Answer:
25.7 mL
Explanation:
Step 1: Given data
Initial volume (V₁): ?Initial concentration (C₁): 0.350 MFinal volume (V₂): 600 mLFinal concentration (C₂): 0.150 MStep 2: Calculate the volume of the initial solution
We have a concentrated solution and we want to prepare a diluted one. We can calculate the initial volume using the dilution rule.
C₁ × V₁ = C₂ × V₂
V₁ = C₂ × V₂ / C₁
V₁ = 0.150 M × 600 mL / 0.350 M
V₁ = 25.7 mL
Complete and balance the molecular equation for the reaction of aqueous sodium sulfate, Na2SO4, and aqueous barium nitrate, Ba(NO3)2. Include physical states.molecular equation:Na_{2}SO_{4}(aq) + Ba(NO_{3})_{2}(aq) ->Na2SO4(aq)+Ba(NO3)2(aq)⟶Enter the balanced net ionic equation for this reaction. Include physical states.net ionic equation:
Answer:
1. The balanced molecular equation is given below:
Na2SO4(aq) + Ba(NO3)2(aq) —> BaSO4(s) + 2NaNO3(aq)
2. The net ionic equation is given below:
SO4^2-(aq) + Ba^2+(aq) —> BaSO4(s)
Explanation:
1. The balanced molecular equation
Na2SO4(aq) + Ba(NO3)2(aq) —> BaSO4(s) + NaNO3(aq)
The above equation can be balance as follow:
Na2SO4(aq) + Ba(NO3)2(aq) —> BaSO4(s) + NaNO3(aq)
There are 2 atoms of Na on the left side and 1 atom on the right side. It can be balance by putting 2 in front of NaNO3 as shown below:
Na2SO4(aq) + Ba(NO3)2(aq) —> BaSO4(s) + 2NaNO3(aq)
Now, the equation is balanced.
2. The bal net ionic equation.
This can be obtained as follow:
Na2SO4(aq) + Ba(NO3)2(aq) —>
In solution, Na2SO4 and Ba(NO3)2 will dissociate as follow:
Na2SO4(aq) —> 2Na^+(aq) + SO4^2-(aq)
Ba(NO3)2(aq) —> Ba^2+(aq) + 2NO3^-(aq)
Na2SO4(aq) + Ba(NO3)2(aq) —>
2Na^+(aq) + SO4^2-(aq) + Ba^2+(aq) + 2NO3^-(aq) —> BaSO4(s) + 2Na^+(aq) + 2NO3^-(aq)
Cancel the spectator ions i.e Na^+ and NO3^- to obtain the net ionic equation.
SO4^2-(aq) + Ba^2+(aq) —> BaSO4(s)
People take antacids, such as milk of magnesia, to reduce the discomfort of acid stomach or heartburn. The recommended dose of milk of magnesia is 1 teaspoon, which contains 500 mg of Mg(OH)2. What volume of HCl solution with a pH of 1.25 can be neutralized by 1 dose of milk of magnesia
Answer:
[tex]V_{HCl}=0.208L=208mL[/tex]
Explanation:
Hello,
In this case, since the chemical reaction is:
[tex]2HCl+Mg(OH)_2\rightarrow MgCl_2+2H_2O[/tex]
We can see that hydrochloric acid and magnesium hydroxide are in a 2:1 mole ratio, which means that the neutralization point, we can write:
[tex]n_{HCl}=2*n_{Mg(OH)_2}[/tex]
In such a way, the moles of magnesium hydroxide (molar mass 58.3 g/mol) in 500 mg are:
[tex]n_{Mg(OH)_2}=500mg*\frac{1g}{1000mg}*\frac{1mol}{58.3g} =0.00858mol[/tex]
Next, since the pH of hydrochloric acid is 1.25, the concentration of H⁺ as well as the acid (strong acid) is:
[tex][H^+]=[HCl]=10^{-pH}=10^{-1.25}=0.0562M[/tex]
Then, since the concentration and the volume define the moles, we can write:
[tex][HCl]*V_{HCl}=2*n_{Mg(OH)_2}[/tex]
Therefore, the neutralized volume turns out:
[tex]V_{HCl}=\frac{2*0.00858mol}{0.0562\frac{mol}{L} }\\ \\V_{HCl}=0.208L=208mL[/tex]
Best regards.
