Answer:
30.12 mL.
Explanation:
We'll begin by calculating the molarity of the phosphoric acid. This can be obtained as follow:
Phosphoric acid H3PO4 will dissociate in water as follow:
H3PO4(aq) <==> 3H^2+(aq) + PO4^3-(aq)
From the balanced equation above,
1 mole of H3PO4 produces 3 moles of H+.
Therefore, XM H3PO4 will produce 8.70 M H+ i.e
XM H3PO4 = 8.70/3
XM H3PO4 = 2.9 M.
Therefore, the molarity of the acid solution, H3PO4 is 2.9 M.
Next, we shall write the balanced equation for the reaction. This is illustrated below:
2H3PO4 + 3Ba(OH)2 —> Ba3(PO4)2 + 6H2O
From the balanced equation above, we obtained the following:
Mole ratio of the acid, H3PO4 (nA) = 2
Mole ratio of the base, Ba(OH)2 (nB) = 3
Data obtained from the question include the following:
Molarity of base, Ba(OH)2 (Mb) = 6.50 M
Volume of base, Ba(OH)2 (Vb) =.?
Molarity of acid, H3PO4 (Ma) = 2.9 M
Volume of acid, H3PO4 (Va) = 45 mL
The volume of the base, Ba(OH)2 Needed for the reaction can be obtained as follow:
MaVa /MbVb = nA/nB
2.9 x 45 / 6.5 x Vb = 2/3
Cross multiply
2 x 6.5 x Vb = 2.9 x 45 x 3
Divide both side by 2 x 6.5
Vb = (2.9 x 45 x 3) /(2 x 6.5)
Vb = 30.12 mL
Therefore, the volume of the base, Ba(OH)2 needed for the reaction is 30.12 mL
While balancing a chemical equation, we change the _____ to balance the number of atoms on each side of the equation.
Answer:
While balancing a chemical equation, we change the coefficient to balance the number of atoms on each side of the equation
Explanation:
While balancing a chemical equation, we change the coefficient to balance the number of atoms on each side of the equation.
What is chemical equation?To summarize in chemistry terms, a chemical equation depicts the initial chemicals, or reactants, on the left-hand side and the final compounds, or products, just on right-hand side, divided by an arrow. In the chemical equation, the number of atoms in each element as well as the total charge are the same on opposite of the equation's sides.
Chemical equations are used in chemistry to depict chemical processes by writing the reactants and products in terms of their corresponding chemical formulas. While balancing a chemical equation, we change the coefficient to balance the number of atoms on each side of the equation.
Therefore, while balancing a chemical equation, we change the coefficient to balance the number of atoms on each side of the equation.
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Question 8 of 30
When is a redox reaction spontaneous?
A. When a metal electrode is in contact with an electrolyte
B. When a power source supplies an electrical current
C. When the cell potential is positive
D. When the cell potential is negative
According to the concept of redox reactions, the answer to this question is option C.
When the cell potential is positive. A redox reaction is spontaneous when the cell potential is positive.The cell potential is the measure of the driving force of the chemical reaction occurring in the electrochemical cell. In an electrochemical cell, a redox reaction occurs, which leads to the production of an electric potential.
If this potential is positive, then the redox reaction is considered spontaneous. However, if the potential is negative, then the reaction is non-spontaneous.In general, a redox reaction is spontaneous if the potential difference between the two electrodes of the cell is positive. This means that the reaction will occur spontaneously without any external energy input.
Thus, the correct option is C.
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Using these metal ion/metal standard reduction potentials Cd2+(aq)|Cd(s) Zn2+(aq)|Zn(s) Ni2+(aq)|Ni(s) Cu2+(aq)/Cu(s) -0.40 V -0.76 V ‑0.25 V +0.34 V Calculate the standard cell potential for the cell whose reaction is Ni2+(aq) + Zn(s) →Zn2+(aq)+ Ni(s)
Answer: The standard cell potential for the cell is +0.51 V
Explanation:
Given : [tex]E^0_{Ni^{2+}/Ni}=-0.25V[/tex]
[tex]E^0_{Zn^{2+}/Zn}=-0.76V[/tex]
The given reaction is:
[tex]Ni^{2+}(aq)+Zn(s)\rightarrow Zn^{2+}(aq)+Ni(s)[/tex]
As nickel is undergoing reduction, it acts as cathode and Zinc is undergoing oxidation, so it acts as anode.
