Explanation:
a) IE1 of Germenium. ionization energy decreases down the group.
b) Here IE2 of Ga is higher because second valence electron of gallium is in s orbital whereas second valence electron of germenium is in p orbital.
c) IE3 of Ge is higher as this follows the trend.
d)IE4 of Ga is higher because the 4th valence electron of Ga is a core electron and the for Ge it is in s orbital and it takes higher energy to break a core electron than the orbital ones.
When comparing Ga and Ge, IE₁ and IE₂ of Ga are lower due to being earlier in the periodic table, but IE₄ of Ga is expected to be higher as it involves removing an electron from a full orbital. IE₃ for both may be similar due to both elements having three valence electrons. The correct answer is d) IE₄ of Ga or IE₄ of Ge.
The subject in question involves comparing first and successive ionization energies (IE) for elements in the periodic table, particularly gallium (Ga) and germanium (Ge). Ionization energy refers to the energy required to remove an electron from a gaseous atom or ion.
As general rules, (a) the first IE tends to increase across a period as the nuclear charge increases, and (b) IE increases for successive ionizations as the atom or ion becomes progressively more positively charged.
(a) IE₁ of Ga would be lower than IE₁ of Ge because Ga is to the left of Ge in the periodic table, and thus Ge has a higher nuclear charge and a stronger attraction to its valence electron.
(b) IE₂ of Ga would also be lower than IE₂ of Ge because the additional electron from Ga would come from the same valence shell as the first electron, while Ge's second ionization would remove an electron that experiences a stronger nuclear charge.
(c) As both Ga and Ge have three valence electrons, the IE₃ of Ga and Ge will both involve removing an electron from a similarly charged ion, but because Ga has a higher energy 4p subshell, its IE₃ might be slightly lower than Ge's, which is removing an electron from the 4s subshell (which after removing two electrons is now full and lower in energy).
(d) At the IE₄ level, we see a large jump in ionization energy for both elements as electrons are being removed from an energy level closer to the nucleus; however, IE₄ of Ga will generally be higher than IE₄ of Ge due to of an electron from a full orbital in Ga, which has higher energy compared to the removal from a half-filled orbital in Ge.
These comparisons are based on the understanding that ionization energies increase both with increasing positive charge and with the removal of electrons closer to the nucleus.
The density of mercury is 13.6 g/cm3 . What volume (in quarts) is occupied by 100. g of Hg? (1 L = 1.06 qt)
Answer:
0.00077 qt
Explanation:
Density -
Density of a substance is given by the mass of the substance divided by the volume of the substance .
Hence , d = m / V
V = volume
m = mass ,
d = density ,
From the question ,
The mass mercury = 100 g
Density of mercury = 13.6 g/cm³ .
Hence , by using the above formula ,and putting the corresponding values , the volume of mercury is calculated as -
d = m / V
13.6 g/cm³ = 100 g / V
V = 7.35 cm³
1 cm³ = 0.001 L
V = 7.35 * 0.001 L = 0.0073 L
Since ,
1 L = 1.06 qt
V = 0.0073* 1.06 qt = 0.0077 qt
A buffer solution is made by mixing a weak acid with its conjugate base. If the ratio of conjugate base to acid is 4, and the pH of the buffer is 7.2, what is the pKa of the weak acid? Round the answer to one decimal place.
Answer:
6.6 is the [tex]pK_a[/tex] of the weak acid.
Explanation:
To calculate the pH of acidic buffer, we use the equation given by Henderson Hasselbalch:
[tex]pH=pK_a+\log(\frac{[salt]}{[acid]})[/tex]
We are given:
[tex]pK_a[/tex] = negative logarithm of acid dissociation constant =?
The ratio of conjugate base to acid is = [tex]\frac{[salt]}{acid}=4[/tex]
pH = 7.2
Putting values in above equation, we get:
[tex]7.2=pK_a+\log(4)[/tex]
[tex]pK_a=7.2-\log(4)=6.598\approx 6.6[/tex]
6.6 is the [tex]pK_a[/tex] of the weak acid.
A solution of chloroform (CHCl3) and acetone((CH3)2CO) exhibits a negative deviation from Raoult's law. This result implies that: W. chloroform-chloroform and acetone-acetone interactions are stronger than chloroform-acetone interactions. X. chloroform-chloroformand acetone-acetone interactions are weaker than chloroform-acetone interactions. Y. acetone-acetone interactions are stronger than chloroform-chloroform interactions. Z. acetone-acetone interactions are weaker than chloroform-chloroform interactions.
Chloroform-Chloroform and Acetone-Acetone interactions are weaker than chloroform-acetone interactions.
Explanation:
Raoult's law states, In a solution, vapor pressure is equal to the sum of the vapor pressures of individual component and this implies when it is multiplied by the mole fraction of that component in the solution. Raoult's Law is expressed by the formula:
P solution = Χ solvent [tex]\times[/tex] P0 solvent.
Raoult's law assumes that the components in the mixture are ideal, it assumes that the Chloroform-Chloroform interactions, Chloroform-Acetone, and Acetone-Acetone are the same. A negative deviation is shown by the measurement of observed vapor pressure which is less than that of calculated. Hence the interaction of Chloroform-Chloroform is weaker than the Chloroform-Acetone interactions.