A buffer is prepared such that [H2PO4-] = 0.095M and [HPO42-] = 0.125M? What is the pH of this buffer solution? (pKa = 7.21 for H2PO4-)
Answer:
pH of the buffer is 7.33
Explanation:
The mixture of the ions H₂PO₄⁻ and HPO₄²⁻ produce a buffer (The mixture of a weak acid, H₂PO₄⁻, with its conjugate base, HPO₄²⁻).
To find pH of a buffer we use H-H equation:
pH = pka + log [A⁻] / [HA]
Where A⁻ is conjugate base and HA weak acid.
For the H₂PO₄⁻ and HPO₄²⁻ buffer:
pH = pka + log [HPO₄²⁻] / [H₂PO₄⁻]
Computing values of the problem:
pH =7.21 + log [0.125M] / [0.095M]
pH = 7.33
pH of the buffer is 7.33
PLEASE HELP!!
You are performing an experiment that involves the electrolysis of gold (I) bromide, also know as AuBr. You know that gold is less reactive than hydrogen. Which of the following would be the product of the reaction?
A. Hydrogen gas
B. Gold bromide
C. Oxygen gas
D. Pure gold
Answer:
D. Pure gold
Explanation:
Hello,
In this case, since gold, as a heavy metal, is said to be less reactive than hydrogen, when it undergoes electrolysis process when forming a salt, due to the action of the electric current, we can appreciate the formation of a layer of gold on the surface of the cathode via a reduction half-reaction from gold (I) to metallic gold:
[tex]Au^++1e^-\rightarrow Au^0[/tex]
Thereby, D. Pure gold is formed as the product of the reaction.
In contrast, more reactive metals than hydrogen such as sodium or potassium, will remain in solution so the hydrogen converted to hydrogen gas.
Best regards-
What would be the voltage (Ecell) of a voltaic cell comprised of Cd(s)/Cd2+(aq) and Zr(s)/Zr4+(aq) if the concentrations of the ions in solution were [Cd2+] = 0.5 M and [Zr4+] = 0.5 M at 298K?
Answer:
1.05 V
Explanation:
Since;
E°cell= E°cathode- E°anode
E°cathode= -0.40 V
E°anode= -1.45 V
E°cell= -0.40-(-1.45) = 1.05 V
Equation of the process;
2Zr(s) + 4Cd^2+(aq) ---->2Zr^4+(aq) + 4Cd(s)
n= 8 electrons transferred
From Nernst's equation;
Ecell = E°cell - 0.0592/n log Q
Ecell= E°cell - 0.0592/8 log [0.5]/[0.5]
Since log 1=0
Ecell= E°cell= 1.05 V
What happens when an atom of sulfur combines with two atoms of chlorine to produce SCl2?
Answer:
Each chlorine atom shares a pair of electrons with the sulfur atom. ... Each chlorine atom shares all its valence electrons with the sulfur atom
1. Explain what the police siren sounds like to Jane:
2. Explain what the police siren sounds like to John:
3. Explain why the police siren sounds different between Jane and John:
Answer:
1. the siren has a lower pitch to Jane
2. the siren has a higher pitch to John
3. sound different due to moving away from Jane making the sound wave lengths longer and moving toward John making the wave lengths shorter
Explanation:
The Doppler effect expresses that sound is comparative with the spectator or observer. This is demonstrated valid by the model given with Jane and John. To one individual it could sound low and to someone else it could sound high, in light of where they are tuning in from. To John, the police alarm playing is a higher pitch. Be that as it may, to Jane this equivalent alarm is a totally extraordinary pitch and is heard lower than in comparison to the john.
