Answer:
[tex]\triangle G= -6.7 KJ/mol[/tex]
Explanation:
From the question we are told that:
Chemical Reactions:
X=A⇌B,ΔG= 14.8 kJ/mol
Y=B⇌C,ΔG= -29.7 kJ/mol
Z=C⇌D,ΔG= 8.10 kJ/mol
Since
Hess Law
The law states that the total enthalpy change during the complete course of a chemical reaction is independent of the number of steps taken.
Therefore
Generally the equation for the Reaction is mathematically given by
[tex]T = +1 * X +1 * Y +1 *Z[/tex]
Therefore the free energy, ΔG is
[tex]\triangle G=1 * \triangle G*X +1 * \triangle G*Y +1 * \triangle G *Z[/tex]
[tex]\triangle G= +1 * (14.9) +1 * (-29.7) +1 * (8.10)[/tex]
[tex]\triangle G= -6.7 KJ/mol[/tex]
For each of the sites specified in the molecules, select whether the site is nucleophilic, electrophilic, or neither. Compound A: The indicated site is a carbon in cyclohexane which is bonded to a bromine and a hydrogen. The indicated carbon in compound A is nucleophilic. neither electrophilic nor nucleophilic. electrophilic. Compound B: The indicated site is the double bond in cyclohexene, a 6 carbon ring with an internal alkene. The indicated bond in compound B is nucleophilic. electrophilic. neither electrophilic nor nucleophilic. Compound C: The indicated site is a carbon double bonded to oxygen, and bonded to O C H 3 and ethyl. The indicated carbon in compound C is neither electrophilic nor nucleophilic. nucleophilic. electrophilic. Compound D: THe indicated site is a carbon bonded to a methyl, two hydrogens and a carbon. There is a nitrogen atom two bonds away. The indicated carbon in compound D is neither electrophilic nor nucleophilic. electrophilic. nucleophilic. Compound E: The indicated site is an oxygen bonded to a carbon and a hydrogen. The indicated oxygen in compound E is neither electrophilic nor nucleophilic. electrophilic. nucleophilic.
The nature of attack on sites in a molecule depends on the nature of such sites. The following are the nature of the sites mentioned in the question:
1) The indicated carbon in compound A is electrophilic.
2) The indicated bond in compound B is nucleophilic.
3) The indicated carbon in compound C is electrophilic.
4) The indicated carbon in compound D is neither electrophilic nor nucleophilic.
5) The indicated oxygen in compound E is nucleophilic.
The terms "electrophilic" and "nucleophilic" are very common in chemistry.
An electrophilic center is usually positively charged, has a positive dipole or is electron deficient hence it attacks negative centers. The term itself means "electron loving". That actually means that it has an affinity for negative charges.
The -I inductive effect of the bromine atom in the carbon in compound A makes that carbon atom to be electrophilic. Also, the carbonyl bond and the O C H 3 attached to the carbon in compound C also makes it electrophilic.
The term "nucleophilic" literately means "nucleus loving". That means a specie that has a high affinity for positive charges. This specie must be electron rich.
The carbon atom in compound B has a double bond which is electron rich and can attack any positive center hence it is nucleophilic. Also, the oxygen atom in E bears two lone pairs of electrons which can attack any positive center in a molecule hence the oxygen atom is also nucleophilic.
In compound D, the carbon atom is bonded to a methyl, two hydrogens and a carbon. There is a nitrogen atom two bonds away. There is no +I or -I inductive effect on this carbon atom because the nitrogen atom is far away. Therefore, the indicated carbon in compound D is neither electrophilic nor nucleophilic.
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For the following reaction, 5.29 grams of water are mixed with excess diphosphorus pentoxide. The reaction yields 13.3 grams of phosphoric acid . diphosphorus pentoxide(s) + water(l) phosphoric acid(aq). What is the theoretical yield of phosphoric acid?
Answer:
19.2 g
Explanation:
Step 1: Write the balanced equation
P₂O₅(s) + 3 H₂O(l) ⇒ 2 H₃PO₄
Step 2: Calculate the moles corresponding to 5.29 g of H₂O
The molar mass of H₂O is 18.02 g/mol.