[tex]E^0_{cell}=E^0_{cathode}-E^0_{anode}[/tex]
where both [tex]E^0[/tex] are standard reduction potentials.
Thus putting the values we get:
[tex]E^0_{cell}=-0.25-(-0.76)[/tex]
[tex]E^0_{cell}=0.51V[/tex]
Thus the standard cell potential for the cell is +0.51 V
Nitric oxide (NO) can be formed from nitrogen, hydrogen and oxygen in two steps. In the first step, nitrogen and hydrogen react to form ammonia: (g) (g) (g) In the second step, ammonia and oxygen react to form nitric oxide and water: (g) (g) (g) (g) Calculate the net change in enthalpy for the formation of one mole of nitric oxide from nitrogen, hydrogen and oxygen from these reactions. Round your answer to the nearest .
Answer: [tex]\Delta H = -272.25kJ[/tex] for 1 mole of NO.
Explanation: Hess' Law of Constant Summation or Hess' Law states that the total enthalpy change of a reaction with multiple stages is the sum of the enthalpies of all the changes.
For this question:
1) [tex]N_{2}_{(g)} + 3H_{2}_{(g)}[/tex] => [tex]2NH_{3}_{(g)}[/tex] [tex]\Delta H=-92kJ[/tex]
2) [tex]4NH_{3}_{(g)}+5O_{2}_{(g)}[/tex] => [tex]4NO_{(g)}+6H_{2}O_{(g)}[/tex] [tex]\Delta H=-905kJ[/tex]
Amonia ([tex]NH_{3}_{(g)}[/tex]) appeares as product in the first equation and as reagent in the 2 reaction, so when adding both, there is no need to inverse reactions. However, in the 2nd, there are 4 moles of that molecule, so to cancel it, you have to multiply by 2 the first chemical equation and enthalpy:
[tex]2N_{2}_{(g)} + 6H_{2}_{(g)}[/tex] => [tex]4NH_{3}_{(g)}[/tex] [tex]\Delta H=-184kJ[/tex]
Now, adding them:
[tex]2N_{2}_{(g)} + 6H_{2}_{(g)}[/tex] => [tex]4NH_{3}_{(g)}[/tex] [tex]\Delta H=-184kJ[/tex]
[tex]4NH_{3}_{(g)}+5O_{2}_{(g)}[/tex] => [tex]4NO_{(g)}+6H_{2}O_{(g)}[/tex] [tex]\Delta H=-905kJ[/tex]
[tex]2N_{2}_{(g)}+6H_{2}_{(g)}+5O_{2}_{(g)}=>4NO_{(g)}+6H_{2}O_{(g)}[/tex] [tex]\Delta H = -185-905[/tex]
[tex]2N_{2}_{(g)}+6H_{2}_{(g)}+5O_{2}_{(g)}=>4NO_{(g)}+6H_{2}O_{(g)}[/tex] [tex]\Delta H = -1089kJ[/tex]
Note net enthalpy is for the formation of 4 moles of nitric oxide.
For 1 mole:
[tex]\Delta H = \frac{-1089}{4}[/tex]
[tex]\Delta H=-272.25kJ[/tex]
To form 1 mol of nitric oxide from nitrogen, oxygen and hydrogen, net change in enthalpy is [tex]\Delta H=-272.25kJ[/tex].
If 2.9g of water is heated from 23.9C to 98.9C, how much heat (in calories) was added to the water?
Answer:
Explanation:
we know that
ΔH=m C ΔT
where ΔH is the change in enthalpy (j)
m is the mass of the given substance which is water in this case
ΔT IS the change in temperature and c is the specific heat constant
we know that given mass=2.9 g
ΔT=T2-T1 =98.9 °C-23.9°C=75°C
specific heat constant for water is 4.18 j/g°C
therefore ΔH=2.9 g*4.18 j/g°C*75°C
ΔH=909.15 j
what is a chemical that is safe to use in food but in small amounts?
