Answer:
D)the first ionization energy (IE1) of B is most probably lower than the first ionization energy (IE1) of A
Explanation:
As mentioned the possible chemical formula for a compound can be predicted from electron configurations of the elements.
The first element 'A' has electronic configuration as "[core]ns²
It means it has two valence electrons and it can give those valence electrons to attain noble gas or full filled stability.
So the element is most likely to form a dipositive ion.
Now for element B the configuration is :[Core]ns²np⁵
thus it has seven valence electrons and it will most likely to accept an electron to gain full filled stability. So B will form a mono negative ion.
The possible formula will be AB₂ [Like MgCl₂]
So the false statements are:
D)the first ionization energy (IE1) of B is most probably lower than the first ionization energy (IE1) of A : the first ionization energy of A will be lower than B as B cannot give an electron easily.
The other statements are correct:
A) element A serves as the reducing agent in the reaction to form AxBy : As A will lose electrons so it will acts as reducing agent and itself gets oxidized
B) yes it is whole number as calculated above.
C) x + y = 1+2 = 3
E) Already calculated that B has oxidation state of -1.
The false statement is that; "the first ionization energy (IE1) of B is most probably lower than the first ionization energy (IE1) of A"
The first ionization energy is the energy required to remove the first electron from an atom. This is a periodic trend that increases across the period but decreases down the group.
Looking at the electron configurations of A and B it is clear that A is an alkaline earth metal while B is a halogen. As such, A is a reducing agent and B will have an oxidation state of -1 as expected of halogens.
The compound will have the formula AX2 hence x + y = 3. However, since ionization energy increases across the period, the first ionization energy of B must certainly be greater than that of A.
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A solution has a pOH of 7.1 at 10∘C. What is the pH of the solution given that Kw=2.93×10−15 at this temperature? Remember to report your answer with the correct number of significant figures
Answer : The pH of the solution is, 7.4
Explanation : Given,
pOH = 7.1
[tex]K_w=2.93\times 10^{-15}[/tex]
First we have to calculate the value of [tex]pK_w[/tex].
The expression used for the calculation of [tex]pK_w[/tex] is,
[tex]pK_w=-\log [K_w][/tex]
Now put the value of [tex]K_w[/tex] in this expression, we get:
[tex]pK_w=-\log (2.93\times 1-^{-15})[/tex]
[tex]pK_w=15-\log (2.93)[/tex]
[tex]pK_w=14.5[/tex]
Now we have to calculate the pH of the solution.
As we know that,
[tex]pH+pOH=pK_w[/tex]
Now put all the given values in this formula, we get:
[tex]pH+7.1=14.5[/tex]
[tex]pH=7.4[/tex]
Therefore, the pH of the solution is, 7.4
A solution has a pOH of 7.1 and a pH of 7.4 at 10 °C, being Kw = 2.93 × 10⁻¹⁵ at that temperature.
What is pH?pH is a figure expressing the acidity or alkalinity of a solution on a logarithmic scale on which 7 is neutral, lower values are more acid and higher values are more alkaline.
A solution has a pOH of 7.1 at 10 °C. We can calculate the pH at this temperature using the following expression.
pH + pOH = pKw
pH = pKw - pOH
pH = -log (2.93 × 10⁻¹⁵) - 7.1 = 7.4
where,
pH = -log [H⁺].pOH = -log [OH⁻].Kw is the ionic product of water.A solution has a pOH of 7.1 and a pH of 7.4 at 10 °C, being Kw = 2.93 × 10⁻¹⁵ at that temperature.
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Estimate the vapor pressure of methanol at 25°C. The heat of vaporization for methanol is 37,400 J/mol and the boiling point is 65°C
The vapor pressure of methanol at 25°C can be estimated using the Clausius-Clapeyron equation. By plugging in the values for methanol's boiling point, enthalpy of vaporization, and the desired temperature, we can solve for the vapor pressure. The estimated vapor pressure of methanol at 25°C is approximately 0.147 atm.
Explanation:In order to estimate the vapor pressure of methanol at 25°C, we can use the Clausius-Clapeyron equation. This equation relates the vapor pressure, temperature, and enthalpy of vaporization. The equation is:
ln(P2/P1) = -ΔHvap/R × (1/T2 - 1/T1)
Where P1 is the known vapor pressure at a known temperature (T1), P2 is the vapor pressure at the desired temperature (T2), ΔHvap is the enthalpy of vaporization, and R is the ideal gas constant. Plugging in the values for methanol:
P1 = vapor pressure at boiling point = 1 atm, T1 = boiling point = 65°C = 338 K, T2 = desired temperature = 25°C = 298 K, ΔHvap = 37,400 J/mol, R = 8.314 J/(mol·K)
the equation becomes:
ln(P2/1) = -37,400 J/mol / (8.314 J/(mol·K)) × (1/298 K - 1/338 K)
Solving for P2:
P2 = 1 × e ^ (-37,400 J/mol / (8.314 J/(mol·K)) × (1/298 K - 1/338 K))
P2 ≈ 0.147 atm
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A tank has a volume of 0.1 m3 and is filled with He gas at a pressure of 5 x 106 Pa. A second tank has a volume of 0.15 m' and is filled with He gas at a pressure of 6 x 106 Pa. A valve connecting the two tanks is opened. Assumıng He to be a monatomic ideal gas and the walls of the tanks to be adiabatic and rigıd, find the final pressure of the system Hint: Note that the internal energy is constant.
Answer:
[tex]P=5.6*10^{6} Pa[/tex]
Explanation:
Consider that, as the system is adiabatic, [tex]U_{1}= U_{2}[/tex] where U1 and U2 are the internal energies before the process and after that respectively.
Consider that: [tex]U=H-PV[/tex], and that the internal energy of the first state is the sum of the internal energy of each tank.
So, [tex]H_{1}-P_{1}V_{1}=H_{2}-P_{2}V_{2}\\H_{1}^{A} -P_{1}^{A}V_{1}^{A}+H_{1}^{B} -P_{1}^{B}V_{1}^{B}=H_{2}-P_{2}V_{2}[/tex]
Where A y B are the tanks. The enthalpy for an ideal gas is only function of the temperature, as the internal energy is too; so it is possible to assume: [tex]H_{1}=H_{2}\\H_{1}^{A}+H_{1}^{B} =H_{2}[/tex]
So, [tex]P_{1}^{A}V_{1}^{A}+P_{1}^{B}V_{1}^{B}=P_{2}V_{2}[/tex]
Isolating [tex]P_{2}[/tex],
[tex]P_{2}=\frac{P_{1}^{A}V_{1}^{A}+P_{1}^{B}V_{1}^{B}}{V_{2}}[/tex]
[tex]V_{2}=V_{1}^{A}+V_{1}^{B}=0.25m^{3}[/tex]
So,
[tex]P_{2}=\frac{5000000*0.1+6000000*0.15}{0.25}=5600000Pa=5.6*10^{6} Pa[/tex]
In the Haber reaction, patented by German chemist Fritz Haber in 1908, dinitrogen gas combines with dihydrogen gas to produce gaseous ammonia. This reaction is now the first step taken to make most of the world's fertilizer.