5.29 g × 1 mol/18.02 g = 0.294 mol
Step 3: Calculate the theoretical yield of phosphoric acid, in moles
The molar ratio of H₂O to H₃PO₄ is 3:2. The theoretical yield of H₃PO₄ is 2/3 × 0.294 mol = 0.196 mol
Step 4: Calculate the mass corresponding to 0.196 moles of H₃PO₄
The molar mass of H₃PO₄ is 97.99 g/mol.
0.196 mol × 97.99 g/mol = 19.2 g
Select True or False: The equilibrium constant for the chemical equation 2NO(g) O2(g) 2NO2(g) is two times the equilibrium constant for the chemical equation NO(g) 1/2O2(g) NO2(g).
Answer:
False
Explanation:
The first reaction is;
NO(g) + 1/2O2(g) ---->NO2(g)
K= [NO2]/[NO] [ O2]^1/2
The second reaction is;
2NO(g) + O2(g) ---->2NO2(g)
K'= [NO2]^2/[NO]^2 [O2]
It now follows that;
K'= K^2
Hence the statement in the question is false
The standard redox potentials of isolated components of an electron transport chain in a cyanobacterium are found to be as follows:
Complex A: standard redox potential: -100 mV
Complex B: standard redox potential: -780 mV
Complex C: standard redox potential: +510 mV
Complex D: standard redox potential: +310 mV
Plastocyanin: standard redox potential: +360 mV
Which complex will likely have a binding site with high affinity for reduced plastocyanin?
A. Complex A.
B. Complex B.
C. Complex C.
D. Complex D.
Answer:
B. Complex B.
Explanation:
Complex B will have binding site with high affinity for reduced plastocyanin due to greater redox potential. The high number of redox potential will will transport electron chain in cyanobacteria.
Redox potential is the measure of the electron gain or loss to the electrode. For reduced plastocyanin complex B will have the highest affinity.
What is electron affinity?Electron affinity is the energy released when the atom gets attached to the atom or other molecule. The high number of redox potential increases the electron transport in the cell.
The greater the redox potential more will be its tendency to show electron affinity. To bond with reduced species, the oxidized species must have greater redox potential.
Therefore, option B. complex b will have the highest affinity.
Learn more about reduction potential here:
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How many grams of CO2 are formed if 44.7 g C5H12 is mixed with 108 g O2?
Explanation:
here's the answer to your question
A mixture of argon and neon gases at a total pressure of 874 mm Hg contains argon at a partial pressure of 662 mm Hg. If the gas
mixture contains 12.0 grams of argon, how many grams of neon are present?
Answer:
6.684g
Explanation:
Here, we can use the mole ratio of the gases to calculate.
We know that the mole ratio of the gases equate to their number of moles.
Firstly, we calculate the number of moles of the oxygen gas. The number of moles of the oxygen gas is equal to the mass of the oxygen gas divided by the molar mass of the oxygen gas. The molar mass of the oxygen gas is 32g/mol
Thus, the number of moles produced is 5.98/32 = 0.186875
Where do we move from here?
We know that if we place the partial pressure of oxygen over the total pressure, this would be equal to the number of moles of oxygen divided by the total number of moles. Now let’s do this.
449/851 = 0.186875/n
n =(0.186875 * 851)/449
n = 0.3542
Now we do the same for argon to get the number of moles of argon.
Firstly, we use dalton’s partial pressure law to get the partial pressure of argon. In the simplest form, the partial pressure of argon is the total pressure minus the partial pressure of oxygen.
P = 851 - 449 = 402 mmHg
We now use the mole ratio relation.
402/851 = n/0.3542
n = (402 * 0.3542) / 851
n = 0.1673
Since we now know the number of moles of argon, we can use this multiplied by the atomic mass of argon to get the mass.
The atomic mass of argon is 39.948 amu
The mass is thus 39.948 * 0.1673 = 6.684g
does anyone know how to solve this and what the answer would be?