This is a prime case of the Doppler Effect. They sound distinctive on the grounds that the sound is moving far from Jane making the sound frequencies longer and it is advancing toward John making the frequencies shorter. This impacts how the sound is heard by the human ear.
What are the conditions that are favorable for extensive solid solubility of one element in another (Hume-rothery rules)
Answer:
Atomic radius less than 15%, similar structure and same valency.
Explanation:
The conditions that are favorable for extensive solid solubility of one element in another are the following.
The atomic radius of the solute and solvent atoms must be less than 15%. The structure of both solute and solvent are similar. Solubility completes when both have same valency. Valency means number of electrons in the outermost shell. If both solute and solvent has same number of electrons so it will be completely soluble in each other.
The conditions that are favorable for extensive solid solubility of one element in another is the same size, electrongativity and valency.
What is Hume - Rothery rules?Hume - Rothery rules are the sets of some important rules which gives idea about the desired condition for the formation of solid solution.
Following main points are described in this rule:
Difference between the size of the solute and the solvent should be less than 15%.Electronegativity difference between the solute and solvents should be small.And they both should have same valency, means same no. of electrons in the outermost shell.Hence size, electronegativity and valency are the conditions.
To know more about Hume - Rothery rules, visit the below link:
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Write the Arrhenius Base reaction for the following:
Sr(OH)2
Answer:
Explanation:
Sr(OH)2 (aq) ⇔ Sr+2 (aq) + 2OH- (aq)
Describe how you would prepare 500ml of 40% (w/v) aqueous iodine solution.
[Atomic mass of iodine =127g/mol].
Answer:
- Weight 333.3 grams of iodine.
- Measure 500 mL of water.
- Vigorously mix the resulting solution.
Explanation:
Hello,
In this case, since 500 mL of a 40% (w/v) aqueous solution iodine is required, we can compute the required mass of iodine by defining the given mass-volume percent:
[tex]\% w/v=\frac{m_{iodine}}{m_{solution}} *100%=\frac{m_{iodine}}{m_{water}+m_{iodine}} *100%[/tex]
In such a way, we need to find mass of iodine, which is computed as:
[tex]m_{iodine}=\frac{\%w/v*m_{water}}{100w/v-\%} \\\\m_{iodine}=\frac{40*500}{100-40}\\ \\m_{iodine}=333.3g\\[/tex]
Thereby, the procedure will be:
- Weight 333.3 grams of iodine.
- Measure 500 mL of water.
- Vigorously mix the resulting solution.
Best regards.
Which of the following is a covalent bond? A NaCl B K2O C H2O D MgO
Answer:
[tex]H_2O[/tex]
Explanation: Research has proven that ;
Water is a Polar Covalent Molecule
It consists of 2 Hydrogen molecules bonded to one Oxygen molecule and the two hydrogen atoms are not evenly distributed around the oxygen atom.
Predict the order of acid strengths in the following series of cationic
species: CH3CH2NH3
+, CH3CH=NH2
+ and CH3C-NH+
Answer:
CH3C-NH+> CH3CH=NH2 >CH3CH2NH3+
Explanation:
The acid strength has to do with the ease of the loss of hydrogen ion from the cationic specie. Hydrogen ion will easily be lost from any specie which contains an atom, group of atoms or bond which withdraws electrons along the chain of the N-H bond.
The pi bond system is known to be highly electronegative and withdraws electrons along the chain hence a withdrawal of electron density along the chain which makes the hydrogen ion to be easily lost from a system which contains a pi bond along the chain. A triple bond is more electronegative than a double bond, hence the answer above.
In the experiment, you heated a sample of metal in a water bath and quickly removed and dried the sample prior to inserting it into the cold water "system". Why did you need to dry the sample prior to placing it into the cold water "system"
a. The extra water might react with the metal which would ruin the sample.
b. Any residual hot water would have added to the heat transferred from the metal to the cold water "system".
c. The wet metal would not transfer any heat thus causing inaccuracy in you measurements.
d. The metal would oxidize in the presence of water thus ruining the sample
Answer:
b
Explanation:
Provided the experiment is about the transfer of heat from one medium to another, any residual hot water on the metal would have added to the heat transferred from the metal to the cold water system.