Suppose a chemical engineer studying a new catalyst for the Haber reaction finds that 505 liters per second of dinitrogen are consumed when the reaction is run at 172 oC and 0.88 atm. Calculate the rate at which ammonia is being produced. Give your answer in kilograms per second. Be sure your answer has the correct number of significant digits.
Answer:
The rate at which ammonia is being produced is 0.41 kg/sec.
Explanation:
[tex]N_2+3H_2\rightarrow 2NH_3[/tex] Haber reaction
Volume of dinitrogen consumed in a second = 505 L
Temperature at which reaction is carried out,T= 172°C = 445.15 K
Pressure at which reaction is carried out, P = 0.88 atm
Let the moles of dinitrogen be n.
Using an Ideal gas equation:
[tex]PV=nRT[/tex]
[tex]n=\frac{PV}{RT}=\frac{0.88 atm\times 505 L}{0.0821 atm l/mol K\times 445.15 K}=12.1597 mol[/tex]
According to reaction , 1 mol of ditnitrogen gas produces 2 moles of ammonia.
Then 12.1597 mol of dinitrogen will produce :
[tex]\frac{2}{1}\times 12.1597 mol=24.3194 mol[/tex] of ammonia
Mass of 24.3194 moles of ammonia =24.3194 mol × 17 g/mol
=413.43 g=0.41343 kg ≈ 0.41 kg
505 L of dinitrogen are consumed in 1 second to produce 0.41 kg of ammonia in 1 second. So the rate at which ammonia is being produced is 0.41 kg/sec.
True or False Materials that have predominantly ionic or covalent bonds have many free electrons.
Answer:
False
Explanation:
Atomic bonding can be of two types:
Ionic bonds: The bond formed between two opposite charged ions. The strong electrostatic force acting between the ions make them compounds of high stability . Example: NaCl
Covalent bonds: The bond formed by sharing of the electrons between the atoms. The balance between the attractive force (nucleus, electron) and repulsive force (electron, electron or nucleus, nucleus) makes the compound stable. Example: [tex]CH_4[/tex]
The bonds are either formed by transferring (losing and gaining) or sharing of electrons. Thus, compound forming these bonds share or transfer electrons to attain noble gas configuration. Thus, they lack free electrons.
The pKa values for the carboxy and ammonium protons of phenylalanine are 2.58 and 9.24, respectively. What is the isoelectric point of phenylalanine? Draw the structure of phenylalanine at its isoelectric point.
Answer:The isoelctric point of phenylalanine is 5.91. Kindly refer the attachment for structure.
Explanation:
Isoeectronic point is the pH at which the amino acid do not migrate in presence of an electric field.Hence at isoelectric point the molecule exists as a neutral molecule.
So for a molecule to not migrate in electric field it must be neutral. At isoelectric point the net charge on the amino acid molecule is zero.
The amino acid at isoelectric point exists as a zwitter ion which has a equal positive and equal negative charge making the net charge of the amino acid to be zero.
since the amino acid molecule has two kinds of functional group present which are an amine group and a carboxylic acid group.
So to calculate the isoelectric point we must know the pKa of these functional groups.
The formula for calculating the isoelectic point (pI) is :
[tex]pI=[pKa1+pKa2]\times 1/2[/tex]
pI=Isoelectric point
pKa1=pKa of carboxy group
pKa2=pKa of amine group
we have the pKa1 and pKa2 given in the question so putting these values in above formula we can calculate the isoelectric point.
pKa1=2.58
pKa2=9.24
pI=1/2×[pKa1+pKa2]
pI=1/2×[2.58+9.24]
pI=5.91
The isoelectric point is 5.91 hence at a pH of 5.91 phenylalanine would exist as a zwitterion and would not migrate in presence of a electric field.
Consider the reaction: HCN (aq) + H2O (l) ⇄ CN- (aq) + H3O + (aq) Which of the following statements will decrease the amount of work the system could perform? (a) Boil off water from the container (increasing the concentration of all species) (b) Add solid NaOH to the reaction (assume no volume change) (c) Selectively remove CNfrom the solution (d) Add water to the reaction vessel (e) Increase the concentration of the HCN
Answer: Option (b) is the correct answer.
Explanation:
The given chemical reaction shows that hydrogen cyanide acid has been added to water which results in the formation of hydronium ion and cyanide ion.
Also, when we add a base like sodium hydroxide (NaOH) to HCN then it will help in accepting a proton ([tex]H^{+}[/tex]) from hydrogen cyanide. As a result, formation of [tex]CN^{-}[/tex] anion will be rapid and easy.
This will make the system not to do any extra work. So, amount of work done by system will decrease.
Thus, we can conclude that out of the given options, add solid NaOH to the reaction (assume no volume change) will decrease the amount of work the system could perform.
Boiling off water from the container will increase the concentration of all species and shift the equilibrium toward the reactants, thus decreasing the potential work the system could perform, aligning with Le Chatelier's Principle. Option A is correct.
Which, action would decrease the amount of work a system, involving the equilibrium reaction of HCN (aq) with water to produce CN⁻ (aq) and H₃O⁺ (aq), could perform. Various actions can shift the equilibrium of this reaction, affecting the system's potential work. According to Le Chatelier's Principle, the system will respond to minimize the effect of any change.
Boiling off water ((a)) from the container will increase the concentration of all species, shifting the equilibrium to the left, thus diminishing the potential work by favoring reactants.Adding solid NaOH ((b)) will increase the hydroxide ion concentration, which consumes H₃O⁺ ions, driving the reaction to the right and potentially increasing work, contradicting the asked condition.Removing CN- ((c)) selectively from the solution will drive the reaction to the right, to replenish CN⁻ ions, potentially increasing the system's capacity to do work, which is against the question's condition.Adding water ((d)) to the reaction vessel will dilute all reactive species, shifting the equilibrium towards CN⁻ and H₃O⁺ formation, which doesn't directly answer the question of decreasing work.Increasing the concentration of HCN ((e)) will push the equilibrium to the right, thus increasing the amount of work the system could perform by forming more products.The action that will decrease the amount of work the system could perform is (a) boiling off water from the container, as it favors the formation of reactants over products.
