Answer : The final concentration of KOH, HF and KF are 0 M, 1.0 M and 1.0 M respectively.
Explanation :
The given chemical reaction is:
[tex]HF(aq)+KOH(aq)\rightarrow KF(aq)+H_2O(l)[/tex]
Initial conc. 2.0 M 1.0 M 0 M
Final conc. 1.0 M 0 M 1.0 M
When 1.0 M of KOH react with 2.0 M HF then 1.0 M KOH will react with 1.0 M HF to form 1.0 M KF and 1.0 M HF remain unreacted.
So, the final solution contains 1.0 M HF and 1.0 M KF that means the solution contains equal amount of weak acid and salt.
Therefore, the final concentration of KOH, HF and KF are 0 M, 1.0 M and 1.0 M respectively.
The final concentration of KOH, HF and KF are 0 M, 1.0 M and 1.0 M respectively.
Chemical reaction:HF ( aq ) + KOH ( aq ) ⟶ KF ( aq ) + H₂O ( l )
Initial conc. 2.0 M 1.0 M 0 M
Final conc. 1.0 M 0 M 1.0 M
When 1.0 M of KOH react with 2.0 M HF then 1.0 M KOH will react with 1.0 M HF to form 1.0 M KF and 1.0 M HF remain unreacted. So, the final solution contains 1.0 M HF and 1.0 M KF that means the solution contains equal amount of weak acid and salt.
Therefore, the final concentration of KOH, HF and KF are 0 M, 1.0 M and 1.0 M respectively.
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The rate expression for a first order reaction could be
A.) Rate=k[A]
B.) Rate=k[A]^2[B]
C.) Rate=k[A][B]
D.) Rate=k[A]^2[B]^2
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
The value of Kw at several different temperatures in given in the table below. What conclusion can be drawn on the basis of this information?
Temperature
0ºC
25ºC
45ºC
Kw
1.14 x 10-15
1.00 x 10-14
5.48 x 10-14
Pure water becomes more alkaline as the temperature is increased.
The ionization of water is an exothermic process.
Pure water becomes more acidic as the temperature is increased.
The pH of pure water decreases as the temperature is increased.
Answer:
The correct answer is:
The pH of pure water decreases as the temperature is increased.
Explanation:
Ionization constant of water :
[tex]2H_2O(l)\rightleftharpoons H_3O^+(aq) + OH^-(aq)[/tex]
Water will dissociate into equal amount of hydronium and hydroxide ions
[tex]K_w=[H_3O^+]\times [OH^-][/tex]
As we can see that from the given information that ionic product of water increase with increase in temperature which means concentration of hydronium ion concentration increases.
And according to definition of pH of the solution is a negative logarithm of hydronium ions concentration.
[tex]pH=-\log [H_3O^+][/tex]
The pH of the solution of solution is inversely proportional to the concentration of hydronium ions.
As the hydronium ion concentration of water increases with increase in temperature , the pH of the water will decrease.
Methanol, CH 3 OH CH3OH , is formed from methane and water in a two‑step process. Step 1: CH4 ( g ) + H2O ( g ) ⟶ CO ( g ) + 3H2 ( g ) Δ S ∘ = 214.7 J / K Step 2: CO ( g ) + 2 H 2 ( g ) ⟶ CH 3 OH ( l ) Δ S ∘ = − 332.3 J / K Calculate Δ H∘ and ΔG∘ at 298 K for step 1.
Answer:
ΔH° = 206.1 kJ
ΔG° = 142.1 kJ
Explanation:
Let's consider the first step in the synthesis of methanol.
Step 1: CH₄(g) + H₂O(g) ⟶ CO(g) + 3 H₂(g) ΔS° = 214.7 J / K
We can calculate the standard enthalpy of the reaction (ΔH°) using the following expression.
ΔH° = ∑np . ΔH°f(p) - ∑nr . ΔH°f(r)
where,
ni are the moles of reactants and products
ΔH°f(p) are the standard enthalpies of formation of reactants and products
ΔH° = [1 mol × ΔH°f(CO(g)) + 3 mol × ΔH°f(H₂(g))] - [1 mol × ΔH°f(CH₄(g)) + 1 mol × ΔH°f(H₂O(g))]
ΔH° = [1 mol × (-110.5 kJ/mol) + 3 mol × (0 kJ/mol)] - [1 mol × (-74.81 kJ/mol) + 1 mol × (-241.8 kJ/mol)]
ΔH° = 206.1 kJ
We can calculate the standard Gibbs free energy (ΔG°) using the following expression.