Dynamic equilibrium is showed at the point at which solid liquid and gas intersect.
At the point at which solid liquid and gas intersect represents a system that shows dynamic equilibrium. There is equal amount of reactants and products at the point of dynamic equilibrium because the transition of substances occur between the reactants and products at equal rates, means that there is no net change. Reactants and products are formed at the rate that no change occur in their concentration.
https://brainly.com/question/24310467
Which type of organic compound is shown below?
A. Carboxylic acid
B. Ester
C. Amine
D. Alcohol
Answer:
I think its A maybe am not sure
Match each land resource to its use.
clay - used to make steel
iron ore - used to make batteries
salt - used to make pottery and tiles
aggregate - used in construction
graphite - used as a flavoring in food
i will give 10 points and brainliest!!!
Answer:
answer in picture
Explanation:
How much heat energy is required to raise the temperature of 50g of bromine from 25°C to 30°C? [Specific heat capacity of bromine = 0.226 J/(g °C]
Answer:
56.5J
Explanation:
To find the heat energy required use the formula for the specific heat capacity which is
c=quantity of heat/mass×change in temperature
in this question c is 0.226j/g,the mass is 50g and the change in temperature is 30-25=5
therefore
0.226=Q/50×5
Q=0.226×250
=56.5J
I hope this helps
list two uses of H2SO4
3. Oxalic acid, C H20, is a toxic substance found in rhubarb leaves. When mixed with sufficient
quantities of a strong base, this weak diprotic acid loses two protons to form a polyatomic ion
called oxalate, C2022. Write a balanced equation that describes the reaction between oxalic acid
and sodium hydroxide.
Answer:
COOHCOOH + 2OH⁻ ⇄ C₂O₄²⁻ + 2H₂O
Explanation:
The reaction of oxalic acid with a strong base like sodium hydroxide is the following:
COOHCOOH + OH⁻ ⇄ COOHCOO⁻ + H₂O (1)
In this first reaction, the oxalic acid loses one proton. In a second reaction with NaOH, the ion COOHCOO⁻ loses its second proton to form ion oxalate as follows:
COOHCOO⁻ + OH⁻ ⇄ C₂O₄²⁻ + H₂O (2)
The general reaction between oxalic acid and NaOH is (eq 1 + eq 2):
COOHCOOH + 2OH⁻ ⇄ C₂O₄²⁻ + 2H₂O
I hope it helps you!
Calculate the vapor pressure of a solution made by dissolving 550 grams of glucose (molar mass = 180.2 g/mol) in 1020.0 ml of water at 25°C. The vapor pressure of pure water at 25°C is 23.76 mm Hg. Assume the density of the solution is 1.00 g/ml. (760 torr = 760 mmHg = 1 atm)
Answer:
22.55 mmHg (0.03 atm)
Explanation:
According to Raoult's law, the vapor pressure (Psolution) of a solution is given by:
Psolution = Xsolvent x Psolvent
Where Xsolvent is the mole fraction of the solvent in the solution and Psolvent is the vapor pressure of the pure solvent.
From the data, we have: Psolvent = 25.76 mmHg
We have to calculate Xsolvent, which is equal to the moles of solvent divided into the total number of moles.
The solution is composed of the solute (glucose) dissolved in the solvent (water). So, the total number of moles is calculated from the moles of solute and solvent.
To calculate the moles of solute (glucose), we divide the mass of glucose into its molar mass:
moles of glucose = mass/molar mass = 550 g/(180.2 g/mol) = 3.05 mol
The same for the moles of solvent (water). The mass of water is obtained from the product of the volume and density:
mass of water = volume x density = 1020.0 mL x 1.00 g/mL = 1020.0 g
molar mass H₂O = (1 g/mol x 2) + 16 g/mol = 18 g/mol
moles of water = mass water/molar masss = 1020.0 g/(18 g/mol) = 56.67 mol
Now, we can calculate Xsolvent:
Xsolvent = moles of water/total moles
total moles = moles glucose + moles water = 3.05 mol + 56.67 mol = 59.72 mol
⇒ Xsolvent = 56.67 mol/(59.72 mol) = 0.9489
Finally, we calculate the vapor pressure of the solution:
Psolution = 0.9489 x 23.76 mmHg = 22.55 mmHg
22.55 mmHg x 1 atm/760 mmHg = 0.03 atm
How are all compounds similar?