Hence, A. is the correct option.
You are holding a container of 28.7L of dinitrogen tetroxide. How many grams of gas are inside? Step by step.
Answer:
=71.76 grams
Explanation:
At Room temperature and pressure, 1 mole of an ideal gas occupies a volume of 24 liters.
Therefore, the number of moles occupied by 28.7 L is:
(28.7×1)/24=1.196 moles.
Mass=Number of moles× RMM
RMM of N₂O₂ is 60
mass=1.196 moles× 60 grams/mol
=71.76 grams
Commercial grade HCl solutions are typically 39.0% (by mass) HCl in water. Determine the molality of the HCl, if the solution has a density of 1.20 g/mL
To determine the molality of the commercial grade HCl solution, we calculate the mass of HCl and water in the solution, and the moles of HCl. We plug these values into the molality formula, which gives us a molality of approximately 17.5 m (mol/kg).
Explanation:To determine the molality of the HCl solution, we first need to understand that molality (m) is defined as the number of moles of solute per kg of solvent. Therefore we need to know the amount of HCl in moles and the weight of water in kg.
Commercial grade HCl solutions are 39.0% HCl by mass. This means that in 1000g of solution there is 390g of HCl. Since the molar mass of HCl is roughly 36.5 g/mol, this means we have about 10.68 moles of HCl in 1000g solution.
The HCl solution has a density of 1.20 g/mL. This means that 1000g (1kg) solution occupy 1000/1.20 = 833.3 mL. Since the solution is made up of HCl and water, we can say that the mass of water = total mass of solution - mass of HCl = 1000g - 390g = 610g (or 0.61 kg).
Then we find the molality of HCl solution by using the formula: Molality (m) = moles of HCl/kg of water = 10.68 moles/0.61 kg = 17.5 m (mol/kg).
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Copy of A buffer consists of 0.33 M Na2HPO4 and 0.28 M Na3PO4. Given that the K values for H3PO4 are, Ka1 = 7.2 x 10-3, Ka2 = 6.3 x 10-8, and Ka3 = 4.2 x 10-13, calculate the pH for this buffer
Answer:
Use Ka3
henderson-hasselbach equation: pH = pKa + log [base]/[acid]
pKa3 = - log Ka3 = - log 4.2 x 10^-13 = 12.38
therefore: pH = 12.38 + log (0.28/0.33) = 12.30
the pH is 12.30
Explanation:
phosphoric acid is a polyprotic acid meaning it donates more than one proton
weak Acid ↔ conjugate Base
H3PO4 ↔ H2PO4^- corresponding to Ka1
H2PO4^- ↔ HPO4^2- corresponding to Ka2
HPO4^2- ↔ PO4^3- corresponding to Ka3
A buffer consist of a weak acid and its conjugate base, the given buffer has the combination of HPO4^2- and PO4^3- thud we used Ka3
knowing that we used henderson-hasselbach equation to get the pH which is 12.30
For a stationary fluid, the pressure will vary in the x, y and z directions.T/F
Answer:
True
Explanation:
For a stationary fluid, the pressure will vary in the x, y and z directions.
This statement is true.
The element chlorine (Cl) has two isotopes: chlorine‑35 and chlorine‑37. Approximately 75.5% of chlorine atoms have 18 neutrons and 17 protons, and the other 24.5% have 20 neutrons and 17 protons. Using the isotopic composition provided, calculate the average atomic mass of chlorine. Round your answer to the tenths place.
Answer:
35.5 amu is the average atomic mass of chlorine
Explanation:
Fractional abundance of chlorine‑35 = 75.5%
Fractional abundance of chlorine‑37 = 24.5%
Average atomic mass is equal to summation of products of all isotopes masses into their fractional abundance.
Average atomic mass = Σ(Mass of an isotope × fractional abundance)
Average atomic of chlorine :
[tex]35 amu\times 0.755+37 amu\times 0.245=35.49 amu\approx 35.5 amu[/tex]
35.5 amu is the average atomic mass of chlorine
A 50% antifreeze solution is to be mixed with a 90% antifreeze solution to get 200 liters of a 80% solution. How many liters of the 50% solution and how many liters of the 90% solution will be used?
Answer:
50 ltr 150 ltr
Explanation:
this problem can be solved by the mixture and allegation concept which can be clearly understand from bellow figure in which the concentration of solution 1 is 50% and concentration of solution 2 is 90% before mixing after mixing with help bellow concept the ratio of concentration become 10:30
ratio of solution 1 and solution 2 =10:30
=1:3
total mixture is 200 liters
part of solution 1=[tex]\frac{1}{4}[/tex] ×200
=50 liters
part of solution 2=[tex]\frac{3}{4}[/tex] ×200
=150 liters
Which of the following four statements is/are accurate with respect to glycolysis?
A. Glycolysis involves the conversion of monosaccharides into glucose.
B. Glycolysis involves the breakdown of glucose into pyruvate.
C. Glycolysis involves the conversion of pyruvate into glucose to glyceraldehyde-3-phosphate.
D. Glycolysis involves the conversion of pyruvate into acetyl CoA.
Answer:
B. Glycolysis involves the breakdown of glucose into pyruvate.
Explanation:
Glycolysis is the biological process whereby one glucose molecule is broken down to two pyruvate molecules. There are a lot of enzymatic processes that are involved in this. It is one of the most important reactions in the world because it allows living cells to harness chemical energy from organic molecules
As part of her nuclear stress test, Simone starts to walk on the treadmill. When she reaches the maximum level, Paul injects a radioactive dye with201Tl with an activity of 72 MBq . The radiation emitted from areas of her heart is detected by a scanner and produces images of her heart muscle. After Simone rests for 3 h, Paul injects more dye with 201Tl and she is placed under the scanner again. A second set of images of her heart muscle at rest is taken. When Simone's doctor reviews her scans, he assures her that she had normal blood flow to her heart muscle, both at rest and under stress.
Part A
If the half-life of 201Tl is 3.0 days, what is its activity, in megabecquerels after 3.0 days?
Express the activity to two significant figures and include the appropriate units.
Part B
What is its activity in megabecquerels after 6.0 days?