A solution of chloroform and acetone exhibits a negative deviation from Raoult's law, which means that the interactions between chloroform and acetone are stronger than those of the individual components
Explanation:A solution of chloroform (CHCl3) and acetone((CH3)2CO) exhibits a negative deviation from Raoult's law. This means the solution has a lower vapor pressure than predicted by Raoult's Law due to the strength of the new intermolecular interactions being stronger than what was present in the pure components. The correct answer is X. chloroform-chloroform and acetone-acetone interactions are weaker than chloroform-acetone interactions. This is because in a solution exhibiting negative deviation, the interactions between the different molecules in the mixture are stronger than those of the pure components. This stronger interaction between chloroform and acetone molecules reduces the overall vapor pressure, leading to a negative deviation from Raoult's Law.
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You need to prepare a solution with a specific concentration of Na+Na+ ions; however, someone used the end of the stock solution of NaClNaCl, and there isn’t any NaClNaCl to be found in the lab. You do, however, have some Na2SO4Na2SO4. Can you substitute the same number of grams of Na2SO4Na2SO4 for the NaClNaCl in a solution? Why or why not?
Explanation:
Ionic equation
NaCl(aq) --> Na+(aq) + Cl-(aq)
Na2SO4(aq) --> 2Na+(aq) + SO4^2-(aq)
In NaCl solution, 1 mole of Na+ is dissociated in 1 liter of solution while in Na2SO4, 2 moles of Na+ is dissociated in 1 liter of solution.
Molecular weight of NA2SO4 = (23*2) + 32 + (16*4)
= 142 g/mol
Molecular weight of NaCl = 23 + 35.5
= 58.5 g/mol
Masses
% Mass of NA+ in Na2SO4 = mass of Na+/total mass of Na2SO4 * 100
= 46/142 * 100
= 32.4%
% Mass of NA+ in NaCl = mass of Na+/total mass of NaCl * 100
= 23/58.5 * 100
= 39.3%
Therefore, the % mass of Na+ in NaCl and Na2SO4 are different so it cannot be used.
You cannot substitute Na2SO4 directly for NaCl based on mass since they have different molar masses. The same mass of Na2SO4 will provide more Na+ ions than NaCl, leading to a change in the Na+ ion concentration.
Explanation:No, you cannot substitute the same number of grams of Na2SO4 for the NaCl in a solution. This is because NaCl and Na2SO4 have different molar masses and therefore different numbers of moles per gram. The concentration of a solution is determined by the number of moles of solute per unit volume of solvent, not the mass. Hence, using the same mass of a different compound would alter the concentration of Na+ ions in the solution.
For instance, if one mole of NaCl gives us one mole of Na+, one mole of Na2SO4 will provide two moles of Na+. In other words, the same mass of Na2SO4 contains more Na+ ions than the same mass of NaCl. So using the same mass of Na2SO4 in place of NaCl will result in a solution with a higher Na+ ion concentration.
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According to the Bohr model of the atom, when an electron goes from a higher-energy orbit to a lower-energy orbit, it ________ electromagnetic energy with an energy that is equal to the ________ between the two orbits.
Answer:
emits (radiates) , energy difference
Explanation:
According to the Bohr theory, when an electron jumps from higher orbital to the lower orbital, it radiates energy which is equal to the energy difference between the orbitals.
Mathematically, it can be shown as:-
The expression for Bohr energy is shown below as:-
[tex]E_n=-2.179\times 10^{-18}\times \frac{1}{n^2}\ Joules[/tex]
For transitions:
[tex]Energy\ Difference,\ \Delta E= E_f-E_i =-2.179\times 10^{-18}(\frac{1}{n_f^2}-\frac{1}{n_i^2})\ J=2.179\times 10^{-18}(\frac{1}{n_i^2} - \dfrac{1}{n_f^2})\ J[/tex]
[tex]\Delta E=2.179\times 10^{-18}(\frac{1}{n_i^2} - \dfrac{1}{n_f^2})\ J[/tex]
Also, [tex]\Delta E=\frac {h\times c}{\lambda}[/tex]
Where,
h is Plank's constant having value [tex]6.626\times 10^{-34}\ Js[/tex]
c is the speed of light having value [tex]3\times 10^8\ m/s[/tex]
According to the Bohr model of the atom, when an electron transitions from a higher-energy orbit to a lower-energy orbit, it emits electromagnetic energy equal to the energy difference between the two orbits.
Explanation:The Bohr model of the atom states that when an electron transitions from a higher-energy orbit to a lower-energy orbit, it emits electromagnetic energy with an energy equal to the difference between the two orbits. This energy is released in the form of a photon.
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How many electrons are necessary to produce 1.0 C of negative charge? (e = 1.60 × 10-19 C) Group of answer choices
Answer:
6.25×10¹⁹ e⁻
Explanation:
Let's apply a rule of three:
1 e⁻ has 1.60×10⁻¹⁹ C
There, we can think:
1.60×10⁻¹⁹ C of charge are made by 1 e⁻
If we want to produce 1 C, we would need ( 1 . 1) / 1.60×10⁻¹⁹
= 6.25×10¹⁹ e⁻
A sample of an alloy of aluminum contains 0.0898 mol Al and 0.0381 mol Mg. What are the mass percentages of Al and Mg in the alloy?
Answer:
Al 72.61%
Mg 27.39%
Explanation:
To obtain the mass percentages, we need to place the individual masses over the total mass and multiply by 100%.
If we observe clearly, we can see that the parameters given are the moles. We need to convert the moles to mass.
To do this ,we need to multiply the moles by the atomic masses. The atomic mass of aluminum is 27 while that of magnesium is 24.
Now, the mass of aluminum is thus = 27 * 0.0898 = 2.4246g
The mass of magnesium is 0.0381 * 24 = 0.9144g
We can now calculate the mass percentage.