The residual hot water on the metal posses its own heat and when such is transferred along with the metal into the cold water, the heat from the residual hot water will interfere with the measurement of the actual heat transferred to the cold water by the metal. Hence, the accuracy of the result would be impacted.
The correct option is b.
Because of movements at the Mid-Atlantic Ridge, the Atlantic Ocean widens by about 2.5 centimeters each year. Explain which type of plate boundary causes this motion.
Answer:
A divergent plate boundary
Explanation:
At a divergent boundary, the plates pull away from each other and generate new crust.
Answer:
Because the ocean becomes larger, this is a divergent plate boundary. Divergent plates cause the ocean floor to expand, making the ocean larger.
Explanation:
PLATO ANSWER
How many grams of PtBr4 will dissolve in 250.0 mL of water that has 1.00 grams of KBr dissolved in it
Answer:
[tex]m_{PtBr_4}=0.306gPtBr_4[/tex]
Explanation:
Hello,
In this case, since the solubility product of platinum (IV) bromide is 8.21x10⁻⁹, and the dissociation is:
[tex]PtBr_4(s)\rightleftharpoons Pt^{4+}(aq)+4Br^-(aq)[/tex]
The equilibrium expression is:
[tex]Ksp=[Pt^{4+}][Br^-]^4[/tex]
Thus, since the salt is added to a solution initially containing 1.00 grams of potassium bromide, there is an initial concentration of bromide ions:
[tex][Br^-]_0=\frac{1.00gKBr*\frac{1molKBr}{119gKBr}*\frac{1molBr^-}{1molKBr} }{0.250L}=0.0336M[/tex]
Hence, in terms of the molar solubility [tex]x[/tex], we can write:
[tex]8.21x10^{-9}=(x)(0.0336+4x)^4[/tex]
In such a way, solving for [tex]x[/tex], we obtain:
[tex]x=0.00238M[/tex]
Which is the molar solubility of platinum (IV) bromide. Then, since its molar mass is 514.7 g/mol, we can compute the grams that get dissolved in the 250.0-mL solution:
[tex]m_{PtBr_4}=0.00238\frac{molPtBr_4}{1L}*0.250L *\frac{514.7gPtBr_4}{1molPtBr_4} \\\\m_{PtBr_4}=0.306gPtBr_4[/tex]
Best regards.
A compound is found to contain 29.68 % sulfur and 70.32 % fluorine by mass. What is the empirical formula for this compound?
Answer:
[tex]SF_4[/tex]
Explanation:
The first thing would be to calculate the number of moles of each element in the compound.
No of moles of sulfur (S) = mass/molar mass
= 29.68/32.065 = 0.9256
No of moles of fluorine (F) = mass/molar mass
= 70.32/18.998 = 3.7014
Then, let us find the atomic ratio of each of the element in the compound by dividing by the no of moles by the smallest no of mole:
S : F
[tex]\frac{0.9256}{0.9256}[/tex] = 1 : [tex]\frac{3.7014}{0.9256}[/tex] = 4
Therefore, the empirical formula for the compound is [tex]SF_4[/tex]
What type of bond would form between two atoms of phosphorus? A. Single covalent bond B. Single ionic bond C. Triple covalent bond D. Double covalent bond
Answer:
A double bond is formed when two pairs of electrons are shared between the two participating atoms. It is represented by two dashes (=). It is represented by two dashes (=). Double covalent bonds are much stronger than a single bond, but they are less stable
Explanation:
Predict the order of acid strengths in the following series of cationic species: CH3CH2NH3 +, CH3CH=NH2
Answer:
CH3CH=NH2+>CH3CH2NH3 +
Explanation:
There are certain structural features that determine the stability of cationic species. These features that lead to the stability and higher acid strength of cations are those features that stabilize the cation.