ΔG° = ΔH° - T.ΔS°
ΔG° = 206.1 kJ - 298 K × (214.7 × 10⁻³ kJ/K)
ΔG° = 142.1 kJ
The values of ΔH∘ and ΔG∘ for the reaction can be calculated using their respective formulas and knowing the standard enthalpy and free energy of formation of all the reactants and products. The data is usually found in Chemistry textbooks, specifically in Appendix G.
Explanation:To calculate ΔH∘ and ΔG∘ at 298K for step 1, we use the Gibb's Helmholtz equation, ΔG = ΔH - TΔS. However, the problem doesn't provide enough information to calculate these directly. Let's presume, however, that we know the values of ΔH∘ for methane and water. Using Appendix G data (usually found in Chemistry textbooks), we could apply the formula ΔH = [ΔH products] - [ΔH reactants] to get ΔH∘.
Similarly, for ΔG∘, if the standard free energy of formation of the compounds is known, we could apply the formula ΔG = [ΔG products] - [ΔG reactants].
Since these values are usually tabulated, this could give us the values of ΔH∘ and ΔG∘ for step 1 of the reaction.
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Silane is a gas that is used to prepare extremely pure silicon for use in semiconductors, such as computer chips. A sample of silane has a mass of 15.0 g and occupies a volume of 4.81 L at a pressure of 2.35 atm and a temperature of 22 ºC. What is the molar mass of silane?
Answer:
Molar mass of silane = 32.15 g/mol
Explanation:
Given:
Pressure = 2.35 atm
Temperature = 22 °C
The conversion of T( °C) to T(K) is shown below:
T(K) = T( °C) + 273.15
So,
T₁ = (22 + 273.15) K = 295.15 K
Volume = 42.9 L
Using ideal gas equation as:
PV=nRT
where,
P is the pressure
V is the volume
n is the number of moles
T is the temperature
R is Gas constant having value = 0.0821 L.atm/K.mol
Applying the equation as:
2.35 atm × 4.81 L = n × 0.0821 L.atm/K.mol × 295.15 K
⇒n = 0.4665 moles
Given, Mass = 15.0 g
The formula for the calculation of moles is shown below:
[tex]moles = \frac{Mass\ taken}{Molar\ mass}[/tex]
Thus,
[tex]0.4665\ mole= \frac{15.0\ g}{Molar\ mass}[/tex]
Molar mass of silane = 32.15 g/mol
A student finds a container of gases in the laboratory classroom with a total pressure reading of 655 mmHg. The label on the container describes the contents as 5.00 g of carbon dioxide gas and 3.75 g of helium gas. Calculate the partial pressure of carbon dioxide in the container in mmHg.
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
a) Draw three structures of 3 sulfur containing acids. Minimize formal charges by expanding sulfur's octet . Include any nonbonded electrons on sulfur only.
b) Indicate the oxidation state of sulfur in each compound. Treat sulfur as more electronegative than carbon.
I. methyl sulfenic acid
II. methyl sulfinic acid
III. methyl sulfonic acid
You may find the structure of compounds as well the oxidation number of sulfur in each compound written with blue color. The nonbonded electrons are represented by black dots.
Explanation:
In the methyl sulfenic acid sulfur have a oxidation number of 0, because we have +1 from carbon and -1 from hydroxil group.
In the methyl sulfinic acid sulfur have a oxidation number of +2 , because we have +1 from carbon, -1 from hydroxil group and -2 from oxygen.
In the methyl sulfonic acid sulfur have a oxidation number of +4 , because we have +1 from carbon, -1 from hydroxil group, -2 from one oxygen and -2 form the other oxygen.
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Final answer:
The structures and oxidation states of the three sulfur containing acids are: I. Methyl sulfenic acid with sulfur's oxidation state of +2, II. Methyl sulfinic acid with sulfur's oxidation state of +4, and III. Methyl sulfonic acid with sulfur's oxidation state of +6.