Express the activity to two significant figures and include the appropriate units.
Answer:
Part A: 36 MBq; Part B: 18 MBq
Explanation:
The half-life is the time it takes for half the substance to disappear.
The activity decreases by half every half-life
A =Ao(½)^n, where n is the number of half-lives.
Part A
3.0 da = 1 half-life
A = Ao(½) = ½ × 72 MBq = 36 MBq
Part B
6.0 da = 2 half-lives
A = Ao(½)^2 = ¼ × 72 MBq = 18 MBq
Final answer:
Part A: The activity of 201Tl after 3.0 days is 36 MBq. Part B: The activity of 201Tl after 6.0 days is 18 MBq.
Explanation:
Part A: To calculate the activity of 201Tl after 3.0 days, we need to determine the number of half-lives that have passed. Since the half-life of 201Tl is 3.0 days, dividing 3.0 days by the half-life gives us 1 half-life. Each half-life reduces the activity of the isotope by half, so after 1 half-life, the activity would be half of the initial activity. Therefore, the activity after 3.0 days would be 72 MBq / 2 = 36 MBq.
Part B: To calculate the activity of 201Tl after 6.0 days, we can use the same method as in Part A. Since the half-life is 3.0 days, 6.0 days is equivalent to 2 half-lives. Each half-life reduces the activity by half, so after 2 half-lives, the activity would be 72 MBq / 2 / 2 = 18 MBq.
Analysis of an unknown sample indicated the sample contained 0.140 grams of N and 0.320 grams of O. The molecular mass of the compound was determined to be 92.02 amu. What is the molecular formula of the compound?
Answer:
the molecular formula of the compound is N2O4
Explanation:
- Find the empirical formula
mole of N present = mass of N divided by molar mass of N = 0.140/14 = 0.01 mole
mole of O present = mass of O divided by molar mass of O = 0.320/16 = 0.02 mole
Divide both by the smallest number of mole to determine the coefficient of each, the smallest number of mole is 0.01 thus:
quantity of N = 0.01/0.01 = 1
quantity of O = 0.02/0.01= 2
thus the empirical formula = NO2
- Now determine the molecular formula by finding the ratio of molecular formula and empirical formula
Molar mass of molecular formula = 92.02 amu = 92.02 g/mole
Molar mass of empirical formula NO2 = (14 + (16 x 2)) = 46 g/mole
the x factor = 92.02/46 = 2
Molecular formula = 2 x NO2 = N2O4
Final answer:
The molecular formula of the compound with 0.140 grams of N and 0.320 grams of O and a molecular mass of 92.02 amu is N₂O₄.
Explanation:
To determine the molecular formula of the compound, we first need to calculate the number of moles of nitrogen (N) and oxygen (O) in the sample. Using the atomic masses of N (14.0 amu) and O (16.0 amu), we divide the mass of each element in the sample by its respective atomic mass to find the moles:
Moles of O = 0.320 g / 16.0 g/mol = 0.02 mol
Next, we determine the simplest whole number ratio of the moles of each element, which gives us the empirical formula. In this case, the ratio of N to O is 1:2, so the empirical formula is NO₂.
Since we know the molecular mass of the compound is 92.02 amu, we can compare it to the mass of the empirical formula to find the molecular formula. The mass of NO₂ is 14.0 + (2 x 16.0) = 46.0 amu. Thus, the molecular mass of our compound (92.02 amu) is approximately twice that of NO₂, so the molecular formula is N₂O₄.
Which group 1 element exhibits slightly different chemistry from the others?
Li
K
Rb
Na
Cs
Answer:
Li
Explanation:
The group 1 elements are the alkali elements, and they are the ones that have one valence electron int heir outer shell, along wiht them Hydrogen is also an element that has one valence electron but is not considered group 1 because it is not a solid in temperature room, but a gas and has different properties, from the options the most different from the others is litium, because it can react with the same elements than the other elements in group one, plus nitrogen in a very violent way.
The group 1 element that exhibits slightly different chemistry from the others is: Lithium (Li).
The chemical elements found in group 1 of the periodic table are highly electro positive metals and have a single (1) valence electron in its outermost shell.
Some examples of these chemical elements (alkali metals) are;
Sodium (Na).Hydrogen (H).Potassium (K).Lithium (Li).Li is the symbol for the chemical element referred to as Lithium.
Lithium (Li) is an alkali metal and as such it is found in group 1 of the periodic table with a single (1) valence electron in its outermost shell.
Also, the electronic configuration of Lithium (Li) is written as;
1s²2s¹As a result of the small size of Lithium (Li), it is exhibits slightly different chemistry from the others such as:
High melting and boiling point.Harder than other alkali metals.Less reactive with water and oxygen.High ionization energy.Read more: https://brainly.com/question/20346057
Ritonavir oral solution contains, in addition to ritonavir, 43.2% alcohol and 26.57% propylene glycol. Calculate the quantity of alcohol in a 240 mL bottle of the oral solution. (Round to two decimal places.)
Hey there!:
240 mL bottle of the oral solution , so 43.2% alcohol
Therefore:
240 mL * 43.2 / 100
= 2.4 * 43.2
=> 103.68 mL of alchohol
Hope this helps!
Ritonavir oral solution contains, in addition to ritonavir, 43.2% alcohol and 26.57% propylene glycol. The quantity of alcohol in a 240 ml bottle of the oral solution is 103.68 ml of alcohol.
What is alcohol?Alcohol is a drink that is made up of rotten fruits and vegetables, It is made by fermentation. They cause unconsciousness to the brain and the body. It creates hallucinations.
Ritonavir is taken with meals twice a day. Ask your doctor or pharmacist to explain any instructions on your prescription label that you are unsure about following.
Given the 43.2% alcohol and 26.57% propylene glycol. The bottle is of 240 ml.
240 mL x 43.2 / 100
2.4 x 43.2 = 103.68 ml of alcohol.
Thus, the quantity of alcohol in a 240 ml bottle is 103.68 ml of alcohol.
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Determine the calcium carbonate equivalent (CCE) of the following compounds: (amount that has the same neutralizing value as 100 g pure CaCO3) (a) MgO (b) Mg(OH)2 (c) and CaMg(CO3)2.
Answer:
a) 40 g
b) 58 g
c) 184 g
Explanation:
The calcium carbonate equivalent of any compound is calculated by calculating it molar mass.