The total mass is 0.9144 + 2.4246 = 3.339g
% mass of Al = 2.4246/3.339 * 100 = 72.61%
% mass of Mg = 0.9144/3.39 * 100 = 27.39%
Answer:
Mass % Al = 72.3 %
Mass % Mg = 27.7 %
Explanation:
Step 1: Data given
Number of moles Al = 0.0898 moles
Number of moles Mg = 0.0381 moles
Molar mass Al = 26.98 g/mol
Molar mass Mg = 24.3 g/mol
Step 2: Calculate mass Al
Mass Al = moles Al * molar mass Al
Mass Al = 0.0898 moles * 26.98 g/mol
Mass Al = 2.42 grams
Step 3: Calculate mass Mg
Mass Mg = 0.0381 moles * 24.3 g/mol
Mass Mg = 0.926 grams
Step 4: Calculate total mass
Total mass = mass Al + mass Mg
Total mass = 2.42 grams + 0.926 grams
Total mass = 3.346 grams
Step 5: Calculate mass %
Mass % Al = (mass Al/ total mass) * 100%
Mass % Al = (2.42 grams / 3.346 grams ) *100%
Mass % Al = 72.3 %
Mass % Mg = (mass Mg/ total mass)*100%
Mass % Mg = (0.926 / 3.346) *100 %
Mass % Mg = 27.7 %
Read the "Chemical Insights: Fireworks" essay within Ch 12 of the Zumdahl textbook. Which of the following substances produce bright yellow emissions that can mask other emission colors? Two of the answer choices below are correct...select the two correct answers.(Grading Note: for this type of question, Canvas will award points for correct selections and deduct points for incorrect selections.)A. sodium saltsB. carbon-based fuelsC. magnesium saltsD. aluminum salts
Answer:
Options A and B are correct.
Sodium salts and Carbon based fuels satisfy the criteria.
Explanation:
Sodium salts and Carbon Based fuels produce bright yellow emissions that can mask other emission colors.
Magnesium Salts produce no emissions, although, Magnesium metal burns brightly white while Aluminium salts produce white emissions.
Hope this Helps!!!
Options A and B are correct.
Sodium salts and Carbon based fuels satisfy the criteria.
The following information should be considered:
Sodium salts and Carbon Based fuels generated bright yellow emissions that can mask other emission colors.Magnesium Salts generated no emissions, however, Magnesium metal burns brightly white while on the other hand Aluminium salts produce white emissions.Learn more: https://brainly.com/question/2386757?referrer=searchResults
Acetic acid, CH3CO2HCH3CO2H, is the main organic constituent of vinegar. Draw an electron-dot structure for acetic acid. (The two carbons are connected by a single bond, and both oxygens are connected to the same carbon.)
Acetic acid, CH3CO2H, is composed of carbon, hydrogen, and oxygen atoms. The electron-dot structure is represented by placing valence electrons as dots around the symbol of each atom, showing their arrangement and the bonding between them.
Explanation:Acetic acid, CH3CO2H, is composed of carbon, hydrogen, and oxygen atoms. The two carbon atoms are connected by a single bond, and both oxygen atoms are connected to the same carbon atom. To draw the electron-dot structure for acetic acid, we represent each atom using its symbol and show the valence electrons as dots surrounding the symbol.
In the case of acetic acid, the carbons are each connected to three hydrogens and one oxygen, and the oxygen is also connected to another carbon. Therefore, for each carbon atom, one dot is placed next to each hydrogen atom and three dots are placed next to the oxygen atom. Two additional dots are placed next to the oxygen atom connected to the other carbon. The arrangement of the dots represents the bonding between the atoms.
The resulting electron-dot structure for acetic acid would be represented as CH3CO2H.
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1.00 M CaCl2 Density = 1.07 g/mL
% (m/m) CaCl2 _______
% (m/v) CaCl2 _______
N Ca+2 ______
N Cl– ______
m CaCl2 ______
ΧCaCl2 _______
χH2O _______
mass of 100. mL of this solution _______ g solution
H2O in 100. mL of this solution _______ g H2O
Explanation:
Molarity of solution = 1.00 M = 1.00 mol/L
In 1 L of solution 1.00 moles of calcium chloride is present.
Mass of solute or calcium chloride = m
[tex]m = 1 mol\times 111 g/mol = 111 g[/tex]
Mass of solution = M
Volume of solution = V = 1L = 1000 mL
Density of solution , d= 1.07 g/mL
[tex]M=d\times V=1.07 g/mL\times 1000 mL=1,070 g[/tex]
1) The value of %(m/M):
[tex]\frac{m}{M}\times 100=\frac{111 g}{1,070 g}\times 100=10.37\%[/tex]
2) The value of %(m/V):
[tex]\frac{m}{V}\times 100=\frac{111 g}{1000 L}\times 100=11.1\%[/tex]
[tex]Molality = \frac{\text{Moles of compound }}{\text{mass of solvent in kg}}[/tex]
[tex]Normality=\frac{\text{Moles of compound }}{n\times \text{volume of solution in L}}[/tex]
n = Equivalent mass
n = [tex]\frac{\text{molar mass of ion}}{\text{charge on an ion}}[/tex]
3) Normality of calcium ions:
Moles of calcium ion = 1 mol (1 [tex]CaCl_2[/tex] mole has 1 mole of calcium ion)
[tex]n=\frac{40 g/mol}{2}=20 [/tex]
[tex]=\frac{1 mol}{20 g/mol\times 1L}=0.050 N[/tex]
4) Normality of chlorine ions:
Moles of chlorine ion = 2 mol (1 [tex]CaCl_2[/tex] mole has 2 mole of chlorine ion)
[tex]n=\frac{35.5 g/mol}{1}=35.5[/tex]
[tex]=\frac{2 mol}{35.5 g/mol\times 1L}=0.056 N[/tex]
Moles of calcium chloride = 1.00 mol
Mass of solvent = Mass of solution - mass of solute
= 1,070 g - 111 g = 959 g = 0.959 kg ( 1 g =0.001 kg)
5) Molality of the solution :
[tex]\frac{1 mol}{0.959 kg}=1.043 mol/kg[/tex]
Moles of calcium chloride = [tex]n_1=1mol[/tex]
Mass of solvent = 959 g
Moles of water = [tex]n_2=\frac{959 g}{18 g/mol}=53.28 mol[/tex]
Mass of solvent = 959 g
6) Mole fraction of calcium chloride =
[tex]\chi_1=\frac{n_1}{n_1+n_2}=\frac{1mol}{1 mol+53.28 mol}=0.01842[/tex]
7) Mole fraction of water =
[tex]\chi_2=\frac{n_2}{n_1+n_2}=\frac{53.28 mol}{1mol+53.28 mol}=0.9816[/tex]
8) Mass of solution = m'
Volume of the solution= v = 100 mL
Density of solution = d = 1.07 g/mL
[tex]m'=d\times v=1.07 g/ml\times 100 g= 107 g[/tex]
Mass of 100 mL of this solution 107 grams of solution.