Answer:
Toxins
Explanation:
candium forms the ion Sc3+. How many bromite ions could bond with Sc3+, and what would be the chemical formula?
a.3 bromite ions, Sc(Broa)2
b.2 bromite ions, Sc(BrO4)3
c.3 bromite ions, Sc(Broz)
d.2 bromite ions, Sc (BrO2)2
Answer: 3 bromite ions and [tex]Sc(BrO_2)_3[/tex]
Explanation:
For formation of a neutral ionic compound, the charges on cation and anion must be balanced. The cation is formed by loss of electrons by metals and anions are formed by gain of electrons by non metals.
Here scandium is having an oxidation state of +3 called as [tex]Sc^{3+}[/tex] cation and bromite is an anion with oxidation state of -1 called as [tex]BrO_2^-[/tex]. Thus 1 Scandium ion combines with three bromite ions and their oxidation states are exchanged and written in simplest whole number ratios to give neutral [tex]Sc(BrO_2)_3[/tex]
Answer:
3 bromite ions and
Explanation:
When the following molecular equation is balanced using the smallest possible integer coefficients, the values of these coefficients are:
________hydrochloric acid (aq) + ___________oxygen (g) → _________water (l) + ________chlorine (g)
Answer:
The coefficients are; 4, 0, 2, 2
Explanation:
The equation is given as;
HCl + O2 --> H2O + Cl2
Upon balancing the equation, we have;
4HCl + O2 --> 2H2O + 2Cl2
You wish to construct a galvanic cell with the anode consisting of a Ni electrode in a 1.0 M Ni(NO3)2 solution. What would be the highest standard cell potential if used as the cathode in this galvanic cell?
Answer:
Au^3+(aq) +3e ------> Au(s). 1.50 V
Explanation:
When we construct the galvanic cell, our intention is to produce energy by spontaneous electrochemical reactions. In order to have a spontaneous electrochemical reaction, E°cell must be positive. The more positive the value of E°cell, the more spontaneous the reaction is.
E°cell= E°cathode - E°anode
If E°cathode= 1.50 V
E°anode= -0.25 V
E°cell= 1.50 -(-0.25)
E°cell= 1.75 V
Hence the process; Au^3+(aq) +3e ------> Au(s) yields the highest standard cell potential
at 89 ∘C∘C , where [Fe2+]=[Fe2+]= 3.60 MM and [Mg2+]=[Mg2+]= 0.310 MM . Part A What is the value for the reaction quotient, QQQ, for the cell?
Answer:
8.6×10^-2
Explanation:
The reaction is;
Mg(s) + Fe^2+(aq) -----> Mg^2+(aq) + Fe(s)
This implies that;
Q = [Mg^2+]/[Fe^2+]
But;
[Fe2+]= 3.60 M
[Mg2+]= 0.310 M
Q= [0.310 M]/[3.60 M]
Q= 0.086
Q= 8.6×10^-2
f the Ksp for HgBr2 is 2.8×10−14, and the mercury ion concentration in solution is 0.085 M, what does the bromide concentration need to be for a precipitate to occur?
Answer:
0.057 M
Explanation:
Step 1: Given data
Solubility product constant (Ksp) for HgBr₂: 2.8 × 10⁻⁴
Concentration of mercury (II) ion: 0.085 M
Step 2: Write the reaction for the solution of HgBr₂
HgBr₂(s) ⇄ Hg²⁺(aq) + 2 Br⁻
Step 3: Calculate the bromide concentration needed for a precipitate to occur
The Ksp is:
Ksp = 2.8 × 10⁻⁴ = [Hg²⁺] × [Br⁻]²
[Br⁻] = √(2.8 × 10⁻⁴/0.085) = 0.057 M
what is ammonium nitrate
Answer:
Ammonium nitrate is a chemical compound with the chemical formula NH₄NO₃. It is a white crystalline solid consisting of ions of ammonium and nitrate.
The Lewis structure of N2H2 shows ________. Group of answer choices a nitrogen-nitrogen single bond each hydrogen has one nonbonding electron pair each nitrogen has one nonbonding electron pair each nitrogen has two nonbonding electron pairs a nitrogen-nitrogen triple bond
Answer:
one bond between nitrogen and hydrogen and a double bond between the nitrogen atoms.