CH3CH=NH2+ is more acidic than CH3CH2NH3 + owing to the fact that CH3CH=NH2+ contains a double bond in close proximity with the hydrogen that can be lost as a proton. Electron withdrawal by the double bond (greater s character) makes it easier for this hydrogen to be lost as a proton compared to CH3CH2NH3 +.
The pH of an acid solution is 5.82. Calculate the Ka for the monoprotic acid. The initial acid concentration is 0.010 M.
Answer:
The answer is
[tex]Ka = 2.29 \times {10}^{ - 14} moldm^{ - 3} [/tex]Explanation:
The Ka of an acid when given the pH and concentration can be found by
[tex]pH = - \frac{1}{2} log(Ka) - \frac{1}{2} log(c) [/tex]where
c is the concentration of the acid
From the question
pH = 5.82
c = 0.010 M
Substitute the values into the above formula and solve for Ka
We have
[tex]5.82 = - \frac{1}{2} log(Ka) - \frac{1}{2} log(0.010) [/tex][tex] - \frac{1}{2} log(Ka) = 5.82 + 1[/tex][tex] - \frac{1}{2} log(Ka) = 6.82[/tex]Multiply through by - 2
[tex] log(Ka) = - 13.64[/tex]Find antilog of both sides
We have the final answer as
[tex]Ka = 2.29 \times {10}^{ - 14} moldm^{ - 3} [/tex]Hope this helps you
The half-life of radium-226 is 1620 years. What percentage of a given amount of the radium will remain after 900 years
Answer:
68%
Explanation:
Since we need a percentage we can use any number we want for our initial value.
5(1/2)^900/1620 = 3.40
(3.40 / 5)*100 = 68%
To make sure lets use a different initial amount
1(1/2)^900/1620 = 0.68
(0.68/1) * 100 = 68%
To solve this question, we'll assume the initial amount of radium-226 to be 1.
Now, we shall proceed to obtaining the percentage of radium-226 that will after 900 years. This can be obtained as illustrated below:
Step 1Determination of the number of half-lives that has elapsed.
Half-life (t½) = 1620 years
Time (t) = 900 years
Number of half-lives (n) =?[tex]n = \frac{t}{t_{1/2}}\\\\n = \frac{900}{1620}\\\\n = \frac{5}{9}[/tex]
Step 2:Determination of the amount remaining
Initial amount (N₀) = 1
Number of half-lives (n) = 5/9
Amount remaining (N) =?[tex]N = \frac{N_{0} }{2^{n}}\\\\N = \frac{1}{2^{5/9}}[/tex]
N = 0.68Step 3Determination of the percentage remaining.
Initial amount (N₀) = 1
Amount remaining (N) = 0.68
Percentage remaining =?Percentage remaining = N/N₀ × 100
Percentage remaining = 0.68/1 × 100
Percentage remaining = 68%Therefore, the percentage amount of radium-226 that remains after 900 years is 68%
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The solubility product for Ag3PO4 is 2.8 × 10‑18. What is the solubility of silver phosphate in a solution which also contains 0.10 moles of silver nitrate per liter?
Answer:
2.8x10⁻¹⁵ M.
Explanation:
Hello,
In this case, the dissociation reaction for silver phosphate is:
[tex]Ag_3PO_4(s)\rightleftharpoons 3Ag^+(aq)+PO_4(aq)[/tex]
Therefore, the equilibrium expression is:
[tex]Ksp=[Ag^+]^3[PO_4^-][/tex]
In such a way, since the initial solution contains an initial concentration of silver ions (from silver nitrate) of 0.10M, we can write the equilibrium expression in terms of the reaction extent [tex]x[/tex]:
[tex]2.8x10^{-18}=(0.10+3x)^3*(x)[/tex]
Thus, solving for [tex]x[/tex] we have:
[tex]x=2.8x10^{-15}M[/tex]
Thus, the molar solubility of silver phosphate is 2.8x10⁻¹⁵ M.