9) Volume of solution = V = 100 mL
Mass of solution = M'' = 107 g
Mass of solute = m
The value of %(m/V) of solution = 11.1%
[tex]11.1\%=\frac{m}{100 mL}\times 100[/tex]
m = 11.1 g
Mass of solvent = M''- m = 107 g -11.1 g = 95.9 g
95.9 grams of water was present in 100 mL of given solution.
Pure gold is defined as having 24 carats. When mixed in an alloy, the carats of gold are given as a percentage of this value. For example, a piece of jewelry made with 50% gold has 12 carats. State the purity of this piece of red gold jewelry in carats.
Answer:
18 Carats because of 75% Purity...
Explanation:
Red Gold Jewelry always contains a mixing of 25% copper by mass to make it durable and strong. That according to the simple parallel ratio rule, gives us a purity level of gold to be exactly 18 Carats. In other words the Red Gold is 75% pure...
The question pertains to the calculation of gold purity in carats. Without an exact gold percentage in the red gold jewelry, a precise purity or carat value cannot be identified.
Explanation:The question relates to the understanding of carats and the purity of gold used in jewellery. Generally, 24 carats is defined as 100% pure gold. Hence, the number of carats indicates the proportion of gold in an alloy.
The question asks about a piece of red gold jewellery but lacks the necessary percentage to determine its carat value. If we had an exact percentage of gold in the item, we could calculate the carat value by multiplying the given percentage by 24.
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(a) Find the concentration of electrons and holes in a sample of germanium that has a concentration of donor atoms equal to 1015 cm−3 . Is the semiconductor n-type or p-type? (b) Repeat part (a) for silicon.
Answer:
a) Germanium = 5.76 x 〖10〗^11 〖cm〗^(-3) , Semiconductor is n-type.
b) Silicon = 2.25 x 〖10〗^5 〖cm〗^(-3) , Semiconductor is n-type.
For clear view of the answers: Please refer to calculation 5 in the attachments section.
Explanation:
So, in order to find out the concentration of holes and electrons in a sample of germanium and silicon which have the concentration of donor atoms equals to 〖10〗^15 〖cm〗^(-3). We first need to find out the intrinsic carrier concentration of silicon and germanium at room temperature (T= 300K).
Here is the formula to calculate intrinsic carrier concentration: For calculation please refer to calculation 1:
So, till now we have calculated the intrinsic carrier concentration for germanium and silicon. Now, in this question we have been given donor concentration (N_d) (N subscript d), but if donor concentration is much greater than the intrinsic concentration then we can write: Please refer to calculation 2.
So, now we have got the concentration of electrons in both germanium and silicon. Now, we have to find out the concentration of holes in germanium and silicon (p_o). (p subscript o)
Equation to find out hole concentration: Please refer to calculation 3. and Calculation 4. in the attachment section.
Good Luck Everyone! Hope you will understand.
(a) The germanium with 10^15 cm^-3 donor atoms: Concentration of electrons = 10^15 cm^-3, Concentration of holes ≈ 0, Semiconductor is n-type. (b) For silicon with 10^15 cm^-3 donor atoms: Concentration of electrons = 10^15 cm^-3, Concentration of holes ≈ 0, Semiconductor is n-type.
(a) In n-type semiconductors like germanium, donor atoms introduce excess electrons into the crystal lattice. To find the concentration of electrons and holes, we can assume that all the donor atoms ionize and release one extra electron each. Therefore, the concentration of electrons will be equal to the donor atom concentration, which is 10^15 cm^-3. In this case, there are very few holes since most of the electrons are engaged in conduction. So, the semiconductor is n-type because it has an excess of electrons.(b) Silicon behaves similarly to germanium but has a larger energy gap. In silicon with a donor atom concentration of 10^15 cm^-3, the concentration of electrons will also be 10^15 cm^-3 because each donor atom donates one electron. As in germanium, there are very few holes compared to electrons, making it an n-type semiconductor due to the excess of electrons generated by the donor atoms.For more such questions on atoms
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Data has been collected to show that at a given wavelength in a 1 cm pathlength cell, Beer's Law for the absorbance of Co2 is linear. If a 0.135 M solution of Co2 has an absorbance of 0.350, what is the concentration of a solution with an absorbance of 0.420?
