Answers
Answer:
First, Second, Second-to-last, and last choice
Explanation:
Chemistry student. The other options are incorrect. If you would like a more thorough explanation, please reply to this comment.
pH according to Arrhenius definition is the measure of hydrogen ion concentration in solution:
[tex]pH=-log[H^{+}][/tex] Acids release H+ ions into solution, while bases release OH- ions into solution. This explains why second-to-last choice is correct.
Litmus paper are thin strips of paper that have been manufactured with indicators, examples being red cabbage or phenolphthalein. Indicators react to changes in pH with an according color change. This explains why the last choice is also correct.
Acid strength is better the lower the number it is, hence why pH 2 is stronger acid than 5, but in turn is a weaker base. This explains why second choice is correct.
The first option can be explained by the pH diagram. HCl is acidic, and will turn blue (basic) litmus paper more acidic, and move towards an acidic pH, hence more "red."
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thanks,
Answer:
I dont know if im correct number 1 is 2
Explanation:
ps im not good at this subject
Related Questions
H2SO4 is a strong acid because the first proton ionizes 100%. The Ka of the second proton is 1.1x10-2. What would be the pH of a solution that is 0.100 M H2SO4? Account for the ionization of both protons.
Answers
Question options:
a) 2.05
b) 0.963
c) 0.955
d) 1.00
Answer:
b) 0.963
Explanation:
H2SO4→ HSO4- + H3O+
HSO4- + H2O ⇌ SO42- + H3O+
Construct ICE table:
HSO4- (aq) + H2O ⇌ SO42- (aq) + H3O+ (aq)
I 0.1 solid & 0 0.1
C -x liquid + x + x
E 0.1 - x are ignored x 0.1 + x
Calculate x
Ka = products/reactants
= [tex]\frac{[SO42-] [H3O+]}{[HSO4-]}[/tex]
0.011 = [tex]\frac{x (0.1 + x)}{0.1 - x}[/tex]
0.011 x (0.1 -x) = o.1x + x^2
0.0011 - 0.011 x - o.1x - x^2 = 0
0.0011 - 0.011 x - x^2 = 0
Use formula to solve for quadratic equation
x = [tex]{ -b +,-\sqrt{b^2 - 4ac[/tex] / 2a
a = -1, b = -0.111, c = 0.001
Solve for x
x = [tex]\sqrt[-(-o.111)]{(-0.111)^2 - 4(-1) (0.0011) }[/tex] / 2(-1)
x = 0.111 +,- [tex]\sqrt{0.012321 + 0.0044}[/tex] / -2
x = 0.111 +,- [tex]\sqrt{0.016721}[/tex] / -2
x = [tex]\frac{0.111 +, - 0.1293}{-2}[/tex]
x = [tex]\frac{0.111 + 0.1293}{-2}[/tex] , x = [tex]\frac{0.111 - 0.1293}{-2}[/tex]
x = [tex]\frac{0.2403}{-2}[/tex] , x = [tex]\frac{0.0183}{-2}[/tex]
x = - 0.12015 , x = 0.00915
x cannot be negative, so
x = 0.00915 M
Calculate [H3O+]
[H3O+] = 0.1 M + x
[H3O+] = 0.1 M + 0.00915 M
[H3O+] = 0.10915 M
Clculate pH
pH = - log [ H3O+]
pH = - log [ 0.10915]
pH = 0.963
Final answer:
The pH of a 0.100 M H2SO4 solution is approximately 1.00, considering the complete ionization of the first proton and the partial ionization of the second proton, which is less significant due to its lower Ka value.
Explanation:
The question asks about the pH of a 0.100 M H2SO4 solution, taking into account the ionization of both protons. Sulfuric acid (H2SO4) is a strong diprotic acid that dissociates completely for the first proton, yielding a concentration of 0.100 M H+ and 0.100 M HSO4-. The second proton dissociation is less extensive, with a given Ka of 1.1x10-2, which we need to include in our calculation of pH.
First step ionization (complete dissociation):
H2SO4 → H+ + HSO4-
Second step ionization (partial dissociation):
HSO4- ↔ H+ + SO42- (Ka = 1.2 x 10-2)
To calculate the pH, we first consider the complete ionization of the first proton, which directly gives us 0.100 M of H+. The pH contribution from this ionization is pH = -log(0.100) = 1.00. Then we consider the second ionization of HSO4-. Given the Ka and the initial HSO4- concentration of 0.100 M, we can set up an equilibrium expression to find the additional contribution of H+ from the second ionization. However, because the ionization is low, the change in concentration of H+ due to the second ionization can be negligible for this approximate calculation. Therefore, we can assume the pH of the solution largely results from the first dissociation.
Therefore, the answer is that the pH of the 0.100 M H2SO4 solution is approximately 1.00.