A. They are all made up of ions that are held together by attractions.
B. They are all made up of the same few elements.
C. They are all made up of atoms of two or more different elements.
D. They are all made up of atoms that share electrons.
Answer:
the answer is C
Explanation:
a molecule can be made up of two atoms of the same kind, as when two oxygen atoms bind together to make an oxygen molecule
I'd really appreciate a brainleast
Nitrous oxide, NO, decomposes exothermically into nitrogen gas and oxygen gas. When graphing [NO] versus time, a straight line can be drawn through the experimental points. From this information, determine the reaction order.
Answer:
Zero-Order
Explanation:
The exothermic decaying of nitrous oxide at 575° C will lead to [tex]N_{2} and O_{2}[/tex] as follows:
[tex]2N_{2}O[/tex] → [tex]2N_{2}(g) + O_{2} (g)[/tex]
Hot platinum wire in the above reaction would function as a catalyst in the zero-order. However, if the reaction is considered in the gaseous phase, it will be more inclined towards second-order.
In the given scenario([tex]2N_{2}O[/tex] → [tex]2N_{2}(g) + O_{2} (g)[/tex]), the reactant molecules of Nitrous oxide are restricted to the ones which have linked themselves to the catalyst's surface. Once this limited surface is filled, the extra molecules of gas would remain vacant until the previously attached molecules with the surface are decayed entirely.
Gallium is produced by the electrolysis of a solution made by dissolving gallium oxide in concentrated NaOH(aq). Calculate the amount of Ga(s) that can be deposited from a Ga(III) solution using a current of 0.680 A that flows for 80.0 min.
Answer:
Mass gallium (Ga°(s)) produced ≅ 0.800 grams (1 sig. fig.)
Explanation:
Ga(OH)₃ => Ga⁺³ + 3OH⁻
Ga⁺³ + 3e⁻ => Ga°(s)
? grams Ga°(s) = 0.680 Amps x 1 mole e⁻/1 Faraday x 1 Faraday/96,500 Amp·sec x 1 mole Ga°/3 moles e⁻ x 69.723 grams Ga°/mole Ga° x 60 sec/1 min x 80 min = [(0.680)(69.723)(60)(80)/(96,500)(3)] grams Ga° = 0.786099731 grams Ga° (calc. ans.) ≅ 0.800 grams Ga° (1 sig. fig.)
work 10
HOMEWORK ASSIGNMENTS Content
Detai
Question 1
6.25 Points
р А.
71
Calculate AE of a gas for a process in which the gas evolves 24 J of heat and has 9 of work done on it.
A
✓
A -33)
B +33)
Gradin
-220)
D) +15)
E -15)
Question 2
6.25 Points
Answer
A
Explanation:
due to high specific heat capacity it loses heat and has low temperature
Select the correct relationship among the concentrations of species present in a 1.0 M aqueous solution of the weak acid represented by HA. A. [H2O] > [HA] > [A-] > [H3O ] > [OH-] B. [H2O] > [A-] ~ [H3O ] > [HA] > [OH-] C. [HA] > [H2O] > [A-] > [H3O ] > [OH-] D. [H2O] > [HA] > [A-] ~ [H3O ] > [OH-] E. [HA] > [H2O] > [A-] ~ [H3O ] > [OH-]
Answer:
D
Explanation:
We have to bear in mind that the acid is a weak acid. A weak acid does not dissociate completely in solution. We will have more concentration of undissociated acid than A^- and H3O^+ and OH^- in the system at equilibrium.
Being a weak acid, there is maximum concentration of water molecules followed by that of undissiociated acid.