Explanation:
I. Methyl sulfenic acid:
Structure: CH3-S(=O)(=O)H
Oxidation state of sulfur: +2
II. Methyl sulfinic acid:
Structure: CH3-S(=O)(=O)-H
Oxidation state of sulfur: +4
III. Methyl sulfonic acid:
Structure: CH3-S(=O)(=O)-OH
Oxidation state of sulfur: +6
Based on Le Chatelier's Principle, under what conditions would the yield of CH4(g) be maximized?
CO2(g) + 4 H2(g) ----> CH4(g) + 2 H2O(g)
ΔHº = -165 kJ/molrxn
high P and low T
low P and high T
high P and high T
low P and low T
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.
The following reaction is exothermic.
2 S(s) + 3 O2(g) → 2 SO3(g)
What can we say about the spontaneity of this reaction?
(A) spontaneous at all temperatures
(B) spontaneous only at high temperatures
(C) spontaneous only at low temperatures
(D) non spontaneous at all temperatures
(E) more information is need to predict if the reaction is spontaneous
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.
At 1000 K, a sample of pure NO2 gas decomposes. 2 NO2(g) equilibrium reaction arrow 2 NO(g) + O2(g) The equilibrium constant KP is 158. Analysis shows that the partial pressure of O2 is 0.29 atm at equilibrium. Calculate the pressure of NO and NO2 in the mixture.
Answer: The pressure of NO and [tex]NO_2[/tex] in the mixture is 0.58 atm and 0.024 atm respectively.
Explanation:
We are given:
Equilibrium partial pressure of [tex]O_2[/tex] = 0.29 atm
For the given chemical equation:
[tex]2NO_2(g)\rightleftharpoons 2NO(g)+O_2(g)[/tex]
Initial: a
At eqllm: a-2x 2x x
Calculating for the value of 'x'
[tex]\Rightarrow x=0.29[/tex]
Equilibrium partial pressure of NO = 2x = 2(0.29) = 0.58 atm
Equilibrium partial pressure of [tex]NO_2[/tex] = a - 2x = a - 2(0.29) = a - 0.58
The expression of [tex]K_p[/tex] for above equation follows:
[tex]K_p=\frac{p_{O_2}\times (p_{NO})^2}{(p_{NO_2})^2}[/tex]
We are given:
[tex]K_p=158[/tex]
Putting values in above expression, we get:
[tex]158=\frac{0.29\times (0.58)^2}{(a-0.58)^2}\\\\a=0.555,0.604[/tex]
Neglecting the value of a = 0.555 because it cannot be less than the equilibrium concentration.
So, [tex]a=0.604[/tex]
Equilibrium partial pressure of [tex]NO_2[/tex] = (a - 0.58) = (0.604 - 0.58) = 0.024 atm
Hence, the pressure of NO and [tex]NO_2[/tex] in the mixture is 0.58 atm and 0.024 atm respectively.
A 25.0 g piece of aluminum (which has a molar heat capacity of 24.03 J/mol°C) is heated to 86.4°C and dropped into a calorimeter containing water (the specific heat capacity of water is 4.18 J/g°C) initially at 21.1°C. The final temperature of water is 26.8°C. Calculate the mass of water in the calorimeter.
a. 1.7 g
b. 0.51 g
c. 150 g
d. 56 g
e. 61 g
Answer:
m H2O = 56 g
Explanation:
Q = mCΔT∴ 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.
Explanation: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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Draw the structure of a soap surfactant molecule. Label the relevant features that enable it to be a good cleaning agent. Explain why surfactants dissolve in both water and water/oil mixtures. Use your knowledge of intermolecular forces in your explanation.
Explanation:
In the Figure can be seen the structure of a soap surfactant. It has two key parts: the hydrophobic tail and the hydrophilic head.
The first one is non-polar and formed only by H and C. The other, is polar due to the O atoms.Surfactant molecules bond with each other forming miscelles, expossing the part that is compatible with the solvent and protecting the other one.
This means thar the polar head enables the surfactant to dissolve in water and the non-polar tail enables it to dissolve in non polar solvents such as oil mixtures.
A 0.50 M solution of an unknown acid has a pH = 4.0. Of the following, which is the acid in the solution?
HOCl (Ka = 2.0 x 10-8)
HBr (strong acid)
HF (Ka = 6.8 x 10-4)
C6H5OH (Ka = 1.0 x 10-10)
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.