The molar mass of the compound is the calcium carbonate equivalent as it corresponds to 1 mole of the compound and equals to one mole of calcium carbonate then.
a) MgO
Atomic mass of Mg = 24
Atomic mass of O = 16
Molar mass = CCE = 24+16 = 40 g
b) Mg(OH)₂
Atomic mass of Mg = 24
Atomic mass of O = 16
Atomic mass of H = 1
Molar mass of Mg(OH)₂= CCE = 24 + (2X16)+2(X1) = 58g
c) CaMg(CO₃)2
Atomic mass of Mg = 24
Atomic mass of O = 16
Atomic mass of C = 12
Atomic mass of Ca = 40
Molar mass = CCE = 40 + 24 + (2X12) + (6X16) = 184 g.
A saturated solution of magnesium fluoride, MgF2, was prepared by dissolving solid MgF2 in water. The concentration of Mg2+ ion in the solution was found to be 1.18×10−3 M . Calculate Ksp for MgF2.
Hey there!:
Ksp =(Mg⁺²)(2F⁻)²
Ksp = (1.18*10⁻³)(2*1.18*10⁻³)²
Ksp = 6.57*10⁻⁹
Hope that helps!
The standard cell potential (E°) of a voltaic cell constructed using the cell reaction below is 0.76 V: Zn (s) + 2H+ (aq) → Zn2+ (aq) + H2 (g) With PH2 = 1.0 atm and [Zn2+] = 1.0 M, the cell potential is 0.53 V. The concentration of H+ in the cathode compartment is ________ M.
Using the Nernst equation with given concentrations and standard cell potential, the [tex]\(H^+\)[/tex] concentration in the cathode compartment is approximately 7916 M.
To solve this problem, we can use the Nernst equation, which relates the cell potential[tex](\(E_{\text{cell}}\))[/tex] to the standard cell potential[tex](\(E^\circ\)),[/tex] the reaction quotient ((Q)), the temperature ((T)), and the gas constant ((R)).
The Nernst equation is given by:
[tex]\[E_{\text{cell}} = E^\circ - \frac{0.0592}{n} \log(Q)\][/tex]
Where:
[tex]- \(E_{\text{cell}}\)[/tex]= cell potential under non-standard conditions
[tex]- \(E^\circ\)[/tex]= standard cell potential
- (n) = number of moles of electrons transferred in the balanced redox reaction
- (Q) = reaction quotient
- (R) = gas constant[tex](\(8.314 \, \text{J/mol} \cdot \text{K}\))[/tex]
- (T) = temperature in Kelvin
Given:
[tex]- \(E^\circ = 0.76 \, \text{V}\)[/tex]
[tex]- \(E_{\text{cell}} = 0.53 \, \text{V}\)[/tex]
[tex]- \(P_{\text{H}_2} = 1.0 \, \text{atm}\)[/tex]
[tex]- \([\text{Zn}^{2+}] = 1.0 \, \text{M}\)[/tex]
- We know that the number of moles of electrons transferred (\(n\)) is 2 because of the balanced equation.
First, let's find the reaction quotient (\(Q\)) using the given concentrations:
[tex]\[Q = \frac{[\text{Zn}^{2+}][\text{H}_2]}{[\text{H}^+]^2}\][/tex]
Given [tex]\([\text{Zn}^{2+}] = 1.0 \, \text{M}\), \([\text{H}_2] = 1.0 \, \text{atm}\)[/tex], and[tex]\([\text{H}^+]\)[/tex]as the unknown concentration in the cathode compartment, we can substitute these values into the equation.
[tex]\[Q = \frac{(1.0 \, \text{M})(1.0 \, \text{atm})}{[\text{H}^+]^2}\][/tex]
Now, we can use the given cell potentials and the Nernst equation to solve for [tex]\([\text{H}^+]\).[/tex]
[tex]\[0.53 \, \text{V} = 0.76 \, \text{V} - \frac{0.0592}{2} \log\left(\frac{(1.0 \, \text{M})(1.0 \, \text{atm})}{[\text{H}^+]^2}\right)\][/tex]
Let's solve this equation step by step:
[tex]\[0.53 \, \text{V} = 0.76 \, \text{V} - \frac{0.0592}{2} \log\left(\frac{(1.0 \, \text{M})(1.0 \, \text{atm})}{[\text{H}^+]^2}\right)\][/tex]
[tex]\[0.53 \, \text{V} = 0.76 \, \text{V} - \frac{0.0296}{2} \log\left(\frac{(1.0 \, \text{M})(1.0 \, \text{atm})}{[\text{H}^+]^2}\right)\][/tex]
[tex]\[0.53 \, \text{V} = 0.76 \, \text{V} - 0.0148 \log\left(\frac{1.0}{[\text{H}^+]^2}\right)\][/tex]
[tex]\[0.53 \, \text{V} = 0.76 \, \text{V} - 0.0148 \left(\log(1.0) - \log([\text{H}^+]^2)\right)\][/tex]
[tex]\[0.53 \, \text{V} = 0.76 \, \text{V} - 0.0296 \log([\text{H}^+]^2)\][/tex]
Now, let's simplify and solve for [tex]\([\text{H}^+]\):[/tex]
[tex]\[0.53 \, \text{V} = 0.76 \, \text{V} - 0.0296 \log([\text{H}^+]^2)\][/tex]
[tex]\[0.53 \, \text{V} = 0.76 \, \text{V} - 0.0592 \log([\text{H}^+])\][/tex]
Now, let's isolate \(\log([\text{H}^+])\):
[tex]\[0.53 \, \text{V} - 0.76 \, \text{V} = - 0.0592 \log([\text{H}^+])\][/tex]
[tex]\[-0.23 \, \text{V} = - 0.0592 \log([\text{H}^+])\][/tex]
Now, divide by (-0.0592):
[tex]\[\frac{-0.23 \, \text{V}}{-0.0592} = \log([\text{H}^+])\][/tex]
[tex]\[3.89189 \approx \log([\text{H}^+])\][/tex]
Now, we can find[tex]\([\text{H}^+]\)[/tex] by taking the antilog of (3.89189):
[tex]\([\text{H}^+] = 10^{3.89189}\)[/tex]
[tex]\([\text{H}^+] \approx 7915.82 \, \text{M}\)[/tex]
So, the concentration of [tex]\(H^+\)[/tex] in the cathode compartment is approximately[tex]\(7915.82 \, \text{M}\).[/tex]
According to Raoult's law, which statement is FALSE? a) The vapor pressure of a solvent over a solution decreases as its mole fraction increases. b) The vapor pressure of a solvent over a solution is less than that of pure solvent. c) The greater the pressure of a gas over a solution the greater its solubility. d) The solubility of a gas increases as the temperature decreases. e) Ionic solutes dissociate in solution causing an enhancement of all colligative properties.