Now, let's finish the calculation and the determination of the formula of the iron compound: Calculate the % water of hydration : Tries 0/3 Calculate the following for Fe3 : g in 100 g sample mol in 100 g sample mol/mol Fe (3 sig figs) mol/mol Fe (whole number) Tries 0/3 Calculate the following for K : g in 100 g sample mol in 100 g sample mol/mol Fe (3 sig figs) mol/mol Fe (whole number) Tries 0/3 Calculate the following for C2O42-: g in 100 g sample mol in 100 g sample mol/mol Fe (3 sig figs) mol/mol Fe (whole number) Tries 0/3 Calculate the following for H2O g in 100 g sample mol in 100 g sample mol/mol Fe (3 sig figs) mol/mol Fe (whole number)
Answers
Complete Question
The complete question is shown on the first and second uploaded image
Answer:
A
The percentage of water of hydration is [tex]P_h= 11.01[/tex]%
Mass of [tex]Fe^{3+}[/tex] in 100mg is 10.60mg
Moles of [tex]Fe^{3+}[/tex] in 100mg is [tex]n_i= 0.19[/tex]
mol / mol Fe (3 sig figs) is [tex]= 1.00[/tex]
mol / mol Fe (whole number) is = 1
B
Mass of [tex]K^{+}[/tex] in 100mg is 27.70mg
Moles of [tex]K^{+}[/tex] in 100mg is [tex]n_i= 0.581 moles[/tex]
mol of K / mol of Fe (3 sig figs) is [tex]= 3.05[/tex]
mol of K / mol of Fe (whole number) is [tex]=3[/tex]
C
Mass of [tex]C_2O_4^{-2}[/tex] in 100mg is 55.69 mg
Moles of [tex]C_2O_4^{-2}[/tex] in 100mg is [tex]n_i= 0.633 moles[/tex]
mol of [tex]C_2O_4^{-2}[/tex] / mol of Fe (3 sig figs) is [tex]= 3.33[/tex]
mol of [tex]C_2O_4^{-2}[/tex] / mol of Fe (whole number) is [tex]=3[/tex]
D
Mass of water in 100mg is 11.01 mg
Moles of water in 100mg is [tex]n_i= 0.611 moles[/tex]
mol of water / mol of Fe (3 sig figs) is [tex]= 3.21[/tex]
mol of water / mol of Fe (whole number) is [tex]=3[/tex]
Explanation:
The percentage of water of hydration is mathematically represented as
[tex]P_h = 100 - (Pi + P_p + P_o)[/tex]
Now substituting 10.60% for [tex]P_i[/tex] (percentage of iron ) , 22.70% for [tex]P_p[/tex](Percentage of potassium) , 55.69% for [tex]P_o[/tex] (percentage of Oxlate)
[tex]P_h =100 - (10.60 + 22.70+55.69)[/tex]
[tex]P_h= 11.01[/tex]%
For IRON
Since the percentage of [tex]Fe^{3+}[/tex] is 10.60% then in a 100 mg of the sample the amount of [tex]Fe^{3+}[/tex] would be 10.60 mg
Now the no of moles is mathematically denoted as
[tex]n = \frac{mass}{molar \ mass }[/tex]
The molar mass of [tex]Fe[/tex] is 55.485 g/mol
So the number of moles of [tex]Fe^{3+}[/tex] in 100mg of he sample is
[tex]n_i = \frac{10.60}{55.485}[/tex]
[tex]n_i= 0.19[/tex]
mol / mol Fe (3 sig figs) is [tex]= \frac{0.19}{0.19} = 1.00[/tex]
FOR POTASSIUM
Since the percentage of [tex]K^{+}[/tex] is 22.70% then in a 100mg of the sample the amount of [tex]K^{+}[/tex] would be 22.70mg
The molar mass of [tex]K[/tex] is 39.1 g/mol
So the number of moles of [tex]K^{+}[/tex] in 100mg of he sample is
[tex]n_i = \frac{22.70}{39.1}[/tex]
[tex]=0.581 moles[/tex]
mol of K / mol of Fe (3 sig figs) is [tex]= \frac{0.581}{0.19} = 3.05[/tex]
FOR OXILATE [tex]C_2O_4^{-2}[/tex]
Since the percentage of [tex]C_2O_4^{-2}[/tex] is 55.69% then in a 100mg of the sample the amount of [tex]C_2O_4^{-2}[/tex] would be 55.69 mg
The molar mass of [tex]C_2O_4^{-2}[/tex] is 88.02 g/mol
So the number of moles of [tex]C_2O_4^{-2}[/tex] in 100mg of he sample is
[tex]n_i = \frac{55.69}{88.02}[/tex]
[tex]=0.633 moles[/tex]
mol of [tex]C_2O_4^{-2}[/tex] / mol of Fe (3 sig figs) is [tex]= \frac{0.633}{0.19} = 3.33[/tex]
FOR WATER OF HYDRATION
Since the percentage of water is 11.01% then in a 100mg of the sample the amount of water would be 11.0 mg
The molar mass of water is 18.0 g/mol
So the number of moles of water in 100mg of he sample is
[tex]n_i = \frac{11.01}{18.0}[/tex]
[tex]=0.611 moles[/tex]
mol of water / mol of Fe (3 sig figs) is [tex]= \frac{0.611}{0.19} = 3.21[/tex]
Predict the products of the reaction below. That is, complete the right-hand side of the chemical equation. Be sure your equation is balanced and contains state symbols after every reactant and product.
Answers
The given question is incomplete. The complete question is :
Predict the products of the reaction below. That is, complete the right-hand side of the chemical equation. Be sure your equation is balanced and contains state symbols after every reactant and product
[tex]HNO_3(aq)+H_2O(l)\rightarrow[/tex]
Answer: The complete equation is [tex]HNO_3(aq)+H_2O(l)\rightarrow NO_3^-(aq)+H_3O^+(aq)[/tex]
Explanation:
According to the Bronsted-Lowry conjugate acid-base theory, an acid is defined as a substance which looses donates protons and thus forming conjugate base and a base is defined as a substance which accepts protons and thus forming conjugate acid.
[tex]HNO_3[/tex] being a strong acid dissociates to give [tex]H^+[/tex] ions an [tex]H_2O[/tex] will act as base and accept [tex]H^+[/tex] to form [tex]H_3O^+[/tex]
Thus the complete equation is [tex]HNO_3(aq)+H_2O(l)\rightarrow NO_3^-(aq)+H_3O^+(aq)[/tex]
Consider the following reaction NaOH (s) --> Na+ (aq) + OH- (aq)
Is the reactant an __________.
Answers
Answer:
base and it's in solution
Explanation:
hope this helps
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The decomposition of nitramide, O 2 NNH 2 , O2NNH2, in water has the chemical equation and rate law O 2 NNH 2 ( aq ) ⟶ N 2 O ( g ) + H 2 O ( l ) rate = k [ O 2 NNH 2 ] [ H + ] O2NNH2(aq)⟶N2O(g)+H2O(l)rate=k[O2NNH2][H+] A proposed mechanism for this reaction is O 2 NNH 2 ( aq ) k 1 ⇌ k − 1 O 2 NNH − ( aq ) + H + ( aq ) ( fast equilibrium ) O2NNH2(aq)⇌k−1k1O2NNH−(aq)+H+(aq)(fast equilibrium) O 2 NNH − ( aq ) k 2 −→ N 2 O ( g ) + OH − ( aq ) ( slow ) O2NNH−(aq)→k2N2O(g)+OH−(aq)(slow) H + ( aq ) + OH − ( aq ) k 3 −→ H 2 O ( l ) ( fast ) H+(aq)+OH−(aq)→k3H2O(l)(fast) What is the relationship between the observed value of k k and the rate constants for the individual steps of the mechaanism?