Hence, for this solution, the concentration of ions in solution follows the order;
[H2O] > [HA] > [A-] ~ [H3O ] > [OH-]
The following reaction is not an oxidation-reduction reaction: Fe(s) + 2Hl(aq) --- Fel (aq) + H_(8) Select one: O True O False
Explanation:
the reaction is indeed an oxidation reduction reaction
For the following reaction, 4.77 grams of carbon (graphite) are allowed to react with 16.4 grams of oxygen gas.
carbon (graphite) (s) + oxygen (g) → carbon dioxide (g)
1. What is the maximum amount of carbon dioxide that can be formed?
2. What is the FORMULA for the limiting reagent?
3. What mass of the excess reagent remains after the reaction is complete?
Answer:
1. 17.5 g of CO₂
2. The limiting reactant is carbon (graphite), and its formula is C(graphite)
3. 3.7 g of O₂
Explanation:
First, we have to write the chemical equation for the reaction. For this, we have to know the chemical formula of each reactant and product:
Reactants: carbon(graphite) ⇒ C(graphite) ; oxygen gas ⇒ O₂(g)Products: carbon dioxide ⇒ CO₂(g)Thus, we write the chemical equation:
C(graphite) + O₂(g) → CO₂(g)
The equation is already balanced because it has the same number of C and O atoms on both sides. Thus, we can see that 1 mol of C(graphite) reacts with 1 mol of O₂ and produce 1 mol of CO₂ (mole-to-mole reaction).
Now we convert the grams of reactants to moles by using the molecular weight (Mw) of each compound:
Mw(C) = 12 g/mol
moles of C(graphite) = 4.77g/(12 g/mol) = 0.3975 mol
Mw(O₂) = 16 g/mol x 2 = 32 g/mol
moles of O₂ = 16.4 g/(32 g/mol) = 0.5125 mol
Now, we can compare the stoichiometric ratio (given by the moles of reactants in the equation) with the actual ratio (given by the mass of reactants we have):
stoichiometric ratio ⇒ 1 mol C(graphite)/mol O₂
actual ratio ⇒ 0.3975 mol C(graphite)/0.5125 mol O₂
We can see that we need 0.3975 moles of O₂ to react with C(graphite) and we have more moles (0.5125 mol) so the excess reactant is O₂. Thus, the limiting reactant is C(graphite).
The amount of product (CO₂) that is formed is calculated from the amount of limiting reactant. We can see in the chemical equation that 1 mol of CO₂ is produced from 1 mol of C(graphite) ⇒ stoichiometric ratio = 1 mol CO₂/mol C(graphite).
Thus, we multiply the moles of C(graphite) we have by the stoichiometric ratio to calculate the moles of CO₂ produced:
moles of CO₂ = 0.3975 mol C(graphite) x 1 mol CO₂/mol C(graphite) = 0.3975 mol CO₂
Now, we convert the moles of CO₂ to mass by using the Mw:
Mw(CO₂) = 12 g/mol + (16 g/mol x 2) = 44 g/mol
mass of CO₂ = 0.3975 mol CO₂ x 44 g/mol CO₂ = 17.5 g
Therefore, the maximum amount of carbon dioxide (CO₂) formed is 17.5 g.
Since this is a mole-to-mole reaction, the moles of excess reactant that remains after the reaction is complete is calculated as the difference between the moles of excess reactant and limiting reactant:
remaining moles of O₂ = 0.5125 mol - 0.3975 mol = 0.115 mol O₂
Finally, we convert the moles of O₂ to mass with the Mw (32 g/mol) :
mass of O₂ = 0.115 mol O₂ x 32 g/mol = 3.68 g
Therefore, the mass of the excess reagent that remains after the reaction is complete is 3.7 g.
State the different radiations emitted by radioactive elements.
Answer:
gamma rays , alpha particles , beta particles , neutrons
write down the different uses of water that you know about
Answer:
The various uses of water :
1. Water is used for daily purpose like cooking , bathing , cleaning and drinking.
2. Water is used as a universal solvent.
3. water maintains the temperature of our body.