Answer:
a) The vapor pressure of a solvent over a solution decreases as its mole fraction increases.
According to Raoult's law, the statement which is FALSE is:
A) The vapor pressure of a solvent over a solution decreases as its mole fraction increases.Raoult's law states that if the pressure of the content of a liquid remains constant to the temperature, then it is proportional to the mole fraction of the mixture.
As a result of this, the vapor pressure of a solvent over a solution is less than the pure solvent and the greater the pressure of a gas over a solution, the greater the solubility.
Therefore, the correct answer is option A because it is false.
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What concentration of Ni2+ ion remains in solution after electrolysis of 100. mL of 0.250 M NiSO4 solution when using a current of 2.40 amperes for 30.0 minutes? Assume Ni metal is plated out.
Answer:
the concentration remains = 0.138 M
Explanation:
Let us calculate the charge passed:
The charge passed during the electrolysis = current X time(s)
charge passed = 2.40X30X60= 2160 C
we know that
96485 C = 1 F
Therefore 2160 C = 0.0224 F
according to Faraday's law of electrolysis if we pass one Faraday of charge through an electrolytic solution, 1 gram equivalent of metal will be deposited.
the gram equivalent of nickel deposited = 0.0224
The moles of Nickel deposited = 0.0224/2 = 0.0112 mol
the initial moles of Nickel ions present in solution is
[tex]moles=molarityXvolume(L)=0.250X0.1=0.0250mol[/tex]
The moles of nickel ion gets consumed = 0.0112
So moles of nickel ion left after electrolysis = 0.0138
[tex][Ni^{+2}]=\frac{moles}{volume(L)}\frac{0.0138}{0.1}=0.138M[/tex]
The concentration of Ni2+ ions remaining in the solution can be found by applying Faraday's first law of electrolysis and subtracting the moles of Ni2+ discharged from the initial moles present in the solution.
Explanation:The electrolysis process involves a chemical reaction in which electricity is used to break down a substance into its constituent elements. In this scenario, we are determining the concentration of Ni2+ in the solution after the electrolysis process of NiSO4.
As per Faraday's first law of electrolysis, the mass of any element discharged during electrolysis is directly proportional to the quantity of electricity (current) passed. Thus, using Faraday's constant (96485 C/mol e-) and the given current and time, we can calculate how many moles of Ni2+ are plated out.
The number of moles of Ni2+ discharged would be equal to the number of moles initially present in the 100.0 mL of 0.250 M NiSO4 solution. By subtracting the moles of Ni2+ that were discharged from the total initial moles of Ni2+, we can obtain the number of moles of Ni2+ remaining in the solution. This value can be converted back into molarity by dividing the moles of Ni2+ remaining by the volume of the solution in liters.
In this way, we can find the concentration of Ni2+ ions that remain in the solution after electrolysis.Learn more about Electrolysis here:https://brainly.com/question/33301115
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Based on Coulomb's Law of electrostatic attraction of oppositely-charged species, which of the following chlorine-containing compounds would be predicted to have the greatest (most exothermic) lattice energy? CaCl2 NaCl MgCl2 KCl CCl4
Based on Coulomb's Law of electrostatic attraction of oppositely-charged species, the chlorine-containing compound that is predicted to have the greatest (most exothermic) lattice energy is; MgCl₂
The lattice energy is defined as the energy required to dissociate one mole of an ionic compound to its constituent gaseous ions.
Coulomb's law of electrostatic attraction states that the force of attraction (F) between two oppositely charged particles is directly proportional to the product of the charges of the particles ( q₁ and q ₂ ) and inversely proportional to the square of the distance between the particles.
Thus; F = (q₁ × q ₂)/r²
Now, Lattice energy is inversely proportional to the size of the ions. This implies that as the size of the ions increases, then the lattice energy will decrease. The size of the ions is also called the atomic radii.
Thus, as atomic radii increases, lattice energy will decrease.
From the formula above and definition, we can tell that Lattice energy decreases down a group. Also, lattice energy will increase as the charges increase.
Now, the compounds we are dealing with are;
CaCl₂ - Calcium Chloride
NaCl - Sodium Chloride
MgCl₂ - Magnesium Chloride
KCl - Potassium Chloride
CCl₄ - Carbon tetrachloride
The charges of the metals that form the chlorides above are;
Ca = +2
Na = +1
Mg = +2
K = +1
The highest charges are Ca and Mg.
Now, they both belong to group two of the periodic table with Ca below Mg in the periodic table and as such Mg will have a greater Lattice energy.
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From the list of chlorine-containing compounds, MgCl₂ would have the greatest lattice energy. Mg²⁺ has a greater charge than the other cations, resulting in stronger electrostatic attraction and most exothermic energy. CCl₄ is a covalently bonded molecule not an ionic compound, so does not possess lattice energy.
Explanation:Based on Coulomb's Law of electrostatic attraction, the strongest (most exothermic) lattice energy would be present in the compound with the highest positively charged ion and smallest ionic radii, which results in a significant increase in lattice energy. This comes from the idea that lattice energy increases with higher charges and decreases with larger ionic radii.
In this case, it would be MgCl₂, because Mg⁺ has a greater charge than the other cations listed (Ca²⁺, Na⁺, K⁺). Greater charge results in stronger electrostatic attraction and hence the most exothermic lattice energy.
It's worth noting that CCl₄ is a covalently bonded molecule and not an ionic compound, thus it does not have lattice energy in the context of ionic compounds.
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Using the Antoine equation, prepare two plots of Psat versus T for Benzene over the range of temperature for which the parameters are valid. One plot should present Psat on a linear scale and the other should present Psat on a log scale. Make these plots using appropriate software of your choice (Excel, Matlab, etc.), not by hand. In both cases, T should be on the horizontal axis (independent variable) and Psat nwww ww www. should be on the vertical axis. Please show all work and use Excel or Matlab. The parameters for Benzene are given below. A 13.7819 B 2726.81 C 217.572 Temp Range( C) 6-104 AHn (kJ/mol) 30.72 Latent heat of Vaporization at normal boiling point a0000nd Normal boiling point Tr (oC) 80.0
Answer:
Here's what I get.
Explanation:
The Antoine equation is
[tex]\log p = A - \dfrac{B }{C+T}[/tex]
A = 13.7819
B = 2726.81
C = 217.572
I did the calculations and the plots in Excel.