How many total moles of ions are released when 0.0859 mol of Rb2SO4 dissolves completely in water?
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.
What are moles?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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Given the standard enthalpy changes for the following two reactions:
(1) N2(g) + 2O2(g)2NO2(g)...... ΔH° = 66.4 kJ
(2) 2N2O(g)2N2(g) + O2(g)......ΔH° = -164.2 kJ
what is the standard enthalpy change for the reaction:
(3) 2N2O(g) + 3O2(g)4NO2(g)......ΔH° = _________?
The standard enthalpy of formation, which has been established for a huge variety of compounds, is enthalpy change. The reactants undergo chemical changes and combine to produce products in any general chemical reaction. Here the enthalpy change is -297 kJ.
The reaction enthalpy is the change in enthalpy, denoted by the symbol ΔrH. By deducting the total enthalpies of all the reactants from the total enthalpies of the products, the reaction enthalpy is determined.
Here we will multiply the equation (1) with "2" and then will subtract it from equation "2".
N₂(g) + 2O₂ (g)→ 2NO₂(g) ...... ΔH° = 66.4 kJ (1) × (2)
2N₂O(g) → 2N₂(g) + O₂(g) .......ΔH° = -164.2 kJ (2)
2N₂(g) + 4O₂ (g)→ 4NO₂(g) ....... ΔH° = 132.8 kJ
2N₂O(g) + 3O₂(g) → 4NO2(g)
ΔH° = ΔH°(eq 2) - 2ΔH°(eq 1) = -164.2 -(2X66.4) = -297 kJ
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Final answer:
Using Hess's Law and manipulating the given reactions to achieve the target reaction, the standard enthalpy change for the reaction 2N2O(g) + 3O2(g) -> 4NO2(g) is found to be -31.4 kJ.
Explanation:
To find the standard enthalpy change ΔH° for the reaction 2N2O(g) + 3O2(g) → 4NO2(g), we need to apply Hess's Law, which states that the total enthalpy change of a reaction is the sum of the enthalpy changes of individual steps that lead to the overall reaction.
The reactions given are:
(1) N2(g) + 2O2(g) → 2NO2(g)...... ΔH° = 66.4 kJ
(2) 2N2O(g) → 2N2(g) + O2(g)......ΔH° = -164.2 kJ
To use these reactions to find the enthalpy change for (3), we need to manipulate them so that when added, they result in the target reaction. We notice that reaction (1) must be doubled to provide the 4NO2(g) as in reaction (3). This also doubles its enthalpy change. Then, we can simply add the first and second reactions together to get the desired reaction.
So, we have:
(1) x 2: 2N2(g) + 4O2(g) → 4NO2(g)...... ΔH° = 2 x 66.4 kJ = 132.8 kJ
(2): 2N2O(g) → 2N2(g) + O2(g)......ΔH° = -164.2 kJ
To find the ΔH° for (3), we add the ΔH° from the doubled reaction (1) to the ΔH° from reaction (2):
132.8 kJ + (-164.2 kJ) = -31.4 kJ
Hence, the standard enthalpy change for the reaction 2N2O(g) + 3O2(g) → 4NO2(g) is -31.4 kJ.
The reaction described by this equation
O3(g)+NO(g)--> O2(g)+NO2(g)
has the following rate law at 310K.
rate of reaction=k[O3][NO] k=3.0*10^6M^-1*s^-1
Given that [O3]=5.0x10^-4M and NO=6.0x10^-5M at t=0 calculate the rate of the reaction at t=0
What is the overall order of this reaction?
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.
Explanation: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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In the tetrachloromethane (carbon tetrachloride) molecule the Cl-C-Cl bond angle is 109.5º. Which distribution of electrons around the central atom provides the best explanation for this bond angle?
3 shared pairs, 1 lone pair
4 shared pairs
2 shared pairs, 2 lone pairs
1 shared pair, 3 lone pairs
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.
Use the Gizmo to answer the following questions.
A. How many electrons are in a neutral atom of lithium? ___________
B. How many neutrons are in an atom of Mg-25? ___________
C. What is the mass number of an atom with 5 protons and 7 neutrons? ________
D. An ion is a charged atom. How many electrons are in O2- ? ___________
E. How many electrons are in Mg2+?
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.