4. Water helps in digestion in our body.
5 .water is used in factories and industries.
6. Water is used to grow plants , vegetables and crops.
Answer:
The various use of water are;
I) Cooking.
ii) Drinking
III) Bathing
iv) Generating hydro- electricity
v) Construction work etc
What is the equilibrium expression for the reaction below?
Caco (s)
Cao(s) + CO (g)
A.
B.
[Cacoz]
[Cao]
[Caco.]
[Cao]+[CO,
[Cao][COL]
[Caco:]
C.
D. [co]
Answer:
D. [CO₂]
Explanation:
Let's consider the following equation at equilibrium.
CaCO₃(s) ⇄ CaO(s) + CO₂(g)
The equilibrium constant, Kc, is the ratio of the equilibrium concentrations of products over the equilibrium concentrations of reactants each raised to the power of their stoichiometric coefficients. It only includes gases and aqueous species.
Kc = [CO₂]
# of protons
# of neutrons
# of electrons
Atomic Number
Mass Number
18
17
35
17
37
6
8
6
6
15
Answer:
35
Explanation:
is the answer for your question
What is Avogadro's number?
O A. 6.02 x 10-23
O B. 6.0223
C. 6.02 x 10
D. 6.02 x 1023
Answer:
E.6.02 10-23Explanation:
Answer:
6.02×10^23 I hope it helps you
Explanation:
I hope it helps you
The half life for the decay of carbon-14 is 5.73 times 10^3 years.
Suppose the activity due to the radioactive decay of the carbon-14 in a tiny sample of an artifact made of wood from an archeological dig is measured to be 77.The activity in a similar-sized sample of fresh wood is measured to be 85.Calculate the age of the artifact. Round your answer to 2 significant digits.
Answer:
790 years
Explanation:
Given that;
0.693/t1/2 = 2.303/t log [A]o/[A]
So;
t1/2 =half life of carbon-14
t= age of the sample
[A]o= activity of the living sampoke
[A] = activity at time t
0.693/5.73 ×10^3 = 2.303/t log 85/77
1.21 × 10^-4 = 2.303/t log 1.1
1.21 × 10^-4 = 0.0953/t
t= 0.0953/1.21 × 10^-4
t= 790 years (to 2sf)
The electronic arrangement of an atom shows how electrons are distributed across the different energy levels. Which of the following elements is represented by the electron arrangement 2, 8, 18, 6?
А Сa
B. Mg
C. S
D Se
E. Ga
Answer:
The answer is D which is Selenium
Benzoyl chloride undergoes hydrolysis when heated with water to make benzoic acid. Reaction scheme of benzoyl chloride with water and heat over the arrow, and benzoic acid and hydrochloric acid as products. Calculate the molar mass of the reactant and product. Report molar masses to 1 decimal place.
Answer:
The molar mass of benzoic acid is 122.1 g/mol
The molar mass of hydrochloric acid = 36.5 g/mol
Explanation:
Benzoyl chloride is an organic compound with the molecular formula C₆H₅COCl. It is an acyl chloride since is it an organic derivative of a carboxylic acid. Acyl chlorides have the general molecular formula, R-COCl, where R is a side chain.
The R group of benzoyl chloride is the benzyl group C₆H₅. It reacts with water (hydrolysis) to produce hydrochloric acid and benzoic acid. The equation of the reaction is given below:
C₆H₅COCl + H₂O → C₆H₅CO₂H + HCl
The molar mass of benzoic acid as well as of hydrochloric acid is calculated from the sum of the masses of the atoms of the elements present in the compound thus:
Molar mass of carbon = 12.0107 g
Molar mass of hydrogen = 1.00784 g
Molar mass of oxygen = 15.999 g
Molar mass of chlorine = 35.453 g
Molar mass of benzoic acid, C₆H₅CO₂H containing 7 moles of atoms of carbon, 6 moles of atoms of hydrogen and 2 moles of atoms of oxygen = 7 × 12.0107 + 6 × 1.00784 + 2 × 15.999 = 122.1 g
Therefore, the molar mass of benzoic acid is 122.1 g/mol
Molar mass of hydrochloric acid, HCl, containing 1 mole of atoms of hydrogen and 1 mole of atoms of chlorine = 1 × 1.00784 + 1 × 35.453 = 36.5 g
Therefore, the molar mass of hydrochloric acid = 36.5 g/mol
A chemical reaction takes place inside a flask submerged in a water bath. The water bath contains 6.90kg of water at 34.7 degrees C . During the reaction 57.1kJ of heat flows out of the bath and into the flask.