Answers
Answer:
Explanation:
The given overall reaction is as follows:
O 2 N N H ₂( a q ) k → N ₂O ( g ) + H ₂ O( l )
The reaction mechanism for this reaction is as follows:
O ₂ N N H ₂ ⇌ k 1 k − 1 O ₂N N H ⁻ + H ⁺ ( f a s t e q u i l i b r i u m )
O ₂ N N H − k ₂→ N ₂ O + O H ⁻ ( s l ow )
H ⁺ + O H − k ₃→ H ₂ O ( f a s t )
The rate law of the reaction is given as follows:
k = [ O ₂ N N H ₂ ] / [ H ⁺ ]
The rate law can be determined by the slow step of the mechanism.
r a t e = k ₂ [ O ₂ N N H ⁻ ] . . . ( 1 )
Since, from the equilibrium reaction
k e q = [ O ₂ N N H ⁻ ] [ H ⁺ ] /[ O ₂ N N H ₂ ] = k ₁ /k − 1
[ O ₂ N N H ⁻] = k ₁ /k − 1 × [ O ₂ N N H ₂ ] /[ H ⁺ ]. . . . ( 2 )
Substitituting the value of equation (2) in equation (1) we get.
r a t e = k ₂ k ₁/ k − 1 × [ O ₂ N N H ₂ ] /[ H ⁺ ]
Therefore, the overall rate constant is
k = k₂k₁/k-1
The observed rate constant k is related to the individual rate constants of the mechanism by the equation k = k2 (k1/k-1), where k2 is the rate constant of the rate-determining slow step and k1/k-1 is the equilibrium constant of the first fast equilibrium step.
Explanation:The relationship between the observed rate constant k (in the rate law) and the rate constants for the individual steps (k1, k-1, k2, k3) in the proposed mechanism for the decomposition of nitramide can be determined by examining the rate-determining step (RDS). In a reaction mechanism, the slowest step controls the overall reaction rate. For the given mechanism:
O2NNH2(aq) ⇌k1⇌k−1 O2NNH−(aq) + H+(aq) (fast equilibrium)O2NNH−(aq) →k2 N2O(g) + OH−(aq) (slow)H+(aq) + OH−(aq) →k3 H2O(l) (fast)the observed rate law is rate = k [O2NNH2] [H+]. Because the second step is slow, it is the RDS. The equilibrium of the first step means that the concentration of the intermediate O2NNH−(aq) can be expressed in terms of the concentrations of the reactants O2NNH2 and H+. Therefore, the observed rate constant k is a function of the rate constants of the individual steps, particularly k2 and the equilibrium constant (K = k1/k−1) from the first step. Hence, we can conclude that k is equal to k2 multiplied by the equilibrium constant of the first step, where k = k2 (k1/k−1)
A solution was prepared by dissolving 125.0 g of KCl in 275 g of water. Calculate the mole fraction of KCl. (The formula weight of KCl is 74.6 g/mol. The formula weight of water is 18.0 g/mol.)
Answers
Answer:
The mole fraction of [tex]KCl[/tex] is [tex]N_{KCl}=0.099[/tex]
Explanation:
Generally number of mole is mathematically represented as
[tex]n = \frac{mass}{Molar mass }[/tex]
The number of mole of [tex]KCl[/tex] is
[tex]n_{KCl} = \frac{mass \ of \ KCl}{Molar\ mass \ of \ KCl }[/tex]
[tex]n_{KCl} = \frac{125}{74.6}[/tex]
[tex]=1.676 \ moles[/tex]
The number of mole of [tex]H_2O[/tex] is
[tex]n_{H_2O} = \frac{mass \ of \ H_2O}{Molar\ mass \ of \ H_2O }[/tex]
[tex]n_{H_2O} = \frac{275}{18}[/tex]
[tex]= 15.28 \ moles[/tex]
Mole fraction of [tex]N_{KCl}= \frac{n_{KCl}}{n_{KCl} + n_{H_2O}}[/tex]
[tex]= \frac{1.676}{15.28 +1.676}[/tex]
[tex]N_{KCl}=0.099[/tex]
Answer:
The mole fraction of KCl is 0.13
Explanation:
no of moles of KCl (n KCl) = W/G.F.Wt
= 135/74.6
= 1.81moles
no of moles of H2O (nH2O) = W/G.F.Wt
= 225/18 = 12.5 moles
mole fraction of KCl ( XKCl) = nKCl/nKCl + nH2O
= 1.81/(1.81+12.5)
= 1.81/14.31
The mole fraction of KCl is = 0.13
Zinc metal reacts with hydrochloric acid to produce hydrogen gas and an aqueous solution of zinc(II) chloride. What is the reducing agent in this reaction?
Answers
Answer:
Zinc is the reducing agent.
Explanation:
Zn is the reducing agent because its oxidation state increases from 0 to 2 and hydrochloric acid, HCl is the oxidizing agent because it loses hydrogen ions, H+ based on the following balanced equation:
Zn (s) + 2HCl (aq) -----> ZnCl2 (aq) + H2 (g)
In oxidation, there is the loss of electrons. In reduction, there is a gain of electrons. A nemonic device that might help is LEO (the lion said) GER.
Losses
Electrons
Oxidation
Gains
Electrons
Reduction
Answer:
Zinc
The reducing agent in these reaction is Zinc (Zn)
Explanation:
Zn is the reducing agent because its oxidation state increases from 0 to 2 and hydrochloric acid, HCl is the oxidizing agent because it loses hydrogen ions, H+ based on the following balanced equation:
Zn (s) + 2HCl (aq) -----> ZnCl2 (aq) + H2 (g)
In oxidation, there is the loss of electrons.
In reduction, there is a gain of electrons.
The mass of the deuterium molecule D2 is twice that of the hydrogen molecule H2. If the vibrational frequency of H2 is 1.29 × 1014 Hz, what is the vibrational frequency of D2, assuming that the "spring constant" of attracting forces is the same for the two species? Answer in units of Hz.
Answers
Answer:
9.12x10¹³ Hz
Explanation:
The vibrational frequency (ω) of a molecule is given by:
[tex] \omega = \sqrt{\frac{k}{\mu}} [/tex]
Where:
k: is the spring constant
μ: is the reduced mass
The reduced mass of a diatomic molecule is:
[tex] \frac{1}{\mu} = \frac{1}{m_{a}} + \frac{1}{m_{b}} [/tex]
Where ma and mb are the atomic masses of the atoms a and b, respectively, of the diatomic molecule.
Hence, the vibrational frequency of the hydrogen molecule is:
[tex]\omega_{H_{2}} = \sqrt{\frac{k}{\mu_{H_{2}}}}[/tex] (1)
From equation (1) we can find k:
[tex] k = \omega_{H_{2}}^{2}*\mu_{H_{2}} [/tex] (2)
The vibrational frequency of the deuterium molecule is:
[tex] \omega_{D_{2}} = \sqrt{\frac{k}{\mu_{D_{2}}}} [/tex] (3)
By entering equation (2) into equation (3) we can calculate the vibrational frequency of the deuterium molecule:
[tex] \omega_{D_{2}} = \sqrt{\frac{\omega_{H_{2}}^{2}*\mu_{H_{2}}}{\mu_{D_{2}}}} [/tex]
[tex] \omega_{D_{2}} = \sqrt{\frac{\omega_{H_{2}}^{2}*\mu_{H_{2}}}{2*\mu_{H_{2}}}} [/tex]
[tex] \omega_{D_{2}} = \frac{\omega_{H_{2}}}{\sqrt{2}} = \frac{1.29 \cdot 10^{14} Hz}{\sqrt{2}} = 9.12 \cdot 10^{13} Hz [/tex]
Therefore, the vibrational frequency of the deuterium molecule is 9.12x10¹³ Hz.
