Sn}^{2+}]}{2.5 \times 10^{-3}} \right) = \frac{1.42}{0.0296} \][/tex]

[tex]\[ \log \left( \frac{[\text{Sn}^{2+}]}{2.5 \times 10^{-3}} \right) = 47.973 \][/tex]

[tex]\[ \frac{[\text{Sn}^{2+}]}{2.5 \times 10^{-3}} = 10^{47.973} \][/tex]

[tex]\[ [\text{Sn}^{2+}] = (2.5 \times 10^{-3}) \times 10^{47.973} \][/tex]

[tex]\[ [\text{Sn}^{2+}] \approx 3.3 \times 10^{45} \, \text{M} \][/tex]

Therefore, the concentration of [tex]\(Sn^{2+}\)[/tex] is approximately [tex]\(3.3 \times 10^{-2} \, \text{M}\)[/tex].

Answer:

The correct answer is a  coenzyme cosubstrate is loosely bound to an enzyme and dissociates in an altered form as the part of the catalytic cycle.Its original form is regenerated not by the cycle but by other enzyme.

Explanation:

The  prosthetic group of the enzyme pyruvate dehydrogenase is

2 TPP or Thymine pyrophosphate.

3  Lipomide.

Answer:

The balanced reaction is:-

[tex]C_6H_5COO^-_{(aq)} + H_2O_{(l)}\rightleftharpoons C_6H_5COOH_{(aq)} + OH^-_{(aq)}[/tex]

[tex]K_b[/tex] expression is:-

[tex]K_{b}=\frac {\left [ C_6H_5COOH \right ]\left [ {OH}^- \right ]}{[C_6H_5COO^-]}[/tex]

Explanation:

Benzoate ion is the conjugate base of the benzoic acid. It is a Bronsted-Lowry base and the dissociation of benzoate ion can be shown as:-

[tex]C_6H_5COO^-_{(aq)} + H_2O_{(l)}\rightleftharpoons C_6H_5COOH_{(aq)} + OH^-_{(aq)}[/tex]

The expression for dissociation constant of benzoate ion is:

[tex]K_{b}=\frac {\left [ C_6H_5COOH \right ]\left [ {OH}^- \right ]}{[C_6H_5COO^-]}[/tex]

Final answer:

The balanced chemical equation for the reaction of benzoate ion, C6H5COO−, in water is C6H5COO−(aq) + H2O(l) ⇌ C6H5COOH(aq) + OH−(aq), and the Kb expression is Kb = [C6H5COOH][OH−] / [C6H5COO−].

Explanation:

The question involves writing a balanced equation for the reaction of the benzoate ion, C6H5COO−, as a Brønsted-Lowry base in water, and then writing the Kb expression for the reaction. In water, the benzoate ion accepts a proton (H+) from water (‘H2O’), forming benzoic acid (C6H5COOH) and hydroxide ions (OH−). The balanced chemical reaction is as follows:

C6H5COO−(aq) + H2O(l) ⇌ C6H5COOH(aq) + OH−(aq)

The Kb expression, which represents the base ionization constant, is written without including the states of the substances. It is derived from the concentrations of the products over the concentration of the reactant, not including water due to its constant concentration in dilute solutions. The Kb expression is:

Kb = [C6H5COOH][OH−] / [C6H5COO−]

Answer:

2.04 is the pOH of a 0.0092 M CsOH solution.

Explanation:

Molarity of cesium hydroxde = [CsOH]= 0.0092

[tex]CsOH(aq)\rightarrow Cs^+(aq)+OH^-(aq)[/tex]

1 mole of cesium hydroxide gives 1 mole of cesium ion and 1 mole of hydroxide ion.

Then 0.0092 M cesium hydroxide will give :

[tex][OH^-]=1\times [CsOH]=1\times 0.0092 M=0.0092M [/tex]

The pOH of the solution is the negative logarithm of concentration of  hydroxide ions in a solution.

[tex]pOH=-\log [OH^-][/tex]

[tex]pOH=-\log [0.0092 M]=2.0362\approx 2.04 [/tex]

2.04 is the pOH of a 0.0092 M CsOH solution.

Final answer:

The pOH of a 0.0092 [tex]M CsOH[/tex] solution is calculated using the negative log of the hydroxide ion concentration, which results in a pOH of 2.04. Thus, the correct answer is (A) 2.04.

Explanation:

The student is asking about how to calculate the pOH of a [tex]CsOH[/tex] solution. Since CsOH is a strong base, it will dissociate completely in water.