Calculate the new temperature of the water bath. You can assume the specific heat capacity of water under these conditions is 4.18J.g^(-1).K^(-1) . Round your answer to significant digits.
Answer:
[tex]T_2= 36.7 \textdegree C[/tex]
Explanation:
Mass of Water [tex]m_w=6.90kg[/tex]
Temperature [tex]T=34.7 degrees[/tex]
Heat Flow [tex]H=57.1kJ[/tex]
Specific heat capacity of water [tex]\mu= 4.18J.g^(-1).K^(-1)[/tex]
Generally the equation for Final Temperature is mathematically given by
[tex]M*\mu *T_1 + Q = M*\mu *T_2[/tex]
[tex]T_2=\frac{M*\mu *T_1 + Q }{M*\mu}[/tex]
Therefore
[tex]T_2=\frac{6.90*4.18*34.7 + 57.1}{6.90*4.18}[/tex]
[tex]T_2= 36.7 \textdegree C[/tex]
150 mL of 0.25 mol/L magnesium chloride solution and 150 mL of 0.35 mol/L silver nitrate solution are mixed together. After reaction is completed; calculate the concentration of nitrate ions in solution. Assume that the total volume of the solution is 3.0 x 10^2 mL
Answer:
[tex]0.175\; \rm mol \cdot L^{-1}[/tex].
Explanation:
Magnesium chloride and silver nitrate reacts at a [tex]2:1[/tex] ratio:
[tex]\rm MgCl_2\, (aq) + 2\, AgNO_3\, (aq) \to Mg(NO_3)_2 \, (aq) + 2\, AgCl\, (s)[/tex].
In reality, the nitrate ion from silver nitrate did not take part in this reaction at all. Consider the ionic equation for this very reaction:
[tex]\begin{aligned}& \rm Mg^{2+} + 2\, Cl^{-} + 2\, Ag^{+} + 2\, {NO_3}^{-} \\&\to \rm Mg^{2+} + 2\, {NO_3}^{-} + 2\, AgCl\, (s)\end{aligned}[/tex].
The precipitate silver chloride [tex]\rm AgCl[/tex] is insoluble in water and barely ionizes. Hence, [tex]\rm AgCl\![/tex] isn't rewritten as ions.
Net ionic equation:
[tex]\begin{aligned}& \rm Ag^{+} + Cl^{-} \to AgCl\, (s)\end{aligned}[/tex].
Calculate the initial quantity of nitrate ions in the mixture.
[tex]\begin{aligned}n(\text{initial}) &= c(\text{initial}) \cdot V(\text{initial}) \\ &= 0.25\; \rm mol \cdot L^{-1} \times 0.150\; \rm L \\ &= 0.0375\; \rm mol \end{aligned}[/tex].
Since nitrate ions [tex]\rm {NO_3}^{-}[/tex] do not take part in any reaction in this mixture, the quantity of this ion would stay the same.
[tex]n(\text{final}) = n(\text{initial}) = 0.0375\; \rm mol[/tex].
However, the volume of the new solution is twice that of the original nitrate solution. Hence, the concentration of nitrate ions in the new solution would be [tex](1/2)[/tex] of the concentration in the original solution.
[tex]\begin{aligned} c(\text{final}) &= \frac{n(\text{final})}{V(\text{final})} \\ &= \frac{0.0375\; \rm mol}{0.300\; \rm L} = 0.175\; \rm mol \cdot L^{-1}\end{aligned}[/tex].