Regards.
For each bond, show the direction of polarity by selecting the correct partial charges. _________ Si-P _________ _________ Si-Cl _________ _________ Cl-P _________ The most polar bond is _______
Answer:
Siδ⁺ -- Pδ⁻⁻
Clδ⁻⁻ -- Pδ⁺
Siδ⁺ -- Clδ⁻⁻
Of the mentioned bonds the most polar bond is Si -- Cl
The polarity of the bond primarily relies upon the electronegativity difference between the two atoms that forms the bond. Therefore, if the electronegativity difference between the two atoms that forms the bond is more the bond will be more polar, and if it is less then the bond will be less polar. The electronegativity of the atoms mentioned is Si = 1.8 , P = 2.1 and Cl = 3.00.
Therefore, the Si - Cl atoms exhibit more electronegativity difference, thus, the Si - Cl bond will be the most polar bond.
Assuming that you start with 21.4 g of ammonia gas and 18.0 g of sodium metal and assuming that the reaction goes to completion, determine the mass (in grams) of each product.
Answer:
[tex]m_{NaNH_2}=30.42gNaNH_2[/tex]
[tex]m_{H_2}=0.783gH_2[/tex]
Explanation:
Hello,
In this case, the reaction between sodium and ammonia is:
[tex]2Na+2NH_3\rightarrow 2NaNH_2+H_2[/tex]
Thus, as we know the initial masses of both sodium and ammonia, we should first identify the limiting reactant, for which we firstly compute the available moles of sodium:
[tex]n_{Na}=18.0gNa*\frac{1molNa}{23.0gNa}=0.783molNa[/tex]
And the moles of sodium consumed by 21.4 g of ammonia (2:2 mole ratio):
[tex]n_{Na}^{\ consumed}=21.4gNH_3*\frac{1molNH_3}{17gNH_3} *\frac{2molNa}{2molNH_3} =1.26molNa[/tex]
In such a way, since less moles of sodium are available than consumed by ammonia, we can say, sodium is the limiting reactant. Furthermore, the mass of both sodium amide (39 g/mol) and hydrogen gas (2 g/mol) that are produced turn out:
[tex]m_{NaNH_2}=0.783molNa*\frac{2molNaNH_2}{2molNa}*\frac{39gNaNH_2}{1molNaNH_2}=30.42gNaNH_2[/tex]
[tex]m_{H_2}=0.783molNa*\frac{1molH_2}{2molNa}*\frac{2gH_2}{1molH_2}=0.783gH_2[/tex]
Best regards.
11mg of cyanide per kilogram of body weight is lethal for 50% of domestic chickens. How many grams per kilogram of body weight is a lethal dose for 50% of domestic chickens?
Answer:
[tex]0.033g[/tex]
Explanation:
Hello,
In this case, since 11 mg per kilogram of body weight has the given lethality, the mg that turn out lethal for a chicken weighting 3 kg is computed by using a rule of three:
[tex]11mg\longrightarrow 1kg\\\\x\ \ \ \ \ \ \longrightarrow 3kg[/tex]
Thus, we obtain:
[tex]x=\frac{3kg*11mg}{1kg}\\ \\x=33mg[/tex]
That in grams is:
[tex]=33mg*\frac{1g}{1000mg} \\\\=0.033g[/tex]
Regards.
Draw the structure for the organic radical species produced by reaction of the compound with a chlorine atom. Assume reaction occurs at the weakest C-H bond.
Answer:
See explanation
Explanation:
The reaction of chlorine with the pictured compound will occur via free radical mechanism. The stability of the free radical formed will depend on its structure.
The order of stability of free radicals is methyl < primary < secondary < tertiary. Hence a tertiary carbon free radical is the most stable.
Looking at the compound, the radical will form at the position shown in the image attached since it will lead to a secondary free radical which is more stable.