Explanation: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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A 5.00 mL of a salt solution of unknown concentration was mixed with 35.0 mL of 0.523 M AgNO3. The mass of AgCl solid formed was 0.508 g. _ AgNO3(aq) + _ NaCl(aq) → _ AgCl(s) + _ NaNO3(aq) What is the chemical term for the silver chloride solid that is formed?
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
In the electrochemical cell using the redox reaction below, the anode half reaction is ________. Sn4+ (aq) + Fe (s) → Sn2+ (aq) + Fe2+ (aq) In the electrochemical cell using the redox reaction below, the anode half reaction is ________. (aq) + (s) (aq) + (aq) Fe→Fe2++2e− Sn4+→Sn2++2e− Fe+2e−→Fe2+ Sn4++2e−→Sn2+ Fe+2e−→Sn2+ Request Answer
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.
Explanation: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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During an exothermic chemical reaction,
A. a system becomes warmer, and the chemical substances undergo a decrease in potential energy.
B. a system becomes warmer, and the chemical substances undergo an increase in potential energy.
C. a system becomes cooler, and the chemical substances undergo a decrease in potential energy.
D. a system becomes cooler, and the chemical substances undergo an increase in potential energy.
E. a system becomes warmer, and additional heat is gained from the surroundings.
An exothermic chemical reaction occurs when a system becomes warmer and the potential energy of the chemical substances increases.
What is Exothermic chemical reaction?An exothermic reaction is one in which energy is released in the form of light or heat. Thus, in an exothermic reaction, energy is transferred into the surroundings rather than taken from the surroundings as in an endothermic reaction. The change in enthalpy ( ΔH ) in an exothermic reaction is negative. Exothermic reactions characterise all combustion reactions. A substance burns as it combines with oxygen during a combustion reaction. When substances burn, they typically emit energy in the form of heat and light. Wood combustion is an exothermic reaction that releases a lot of energy in the form of heat and light.The net amount of energy required to start an exothermic reaction is less than the net amount of energy released by the reaction. The net amount of heat energy that flows through a calorimeter, a device used to measure the heat released by a chemical reaction, is equal to the negative of the total energy change of the system. The absolute total of energy in a given chemical system is extremely difficult to measure or even calculate. As a result, the energy change (or enthalpy change, denoted by ΔH) is used instead. The following equation describes the relationship between the value of ΔH and the reaction's bond energies.ΔH = (energy used in the bond formation that yields products) – (energy released when the reactant bonds are broken)To learn more about exothermic chemical reaction refer :
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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.
Which compound is most likely to form intermolecular hydrogen bonds?
A.) C4H10
B.) NaH
C.) C2H4OH
D.) C2H5SH
Final answer:
C2H4OH is most likely to form intermolecular hydrogen bonds due to its hydroxyl (-OH) group, which is highly polar and able to engage in hydrogen bonding.
Explanation:
The compound most likely to form intermolecular hydrogen bonds is C2H4OH. This is because C2H4OH has an -OH (hydroxyl) group, which is capable of forming hydrogen bonds due to the high polarity of the O-H bond. Let's analyze why the other options are less likely:
C4H10 (butane) primarily experiences dispersion forces, because it only has C-H bonds, which are minimally polar.NaH (sodium hydride) is an ionic compound and does not form hydrogen bonds; it's held together by strong ionic interactions.C2H5SH (ethyl mercaptan) has an S-H bond, which is less polar than an O-H bond and thus less capable of forming hydrogen bonds compared to hydroxyl groups.Therefore, the answer is C2H4OH, as it is the only one among the given options with the ability to form significant hydrogen bonds due to its -OH group.
When 1.550 g of liquid hexane (C6H14) undergoes combustion in a bomb calorimeter, the temperature rises from 25.87∘C to 38.13∘C. Find ΔErxn for the reaction in kJ/mol hexane. The heat capacity of the bomb calorimeter, determined in a separate experiment, is 5.73 kJ/∘C. Express your answer in kilojoules per mole to three significant figures.2. The combustion of toluene has a ΔErxn of –3.91×103 kJ/mol. When 1.55 g of toluene (C7H8) undergoes combustion in a bomb calorimeter, the temperature rises from 23.12∘C to 37.57∘C. Find the heat capacity of the bomb calorimeter. Express the heat capacity in kilojoules per degree Celsius to three significant
Explanation:
1). The given data is as follows.