Explanation:
H-N=N-H
Advantages of using a resource person in handling the first aid lesson
The advantage of a resource person would be that it will provide a hands-on activity that will allow the students to experience spacing between organs and on the body of the person.
It will also allow them to identify challenges when doing this and will engage them more in the activity and lesson.
Answer:A resource person add knowledge to the course
Explanation:
Will a precipitate of magnesium fluoride form when 300. mL of 1.1 × 10 –3 M MgCl 2 are added to 500. mL of 1.2 × 10 –3 M NaF? [K sp (MgF 2) = 6.9 × 10 –9]
Answer:
No precipitate is formed.
Explanation:
Hello,
In this case, given the dissociation reaction of magnesium fluoride:
[tex]MgF_2(s)\rightleftharpoons Mg^{2+}+2F^-[/tex]
And the undergoing chemical reaction:
[tex]MgCl_2+2NaF\rightarrow MgF_2+2NaCl[/tex]
We need to compute the yielded moles of magnesium fluoride, but first we need to identify the limiting reactant for which we compute the available moles of magnesium chloride:
[tex]n_{MgCl_2}=0.3L*1.1x10^{-3}mol/L=3.3x10^{-4}molMgCl_2[/tex]
Next, the moles of magnesium chloride consumed by the sodium fluoride:
[tex]n_{MgCl_2}^{consumed}=0.5L*1.2x10^{-3}molNaF/L*\frac{1molCaCl_2}{2molNaF} =3x10^{-4}molMgCl_2[/tex]
Thus, less moles are consumed by the NaF, for which the moles of formed magnesium fluoride are:
[tex]n_{MgF_2}=3x10^{-4}molMgCl_2*\frac{1molMgF_2}{1molMgCl_2}=3x10^{-4}molMgF_2[/tex]
Next, since the magnesium fluoride to magnesium and fluoride ions is in a 1:1 and 1:2 molar ratio, the concentrations of such ions are:
[tex][Mg^{2+}]=\frac{3x10^{-4}molMg^{+2}}{(0.3+0.5)L} =3.75x10^{-4}M[/tex]
[tex][F^-]=\frac{2*3x10^{-4}molMg^{+2}}{(0.3+0.5)L} =7.5x10^{-4}M[/tex]
Thereby, the reaction quotient is:
[tex]Q=(3.75x10^{-4})(7.5x10^{-4})^2=2.11x10^{-10}[/tex]
In such a way, since Q<Ksp we say that the ions tend to be formed, so no precipitate is formed.
Regards.
Two elements represents by the letter Q and R atomic number 9 and 12 respectively. Write the electronic configuration of R
Answer:
Atomic no = 12 = Mg
Explanation:
It is given that,
The atomic number of two elements that are represented by letter Q and R are 9 and 12.
We need to write the electronic configuration of R. Atomic number shows the number of protons in atom.
For R, atomic number = 12
Its electronic configuration is : 2,8,2
It has two valance electrons in its outermost shell. The element is Magnesium (Mg).
3,3-dibromo-4-methylhex-1-yne
Explanation:
see the attachment. hope it will help you...Q 11.20: What is the product of the reaction between t-BuCl and MeOH? A : t-BuOH B : MeOCl C : t-BuOMe D : (CH3)2CCH2
Answer:
C : t-BuOMe
Explanation:
The tert -butanol is a tertiary alcohol and when chloride ion attacks the carbocation, it forms t-BuCl.
The reaction of tert-butyl chloride or t-BuCl ((CH3)3C−Cl) with methanol and MeOH (CH3−OH) gives the product tert-Butyl methyl ether or t-BuOMe (CH3)3C−OCH3:
(CH3)3C−Cl + CH3−OH => (CH3)3C−OCH3 + HCl
Hence, the correct asnwer is C : t-BuOMe
Which of the following elements is in the same group as Sulfur (S)?
Answer:
PLEASE SHOW ME THE ELEMENTS OR I WOULD ENLIST ALL THE ELEMENTS.
Explanation:
Group 6A (or VIA) of the periodic table are the chalcogens: the nonmetals oxygen (O), sulfur (S), and selenium (Se), the metalloid tellurium (Te), and the metal polonium (Po)
1.40 m3 is how many mL
[tex] \LARGE{ \boxed{ \rm{ \pink{Solution:}}}}[/tex]
We know, 1 m³ of space can hold 1000 l of the substance.