Answer : The concentration of a solution with an absorbance of 0.420 is, 0.162 M
Explanation :
Using Beer-Lambert's law :
[tex]A=\epsilon \times C\times l[/tex]
As per question, at constant path-length there is a direct relation between absorbance and concentration.
[tex]\frac{A_1}{A_2}=\frac{C_1}{C_2}[/tex]
where,
A = absorbance of solution
C = concentration of solution
l = path length
[tex]A_1[/tex] = initial absorbance = 0.350
[tex]A_2[/tex] = final absorbance = 0.420
[tex]C_1[/tex] = initial concentration = 0.135 M
[tex]C_2[/tex] = final concentration = ?
Now put all the given value in the above relation, we get:
[tex]\frac{0.350}{0.420}=\frac{0.135}{C_2}[/tex]
[tex]C_2=0.162M[/tex]
Thus, the concentration of a solution with an absorbance of 0.420 is, 0.162 M
In the investigation of an unknown alcohol, there was a positive Jones test and a negative Lucas test. What deductions may be made as to the nature of the alcohol? State reasons for your deductions.
Answer:
Primary alcohol.
Explanation:
Jones reagent is mixture of chromium trioxide (CrO3) and sulfuric acid (H2SO4) dissolved in a mixture of acetone and water. Alternatively, potassium dichromate (K2CrO7) can be used in place of chromium trioxide because of its carcinogenic nature.
This oxidation reaction is an organic reaction for the oxidation of primary alcohols to aldehydes then carboxylic acid and secondary alcohols to ketones.
Lucas reagent is a solution of anhydrous zinc chloride (ZnCl) in concentrated hydrochloric acid. The reaction involves substitution reaction in which the chloride replaces a hydroxyl group.
A positive test is indicated by a change in appearance of the solution, from clear and colourless to fog-like, which shows the formation of a chloroalkane. Accurate results for this test are observed in tertiary alcohols, as they form alkyl halides the fastest due to the stability of their intermediate tertiary carbocation.
Therefore, an alcohol in which there was a positive Jones test and a negative Lucas test indicates the presence of primary alcohol.
This is because:
A primary alcohol would test positive to Jones test but in Lucas test, the substitution reaction is the slowest as compared to the secondary and tertiary alcohols.
1° alcohols < 2° alcohols < 3° alcohols
So a primary alcohol will give a negative result to Lucas reagent.
A student reacts 25.0 mL of 0.175 M H3PO4 with 25.0 mL of 0.205 M KOH. Write a balanced chemical equation to show this reaction. Calculate the concentrations of H3PO4 and KOH that remain in solution, as well as the concentration of the salt that is formed during the reaction.
Answer: The concentration of salt (potassium phosphate), phosphoric acid and KOH in the solution is 0.0342 M, 0.0533 M and 0 M respectively.
Explanation:
To calculate the number of moles for given molarity, we use the equation:
[tex]\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}[/tex] .....(1)
For KOH:
Initial molarity of KOH solution = 0.205 M
Volume of solution = 25.0 mL = 0.025 L (Conversion factor: 1 L = 1000 mL)
Putting values in equation 1, we get:
[tex]0.205M=\frac{\text{Moles of KOH}}{0.025L}\\\\\text{Moles of KOH}=(0.205mol/L\times 0.025L)=5.123\times 10^{-3}mol[/tex]
For phosphoric acid:
Initial molarity of phosphoric acid solution = 0.175 M
Volume of solution = 25.0 mL = 0.025 L
Putting values in equation 1, we get:
[tex]0.175M=\frac{\text{Moles of }H_3PO_4}{0.025L}\\\\\text{Moles of }H_3PO_4=(0.175mol/L\times 0.025L)=4.375\times 10^{-3}mol[/tex]
The chemical equation for the reaction of KOH and phosphoric acid follows:
[tex]3KOH+H_3PO_4\rightarrow K_3PO_4+3H_2O[/tex]
By Stoichiometry of the reaction:
3 moles of KOH reacts with 1 mole of phosphoric acid
So, [tex]5.123\times 10^{-3}[/tex] moles of KOH will react with = [tex]\frac{1}{3}\times 5.123\times 10^{-3}=1.708\times 10^{-3}mol[/tex] of phosphoric acid
As, given amount of phosphoric acid is more than the required amount. So, it is considered as an excess reagent.
Thus, KOH is considered as a limiting reagent because it limits the formation of product.
Excess moles of phosphoric acid = [tex](4.375-1.708)\times 10^{-3}=2.667\times 10^{-3}mol[/tex]
By Stoichiometry of the reaction:
3 moles of KOH produces 1 mole of potassium phosphate
So, [tex]5.123\times 10^{-3}[/tex] moles of KOH will produce = [tex]\frac{1}{3}\times 5.123\times 10^{-3}=1.708\times 10^{-3}moles[/tex] of potassium phosphate
For potassium phosphate:Moles of potassium phosphate = [tex]1.708\times 10^{-3}moles[/tex]
Volume of solution = [25.0 + 25.0] = 50.0 mL = 0.050 L
Putting values in equation 1, we get:
[tex]\text{Molarity of potassium phosphate}=\frac{1.708\times 10^{-3}}{0.050}=0.0342M[/tex]
For phosphoric acid:Moles of excess phosphoric acid = [tex]2.667\times 10^{-3}moles[/tex]
Volume of solution = [25.0 + 25.0] = 50.0 mL = 0.050 L
Putting values in equation 1, we get:
[tex]\text{Molarity of phosphoric acid}=\frac{2.667\times 10^{-3}}{0.050}=0.0533M[/tex]
For KOH:Moles of KOH remained = 0 moles
Volume of solution = [25.0 + 25.0] = 50.0 mL = 0.050 L
Putting values in equation 1, we get:
[tex]\text{Molarity of KOH}=\frac{0}{0.050}=0M[/tex]
Hence, the concentration of salt (potassium phosphate), phosphoric acid and KOH in the solution is 0.0342 M, 0.0533 M and 0 M respectively.