The concentration of hydroxide ions (OH-) in a 0.0092 [tex]M CsOH[/tex] solution will be the same as the concentration of CsOH itself, which is 0.0092 M.

To find the pOH, we take the negative logarithm (base 10) of the hydroxide ion concentration:

[tex]pOH = -log[OH-][/tex]
[tex]pOH = -log(0.0092)[/tex]
[tex]pOH = -(-2.036)[/tex]
[tex]pOH = 2.036[/tex]

Therefore, rounding to two decimal places, the pOH of a 0.0092 [tex]M CsOH[/tex] solution is 2.04. Hence, the correct answer is (A) 2.04.

Answer:

Longwave ridges = Global scale

High pressure system persists over the central plains for 4 days = Synoptic scale

Supercell thunderstorm that lasts for over an hour = mesoscale

A turbulent eddy = microscale

Explanation:

Global scale has a range of the entire earth

synoptic scale has a range of about 100-1000km and can last fro days to weeks

mesoscale has a range of 4-100km and lasts for a day maximum

microscale is the least of all atmospheric motions.

Answer:

From Molarity=concentration/molar mass

Concentration=10/0.3dm³

33.33g/dm³

Molar mass=H2O=18g/mol

Molarity=1.852mol/dm³

Answer:

0. 446 mol L−1

Explanation:

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 potassium chloride = 74.55 g  

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

No of moles = 10 / 74.55

                   = 0.134 moles  

Now we know that  molarity is expressed per liter of solution. Since you dissolve 0.134 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 = 0.134 moles

For 1 ml of solution, no of moles are = 0.134/300 moles

For 1(1000) ml of solution, no of moles are=  0.134/300 x 1000

                                                                 = 0.446 moles/ L

Answer is =0. 446 mol L−1

Answer:

A human need to drink 48.93 mL of sprite  to reach a plasma glucose concentration of 80 mg/dL.

Explanation:

Volume of blood in 7 Kg human = 5 L

Percentage of plasma in blood = 55%

Volume of plasma in 5 L blood = [tex]\frac{55}{100}\times 5 L=2.75 L=27.5 dL[/tex] (1 L = 10 dL)

Concentration of glucose in plasma = 80 mg/dL

Amount of glucose present in 27.5 dL : 80 mg /dL × 27.5 dL= 2200 mg

Let the volume of sprite with 2200 mg glucose be x

Concentration of glucose in sprite = 44.96 mg/mL

[tex]x\times 44.96 mg/mL=2200 mg[/tex]

[tex]x=\frac{2200 mg}{44.96 mg/mL}=48.93 mL[/tex]

A human need to drink 48.93 mL of sprite  to reach a plasma glucose concentration of 80 mg/dL.

Answer:

The heat of the given reaction is -23.0 kJ.

Explanation:

We are given with ;

[tex]NO(g) + NO_2(g)\rightarrow N_2O_3(g) \Delta H_{1,rxn}=-39.kJ[/tex]..[1]

[tex]NO(g) + NO_2(g) + O_2(g)\rightarrow N_2O_5(g), \Delta H_{2,rxn} = -112.5 kJ [/tex]..[2]

[tex]2NO_2(g) \rightarrow N_2O_4(g) ,\Delta H_{3,rxn} = -57.2 kJ [/tex]..[3]

[tex]2NO(g) + O_2(g)\rightarrow 2NO_2(g), \Delta H_{4,rxn} = -114.2 kJ[/tex]..[4]

[tex]N_2O_5(s)\rightarrow N_2O_5(g) ,\Delta H_{5,sub}= 54.1 kJ[/tex]..[5]

To find heat of reaction:

[tex]N_2O_3(g) + N_2O_5(s)\rightarrow 2N_2O_4(g),\Delta H_{6,rxn} = ?[/tex]..[6]

Using Hess's law:

[5] +2 × [3] + [4] - [2] - [1] = [6]

[tex]\Delta H_{6,rxn}=\Delta H_{5,sub}+2\times \Delta H_{3,rxn}+\Delta H_{4,rxn}-\Delta H_{2,rxn}-\Delta H_{1,rxn}[/tex]

[tex]\Delta H_{6,rxn}=54.1 kJ+(2\times (-57.2 kJ))+(-114.2 kJ)-(-112.5 kJ)-(-39.0 kJ)[/tex]

[tex]\Delta H_{6,rxn}=-23.0 kJ[/tex]

The heat of the given reaction is -23.0 kJ.