⇛ 1 m³ = 1000 l----(1)
And, 1 l is 1000 times more than 1 ml
⇛ 1 l = 1000 ml------(2)
So, From (1) and (2),
⇛ 1 m³ = 1000 × 1000 ml
⇛ 1m³ = 1000000 ml
We had to find,
⇛ 1.40 m³ = 1.40 × 1000000 ml
⇛ 1.40 m³ = 140/100 × 1000000 ml
⇛ 1.40 m³ = 1400000 ml
⇛ 1.40 m³ = 14,00,000 ml / 14 × 10⁵ ml / 1.4 × 10⁶ ml
☃️ So, 1.40 m³ = 14 × 10⁵ ml / 1.4 × 10⁶ ml.
━━━━━━━━━━━━━━━━━━━━
A strand of messenger RNA coding the synthesis of a protein with 146 amino acids must have a string of___________ bases along its backbone. a. 146 b. 292 c. 438 d. 20 e. 3
Answer:
Since the relationship between amino acid and codon bases would be the values of 3 nitrogenous bases per 1 amino acid.
knowing this relationship what you would do is simply multiply 146 x 3 to find the number of codon bases which would be C. 438.
A strand of messenger RNA coding the synthesis of a protein with 146 amino acids must have a string of 438 bases along its backbone and the correct option is option C.
What is messenger RNA?mRNA or messenger RNA is a single stranded RNA molecule. It is complementary to the DNA and carries genetic information present in the DNA. It is translated to form proteins. The genetic codes (triplet) present on mRNA get translated to amino acids, giving rise to the functional product of a gene.
So mRNA really is a form of nucleic acid, which helps the human genome which is coded in DNA to be read by the cellular machinery. mRNA is actually the translated form of DNA that the machinery can recognize and use to assemble amino acids into proteins.
Each strand has 3 bases, so 146 × 3 = 438 bases
Therefore, A strand of messenger RNA coding the synthesis of a protein with 146 amino acids must have a string of 438 bases along its backbone and the correct option is option C.
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Suppose that a 100 mL sample of ideal gas is held in a piston-cylinder apparatus. Its volume could be increased to 200 mL by
Answer:
e. reducing the pressure from 608 torr to 0.40 atm at constant temperature.
Explanation:
According to Boyle's law when a gas is at the same temperature and there is a mass in a closed container so the pressure and the volume changes in the opposite direction
So here the equation is
[tex]P_1V_1=P_2V_2[/tex]
Now we choose the options
where,
[tex]V_1 = 100\ mL = 0.1\ L\\\\V_2 = 200\ mL = 0.2\ L[/tex]
[tex]P_1 = 608\ torr = 0.8\ atm \\\\P_2= 0.4\ atm[/tex]
Now applying these values to the above equation
So,
P1V1=P2V2
[tex]P_1V_1=P_2V_2[/tex]
[tex]0.8\times0.1 = 0.4\times0.2[/tex]
0.8 = 0.8
Hence, it is proved
An HCl solution has a concentration of 0.09714 M. Then 10.00 mL of this solution was then diluted to 250.00 mL in a volumetric flask. The diluted solution was then used to titrate 250.0 mL of a saturated AgOH solution using methyl orange indicator to reach the endpoint.
Required:
a. What is the concentration of the diluted HCI solution?
b. If 7.93 mL of the diluted HCI solution was required to reach the endpoint, what is the concentration of OH- in solution?
c. What is the concentration of Ag+ in solution?
d. What is the Ksp expression for the dissolution of AgOH?
Answer:
a. 3.8856x10⁻³M HCl
b. 1.23x10⁻⁴M OH⁻
c. 1.23x10⁻⁴M Ag⁺
d. Ksp = [Ag⁺] [OH⁻]
Explanation:
a. The reaction that you are studying is:
HCl(aq) + AgOH(aq) → H₂O(l) + AgCl(s)
The HCl solution is diluted from 10.00mL to 250.00mL, that is:
250.00mL / 10.00mL = 25 -The solution is diluted 25 times-
As original concentration of HCl is 0.09714M, the concentration of the diluted solution is:
0.09714M / 25 =
3.8856x10⁻³M HClb. 1 mole of HCl reacts per mole of AgOH, moles of HCl that reacts are:
7.93mL = 7.93x10⁻³L × (3.8856x10⁻³mol HCl / L) = 3.0813x10⁻⁵ moles of HCl.