The structure that should be drawn is shown below.
The reaction of chlorine:It should be within the pictured compound that will arise via a free radical mechanism. The stability should be based on the structure. The stability of the order of free radicals should be methyl < primary < secondary < tertiary. Thus, a tertiary carbon free radical should be most stable.
Here look at the compound, the radical should form at the position that should be shown in the image that resulted in the secondary free radical i.e. more stable.
What is the electron configuration for the transition metal ion in each of the following compounds?
[Ni(H2O)6]Br2
[Cr(H2O)4(NO2)2]I
Answer:
1)Ni=1s2, 2s2, 2p6, 3s2, 3p6, 4s0, 3d10 called full-filled
2)Cr=1s2, 2s2, 2p6, 3s2, 3p6, 4s1, 3d5 called half-filled
A 1 liter solution contains 0.436 M hypochlorous acid and 0.581 M potassium hypochlorite. Addition of 0.479 moles of barium hydroxide will: (Assume that the volume does not change upon the addition of barium hydroxide.)
Answer:
Exceed the buffer capacity and Raise the pH by several units
Explanation:
Options are:
Raise the pH slightly
Lower the pH slightly
Raise the pH by several units
Lower the pH by several units
Not change the pH
Exceed the buffer capacity
The hypochlorous acid, HClO, is in equilibrium with Hypochlorite ion (From potassium hypochlorite, ClO⁻) producing a buffer. Using H-H equation, pH of initial buffer is:
pH = pKa + log [ClO⁻] / [HClO]
pKa for hypochlorous acid is 7.53
pH = 7.53 + log [0.581M] / [0.436M]
pH = 7.65
Barium hydroxide reacts with HClO producing more ClO⁻, thus:
Ba(OH)₂ + 2HClO → 2ClO⁻ + 2H₂O
As 0.479 moles of Barium hdroxide are added. For a complete reaction you require 0.479mol * 2 = 0.958 moles of HClO
As you have just 0.436 moles (Volume = 1L),
The addition will:
Exceed the buffer capacityThe Ba(OH)₂ that reacts is:
0.436 moles HClO * (1mole (Ba(OH)₂ / 2 mol HClO) = 0.218 moles Ba(OH)₂ and will remain:
0.479 mol - 0.218 mol = 0.261 moles Ba(OH)₂
As 1 mole of Ba(OH)₂ contains 2 moles of OH⁻, moles of OH⁻ and molarity is:
0.261 moles* 2 = 0.522 moles OH⁻ = [OH⁻]
pOH = -log [OH⁻]
pOH = 0.28
And pH = 14 - pOH:
pH = 13.72
Thus, after the addition the pH change from 7.65 to 13.62:
Raise the pH by several units
You have to prepare a pH 3.65 buffer, and you have the following 0.10M solutions available: HCOOH, CH3COOH, H3PO4, HCOONa, CH3COONa, and NaH2PO4. How many mL of HCOOH and HCOONa would you use to make approximately a liter of the buffer?
Answer:
550mL of HCOOH 0.1M and 450mL of HCOONa 0.1M
Explanation:
It is possible to find the pH of a buffer by using H-H equation:
pH = pKa + log [A⁻]/[HA]
For the formic buffer (HCOOH/HCOONa):
pH = 3.74 + log [HCOONa]/[HCOOH]
As you need a buffer of pH 3.65:
pH = 3.74 + log [HCOONa]/[HCOOH]
3.65 = 3.74 + log [HCOONa]/[HCOOH]
0.81283 = [HCOONa]/[HCOOH] (1)Where [HCOONa]/[HCOOH] can be taken as the moles of each specie.