Answer:

a, g, c

Explanation:

The conversion of the stable cyclopentane into Trans-1, 2dibromocyclopentane will require three step reactions.

The first is to convert the compound into a cyclopentene, through the addition of Bromine water under heat and photons (light). So option A is the first in the order. This will generate 1 bromocyclopentane through halogenation of the alkane. Secondly, a hot and strong base should be added like the NaOEt, EtOH to remove the added bromine and one atom of hydrogen from the resulting 1 bromocyclopentane in the previous reaction. This will yield cyclopentene, thus making the compound more electrophilic. So option g is required. Thirdly, bromine molecules will be added (C) to take up their places at the two electrophilic regions of the compound to produce Trans-1, 2dibromocyclopentane.

Final answer:

To synthesize trans-1,2-dibromocyclopentane from cyclopentane, the correct reagent is Br2 alone (option c). Other reagents like heat and light could lead to radical mechanisms, which are not appropriate for this specific transformation.

Explanation:

To prepare trans-1,2-dibromocyclopentane from cyclopentane, you need to add bromine (Br2) across the double bond in a trans configuration. The best way to achieve this without any other rearrangements or products is to use Br2 alone (option c). When Br2 is added to a double bond, it will typically add in an anti-fashion, leading to the trans product. Using Br2 with heat/light (option a) or with solvents like EtOH (option f) would likely lead to radical mechanisms or other side products.

Therefore, the correct order of reagents to prepare trans-1,2-dibromocyclopentane from cyclopentane would be just: c. The other two spaces would be filled with 'na' as no additional reagents are needed for this transformation.

Thus, the list in order would be: c, na, na.

Answer:

All of these indicate plates exist and move.

Explanation:

Evidence supporting the existence of tectonic plates includes geological features such as the Ring of Fire, earthquake locations, and seafloor spreading. Fossil distribution across continents and the formation of mountain ranges also support the theory. Hot spots show how islands form and move, further confirming plate tectonics.

There is substantial evidence to suggest that the Earth is composed of tectonic plates. Some of the most compelling pieces of evidence include the presence of the Ring of Fire, the occurrence of hot spots, and the locations of earthquakes. These features highlight areas where plates interact. Additionally, data from seafloor spreading, such as the formation of mid-ocean ridges and the discovery of symmetrical patterns of magnetic stripes on either side of these ridges, supports this theory.

The presence of fossils provides further proof. For example, identical fossils of the fern Glossopteris have been found on continents that are now widely separated, indicating that these landmasses were once connected. Similarly, mountain ranges that were once part of the same range are now found on different continents. These phenomena can be explained by the movement of tectonic plates.

Moreover, the study of hot spots, such as those under Hawaii, shows how islands are formed and then move away from the hot spot due to plate movement. These various lines of evidence converge to form a comprehensive picture of how tectonic plates shape the Earth's surface.

Answer:

D.) The higher vapor pressure of acetone results in more rapid evaporation and heat loss

Explanation:

Acetone being liquid with very low boiling point and hence can be easily be converted to gaseous state .

And ,

Therefore have higher vapour pressure and as liquid gets converted to gas , the process of evaporation takes place and leads to loss of heat .

Hence , from the given options , the most appropriate option according to given statement of the question is ( d ) .

The cooling sensation felt with acetone is due to its rapid evaporation, which absorbs heat from the skin. This occurs because acetone has a higher vapor pressure and weaker intermolecular forces compared to water. Option D is the correct option.

The phenomenon observed by the student can be explained by the fact that acetone evaporates more rapidly than water. This rapid evaporation leads to a cooling effect because evaporation is an endothermic process that absorbs heat from the surroundings. Option D is the correct explanation: The higher vapor pressure of acetone results in more rapid evaporation and heat loss.

Acetone has weaker intermolecular forces compared to water, primarily dipole-dipole interactions rather than hydrogen bonds. This means acetone molecules can escape into the gas phase much more easily. When acetone evaporates, it requires energy, which it absorbs from the surroundings, leading to a cooling sensation on the skin.