Based on the reaction, you have in solution
3.0813x10⁻⁵ moles of AgOH = Ag⁺ = OH⁻
The AgOH solution was 250.0mL = 0.2500L, its concentration is:
3.0813x10⁻⁵ moles OH⁻ / 0.2500L =
1.23x10⁻⁴M OH⁻c. In solution, AgOH produce Ag⁺ and OH⁻ in equals proportions, that means:
1.23x10⁻⁴M OH⁻ =
1.23x10⁻⁴M Ag⁺d. The solubility product reaction of AgOH(s) is:
AgOH(s) ⇄ Ag⁺(aq) + OH⁻(aq)
Where Ksp for this reaction is defined as:
Ksp = [Ag⁺] [OH⁻]A balloon is filled to a volume of 1.50 L with 3.00 moles of gas at 25 °C. With pressure and temperature held constant, what will be the volume (in L) of the balloon if 0.50 moles of gas are released?
Answer:
Volume : 1.25 L
Explanation:
We are given here that the volume ( V[tex]_1[/tex] ) = 1.50 Liters, the initial moles ( held at 25 °C ) = 3.00 mol, and the final moles ( n[tex]_2[/tex] ) = 3.00 - 0.5 = 2.5 mol. The final mol is calculated given that 0.50 mol of gas are released from the prior 3.00 moles of gas.
Volume ( V[tex]_1[/tex] ) = 1.50 L,
Initial moles ( n[tex]_1[/tex] ) = 3.00 mol,
Final Volume ( n[tex]_2[/tex] ) = 3.00 - 0.5 = 2.5 mol
Applying the combined gas law, we can calculate the final volume ( V[tex]_2[/tex] ).
P[tex]_1[/tex]V[tex]_1[/tex] / n[tex]_1[/tex]T[tex]_1[/tex] = P[tex]_2[/tex]V[tex]_2[/tex] / n[tex]_2[/tex]T[tex]_2[/tex] - we know that the pressure and temperature are constant, and therefore we can apply the following formula,
V[tex]_1[/tex] / n[tex]_1[/tex] = V[tex]_2[/tex] / n[tex]_2[/tex] - isolate V[tex]_2[/tex],
V[tex]_2[/tex] = V[tex]_1[/tex] n[tex]_2[/tex] / n[tex]_1[/tex] = 1.50 L [tex]*[/tex] 2.5 mol / 3.00 mol = ( 1.5 [tex]*[/tex] 2.5 / 3 ) L = 1.25 L
The volume of the balloon will be 1.25 L.
Oxygen condenses into a liquid at approximately 90 K. What temperature, in degrees Fahrenheit, does this correspond to?
Answer:
-297.67 °F
Explanation:
Oxygen condenses into a liquid at approximately 90 K. We can convert any temperature in the Kelvin scale (absolute scale) to the Fahrenheit scale using the following expression.
°F = (K − 273.15) × 9/5 + 32
°F = (90 − 273.15) × 9/5 + 32
°F = (-183.15) × 9/5 + 32
°F = -329.67 + 32
°F = -297.67 °F
The enthalpy change for a chemical reaction is: a. the temperature change b. the amount of heat given off or absorbed c. related to molar volume d. none of the above
Answer:
b. the amount of heat given off or absorbed
Explanation:
Hello,
In this case, we should take into account a formal definition of enthalpy change such as an energetic change that occurs in a system when matter is transformed by a given chemical reaction from reactants to products. Thus, such energetic change is macroscopically exhibited and it is related with either a temperature increase or decrease; it means that if a reaction exhibits a temperature increase, we say that heat was given off and if the temperature exhibits a decrease, we say that heat is absorbed. For that reason, answer is b. the amount of heat given off or absorbed.
Regards.
The direction of the functional group is called?