[tex]T_{i} = 25.87^{o}C[/tex], [tex]T_{f} = 38.13^{o}C[/tex]
C = 5.73 [tex]kJ/^{o}C[/tex]
Hence, calculate the change in enthalpy of the reaction as follows.
[tex]\Delta E_{rxn} = -C \times \Delta T[/tex]
= [tex]-5.73 \times (38.13 - 25.87)^{o}C[/tex]
= -70.25 KJ
As, number of moles = [tex]\frac{mass}{\text{molar mass}}[/tex]
= [tex]\frac{1.55}{(6 \times 12 + 14 \times 1)}[/tex]
= 0.018 mol
Therefore, enthalpy of reaction in kJ/mol hexane is as follows.
[tex]\Delta E_{rxn} = \frac{-70.25 KJ}{0.018 mol}[/tex]
= [tex]-3.90 \times 10^{3}[/tex] kJ/mol
Thus, we can conclude that [tex]\Delta E_{rxn}[/tex] for the reaction in kJ/mol hexane is [tex]-3.90 \times 10^{3}[/tex] kJ/mol .
2). As we know that,
Number of moles = [tex]\frac{mass}{\text{molar mass}}[/tex]
= [tex]\frac{1.55}{(7 \times 12 + 8 \times 1)}[/tex]
= 0.017 mol
[tex]\Delta E_{rxn} = \E_{rxn} per mol \times \text{number of moles}[/tex]
= [tex]-3.91 \times 10^{3} \times 0.017[/tex]
= -65.875 kJ
As, [tex]\Delta E_{rxn} = C \times \Delta T [/tex]
-65.875 kJ = [tex]-C \times (37.57 - 23.12)^{o}C[/tex]
C = 4.56 [tex]kJ/^{o}C[/tex]
Hence, heat capacity of the bomb calorimeter is 4.56 [tex]kJ/^{o}C[/tex].
When 1.550 g of liquid hexane (C₆H₁₄) undergoes combustion in a bomb calorimeter, the temperature rises from 25.87 °C to 38.13 °C. The change in the internal energy of the reaction is -3.90 × 10³ kJ/mol.
When 1.55 g of toluene (C₇H₈) undergoes combustion in a bomb calorimeter, the temperature rises from 23.12 °C to 37.57 °C. The heat capacity of the bomb calorimeter is 4.55 kJ/°C.
1. When 1.550 g of liquid hexane (C₆H₁₄) undergoes combustion in a bomb calorimeter, the temperature rises from 25.87 °C to 38.13 °C. Find ΔErxn for the reaction in kJ/mol hexane. The heat capacity of the bomb calorimeter, determined in a separate experiment, is 5.73 kJ/∘C. Express your answer in kilojoules per mole to three significant figures.First, we will calculate the moles corresponding to 1.550 g of hexane using its molar mass (86.18 g/mol).
[tex]n = 1.550 g \times \frac{1mol}{86.18 g} = 0.01799mol[/tex]
Then, we will calculate the heat (Q) absorbed by the bomb calorimeter using the following expression.
[tex]Q = C \times \Delta T = \frac{5.73kJ}{\° C} \times (38.13\° C - 25.87\° C) = 70.2 kJ[/tex]
According to the law of conservation of energy, the sum of the heat absorbed by the bomb calorimeter (Qb) and the heat released by the combustion (Qc) is zero.
[tex]Qb + Qc = 0\\\\Qc = -Qb = -70.2 kJ[/tex]
Finally, we will calculate the change in the internal energy of the reaction (ΔErxn) using the following expression.
[tex]\Delta Erxn = \frac{Qc}{n} = \frac{-70.2kJ}{0.01799mol} = -3.90 \times 10^{3} kJ/mol[/tex]
2. The combustion of toluene has a ΔErxn of –3.91 × 10³ kJ/mol. When 1.55 g of toluene (C₇H₈) undergoes combustion in a bomb calorimeter, the temperature rises from 23.12 °C to 37.57 °C. Find the heat capacity of the bomb calorimeter. Express the heat capacity in kilojoules per degree Celsius to three significant.First, we will calculate the moles corresponding to 1.55 g of toluene using its molar mass (92.14 g/mol).