Answer:

[tex]150~ppm[/tex]

Explanation:

The first step is to draw the ionization reaction of [tex]NaNO_2[/tex], so:

[tex]NaNO_2~->~Na^+~+~NO_2^-[/tex]

The molar ratio between [tex]NaNO_2[/tex] and [tex]NO_2^-[/tex] is 1:1

The next step is the calculation of the concentration. We have to remember that the formula of ppm is:

[tex]ppm=\frac{mg}{L}[/tex]

In this case we will have a mass of 150 mg and a volume of 1 L so:

[tex]ppm=\frac{150~mg}{1~L}=~150ppm[/tex]

Answer:

This reaction is spontaneous for temperature lower than 3722.1 Kelvin or 3448.95 °C

Explanation:

Step 1: Data given

ΔH = −320.1 kJ/mol

ΔS = −86.00 J/K · mol.

Step 2: Calculate the temperature

ΔG<0 = spontaneous

ΔG= ΔH - TΔS

ΔH - TΔS  <0

-320100 - T*(-86) <0

-320100 +86T < 0

-320100 < -86T

320100/86 > T

3722.1 > T

The temperature should be lower than 3722.1 Kelvin (= 3448.9 °C)

We can prove this with Temperature T = 3730 K

-320100 -3730*(-86) <0

-320100 + 320780  = 680 this is greater than 0 so it's non spontaneous

T = 3700 K

-320100 -3700*(-86) <0

-320100 + 318200  = -1900 this is lower than 0 so it's spontaneous

The temperature is quite high because of the big difference between ΔH and ΔS.

This reaction is spontaneous for temperature lower than 3722.1 Kelvin or 3448.95 °C

Answer:

8.3 kJ

Explanation:

In this problem we have to consider that both water and the calorimeter absorb the heat of combustion, so we will calculate them:

q for water:

q H₂O = m x c x ΔT where m: mass of water = 944 mL x 1 g/mL = 944 g

                                      c: specific heat of water = 4.186 J/gºC

                                     ΔT : change in temperature = 2.06 ºC

so solving for q :

q H₂O = 944 g x 4.186 J/gºC x 2.06 ºC = 8,140 J

For calorimeter

q calorimeter  = C x  ΔT  where C: heat capacity of calorimeter = 69.6 ºC

                                     ΔT : change in temperature = 2.06 ºC

q calorimeter = 69.60J x 2.06 ºC = 143.4 J

Total heat released = 8,140 J +  143.4 J = 8,2836 J

Converting into kilojoules by dividing by 1000 we will have answered the question:

8,2836 J x 1 kJ/J = 8.3 kJ

The heat produced by the combustion of 1.98g of the hydrocarbon is 0.143 kJ.

The volume of water was not needed for this particular calculation.

Where:

Given:

Let's plug in the values:

Answer:

Is not possible to make a buffer near of 7.

Optimal pH for sulfate‑based buffers is 2.

Explanation:

The dissociations of H₂SO₄ are:

H₂SO₄ ⇄ H⁺ + HSO₄⁻ pka₁ = -10

HSO₄⁻ ⇄ H⁺ + SO₄²⁻ pka₂ = 2.

The buffering capacity is pka±1. That means that for H₂SO₄ the buffering capacity is in pH's between -11 and -9 and between 1 and 3, having in mind that pH's<0 are not useful. For that reason, is not possible to make a buffer near of 7.

The optimal pH for sulfate‑based buffers is when pka=pH, that means that optimal pH is 2.

I hope it helps!

Answer:

a) f = 3.02x10¹⁵ s⁻¹, and λ = 99.4 nm.

b) 99.4 nm

Explanation:

a) The energy of radiation is given by:

E = h*f

Where h is the Planck constant (6.626x10⁻³⁴ J.s), and f is the frequency. To have the highest frequency, the energy must be the highest too, because they're directly proportional. So we must use E = -E1 = 20x10⁻¹⁹ J

20x10⁻¹⁹ = 6.626x10⁻³⁴xf

f = 3.02x10¹⁵ s⁻¹

The wavelenght is the velocity of light (3.00x10⁸ m/s) divided by the frequency:

λ = 3.00x10⁸/3.02x10¹⁵

λ = 9.94x10⁻⁸ m = 99.4 nm

b) To have the shortest wavelength, it must be the highest energy and frequency, so it would be the same as the letter a) 99.4 nm.

Answer:

0.0938 W/m.°C

Explanation:

The heat is flowing by the refrigerator door by conduction, and can be calculated by the equation:

q = k*ΔT/L

Where q is the heat flux (25 W/m²), k is the thermal conductivity of the refrigerator, ΔT is the variation of the temperature (inner - outer), and L is the thickness of the door (0.03 m).

25 = k*(15-7)/0.03

8k = 0.75

k = 0.0938 W/m.°C