Figure 1 shows the calculations, Figure 2 is the linear plot, and Figure 3 is the log plot.
Cavendish prepared hydrogen in 1766 by the novel method of passing steam through a red-hot gun barrel: 4H2 O(g) + 3Fe(s) ⟶ Fe3 O4 (s) + 4H2 (g) (a) Outline the steps necessary to answer the following question: What volume of H2 at a pressure of 745 torr and a temperature of 20 °C can be prepared from the reaction of 15.O g of H2O? (b) Answer the question.
Answer : The volume of [tex]H_2[/tex] will be, 0.2690 L
Solution :
(a) Steps involved for this problem are :
First we have to calculate the moles of [tex]H_2O[/tex].
Now we have to calculate the volume of hydrogen gas by using the ideal gas equation.
(b) First we have to calculate the moles of [tex]H_2O[/tex].
[tex]\text{Moles of }H_2O=\frac{\text{Mass of }H_2O}{\text{Molar mass of }H_2O}=\frac{15g}{18g/mole}=0.833moles[/tex]
The balanced chemical reaction is,
[tex]4H_2O(g)+3Fe(s)\rightarrow Fe_3O_4(s)+4H_2(g)[/tex]
From the balanced chemical reaction, we conclude that the moles of hydrogen is equal to the moles of water.
Thus, the moles of hydrogen gas = 0.833 mole
Now we have to calculate the volume of hydrogen gas.
Using ideal gas equation,
[tex]PV=nRT[/tex]
where,
n = number of moles of gas = 0.833 mole
P = pressure of the gas = [tex]745torr=\frac{745}{760}=0.98atm[/tex]
conversion used : (1 atm = 760 torr)
T = temperature of the gas = [tex]20^oC=273+20=293K[/tex]
R = gas constant = 0.0821 Latm/moleK
V = volume of gas = ?
Now put all the given values in the above equation, we get :
[tex](0.98atm)\times V=(0.833mole)\times (0.0821Latm/moleK)\times (293K)[/tex]
[tex]V=0.2690L[/tex]
Therefore, the volume of [tex]H_2[/tex] will be, 0.2690 L
Answer:
V = 20.4 L
Explanation:
Step 1: Write the balanced equation
4 H₂O(g) + 3 Fe(s) ⟶ Fe₃O₄(s) + 4 H₂(g)
Step 2: Find the moles of H₂
We can establish the following relations.
The molar mass of H₂O is 18.0 g/molThe molar ratio of H₂O to H₂ is 4:4.The moles of H₂ obtained from 15.0 g of H₂O is:
[tex]15.0gH_{2}O.\frac{1molH_{2}O}{18.0gH_{2}O} .\frac{4molH_{2}}{4molH_{2}O} =0.833molH_{2}[/tex]
Step 3: Find the volume of H₂
We will use the ideal gas equation.
P = 745 torr × (1 atm/760 torr) = 0.980 atm
V = ?
n = 0.833 mol
R = 0.08206 atm.L/mol.K
T = 20°C + 273 = 293 K
P × V = n × R × T
0.980 atm × V = 0.833 mol × (0.08206 atm.L/mol.K) × 293 K
V = 20.4 L
Glucose, C6H12O6, is used as an energy source by the human body. The overall reaction in the body is described by the equation C6H12O6(aq)+6O2(g)⟶6CO2(g)+6H2O(l) Calculate the number of grams of oxygen required to convert 48.0 g of glucose to CO2 and H2O. mass of O2: g Calculate the number of grams of CO2 produced. mass of CO2: g
Answer:
70.3824 grams of carbondioxide is produced.
Explanation:
[tex]C_6H_{12}O_6(aq)+6O_2(g)⟶6CO_2(g)+6H_2O(l)[/tex]
Moles of glucose = [tex]\frac{48.0 g}{180 g/mol}=0.2666 mol[/tex]
According to reaction 1 mole of glucose reacts with 6 mole of oxygen gas.
Then 0.2666 mol of gluocse will reactwith :
[tex]\frac{6}{1}\times 0.2666 mol =1.5996 mol[/tex] of oxygen
Mass of 1.5996 moles of oxygen gas:
[tex]1.5996 mol\times 32 g/mol = 51.1872 g[/tex]
51.1872 grams of oxygen are required to convert 48.0 grams of glucose to [tex]CO_2[/tex] and [tex]H_2O[/tex].
According to reaction, 1 mol, of glucose gives 6 moles of carbon dioxide.
Then 0.2666 moles of glucose will give:
[tex]\frac{6}{1}\times 0.2666 mol =1.5996 mol[/tex] of carbon dioxide
Mass of 1.5996 moles of carbon dioxide gas:
[tex]1.5996 mol\times 44 g/mol = 70.3824 g[/tex]
70.3824 grams of carbondioxide is produced.
Glucose reacts with 51.2 g of oxygen to produce 70.4 g of carbon dioxide and water.
Let's consider the overall equation for the combustion of glucose in the human body.
C₆H₁₂O₆(aq) + 6 O₂(g) ⟶ 6 CO₂(g) + 6 H₂O(l)
We can calculate the grams of oxygen required to react with 48.0 g of glucose considering the following relations.
The molar mass of glucose is 180.16 g/mol.The molar ratio of glucose to oxygen is 1:6.The molar mass of oxygen is 32.00 g/mol.[tex]48.0gGlucose \times \frac{1molGlucose}{180.16gGlucose} \times \frac{6molO_2}{1molGlucose} \times \frac{32.00 gO_2}{1 molO_2} = 51.2 gO_2[/tex]
We can calculate the grams of carbon dioxide produced from 48.0 g of glucose considering the following relations.
The molar mass of glucose is 180.16 g/mol.The molar ratio of glucose to carbon dioxide is 1:6.The molar mass of carbon dioxide is 44.01 g/mol.[tex]48.0gGlucose \times \frac{1molGlucose}{180.16gGlucose} \times \frac{6molCO_2}{1molGlucose} \times \frac{44.01 gCO_2}{1 molCO_2} = 70.4 gCO_2[/tex]
Glucose reacts with 51.2 g of oxygen to produce 70.4 g of carbon dioxide and water.
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Enter the balanced complete ionic equation for HCl(aq)+K2CO3(aq)→H2O(l)+CO2(g)+KCl(aq). Express your answer as a chemical equation. Identify all of the phases in your answer.