[tex]1.55 g \times \frac{1mol}{92.14g} = 0.0168 mol[/tex]
The combustion of toluene has a ΔErxn of –3.91 × 10³ kJ/mol. The heat released by the combustion of 0.0168 moles of toluene is:
[tex]0.0168 mol \times \frac{-3.91\times 10^{3}kJ }{mol} = -65.7 kJ[/tex]
According to the law of conservation of energy, the sum of the heat absorbed by the bomb calorimeter (Qb) and the heat released by the combustion (Qc) is zero.
[tex]Qb + Qc = 0\\\\Qb = -Qc = 65.7 kJ[/tex]
We can calculate the heat capacity of the bomb calorimeter (C) using the following expression.
[tex]C = \frac{Qb}{\Delta T } = \frac{65.7 kJ}{37.57 \° C - 23.12 \° C } = 4.55 kJ/ \° C[/tex]
When 1.550 g of liquid hexane (C₆H₁₄) undergoes combustion in a bomb calorimeter, the temperature rises from 25.87 °C to 38.13 °C. The change in the internal energy of the reaction is -3.90 × 10³ kJ/mol.When 1.55 g of toluene (C₇H₈) undergoes combustion in a bomb calorimeter, the temperature rises from 23.12 °C to 37.57 °C. The heat capacity of the bomb calorimeter is 4.55 kJ/°C.Learn more: https://brainly.com/question/24245395
A 6.22-kg piece of copper metal is heated from 20.5 °C to 324.3 °C. The specific heat of Cu is 0.385 Jg-1°C-1.a.Calculate the heat absorbed (in kJ) by the metal.b.How close is this heat capacity to the expected heat capacity at the classical limit?
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.
Explanation: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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An acid-base indicator, Hln, dissociates according to the following reaction in an aqueous solution. HInlag) In (aq) H (aq) The protonated form of the indicator, Hln, has a molar absorptivity of 2929 M cm 1 and the deprotonated form, In has a molar absorptivity of 20060 M-1. cm 1 at 440 nm. The pH of a solution containing a mixture of Hin and In s adjusted to 6.12. The total concentration of HIn and In s 0.000127 M. The absorbance of this solution was measured at 440 nm in a 1.00 cm cuvette and was determined to be 0.818. Calculate pKa for HIn.
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.
Explanation: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.
Determine the electron-group arrangement, molecular shape, and ideal bond angle for the following molecule: PH3 Electron-group arrangement: tetrahedral trigonal pyramidal V-shaped trigonal planar Molecular shape: tetrahedral trigonal pyramidal T-shaped bent Ideal bond angle: degrees.
Answer:
The molecular shape and ideal bond angle of the [tex]PH_{3}[/tex] is trigonalbipyramidal and [tex]109.5^{o}[/tex] respectively.
Explanation:
The structure of [tex]PH_{3}[/tex] is as follows.(in attachment)
From the structure,
Phosphor atom has one lone pair and three hydrogens are bonded by six electrons.
Therefore, total electrons invovled in the formation [tex]PH_{3}[/tex] is eight.
Hence, four electron groups which indicate the tetrahedral shape. But one pair is lone pair i.e, present on the phosphor atom.
Therefore, ideal geometry of the [tex]PH_{3}[/tex] molecule is Trigonalbipyramidal.
The ideal angle of trigonalbipyrmidal is [tex]109.5^{o}[/tex].
All three bonds of P-H has [tex]109.5^{o}[/tex].
Therefore, ideal bond angle is [tex]109.5^{o}[/tex].
The electron-group arrangement of PH3 is tetrahedral, the molecular shape is trigonal pyramidal, and the ideal bond angle is 104.5°.
Explanation:In the case of PH3, the electron-group arrangement is tetrahedral and the molecular shape is trigonal pyramidal. The ideal bond angle for PH3 is 104.5°.
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A boy with pneumonia has lungs with a volume of 1.9 L that fill with 0.080 mol of air when he inhales. When he exhales, his lung volume decreases to 1.5 L. Enter the number of moles of gas that remain in his lungs after he exhales. Assume constant temperature and pressure. g
Answer: The number of moles of gas remaining in the lungs is 0.063 moles
Explanation:
The relationship of number of moles and volume at constant temperature and pressure was given by Avogadro's law. This law states that volume is directly proportional to number of moles at constant temperature and pressure.