Write the appropriate symbol for each of the following isotopes: (a) Z = 19, A = 41 (b) Z = 46, A = 106 (c) Z = 52, A = 125 (d) Z = 38, A = 88
Answer:
(a) ⁴¹₁₉K
(b) ¹⁰⁶₄₆Pd
(c) ¹²⁵₅₂Te
(d) ⁸⁸₃₈Sr
Explanation:
The identity of an element is its atomic number, by convention we write the atomic mass as superscript and the and the atomic number as subscript to the left of the element .
(a) Z = 19 A = 41 symbol: ⁴¹₁₉K
(b) Z = 46 A = 106 symbol: ¹⁰⁶₄₆Pd
(c) Z = 52 A = 125 symbol: ¹²⁵₅₂Te
(d) Z = 38 A = 88 symbol: ⁸⁸₃₈Sr
The isotopes correspond to Potassium-41 (41K), Palladium-106 (106Pd), Tellurium-125 (125Te), and Strontium-88 (88Sr). Z and A denote atomic and mass numbers, which help in identifying particular isotopes.
Explanation:The appropriate symbol for each of these isotopes would be:
(a) Z = 19, A = 41, which corresponds to Potassium-41 (41K), (b) Z = 46, A = 106, this stands for Palladium-106 (106Pd), (c) Z = 52, A = 125, this refers to Tellurium-125 (125Te), (d) Z = 38, A = 88, the symbol for this is Strontium-88 (88Sr).
Z and A represent atomic and mass numbers respectively. These isotopes are written with a mass number (A) preceding the symbol and atomic number (Z) is typically understood from the symbol itself.
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For some reaction carried out at constant atmospheric pressure and at a constant temperature of 25◦C, it is found that ∆H = −38.468 kJ/mol and ∆S = +51.4 J mol · K . What is the value of ∆G for this reaction under these conditions?
Answer: ΔG =23.169kJ/mol
Explanation:
Solution
To Calculate Gibbs free energy ΔG for the reaction above we use the equation ΔG=ΔH−TΔS.
Where
ΔH= 38.468 kJ/mol = 38468 J/mol
∆S = +51.4 J mol−1 K−1).
T = 25◦C =298k
ΔG= 38468J/mol−298k(51.4 J mol−1 K−1).
ΔG = 38468 J/mol - 15317.2J/mol
ΔG = 23168.8J/mol
ΔG =23.169kJ/mol
Find the angle between the diagonal of a cube of side length 8 and the diagonal of one of its faces, so that the two diagonals have a common vertex. The angle should be measured in radians. (Hint: we may assume that the cube is in the first octant, the origin is one of its vertices, and both diagonals start at the origin.)
Answer:
35.26 rad
Explanation:
Let's assume the cube in the figure below. If it's in the first octant, then origin (0, 0, 0) is one of the vertices and it's also the common vertex of the diagonals (OB and OE).
The point B is at the y-axis, so since the length is 8, it is (8, 0, 8), and the point E is (8, 8, 8). The vectors of the diagonals are the subtraction of the coordinates of the two points, so OB = <8, 0, 8> and OE = <8, 8, 8>. The angle between two vectors in the tridimensional space is:
θ = cos⁻¹[(OB · OE)/(|OB|·|OE|)]
The module (| |) of a vector <x, y, z> is √(x² + y² + z²)
θ = cos⁻¹[(<8, 0, 8> · <8, 8, 8>)/(√(8² + 0² + 8²) · √(8² + 8² + 8²))]
θ = cos⁻¹[(8*8 + 8*0 + 8*8)/(√128 ·√192)]
θ = cos⁻¹[128/156.77]
θ = cos⁻¹[0.8165]
θ = 35.26 rad
There are several reagents that can be used to effect addition to a double bond including: acid and water, oxymercuration–demercuration reagents, and hydroboration–oxidation reagents. Select all the reasons why hydroboration–oxidation reagents were chosen to effect the following transformation instead of the other reagents?
a. The reaction requires the Markovnikov product without sigmatropic rearrangement.
b. Addition with acid and water as reagents avoids sigmatropic rearrangements.
c. Hydroboration-oxidation reagents yield the anti-Markovnikov product of addition.
d. The reaction requires anti-Markovnikov product without sigmatropic rearrangement.
e. Addition with acid and water as reagents gives the Markovnikov product.
f. Hydroboration-oxidation reagents yield the Markovnikov product of addition.
g. The reaction requires sigmatropic rearrangement.
Answer:
The reaction requires anti-Markovnikov product without sigmatropic rearrangement.
Explanation:
The reaction is known to begin with the concerted syn addition of B and H across the double bond, with the boron adding to the less substituted carbon atom.
The second step of the reaction involves hydrogen peroxide and a base such as NaOH are added, NaOH deprotonates the hydrogen peroxide.
The resulting NaOOH then attacks the boron and sets up the key migration step, where the carbon-boron bond migrates to the oxygen bound to boron, breaking the weak oxygen-oxygen bond . Then the -OH expelled then returns to form a bond on the boron resulting in a deprotonated alcohol (alkoxide). The alkoxide is then protonated by water or some other comparably acidic species.
Hydroboration is a syn addition that gives an anti-Markovnikov product without sigmatropic rearrangement.