The complete net ionic equation for the reaction between HCl(aq) and K₂CO₃(aq) is:
2H⁺(aq) + CO₃²¯(aq) —> H₂O(l) + CO₂(g)We'll begin by writing the dissociation equation for HCl and K₂CO₃. This is illustrated below:
HCl(aq) —> H⁺(aq) + Cl¯(aq)
K₂CO₃(aq) —> 2K⁺(aq) + CO₃²¯(aq)
In solution, the reaction will proceed as follow:
HCl(aq) + K₂CO₃(aq) —>
2H⁺(aq) + 2Cl¯(aq) + 2K⁺(aq) + CO₃²¯(aq) —> H₂O(l) + CO₂(g) + 2Cl¯(aq) + 2K⁺(aq)
Cancel out the spectator ions (i.e Cl¯ and K⁺) to obtain the net ionic equation.
2H⁺(aq) + CO₃²¯(aq) —> H₂O(l) + CO₂(g)Learn more: https://brainly.com/question/24900805
A sample of potassium phosphate octahydrate (K3PO4•8H2O) is heated until 7.93 grams of water are released. How many grams did the original hydrate weigh?
The original sample of potassium phosphate octahydrate had a mass of 19.6 grams. When it was heated, it released 7.93 grams of water.
Further Explanation:
For every mole of the compound potassium phosphate octahydrate, there are 8 moles of water of hydration which can be removed from the crystal by heating without altering the chemical composition of the substance.
To determine how much original sample was used, the amount of water released upon heating may be used as well as the mole ratio of the water of hydration with the compound itself following the steps below:
Convert mass of water released to moles.Use the mole ratio of water of hydration to the compound (8 mol water for every mol of potassium phosphate octahydrate) to get the moles of original sample.Convert the moles of original sample to grams.STEP 1: Convert 7.93 g water to moles.
[tex]moles \ of\ H_{2}O \ = 7.93 \ g \ H_{2}O \ (\frac{1 \ mol \ H_{2}O}{18.00 \ g \ H_{2}O})\\\boxed {moles \ of \ H_{2}O \ = 0.4406 \ mol}[/tex]
STEP 2: Calculate the moles of original sample using the mole ratio: 1 mol K3PO4 8H2O : 8 mol H2O.
[tex]moles \ of \ K_{3}PO_{4}\ 8H_{2}O \ = 0.4406 \ mol \ H_{2}O \ (\frac{1 \ mol \ K_{3}PO_{4}\ 8H_{2}O \ }{8 \ mol \ H_{2}O})\\\\\boxed {moles \ of \ K_{3}PO_{4}\ 8H_{2}O \ = 0.0551 \ mol}[/tex]
STEP 3: Convert the moles of original sample to mass.
[tex]mass \ of \ K_{3}PO_{4}\ 8H_{2}O = 0.0551 \ mol \ K_{3}PO_{4}\ 8H_{2}O \ (\frac{356.3885 \ g}{1 \ mol\ K_{3}PO_{4}\ 8H_{2}O})\\ mass \ of \ K_{3}PO_{4}\ 8H_{2}O \ = 19.637 \ g[/tex]
Following the significant figures of the given, the final answer should be:
[tex]\boxed {mass \ of \ K_{3}PO_{4}\ 8H_{2}O = 19.6 \ g}[/tex]
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Answer: The mass of original hydrate is 19.63 grams.
Explanation:
We are given:
Mass of water released = 7.93 grams
We are given a chemical compound known as potassium phosphate octahydrate having chemical formula of [tex]K_3PO_4.8H_2O[/tex]
Mass of [tex]K_3PO_4.8H_2O[/tex] = 356.4 grams
Mass of 8 water of crystallization = (8 × 18) = 144 grams
By applying unitary method, we get:
144 grams of water is released when 356.4 grams of salt is heated.
So, 7.93 grams of water will be released when = [tex]\frac{356.4g}{144g}\times 7.93g=19.63g[/tex] of salt is heated.
Hence, the mass of original hydrate is 19.63 grams.
You decided to prepare a phosphate buffer from solid sodium dihydrogen phosphate (NaH2PO4) and disodium hydrogen phosphate (Na2HPO4)and you need 1L of the buffer at pH 7.00 with a total phosphate concentration(NaH2PO4+ Na2HPO4) of 0.100 M. Hint:Phosphoric acid (H3PO4), a triprotic acid, has 3 pKa values: 2.14, 6.86, and12.4. Determine the weight in grams of sodium dihydrogen phosphate (NaH2PO4;138g/mol) and disodium hydrogen phosphate (Na2HPO4;142g/mol) needed to prepare this buffer.
Answer:
For disodium hydrogen phosphate:
5.32g Na2HPO4
For sodium dihydrogen phosphate:
7.65g Na2HPO4
Explanation:
First, you have to put all the data from the problem that you going to use:
-NaH2PO4 (weak acid)
-Na2HPO4 (a weak base)
-Volume = 1L
-Buffer pH = 7.00
-Concentration of [NaH2PO4 + Na2HPO4] = 0.100 M
What we need to find the pKa of the weak acid, in this case NaH2PO4, for that you need to find the Ka (acid constant) of NaH2PO4, and for this we use the pKa of the phosphoric acid as follow:
H3PO4 = H2PO4 + H+ pKa1 = 2.14
H2PO4 = HPO4 + H+ pKa2 = 6.86
HPO4 = PO4 + H+ pKa3 = 12.4
So, for the preparation of buffer, you need to use the pKa that is near to the value of the pH that you want, so the choice will be:
pKa2= 6.86
Now we going to use the Henderson Hasselbalch equation for the pH of a buffer solution:
pH = pKa2 + log [(NaH2PO4)/(Na2HPO4)]
The solution of the problem is attached to this answer.
To prepare 1L of a phosphate buffer at pH 7 with a total concentration of 0.1 M, one would need to mix equal molar amounts of NaH2PO4 and Na2HPO4, resulting in 6.9 g of NaH2PO4 and 7.1 g of Na2HPO4.
The relevant pKa value for the pH target of 7.00 is 6.86, which is close to the second pKa of phosphoric acid. The Henderson-Hasselbalch equation is as follows:
pH = pKa + log([A-]/[HA]), where:
At pH 7.00, the ratio [A-]/[HA] is 1:1 because pH = pKa. Thus, we need equal molar amounts of NaH2PO4 and Na2HPO4. Since the total molarity is 0.1 M, this means we need 0.05 M of each component.
Therefore, you would need to weigh 6.9 g of NaH2PO4 and 7.1 g of Na2HPO4 to prepare your buffer.