I hope it helps you!
The vibrational frequency of D₂ is : 9.12 * 10¹³ Hz
Given that:
Vibrational frequency ( w ) = [tex]\sqrt{\frac{k}{u} }[/tex]
u = reduced mass
The reduced mass of a diatomic molecule is expressed as
[tex]\frac{1}{u} = \frac{1}{m_{a} } + \frac{1}{m_{b} }[/tex]
Where : Ma and Mb are the atomic masses of mass A and mass B
First step : expressing the vibrational frequency of the hydrogen molecule
wH₂ = [tex]\sqrt{\frac{k}{uH_{2} } }[/tex] ----- ( i )
from the equation
k = ( wH₂ )² * uH₂ ---- ( ii )
Next step : expressing the vibrational frequency of the deuterium molecule.
wD₂ = [tex]\sqrt{\frac{k}{uD_{2} } }[/tex] ---- ( iii )
Insert equation ( ii ) into equation ( iii )
wD₂ = [tex]\frac{wH_{2} }{\sqrt{2} }[/tex] = ( 1.29 * 10¹⁴ ) / ( √2 ) = 9.12 * 10¹³ Hz
Hence we can conclude that The vibrational frequency of D₂ is : 9.12 * 10¹³ Hz.
Learn more about vibrational frequency : https://brainly.com/question/6075512
100 mL of a strong acid is completely neutralized by 100 mL of a strong base. The observed products are salt and water. Upon further investigation, the solution has a pH > 7. How is this possible?
A) The concentration of acid is the same as the concentration of the base.
B) The concentration of acid is higher than the concentration of the base.
C) The concentration of base is higher than the concentration of the acid.
D) The water autoionizes to give [OH-] ions.
Answers
Answer:
Option C. The concentration of base is higher than the concentration of the acid.
Explanation:
The reaction between a strong acid and a strong base follows the equation:
HA + OH⁻ ⇆ A⁻ + H₂O
The pH is:
[tex] pH = -log [H_3O^{+}] [/tex]
If we have 100 mL of a strong acid and 100 mL of a strong base, for the pH to be more than 7, that means that the concentration of the base is higher than the concentration of the acid. This is because, the number of moles that remains in the solution after the reaction between the acid and the base will be the moles of the base. The number of moles of the reaction above is:
[tex] n_{T} = n_{a} - n_{b} [/tex] (1)
Where na: is the moles of acid, nb: the moles of the base, and nT is the total number of moles.
Case A) If the concentration of acid is the same as the concentration of the base since the volume of the acid and the base are the same, the number of moles of acid is the same as the number of moles of the base, hence, they neutralize, so the pH = 7. This is not the correct option.
Case B) If the concentration of acid is higher than the base since the volume of the acid and the base are the same, the number of moles of acid is also higher than the number of moles of the base and the total moles in equation (1) results in an excess of moles of the acid, so the pH is < 7. This is not the correct option.
Case C) If the concentration of base is higher than the concentration of the acid since the volume of the acid and the base are the same, the number of moles of the base is also higher than the number of moles of the acid and the total moles in equation (1) results in an excess of moles of the base, so the pH is > 7. This is the correct option.
Case D) The water autoionizes to give [OH-] ions. The autoionization of the water produces the same concentration of acid that the base, so this is not the correct option.
From all of the above, the correct option is C. For the pH to be more than 7, the concentration of the base is higher than the concentration of the acid.
I hope it helps you!
2K + 2HBr → 2 KBr + H2
When 5.5moles of K reacts with 4.04moles of HBr, to produce Hydrogen gas(H₂)
●a). What is the limiting reactant?
●b.)What is the excess reactant?
2K + 2HBr → 2 KBr + H2
When 5.5moles of K reacts with 4.04moles of HBr, to produce Hydrogen gas(H₂)
●a). What is the limiting reactant?
●b.)What is the excess reactant?
2K + 2HBr → 2 KBr + H2
When 5.5moles of K reacts with 4.04moles of HBr, to produce Hydrogen gas(H₂)
●a). What is the limiting reactant?
●b.)What is the excess reactant?
●C.)How much product is produced?
Answers
Answer:
[tex]\large \boxed{\text{a) HBr; b) K; c) 0.0503 g}}[/tex]
Explanation:
The limiting reactant is the reactant that gives the smaller amount of product.
Assemble all the data in one place, with molar masses above the formulas and masses below them.
M_r: 39.10 80.41 2.016
2K + 2HBr ⟶ 2KBr + H₂
m/g: 5.5 4.04
a) Limiting reactant
(i) Calculate the moles of each reactant
[tex]\text{Moles of K} = \text{5.5 g} \times \dfrac{\text{1 mol}}{\text{31.10 g}} = \text{0.141 mol K}\\\\\text{Moles of HBr} = \text{4.04 g} \times \dfrac{\text{1 mol}}{\text{80.91 g}} = \text{0.049 93 mol HBr}[/tex]
(ii) Calculate the moles of H₂ we can obtain from each reactant.
From K:
The molar ratio of H₂:K is 1:2.
From HBr:
The molar ratio of H₂:HBr is 3:2.
[tex]\text{Moles of H}_{2} = \text{0.049 93 mol HBr } \times \dfrac{\text{1 mol H}_{2}}{\text{2 mol HBr}} = \text{0.024 97 mol H}_{2}[/tex]
(iii) Identify the limiting reactant
HBr is the limiting reactant because it gives the smaller amount of NH₃.
b) Excess reactant
The excess reactant is K.
c) Mass of H₂
[tex]\text{Mass of H}_{2} = \text{0.024 97 mol H}_{2} \times \dfrac{\text{2.016 g H}_{2}}{\text{1 mol H}_{2}} = \textbf{0.0503 g H}_{2}\\\text{The mass of hydrogen is $\large \boxed{\textbf{0.0503 g}}$}[/tex]
Zinc is to be electroplated onto both sides of an iron sheet that is 20 cm2 as a galvanized sacrificial anode. It is desired to electroplate the zinc to a thickness of 0.025 mm. It is found that a current of 20 A produces a zinc coating of sufficient quality for galvanized iron. Determine the time required to produce the desired coating, assuming 100 % efficiency.