Explanation:
they are called hydrocarbyls
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Answer:
The first carbon atom that attaches to the functional group is referred to as the alpha carbon.
Use the following equation to determine the charge on bromine when it dissociates from sodium and determine whether it is being oxidized or reduced: Cl2 + NaBr -> NaCl + Br2A. It starts with a charge of 0 and is oxidized. B. It starts with a charge of -1 and is reduced. C. It starts with a charge of 0 and is reduced. D. It starts with a charge of -1 and is oxidized.
It starts with a charge of 0 and is oxidized. Hence, option A is correct.
What is a chemical equation?A chemical reaction is a representation of symbols of the elements to indicate the number of substances and moles of reactant and product.
[tex]Cl_2 + NaBr[/tex] ->[tex]NaCl + Br_2[/tex]
The oxidation number of bromine changes from -1 (in NaBr) to 0 (in Br
2). Thus, NaBr is oxidized.
The oxidation number of chlorine changes from 0 to -1. Thus, chlorine is reduced.
Hence, option A is correct.
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A monoprotic weak acid, HA , dissociates in water according to the reaction HA(aq)+H2O(l)↽−−⇀H3O+(aq)+A−(aq) The equilibrium concentrations of the reactants and products are [HA]=0.260 M , [H3O+]=4.00×10−4 M , and [A−]=4.00×10−4 M . Calculate the Ka value for the acid HA.
Answer:
Ka = 6.15x10⁻⁷
Explanation:
Ka is defined as dissociation constant in the equilibrium of a weak acid with water. The general reaction is:
HA(aq) + H₂O(l) ⇆ H₃O⁺(aq) + A⁻(aq)
And Ka is defined as the ratio between molar concentrations in equilibrium of products over reactants as follows:
Ka = [H₃O⁺] [A⁻] / [HA]
You don't take water in the equilibrium beacuse is a pure liquid
Replacing with the concentrations of the problem:
Ka = [H₃O⁺] [A⁻] / [HA]
Ka = [4.00x10⁻⁴] [4.00x10⁻⁴] / [0.260]
Ka = 6.15x10⁻⁷
Calculate the pH of a 0.20 M NH3/0.20 M NH4Cl buffer after the addition of 25.0 mL of 0.10 M HCl to 65.0 mL of the buffer.
Answer:
Explanation:
For pH of a buffer solution , the formula is
pH = pKa + log [ Base ] / [ conjugate acid ]
= pKa + log [ NH₃ ] / [ NH₄⁺ ]
Ka = Kw / Kb
Kb for NH₄OH = 1.8 x 10⁻⁵
Ka = 10⁻¹⁴ / 1.8 x 10⁻⁵
= 5.6 x 10⁻¹⁰
pH = - log ( 5.6 x 10⁻¹⁰ ) + log 0.2 / 0.2
= 10 - log 5.6
= 9.25
Effect of addition of HCl
H⁺ of HCl will react with NH₃ to produce NH₄⁺
25 mL of .1 HCl = 2.5 mM of HCl
25 mL of .1 NH₄⁺ = 2.5 mM of NH₄⁺
65 mL of .2 M NH₃ = 13 mM of NH₃
65 mL of .2 M NH₄⁺ = 13 mM of NH₄⁺
NH₃ + H⁺ = NH₄⁺
NH₄⁺ formed = 2.5 + 13 mM
15.5 mM of NH₄⁺
NH₃ = 13 mM
Concentration of NH₃ = 13 / 90
Concentration of NH₄⁺ = 15.5 / 90
pH of final buffer mixture
= 9.6 + log 13 / 15.5
= 9.25 - .076
= 9.174
The pH value is mathematically given as
pH= -6.332.
What is the pH of a 0.20 M NH3/0.20 M NH4Cl buffer?Question Parameters:
the pH of a 0.20 M NH3/0.20 M NH4Cl buffer
the addition of 25.0 mL of 0.10 M HCl to 65.0 mL of the buffer.
Generally, the equation for the Chemical Reaction is mathematically given as
HCl + NH3 --> NH4^+ + Cl^-
Therefore
pH= pka + log(13/14).
pH= -6.3 + log 0.93.
pH= -6.3+ (-0.032).
pH= -6.332.
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