The equation used to calculate number of moles is given by:
[tex]\frac{V_1}{n_1}=\frac{V_2}{n_2}[/tex]
where,
[tex]V_1\text{ and }n_1[/tex] are the initial volume and number of moles
[tex]V_2\text{ and }n_2[/tex] are the final volume and number of moles
We are given:
[tex]V-1=1.9L\\n_1=0.080mol\\V_2=1.5L\\n_2=?[/tex]
Putting values in above equation, we get:
[tex]\frac{1.9}{0.080}=\frac{1.5}{n_2}\\\\n_2=\frac{1.5\times 0.080}{1.9}=0.063[/tex]
Hence, the number of moles of gas remaining in the lungs is 0.063 moles
A 23.9 g sample of iridium is heated to 89.7°C, and then dropped into 20.0 g of water in a foam-cup calorimeter. The temperature of the water went from 20.1°C to 22.6°C. Calculate the specific heat of iridium (specific heat of water = 4.18 J/g.°C)
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].
CalorimetryCalorimetry 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:
Q is the heat exchanged by a body of mass m.c specific heat substance.ΔT is the temperature variation.Specific heat of iridiumIn this case, you know:
For iridium:Mass of iridium = 13.5 gInitial temperature of gold= 89.3 °CFinal temperature of gold= 22.6 ºCSpecific heat of gold = UnknownFor water:Mass of water = 20 gInitial temperature of water= 20.1 ºCFinal temperature of water= 22.6 ºCSpecific heat of water = 4.18 [tex]\frac{J}{gC}[/tex]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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An experimental spacecraft consumes a special fuel at a rate of 353 L/min . The density of the fuel is 0.700 g/mL and the standard enthalpy of combustion of the fuel is − 57.9 kJ/g . Calculate the maximum power (in units of kilowatts) that can be produced by this spacecraft. 1 kW = 1 kJ/s
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.
The mass of a proton is 1.00728 amu andthat of a neutron is 1.00867 amu. What is the binding energy pernucleon (in J) of a Co nucleus? (The mass of a cobalt-60 nucleus is59.9338 amu.) a. 3.039× 10^-12 b. 2.487 × 10^-12 c. 7.009 × 10^-14 d. 1.368 × 10^-12 e. 9.432 × 10^-13
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.
Explanation: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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Given that H 2 ( g ) + F 2 ( g ) ⟶ 2 HF ( g ) Δ H ∘ rxn = − 546.6 kJ 2 H 2 ( g ) + O 2 ( g ) ⟶ 2 H 2 O ( l ) Δ H ∘ rxn = − 571.6 kJ calculate the value of Δ H ∘ rxn for 2 F 2 ( g ) + 2 H 2 O ( l ) ⟶ 4 HF ( g ) + O 2 ( g )
Answer: - 521.6kJ
Explanation:
According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.
According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.
[tex]H_2(g)+F_2(g)\rightarrow 2HF(g)[/tex] [tex]\Delta H^0_1=-546.6kJ[/tex] (1)
[tex]2H_2(g)+O_2(g)\rightarrow 2H_2O(l)[/tex] [tex]\Delta H^0_2=-571.6kJ[/tex] (2)
The final reaction is:
[tex]2F_2(g)+2H_2O(l)\rightarrow 4HF(g)+O_2(g)[/tex] [tex]\Delta H^0_3=?[/tex] (3)
By multipling (1) by 2
[tex]2H_2(g)+2F_2(g)\rightarrow 4HF(g)[/tex] [tex]\Delta H^0_1'=-2\times 546.6kJ=-1093.2kJ[/tex] (1')
Subtracting (2) from (1')
[tex]2F_2(g)+2H_2O(l)\rightarrow 4HF(g)+O_2(g)[/tex] [tex]\Delta H^0_3=?[/tex] (3)
Hence [tex]\Delta H^0_3=\Delta H^0_1'-\Delta H^0_2=-1093.2-(-571.6)kJ=-521.6kJ[/tex].
The enthalpy for the reaction is -521.6kJ