Answers
Answer:
The time required for the coating is 105 s
Explanation:
Zinc undergoes reduction reaction and absorbs two (2) electron ions.
The expression for the mass change at electrode [tex](m_{ch})[/tex] is given as :
[tex]\frac{m_{ch}}{M} ZF = It[/tex]
where;
M = molar mass
Z = ions charge at electrodes
F = Faraday's constant
I = current
A = area
t = time
also; [tex](m_{ch})[/tex] = [tex](Ad) \rho[/tex] ; replacing that into above equation; we have:
[tex]\frac{(Ad) \rho}{M} ZF = It[/tex] ---- equation (1)
where;
A = area
d = thickness
[tex]\rho[/tex] = density
From the above equation (1); The time required for coating can be calculated as;
[tex][ \frac{20 cm^2 *0.0025 cm*7.13g/cm^3}{65.38g/mol}*2 \frac{moles\ of \ electrons}{mole \ of \ Zn} * 9.65*10^4 \frac{C}{mole \ of \ electrons } ] = (20 A) t[/tex]
[tex]t = \frac{2100}{20}[/tex]
= 105 s
Write a balanced equation for the combustion of gaseous methane (CH4), a majority component of natural gas, in which it combines with gaseous oxygen to form gaseous carbon dioxide and gaseous water.
Answers
Answer:
CH4(g) + 2O2(g) → CO2(g) + 2H2O(g)
Explanation:
Step 1: Data given
gaseous methane = CH4(g)
Combustion reaction is adding O2. The products will be carbondioxide (CO2) and water vapor (H2O)
Step 2: The unbalanced equation
CH4(g) + O2(g) → CO2(g) + H2O(g)
Step 3: Balancing the equation
CH4(g) + O2(g) → CO2(g) + H2O(g)
On the left side we have 4x H (in CH4), on the right side we have 2x H (in H2O). To balance the amount H on both sides, we have to multiply H2O by 2.
CH4(g) + O2(g) → CO2(g) + 2H2O(g)
On the left side we have 2x O (in O2), on the right side we have 4x O (2x in CO2 and 2x in 2H2O). To balance the amount of O on both sides, we have to multiply O2 on the left side, by 2. Now the equation is balanced.
CH4(g) + 2O2(g) → CO2(g) + 2H2O(g)
The balanced chemical equation for the combustion of gaseous methane (CH₄) with gaseous oxygen (O₂) to form gaseous carbon dioxide (CO₂) and gaseous water (H₂O) is:
CH₄ + 2O₂ → CO₂ + 2H₂O
The combustion of gaseous methane (CH₄) with gaseous oxygen (O₂) to form gaseous carbon dioxide (CO₂) and gaseous water (H₂O) is a fundamental chemical reaction that occurs in natural gas combustion and many other combustion processes. To write a balanced chemical equation for this reaction, we must ensure that the number of atoms of each element on both sides of the equation is the same.
The unbalanced equation for the combustion of methane is:
CH₄ + O₂ → CO₂ + H₂O
Now, let's balance the equation:
Balance the carbon (C) atoms:
There is one carbon atom on the left and one on the right, so carbon is already balanced.
Balance the hydrogen (H) atoms:
There are four hydrogen atoms on the left (in CH₄) and two on the right (in H₂O). To balance hydrogen, we need to place a coefficient of 2 in front of H₂O on the right side.
CH₄ + O₂ → CO₂ + 2H₂O
Balance the oxygen (O) atoms:
On the left side, there are two oxygen atoms in CH₄ and two in O₂, making a total of four oxygen atoms. On the right side, there are two oxygen atoms in CO₂ and four in 2H₂O, making a total of six oxygen atoms. To balance the oxygen atoms, we need to adjust the coefficient of O₂ on the left side.
CH₄ + 2O₂ → CO₂ + 2H₂O
Now, the equation is balanced with an equal number of atoms of each element on both sides:
CH₄ + 2O₂ → CO₂ + 2H₂O
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Calculate the volume of 0.684mol of carbon dioxide at s.t.p. show working
Answers
Answer: The volume of 0.684 mol of carbon dioxide at s.t.p. is 15.3 L
Explanation:
According to ideal gas equation:
[tex]PV=nRT[/tex]
P = pressure of gas = 1 atm (at STP)
V = Volume of gas = ?
n = number of moles = 0.684
R = gas constant = [tex]0.0821Latm/Kmol[/tex]
T =temperature =[tex]273K[/tex] (at STP)
[tex]V=\frac{nRT}{P}[/tex]
[tex]V=\frac{0.684\times 0.0821L atm/K mol\times 273K}{1atm}[/tex]
[tex]V=15.3L[/tex]
Thus the volume of 0.684 mol of carbon dioxide at s.t.p. is 15.3 L
Your value for the stoichiometric ration (slope) is most likely slight different than the predicted value of 1. List a reasonable error that could have caused this discrepancy and briefly explain.
Answers
Answer:
The stopper is not fitted on the flask quickly.
Explanation:
The reaction has to do with a gas. The predicted value of the slope is 1. The value of slope obtained from the experiment varies slightly from the predicted value of 1. This discrepancy may be caused by not fitting the stopper on the flask quickly enough. This means that some gas may still be left in the flask leading a discrepancy in the slope obtained.
g Copper (II) Sulfate forms several hydrates with the general formula CuSO4 times xH2O, where x is an integer. If the hydrate is heated, the water can be drive off leaving pure CuSO4 behind. Suppose a sample of a certain hydrate is heated until all water is removed, and its found that the mass of the sample decreases by 31%. Which hydrate is it? THat is, WHAT IS X?
Answers
Answer:
Water of crystallization, X = 4.
Explanation:
Molar mass of [tex]CuSO_{4}.XH_{2} O[/tex]
64 + 32 + (4x18) + x ( 1 × 2 + 16)
= 160 + 18x
Given: % water of crystallization (decrease in mass after heating) = 30%
⇒ [tex]\frac{18x}{160 + 18x} =\frac{31}{100}[/tex]
1800x = 31 (160 + 18x)
58.0645x = 160 + 18x
(58.0645 - 18)x = 160
x = [tex]\frac{160}{40.0645}[/tex] = 3.99 ≅ 4.
Water of crystallization, X = 4.
Qualitative measurements involve numerical measurements while quantitative measurements record descriptions.
Question 5 options:
True
False
Answers
Calculate the solubility of CuX (Ksp=[Cu2+][X2−]=1.27×10−36) in a solution that is 0.200 M in NaCN.
I have already tried to square root the Ksp value to get the answer but it was wrong.
Answers
The solubility of CuX in a 0.200 M NaCN solution cannot be calculated solely based on the Ksp of CuX, because NaCN forms a complex with Cu2+, significantly affecting its solubility. Additional information on the stability constant of the copper-cyanide complex is needed
To calculate the solubility of CuX in a solution that is 0.200 M in NaCN, we must consider the common ion effect due to the presence of the cyanide ion, CN-. The equation for the solubility product (Ksp) is given by Ksp = [Cu2+][X2−]. Since cyanide ions form a complex with copper ions, the direct precipitation of CuX is suppressed, and we cannot simply take the square root of the Ksp to find its molar solubility.
Instead, we would write the reaction of copper with cyanide: CuX(s) + 4CN−
ightleftharpoons [Cu(CN)4]3− + X2−. This complexation reaction would vastly reduce the concentration of free Cu2+ ions in solution, thereby affecting the solubility of CuX. To find the actual solubility, we would need to know the stability constant (Kf) for the copper-cyanide complex, and then set up an equilibrium calculation that includes both the Ksp of CuX and the Kf of [Cu(CN)4]3−. Since the problem doesn't provide Kf, we would not be able to calculate the solubility without additional information
The presence of NaCN significantly decreases the solubility of CuX due to the common ion effect. Therefore, the solubility of CuX in a solution that is 0.200 M in NaCN remains extremely low, approximately 1.27×10 ⁻¹⁸ M.
To calculate the solubility of CuX in a solution that is 0.200 M in NaCN, we need to consider the common ion effect. When NaCN dissolves, it produces CN⁻ ions, which can interact with Cu²⁺ ions, reducing the solubility of CuX.
Given that the solution is 0.200 M in NaCN, we can assume that the concentration of CN⁻ ions ([CN⁻]) is 0.200 M.
Now, let's denote the solubility of CuX as x M. Since CuX dissociates into Cu²⁺ and X²⁻ ions, the concentration of Cu²⁺ ions ([Cu²⁺]) and X²⁻ ions ([X²⁻]) will both be equal to x M at equilibrium.
The solubility product constant (K sp ) expression for CuX is:
K sp =[Cu²⁺ ][X²⁻ ]
Given that K sp =1.27×10⁻³⁶ , we can substitute the concentrations into the expression:
1.27×10⁻³⁶ =(x)(x)
1.27×10⁻³⁶ =x²
Now, we'll solve this equation for x to find the solubility of CuX. Let's proceed with the calculations.
To solve for x, we take the square root of both sides of the equation:
x= 1.27×10⁻³⁶
x=1.127×10⁻¹⁸
So, the solubility of CuX in a solution that is 0.200 M in NaCN is
1.127×10⁻¹⁸ M.
At 500 K the reaction 2 NO(g) + Cl2(g) ⇌ 2 NOCl(g) has Kp = 51 In an equilibrium mixture at 500 K, the partial pressure of NO is 0.125 atm and Cl2 is 0.165 atm. What is the partial pressure of NOCl in the equilibrium mixture?
Answers
Answer:
p3=0.36atm (partial pressure of NOCl)
Explanation:
2 NO(g) + Cl2(g) ⇌ 2 NOCl(g) Kp = 51
lets assume the partial pressure of NO,Cl2 , and NOCl at eequilibrium are P1 , P2,and P3 respectively
[tex]Kp=\frac{[NOCl]^{2} }{[NO]^{2} [Cl_2] }[/tex]
[tex]Kp=\frac{[p3]^{2} }{[p1]^{2} [p2] }[/tex]
p1=0.125atm;
p2=0.165atm;
p3=?
Kp=51;
On solving;
p3=0.36atm (partial pressure of NOCl)
The reaction,
[tex]2 NO(g)+Cl_2(g) \rightleftharpoons 2 NOCl (g)[/tex]Given values,
Pressure,
[tex]P_1 = 0.125 \ atm[/tex][tex]P_2 = 0.165 \ atm[/tex]Value of Kp,
[tex]51[/tex]Now,
→ [tex]K_p = \frac{[NOCl]^2}{[NO]^2[Cl_2]^2}[/tex]
or,
→ [tex]K_p = \frac{[P_3]^2}{[P_1]^2[P_2]}[/tex]
By substituting the values,
[tex]51 = \frac{[P_3]^2}{[0.125]^2[0.165]}[/tex]
[tex]P_3 = 0.36 \ atm[/tex]
Thus the response above is appropriate.
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Name the following compound:
CH3
I
CH = CH2 - CH3
I
CH2
I
CH - CH3
I
CH2
I
CH
I
CH3
2-ethyl-4-methylheptane
2-ethyl-4-methylheptene
3-methyl-5-propyl-2-hexene
3,5-dimethyl-2-octene
Answers
Answer:
3,5-dimethyl-2-octene
Explanation:
Please note that there is no H at carbon 3 less carbon becomes penta hydra.
The compound is:
CH3
I
C = CH2 - CH3
I
CH2
I
CH - CH3
I
CH2
I
CH2
I
CH3
To name the above compound, do the following:
1. Locate the longest continuous chain i.e octene
2. Start counting from the side that gives the double bond the lowest low count since the double bond is the functional group. In doing this, the double bond is at carbon 2.
3. Locate the substituent groups attached and their position in the parent chain. In doing so, you will see that there are two CH3 group attached and they are at carbon 3 and carbon 5. Since the substituents attached are the same, we'll name them as 'dimethyl' indicate that they are two methyl groups
Now, we'll combine the above findings in order to obtain the name. Therefore, the name of the compound is:
3,5-dimethyl-2-octene
Four gases were combined in a gas cylinder with these partial pressures: 3.5 atm N2, 2.8 atm O2, 0.25 atm Ar, and 0.15 atm He.
What is the total pressure inside the cylinder?
6.7
atm
What is the mole fraction of N2 in the mixture?
0.52
atm
2 equations: First, P subscript T equals P subscript 1 plus P subscript 2 plus P subscript 3 plus ellipses plus P subscript n. Second: StartFraction P subscript a over P subscript T EndFraction equals StartFraction n subscript a over N subscript T EndFraction.
What is the mole fraction of O2 in the mixture?
atm
What is the mole fraction of Ar in the mixture?
atm
Answers
Answer:
Mole Fraction of O2 --> 0.42
Mole Fraction of Ar --> 0.037
Explanation:
Answer:
Mole Fraction of O2 --> 0.42
Mole Fraction of Ar --> 0.037
Explanation:
Which statements describes why ethyl methyl ether (b.p. 7.9°C) and 1-propanol (b.p. 97.2°C) have such different boiling points despite identical molar masses? A) Ethyl methyl ether has more hydrogen bonding than 1-propanol. B) 1-propanol has more hydrogen bonding than ethyl methyl ether. C) Ethyl methyl ether has greater dispersion forces than 1-propanol. D) 1-propanol has greater dispersion forces than ethyl methyl ether. E) Ethyl methyl ether has greater molecular volume than 1-propanol.
Answers
Answer:
B) 1-propanol has more hydrogen bonding than ethyl methyl ether.
Explanation:
1-propanol and ethyl methyl ether look very similar but have very different boiling points. They both have three carbon atoms but ethyl methyl ether has an oxygen atom.
Ethyl methyl ether has both London dispersion forces and dipole-dipole interactions.
1-Propanol has London dispersion forces, dipole-dipole interactions and hydrogen bonding.
1-propanol has more hydrogen bonding than ethyl methyl ether which explains why they both have different boiling points.
What is the temperature of CO2 gas if the average speed (actually the root-mean-square speed) of the molecules is 750 m/s?
Answers
Answer:
992.302 K
Explanation:
V(rms) = 750 m/s
V(rms) = √(3RT / M)
V = velocity of the gas
R = ideal gas constant = 8.314 J/mol.K
T = temperature of the gas
M = molar mass of the gas
Molar mass of CO₂ = [12 + (16*2)] = 12+32 = 44g/mol
Molar mass = 0.044kg/mol
From
½ M*V² = 3 / 2 RT
MV² = 3RT
K = constant
V² = 3RT / M
V = √(3RT / M)
So, from V = √(3RT / M)
V² = 3RT / M
V² * M = 3RT
T = (V² * M) / 3R
T = (750² * 0.044) / 3 * 8.314
T = 24750000 / 24.942
T = 992.302K
The temperature of the gas is 992.302K
Note : molar mass of the gas was converted from g/mol to kg/mol so the value can change depending on whichever one you use.
The temperature of a gas can be calculated from the root-mean-square speed of its molecules using the kinetic theory of gases. For CO2 gas with the rms speed of 750 m/s, the derived temperature is approximately 485.4 K.
Explanation:The temperature of a gas can be calculated from the root-mean-square (rms) speed of its molecules using the kinetic theory of gases. This theory develops a relationship between the average kinetic energy of the gas molecules and the temperature of the gas through the equation K = 3/2kBT = mv²/2, where kB is Boltzmann’s constant, T is the temperature, m is the mass of a gas molecule, and v is the rms speed. Thus, temperature can be derived as T = mv² / (3kB).
For CO2, the molar mass is 0.044 kg/mol. Knowing that the number of molecules (n) is the number of moles times Avogadro's number (6.022 × 10²³), we derive the molecular mass m = 0.044 kg/mol / (6.022 × 10²³) = 7.3 × 10^-26 kg. Plugging in the values for v (750 m/s), m, and kB (1.38 × 10^-23 J/K), we get an approximate temperature of 485.4 K.
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Beyond simple thermal decomposition, an even more effective means of obtaining energy from NH4ClO4(s) \rm NH_4ClO_4(s) is to mix this oxidizer with a fuel. When NH4ClO4(s) is employed in solid-fuel booster rockets, it is packed with powdered aluminum. The powdered aluminum is the fuel, and the ammonium perchlorate is the oxidizer. Two of the reactions occurring on ignition are
6NH4ClO4(s)+10Al(s)2NH4ClO4(s)+2Al(s)??5Al2O3(s)+6HCl(g)+9H2O(g)+3N2(g),?H1=?4392.5 kJ
Al2O3(s)+2HCl(g)+3H2O(g)+2NO(g),?H2=?1172.6 k
Which of the following statements are correct?
Check all that apply.
a.All nitrogen atoms lose three electrons in both reactions.
b.Chlorine is reduced from + +7 to ? -1 in both reactions.
c.Reaction 2 produces more energy per gram of Al \rm Al.
d.The thrust produced by the formation of gaseous products is greater per mole of ammonium perchlorate in Reaction 2.
Answers
Final answer:
Only statement b, which says that chlorine is reduced from +7 to -1 in both reactions, is correct. Statements a, c, and d cannot be confirmed as correct based on the given information.
Explanation:
The student has presented two chemical reaction equations involving the thermal decomposition of NH4ClO4 (ammonium perchlorate) and powdered aluminum. Let's address the statements provided:
a. All nitrogen atoms lose three electrons in both reactions. This statement is incorrect. In the provided reactions, nitrogen goes from an oxidation state of -3 in NH4+ to 0 in N2, which means each nitrogen atom gains three electrons.
b. Chlorine is reduced from +7 to -1 in both reactions. This statement is correct. In NH4ClO4, chlorine starts with an oxidation state of +7 and is reduced to -1 in HCl.
c. Reaction 2 produces more energy per gram of Al. Without details of the mass of aluminum involved in Reaction 2, we cannot determine which reaction produces more energy per gram of aluminum. However, if the reactions involve the same mass of aluminum, then Reaction 1 is more energetic since the absolute value of ΔH is greater.
d. The thrust produced by the formation of gaseous products is greater per mole of ammonium perchlorate in Reaction 2. This statement cannot be evaluated without more information about the moles of gaseous products formed in Reaction 2.
Therefore, based on the information provided, the correct statement is b. Chlorine is reduced from +7 to -1 in both reactions.
What is the molarity of a KF solution that contains 116 grams of KF in 2 L of solution?
Answers
Answer : The molarity of KF solution is, 1 M
Explanation : Given,
Mass of [tex]KF[/tex] = 116 g
Volume of solution = 2 L
Molar mass of [tex]KF[/tex] = 58 g/mole
Molarity : It is defined as the number of moles of solute present in one liter of volume of solution.
Formula used :
[tex]\text{Molarity}=\frac{\text{Mass of }KF}{\text{Molar mass of }KF\times \text{Volume of solution (in L)}}[/tex]
Now put all the given values in this formula, we get:
[tex]\text{Molarity}=\frac{116g}{58g/mole\times 2L}=1mole/L=1M[/tex]
Therefore, the molarity of KF solution is, 1 M
A stock solution of FeCl2 is available to prepare solutions that are more dilute. Calculate the volume, in mL, of a 2.0-M solution of FeCl2 required to prepare exactly 100 mL of a 0.630-M solution of FeCl2.
Answers
Answer:
31.5mL
Explanation:
The following were obtained from the question:
C1 (concentration of stock solution) = 2M
V1 (volume of stock solution) =.?
C2 (concentration of diluted solution) = 0.630M
V2 (volume of diluted solution) = 100mL
Using the dilution formula C1V1 = C2V2, the volume of the stock solution needed can be obtained as follow:
C1V1 = C2V2
2 x V1 = 0.630 x 100
Divide both side by 2
V1 = (0.630 x 100) /2
V1 = 31.5mL
Therefore, 31.5mL of 2M solution of FeCl2 required
Answer:
We need 31.5 mL of the 2.0 M FeCl2 solution
Explanation:
Step 1: Data given
Molarity of a FeCl2 solution = 2.0 M
Initial volume of FeCl2 = 100 mL
Initial molarity of FeCl2 = 0.630 M
Step 2: Calculate volume of the stock solution
C1V1 = C2V2
⇒with C1 = the initial molarity FeCl2 = 0.630 M
⇒with V1 = the initial volume = 100 mL = 0.100 L
⇒with C2 = the new molarity FeCl2 = 2.0 M
⇒with V2 = the new volume = TO BE DETERMINED
0.630M * 0.100 L = 2.0 M * V2
V2 = (0.630 * 0.100) / 2.0
V2 = 0.0315 L = 31.5 mL
We need 31.5 mL of the 2.0 M FeCl2 solution
The nucleophilic addition reaction depicted below involves a prochiral ketone carbon atom reacting with a nucleophilic hydride ion source (LiAlH4 or NaBH4) and, subsequently, a proton source (e.g., H2O or dilute aq. HCl). Consequently, the reaction produces a racemic mixture of an alcohol. Finish drawing the structures of the products resulting from nucleophilic attack upon the front and back faces of the carbonyl group, being careful to specify the stereochemistry via wedge-and-dash bonds.
Answers
Answer:
we are given the 3-methyl2 butanone and upon the reduction with LiAIH4 there is formed alcohol, there are two possible side attack,
from the back sidefrom the front side.therefore, whenever the front side attacks then the -CH3 moves back the plain and the H will be above the plane. More so, when attack from the back side the H moves below the plane and -CH3 moves above the plane. -OH is evident in the plane. see the attachment below to view the structure.
Explanation:
When you place a saturated sodium bicarbonate solution in with the mixture in the separatory funnel, a gas should be evolved. Is this evidence your desired product is present?
Answers
Answer:
Yes
Explanation:
It causes excess bromine to be given off as a gas. -It is used to absorb excess heat generated by the exothermic reaction.
Answer
The desired product should contain acidic group like -COOH group.
Explanation:
Saturated sodium bicarbonate solution will react with a acid compound which contain in reaction mixture and produce CO2 gas.
We have the reaction has;
NaHCO3 (aq) + R-COOH (aq) ..............> R-COONa (aq) + H2O (l) + CO2 (g)
Therefore.
The desired product will contain acidic group like -COOH (carboxylic acid group).
A chemist dissolves 327.mg of pure hydrochloric acid in enough water to make up 120.mL of solution. Calculate the pH of the solution. Be sure your answer has the correct number of significant dig
Answers
Answer:
pH → 1.13
Explanation:
Our solution is pure HCl
HCl(aq) + H₂O(l) → H₃O⁺(aq) + Cl⁻(aq)
As a strong acid, it is completely dissociated.
1 mol of HCl, can give 1 mol of H⁺ to the medium. Water does not participate. Let's find out M for the acid.
1st step: We convert the mass from mg to g → 327 mg . 1g /1000mg = 0.327 g
2nd step: We convert the mass(g) to moles: 0.327 g / 36.45 g/mol = 8.97×10⁻³ moles
3rd step: We convert the volume from mL to L → 120mL . 1L /1000 mL = 0.120L
Molarity (mol/L) = 8.97×10⁻³ mol / 0.120L = 0.075M
We propose: HCl(aq) + H₂O(l) → H₃O⁺(aq) + Cl⁻(aq)
0.075M 0.075M
pH = - log [H₃O⁺] → - log 0.075 = 1.13 → pH
Between HClO3 and HIO3, which is stronger and why? Question 16 options: 1) HClO3 is stronger because chlorine is in a higher oxidation state than iodine. 2) HClO3 is stronger because chlorine is more electronegative than iodine. 3) HIO3 is stronger because iodine is in a higher oxidation state than chlorine. 4) HIO3 is stronger because iodine is less electronegative than chlorine.
Answers
Answer:
2) HClO3 is stronger because chlorine is more electronegative than iodine.
Explanation:
The more electronegative the element is the more strong or acidic it becomes.
Chlorine being more electronegative than Iodine makes it easier for it to pull the electron of hydrogen more strongly and hence has a higher tendency to release a H+ unit. Hence that makes it stronger.
HClO3 is stronger than HIO3 because chlorine is more electronegative than iodine.
Chloric acid (HClO3) is more stronger than Iodic acid (HIO3) because chlorine is more electronegative than iodine. As we go from top to bottom in the periodic table, the atomic size increases and electronegativity decreases.
Electronegativity is the ability of an atom to attract electron pair towards itself. Chlorine atom comes on the top whereas iodine is present lower than chlorine in the periodic table so the atomic size of chlorine is smaller and higher value of electronegativity as compared to iodine so we can conclude that Chloric acid (HClO3) is more stronger than Iodic acid (HIO3).
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For the reaction where Δn=−1Δn=−1 , what happens after in increase in volume? ????KQ>K so the reaction shifts toward reactants. For the reaction where Δn=0Δn=0 , what happens after in increase in volume? ????KQ>K so the reaction shifts toward reactants. For the reaction where Δn=+1Δn=+1 , what happens after in increase in volume? ????KQ>K so the reaction shifts toward reactants.
Answers
In chemical reactions, a change in volume affects the position of equilibrium based on stoichiometry. Increasing volume results in a shift towards the side with more gaseous molecules for reactions with Δn not equal to zero, while there is no shift if Δn equals zero. The direction of the shift is to decrease the reaction quotient (Q) in order to re-establish equilibrium with the equilibrium constant (K).
Explanation:
The effect of volume change on the position of equilibrium in chemical reactions is dependent on the stoichiometry and the reaction in question. For a reaction where Δn = -1, increasing the volume would result in a decrease in pressure and a shift towards the side with more moles of gas to re-establish equilibrium, typically the side with more molecules. However, if Δn = 0, an increase in volume has no effect on the equilibrium as there is no change in moles of gaseous substances on either side of the reaction. When Δn = +1, increasing the volume leads the equilibrium to shift towards the products, as this increases the total number of gaseous molecules which tends to lower the pressure.
When volume is increased and the reaction quotient Q becomes greater than the equilibrium constant K (Q > K), the reaction tends to shift towards the reactants to re-establish equilibrium. Conversely, when volume is decreased, and the pressure is increased, the reaction tends to shift towards the side of the reaction with fewer moles of gas.