The reasons why hydroboration–oxidation reagents were chosen to effect the transformation will be:
Oxymercuration - demercuration reagents give the Markovnikov product.Hydroboration-oxidation reagents yield the anti-Markovnikov product of addition.The reaction requires anti-Markovnikov product without sigmatropic rearrangement.Addition with acid and water as reagents gives the Markovnikov product.It should be noted that hydroboration–oxidation is vital for the production of alcohol. It's needed for the transformation rather than using other reagents because hydroboration-oxidation reagents yield the anti-Markovnikov product of addition.
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Draw a mechanism for chlorination of 1,1,1-trichloroethane to produce 1,1,1,2-tetrachloroethane. Do not use abbreviations in your answer.
Answer:
1,1,1-trichloroethane being a saturated halogenated alkane will undergo substitution reaction via free radical mechanism. The mechanism is divided in three steps,
Step 1: Initiation:
In this step the reaction is started by treating the chlorine gas either with UV light or by sunlight. This results in the formation of free radical.
Step 2: Propagation:
In this step the radical formed will react with the hydrogen atom resulting in formation of HCl and generating free radical of corresponding alkane. Hence, the radical will agian react with Cl2 molecule generating another Chlorine radical and corresponding halogenated compound i.e. 1,1,1,2-tetrachloroethane.
Step 3: Termination:
This is the last step. In this step the reaction is stopped/terminated. The free radicals react with each other forming a single bonds and stopping the formation of further radicals.
The mechanism is shown below,
When 2-methylpropane is treated with bromine in the presence of UV light, one product predominates a. Identify the structure of the major productb. Identify the structure of the minor productc. Draw a mechanism for formation of the major product (include all steps:initiation, propagation, and termination)
Answer: Inittiation: Br2 -> Br. + Br.
Propagation: CH3CH(CH3)CH3+Br. -> CH3CH(CH3)CH2. +HBr
CH3CH(CH3)CH2. + Br. -> CH3CH(CH3)CH2Br
CH3CH(CH3)CH2Br +Br. -> CH3C.(CH3)CH2Br +HBr
CH3C.(CH3)CH2Br + Br. -> CH3CBr(CH3)CH2Br
Termination: Br. + Br. ->Br2
The Henry’s law constant for CO2 is 3.4 × 10−2 M/atm at 25 °C. Assuming ideal solution behavior, what pressure of carbon dioxide is needed to maintain a CO2 concentration of 0.10 M in a can of lemon-lime soda?
Answer: The pressure of carbon dioxide needed is 2.94 atm
Explanation:
To calculate the partial pressure of carbon dioxide, we use the equation given by Henry's law, which is:
[tex]C_{CO_2}=K_H\times p_{CO_2}[/tex]
where,
[tex]K_H[/tex] = Henry's constant = [tex]3.4\times 10^{-2}mol/L.atm[/tex]
[tex]C_{CO_2}[/tex] = molar solubility of carbon dioxide gas = [tex]0.10mol/L[/tex]
[tex]p_{CO_2}[/tex] = pressure of carbon dioxide = ?
Putting values in above equation, we get:
[tex]0.10mol/L=3.4\times 10^{-2}mol/L.atm\times p_{CO_2}\\\\p_{CO_2}=\frac{0.10mol/L}{3.4\times 10^{-2}mol/L.atm}=2.94atm[/tex]
Hence, the pressure of carbon dioxide needed is 2.94 atm
If a buffer solution is 0.250 M 0.250 M in a weak base ( K b = 8.0 × 10 − 5 ) Kb=8.0×10−5) and 0.540 M 0.540 M in its conjugate acid, what is the pH ?
Answer:
9.57
Explanation:
Given that:
[tex]pK_{b}=-\log\ K_{b}=-\log(8.0\times 10^{-5})=4.1[/tex]
Considering the Henderson- Hasselbalch equation for the calculation of the pOH of the basic buffer solution as:
[tex]pOH=pK_b+log\frac{[conjugate\ acid]}{[base]}[/tex]
So,
[tex]pOH=4.1+\log\frac{0.540}{0.250}=4.43[/tex]
pH + pOH = 14
So, pH = 14 - 4.43 = 9.57
Write an equation that represents the action in water of formic acid (HCOOH)(HCOOH) as a Brønsted–Lowry acid. Express your answer as a chemical equation.
Answer:
HCOOH + H2O <===> HCOO- + H3O+
Explanation:
A coal-fired power plant emitshot gases at a rate of 50 ft3/sec. The hot gas contains flyash, which is emitted at a rate of 120 lb of flyash per hour. What is the concentration of flyash in micrograms per cubic meter?
Answer:
C flyash = 10677789.55 μg/m³
Explanation:
A coal-fired power plant:
∴ rate = 50 ft³/s = (1.416 m³/s)(3600 s/h) = 5097.6 m³/h
hot gas-flyash:
∴ rf = 120 Lb/h = 54431.1 g/h = 54431100000 μg/h
⇒ C flyash = ? [=] μg/m³
⇒ C flyash = (54431100000 μg/h)/(h/5097.6 m³)
⇒ C flyash = 10677789.55 μg/m³
Summarize the trend in metallic character as a function of position in the periodic table. Is it the same as the trend in atomic size? Ionization energy?
Answer:
The trend in metallic character as a function of position in the periodic table is that the metallic character increases as you go down a group. Since the ionization energy decreases going down a group (or increases going up a group), the increased ability for metals lower in a group to lose electrons makes them more reactive.
This is not the same for the atomic size, as you go down a column of the periodic table, the atomic radii increase. This is because the valence electron shell is getting a larger and there is a larger principal quantum number, so the valence shell lies physically farther away from the nucleus.