As molarity of both solutions is 0.10M (0.10mol / L) and you need 1L of solution, total moles of the buffer are:
0.10 moles = [HCOONa] + [HCOOH] (2)Replacing (2) in (1):
0.81283 = 0.10 - [HCOOH] /[HCOOH]
0.81283[HCOOH] = 0.10 - [HCOOH]
1.81283[HCOOH] = 0.10
[HCOOH] = 0.055 molesAnd moles of HCOONa are:
[HCOONa] = 0.1 mol - 0.055mol =
[HCOONa] = 0.045 molesAs concentration of the solutions is 0.1M, the volume you need to add of both solutions is:
HCOOH = 0.055 mol ₓ (1L / 0.1mol) = 0.55L = 550mL of HCOOH 0.1M
HCOONa = 0.045 mol ₓ (1L / 0.1mol) = 0.45L = 450mL of HCOONa 0.1M
The number should be considered like 550mL of HCOOH 0.1M and 450mL of HCOONa 0.1M.
Calculation of mL:Here we used the H-H equation:
pH = pKa + log [A⁻]/[HA]
Now
For the formic buffer (HCOOH/HCOONa):
So,
pH = 3.74 + log [HCOONa]/[HCOOH]
Now
need a buffer of pH 3.65:
So,
pH = 3.74 + log [HCOONa]/[HCOOH]
3.65 = 3.74 + log [HCOONa]/[HCOOH]
0.81283 = [HCOONa]/[HCOOH] (1)
here [HCOONa]/[HCOOH] can be considered as the moles of each specie.
Now the total moles should be
0.10 moles = [HCOONa] + [HCOOH] (2)
Now
0.81283 = 0.10 - [HCOOH] /[HCOOH]
0.81283[HCOOH] = 0.10 - [HCOOH]
1.81283[HCOOH] = 0.10
[HCOOH] = 0.055 moles
And moles of HCOONa should be
[HCOONa] = 0.1 mol - 0.055mol =
[HCOONa] = 0.045 moles
Now
HCOOH = 0.055 mol ₓ (1L / 0.1mol) = 0.55L = 550mL of HCOOH 0.1M
HCOONa = 0.045 mol ₓ (1L / 0.1mol) = 0.45L = 450mL of HCOONa 0.1M
Learn more about moles here: https://brainly.com/question/2154505
a) Provide equation of K of this reaction, use symbol " ^ " for exponents. That means 1000 = 10^3 and 1/100 is 10^(-2). b) How many moles of compound F will be produced if only 2 moles of compound C is available? describe or show your work. 3 A + 5 B +4 C 5 D +7 E + F
Answer and Explanation:
a. The equation of K of this reaction is shown below:-
3 A + 5 B + 4 C↔5 D + 7 E + F
[tex]K = \frac{(D)^5 (E)^7 (F)}{(A)^3 (B)^5 (C)^4}[/tex]
b. The moles of compound F is shown below:-
3 A + 5 B + 4 C↔5 D + 7 E + F
2 moles
Now, the mole of produced is
[tex]= \frac{1}{4} \times \ moles\ of\ c[/tex]
Now, we will the value of c by using the above equation
[tex]= \frac{1}{4} \times 2[/tex]
After solving the above equation we will get
0.5 moles
Consider the reaction for the dissolution of solid magnesium hydroxide.
Mg(OH)2(s)g2 (a) +2OH (ag)
If the concentration of hydroxide ion in a saturated solution of magnesium hydroxide is 2.24 x 104 M.
what is the molar solubility of magnesium hydroxide? Report your answer in scientific notation with three significant figures.
Answer:
Molar solubility is 1.12x10⁻⁴M
Explanation:
The dissolution of magnesium hydroxide is:
Mg(OH)₂(s) ⇄ Mg²⁺ + 2OH⁻
The molar solubility represents the moles of the solid that the solution can dissolve, that could be written as:
Mg(OH)₂(s) ⇄ X + 2X
Where X is solubility.
If you obtained a [OH⁻] = 2.24x10⁻⁴M and you know [OH⁻] = 2X:
2X = 2.24x10⁻⁴M
X = 2.24x10⁻⁴M/2
X =1.12x10⁻⁴M
Molar solubility is 1.12x10⁻⁴M