Similarly, it is also different for the ionization energy trend, as you go down the periodic table, it becomes easier to remove an electron from an atom (i.e., IE decreases) because the valence electron is farther away from the nucleus.
Final answer:
The metallic character in the periodic table decreases across a period and increases down a group. It trends similarly to atomic size but oppositely to ionization energy.
Explanation:
Metallic character: The metallic trend follows the trend of the atomic radius. It increases within a group of the periodic table from the top to the bottom and decreases within a period from left to right. Metallic character relates to the ease of losing an electron in a chemical reaction and is opposite to the trend of ionization energy.
Atomic size shows a trend that parallels the metallic character. Atomic size increases down a group because of the increase in electron shells, which makes the valence electrons less tightly held. Conversely, atomic size decreases from left to right within a period due to an increasing effective nuclear charge, which draws electrons closer to the nucleus, reducing the size of the atom.
In summary, the metallic character and atomic size increase from right to left in a period and from top to bottom in a group, while ionization energy generally shows the opposite trend. Hence, the trend in metallic character is similar to the trend in atomic size but opposite to the trend in ionization energy.
Draw the triglyceride formed from the esterification of glycerol and three molecules of myristic acid.
A triglyceride formed from glycerol and myristic acid involves esterifying three myristic acid molecules with a glycerol backbone through a dehydration reaction, releasing three water molecules and forming a simple triglyceride.
The triglyceride formed from the esterification of glycerol and three molecules of myristic acid is constructed by attaching each of the three fatty acid molecules to the glycerol backbone through a dehydration reaction. In this process, each fatty acid's carboxyl group (COOH) reacts with one of the hydroxyl groups (OH) on the glycerol molecule, resulting in the formation of an ester bond and the release of water.
The chemical structure of glycerol (H₂C-OH | HC-OH | H₂C-OH) bonded with three myristic acid molecules (which have the formula CH₃(CH₂)₌₁₂COOH) will show three ester linkages replacing the hydroxyl groups of glycerol with the alkyl chains of myristic acid.
Each dehydration synthesis reaction between the carboxyl group of a fatty acid and a hydroxyl group of glycerol results in the release of one molecule of water, leading to a total of three water molecules released when forming the triacylglycerol. If all three OH groups on the glycerol molecule are esterified with myristic acid, the molecule formed is a simple triglyceride because it contains only one type of fatty acid.
How much heat must be absorbed by 125 g of ethanol to change its temperature from 21.5 ∘C to 34.8 ∘C? (Specific heat capacity of ethanol is 2.42 J/g∘C)
a. 86.6 kJ
b. 4.02×103kJ
c. 6.95 kJ
d. 4.02 kJ
Answer:
Option D. 4.02 kJ
Explanation:
A simple calorimetry problem
Q = m . C . ΔT
ΔT = Final T° - Initial T°
C = Specific heat capacity
m = mass
Let's replace the data
Q = 125 g . 2.42 J/g∘C . (34.8°C -21.5 °C)
Q= 4023.25 J
We must convert the answer to kJ
4023.25 J . 1kJ /1000 =4.02kJ
Write the balanced molecular equation for the reaction of sodium metabisulfite (Na2S2O5) with water to produce sodium bisulfite (NaHSO3) and then write the net ionic equation for this reaction. Why is sodium bisulfite prepared using this method?
Answer:
Balanced molecular reaction
Na2S2O5 + H2O ----> 2NaHSO3
Net ionic reaction
(S2O5)^(2-) + H2O ----> 2((HSO3)^(1-))
Explanation:
The sodium bisulfite compound is produced in this manner because Sodium metabisulfite has better preservative properties. It is more stable compared to sodium bisulfite and readily dissolves in water to give the required sodium bisulfite.
Sodium bisulfite sold in the market contains sodium metabisulfite as it is the more stable one of the pair.
And of all the ways of preparing the compound, this is the cheapest and easiest one.
Answer: check explanation.
Explanation:
The balanced molecular equation for the reaction of sodium metabisulfite (Na2S2O5) with water to produce sodium bisulfite (NaHSO3) is given below;
Na2S2O5 + H2O --------> 2NaHSO3.
Therefore, the net ionic equation for this reaction is given below;
S2O5^2- + H2O ------> 2HSO3^-1.
S2O5^2- + H+ ------------> 2HSO3^-1
=====> Why is sodium bisulfite prepared using this method?
sodium bisulfite is prepared using this method because of the following reasons;
(1). It is the most easy way of synthesisizing/producing sodium bisulfite (NaHSO3).
(2). In order to produce sodium bisulfite (NaHSO3) through this method, it does not require high cost,that is to say that it is financial friendly.
(3). Because of high preservative properties of sodium metabisulfite (Na2S2O5).
(4). It reacts with water with ease to produce sodium bisulfite (NaHSO3).
What is the volume of 2 moles of methane (CH4)? (One mole of any gas occupies 22.4 L under certain conditions of temperature and pressure. Assume those conditions for this question.)
a. 44.8 L
b. 22.4 L
c. 20 L
d. 2.0L
Answer:
2 moles of Methane gas will occupy 44.8 L
Explanation:
If one mole of any gas occupies 22.4 L under certain conditions of temperature and pressure, and those conditions are assumed in this question, then we comfortably solve this problem as follows;
1 mole of Methane gas ---------------> 22.4 L
2 moles of Methane gas -------------->?
Cross and multiply, 2 moles of Methane gas = 2 X 22.4 L = 44.8L
Therefore, 2 moles of Methane gas will occupy 44.8 L, if the conditions of temperature and pressure are maintained.
OPTION A IS THE RIGHT SOLUTION.
Answer:
The answer is A
Explanation: