Answer:
False
Explanation:
Magnesium is the element of second group and third period. The electronic configuration of magnesium is - 2, 8, 2 or [tex]1s^22s^22p^63s^2[/tex]
There are 2 valence electrons of magnesium.
Only the valence electrons are shown by dots in the Lewis structure.
As, stated above, there are only two valence electrons of magnesium, so in the Lewis structure, two dots are made around the magnesium symbol.
Given that the electronic configuration is:- [tex]1s^22s^22p^63s^3[/tex] .
Orbital s cannot accommodate 3 electrons and also in magnesium it has [tex]3s^2[/tex] . Hence, the statement is false.
We consider a spherical air bubble rising through a water tank. The pressurepinside the bubble isequal to its surrounding pressure. The volumeVof the bubble is connected to the pressure throughthe isentropic relationWhy does the volume of a water bubble increase as it moves higher to the liquid surface?
The answer is a change in internal energy causes work to be done and heat to flow into the system.
Explanation:
Boyle's law says, PV=RT
Here P represents the pressure, V represents the volume and T represents the temperature. R is a constant. The volume of an ideal gas is inversely proportional to its pressure if the temperature is constant.When a bubble is present in deep water it has water pressure and atmospheric pressure. Then the Volume increases when water pressure raises which is proportional to the depth reduces.But we should not finalize the volume of the bubble will be four-time as great as at the top than the bottom. if the bottom of the lake is at four atmospheres, the temperature will not be equal to the top. If the bubble travels from the bottom to the top or vice-versa, it's going to lose or gain heat in a way that must be quite hard to measure.The solid XY decomposes into gaseous X and Y: XY(s) m X(g) + Y(g) Kp = 4.1 (at 0 °C) If the reaction is carried out in a 22.4 L container, which initial amounts of X and Y will result in the formation of solid XY?a) 5 mol X; 0.5 mol Yb) 2.0 mol X; 2.0 mol Yc) 1 mol X; 1 mol Yd) none of the above
Answer:
None of the conditions will favor either the forward reaction or backward reaction , hence the answer is D
Explanation:
The principle of chemical Equilibrium is applied here, where the concentration of the reactants or the forward reaction is same as the concentration of the products or the backward reaction. The equilibrium constants is also involved here, K can be in terms of pressure (Kp) or concentration (Kc) hence equilibrium constant is the ration of the concentration of the products to the concentration of the reactants raised to the power of the coefficient of the reactants and products.Partial pressure , total pressure and the mole fraction relationship is also appliedThe step by step explanation is as shown in the attachment below.It's challenging to determine the precise initial amounts of gaseous elements X and Y in the reaction, XY(s) → X(g) + Y(g), with the given Kp and the conditions provided. Based on the data provided, the accurate answer would be 'none of the above'.
Explanation:This question involves equilibrium in chemical reactions, specifically the concept of Kp, the equilibrium constant under constant pressure circumstances. In its simplest form, it represents a ratio of the concentrations of products and reactants, each raised to the power of their respective stoichiometric coefficients in the balanced chemical equation. In this context, if the reaction XY(s) → X(g) + Y(g) has a Kp = 4.1, it means that when equilibrium is reached, the ratio [X][Y]/[XY] is 4.1.
Given that one mole of gas at 0°C and 1 atmosphere of pressure occupies a volume of 22.4L (standard molar volume), we can establish the initial conditions of X and Y using the stoichiometric ratio. However, without additional specifics on the initial conditions of the solid XY such as its volume or weight, it's challenging to determine precise initial amounts of X and Y. Therefore, the accurate answer from the given choice would be (d) none of the above.
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CFC-11 (trichlorofluoromethane, CCl3F) has been used for many years as the working fluid in refrigerators. Given its heat of vaporization is 26.88 kJ/mol and its entropy of vaporization is 90.51 J/(mol ⋅ K), what is the boiling point of CFC-11?
Answer:
297.0 K
Explanation:
Given data
Enthalpy of vaporization (ΔHvap): 26.88 kJ/mol = 26.88 × 10³ J/molEntropy of vaporization (ΔSvap): 90.51 J/(mol ⋅ K)Boiling point (Tb): ?We can find the boiling point of CFC-11 using the following expression.
ΔSvap = ΔHvap/Tb
Tb = ΔHvap/ΔSvap
Tb = (26.88 × 10³ J/mol)/(90.51 J/(mol ⋅ K))
Tb = 297.0 K
Chlorofluorocarbons (CFCs) are no longer used as refrigerants because they destroy the ozone layer.Trichlorofluoromethane (CCl3F) boils at 296.95 K, and its molar heat of vaporization is 24.8 kJ/mol.
Explanation:
It is known that entropy is the measure of degree of randomness present in a substance due to the movement of its molecules.
More randomly the atoms are moving from one place to another more will be the entropy of system.
In order to calculate the entropy, we divide the amount of heat transferred by the temperature at which heat transfer occurs.
As the given temperature is 296.95 K and heat of vaporization is 24.8 kJ/mol. Therefore, calculate the molar entropy of vaporization as follows.
Molar entropy of vaporization = [tex]\frac{24.8 kJ/mol}{296.95 K}[/tex]
= 0.08351 kJ/mol K
Thus, we can conclude that molar entropy of vaporization of [tex]CC_{3}F[/tex] is 0.08351 kJ/mol K.
CFCs, including Trichlorofluoromethane, have been banned due to their destructive effect on the ozone layer. UV light causes them to release chlorine atoms, which react with and deplete ozone. The harmful side effect of this is increased solar radiation reaching Earth.
Explanation:Chlorofluorocarbons, or CFCs, like Trichlorofluoromethane, were previously widely used in a variety of industries, such as refrigeration and aerosols. However, the stable nature of these compounds, combined with their ability to break down and release chlorine atoms under ultraviolet light, led to significant depletion of the ozone layer in the stratosphere. This process happens because as CFCs are broken down by UV light, they produce chlorine atoms. These chlorine atoms then react with ozone molecules, leading to a net decrease in stratospheric ozone. This ozone depletion had severe implications, such as increased risks from solar radiation, which prompted the worldwide effort to phase out the use of CFCs under the Montreal Protocol.
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How many grams of magnesium bromide, MgBr2, must be dissolved to prepare 400. mL of a 0.183 M aqueous solution of the salt?
Answer: 13.47g of MgBr2
Explanation:
MM of MgBr2 = 24 + (2 x 80) = 24 + 160 = 184g/mol
Mass conc. Of MgBr2 = 0.183 x184 = 33.672g
33.672g of MgBr2 dissolves in 1000mL
Therefore Xg of MgBr2 will dissolve in 400mL
Xg of MgBr2 = ( 33.672 x 400)/1000
Xg of MgBr2 = 13.47g
N2O5(g) → NO3(g) + NO2(g)
This reaction is 1st order with half-life = 18.0 seconds. How long will it take for 95% of a sample of N2O5(g) to react?
A. 42.2 seconds
B. 132 seconds
C. 109 seconds
D. 77.8 seconds
E. 158 seconds
Answer:
It will take 77.8 seconds for 95% of sample to react
Explanation:
As the given reaction obeys 1st order therefore-
[tex][N_{2}O_{5}]=[N_{2}O_{5}]_{0}\times (\frac{1}{2})^{\frac{t}{t_{\frac{1}{2}}}}[/tex]
Where [tex][N_{2}O_{5}][/tex] is the concentration of [tex]N_{2}O_{5}[/tex] after "t" time, [tex][N_{2}O_{5}]_{0}[/tex] is the initial concentration of [tex]N_{2}O_{5}[/tex] and [tex]t_{\frac{1}{2}}[/tex] is half life
Here, [tex]\frac{[N_{2}O_{5}]}{[N_{2}O_{5}]_{0}}=\frac{100-95}{100}=0.05[/tex] and [tex]t_{\frac{1}{2}}=18.0 seconds[/tex]
So, [tex]0.05=(\frac{1}{2})^{\frac{t}{18.0seconds}}[/tex]
or, [tex]t=77.8 seconds[/tex]
So, it will take 77.8 seconds for 95% of sample to react
A sample of gas occupies a volume of 10.5 L when the pressure is 14.3 psi. Use Boyle’s Law to calculate the pressure (psi) when the volume is reduced to 8.55L. Show the calculation.
Answer: 17.6psi
Explanation:
V1 = 10.5 L
P1 = 14.3 psi
V2 = 8.55L
P2 =?
P1V1 = P2V2
14.3 x 10.5 = P2 x 8.55
P2 = (14.3 x 10.5) / 8.55
P2 = 17.6psi
Answer:
The final pressure is 17.56 psi
Explanation:
Step 1: Data given
Initial volume = 10.5 L
Initial pressure = 14.3 psi
Final volume = 8.55 L
Step 2: Calculate the new pressure
P1V1 = P2V2
⇒ P1 = The initial pressure = 14.3 psi
⇒ V1 = The initial volume = 10.5L
⇒ P2 = The final pressure = TO BE DETERMINED
⇒ V2 = The final volume = 8.55 L
14.3 * 10.5 = P2*8.55
P2 = 17.56 psi
The final pressure is 17.56 psi
Choose the aqueous solution below with the highest freezing point. These are all solutions of nonvolatile solutes and you should assume ideal van't Hoff factors where applicable. Choose the aqueous solution below with the highest freezing point. These are all solutions of nonvolatile solutes and you should assume ideal van't Hoff factors where applicable.
a. 0.200 m Ba(NO3)2
b. 0.200 m Mg(ClO4)2
c. 0.200 m HOCH2CH2OH
d. 0.200 m Na3PO3
Answer: c. 0.200 m HOCH2CH2OH
Explanation:
The collective properties of solvents expressed that the freezing point of water is lowered when solute particles are dissolved in it.
When more particles are dissolved, the more the freezing point is deepened and lowered.
Here, all the molarities are equal, the deciding factor in measuring the lowering will be which substance produces the fewest particles in solution; then we relate it to lowering of the freezing point.
Ba(NO3)2 ---> Ba2+ + 2NO3(1-)
1 mol releases 3 mol of ions.
Mg(ClO4)2 ---> Mg2+ + 2ClO4(1-)
1 mol releases 3 mol of ions.
HOCH2CH2OH is covalent and doesn't ionize.
1 mol in water is just 1 mol of molecules.
Na3PO3 ---> 3Na+ + PO3(1-)
1 mol releases 4 mol ions.
The substance that produces the fewest particles in solution is the 0.200 m HOCH2CH2OH
The solution with the highest freezing point will be the one that forms the fewest particles when dissolved. In this case, that would be the ethylene glycol solution (HOCH2CH2OH), because it does not ionize in solution and therefore has a van't Hoff factor of 1.
Explanation:In order to understand which solution has the highest freezing point, think about a concept called freezing point depression. The freezing point depression is due to the number of solute particles in a given volume of solvent; more particles will result in a lower freezing point. The key to the number of particles is the van't Hoff factor (i), which shows how many particles a solute breaks up into when dissolved.
Selected solution (a) Ba(NO3)2 is a salt which breaks up into three ions in solution: Ba2+ and 2NO3-, so i=3. Solution (b) Mg(ClO4)2 also breaks up into three ions: Mg2+ and 2ClO4-, so i=3. Solution (c) HOCH2CH2OH, ethylene glycol, is a nonelectrolyte and doesn't ionize in solution, so i=1. Solution (d) Na3PO3 breaks up into four ions: 3Na+ and PO33-, so i=4.
Therefore, the solution with the highest freezing point will be the one that forms the fewest particles when dissolved, which is 0.200 m HOCH2CH2OH.
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Mechanism for enzyme catalyzed reactions to explain the kinetics of enzyme-substrate reactions, michaelis final rate from -ra to ________
Answer:
ra to r+a
Explanation: In an enzyme catalyzed reaction, enzyme binds with the active side(3D) of the substrate substrate. It provides an alternative path for the reaction to take place with lower activation energy. in the reaction the kinetic energy of the molecules increases so reaction takes place at a higher rate. when the reaction is completed enzyme leaves the active side.
A reaction has an equilibrium constant of 7.9×10³ at 298 K. At 713 K, the equilibrium constant is 0.77.
Do you predict that the enthalpy of the reaction is positive or negative? Why?
a. Positive, because endothermic reactions shift toward products at higher temperatures
b. Negative, because exothermic reactions shift toward reactants at higher temperatures
c. Positive, because endothermic reactions shift toward reactants at higher temperatures
d. Negative, because exothermic reactions shift toward products at higher temperatures
Answer:
option b is correct
Explanation:
Taking the Van't Hoff equation
d ln Keq / dT = ΔH/(RT²)
then
Keq increases with increasing temperature (d ln Keq / dT>0) when ΔH>0 and decreases with increasing temperature (d ln Keq / dT<0) when ΔH<0
if Keq decreases (reactions shift toward reactants) from 7.9×10³ to 0.77 when temperature increases from 298 K to 713 K (d ln Keq / dT<0) → ΔH<0 ( exothermic reaction)
therefore option b is correct
The enthalpy of the reaction is predicted to be negative, indicating it is an exothermic reaction, because the equilibrium constant decreases with an increase in temperature, a characteristic of exothermic reactions. Therefore, the correct answer is b.
The question asks if the enthalpy of the reaction is positive or negative based on changes in the equilibrium constant at different temperatures. The equilibrium constant decreases from 7.9×10³ at 298 K to 0.77 at 713 K. According to the principle that for an endothermic reaction (ΔH° > 0), the magnitude of K increases with increasing temperature, and conversely, for an exothermic reaction (ΔH° < 0), the magnitude of K decreases with increasing temperature, we can deduce the nature of the reaction. As the equilibrium constant decreases with an increase in temperature, the reaction must be exothermic. Therefore, the correct answer is b. Negative, because exothermic reactions shift toward reactants at higher temperatures.
Raoult's law accounts for the fact that the vapor pressure of a solvent will decrease as the mole fraction of the solvent is decreased. In considering the mole fraction, it is important to consider the total moles of dissolved particles. Remember: a particle can be a dissolved molecule or ion. Which aqueous solutions would have the lowest vapor pressure
Raoult's law accounts for the fact that the vapor pressure of a solvent will decrease as the mole fraction of the solvent is decreased. In considering the mole fraction, it is important to consider the total moles of dissolved particles. Remember: a particle can be a dissolved molecule or ion. Which aqueous solutions would have the lowest vapor pressure.
0.1 M [tex]NH_4NO_3(aq)[/tex] , 0.1 M [tex]NaF(aq)[/tex], 0.1 M [tex]LiNO_3(aq)[/tex], 0.1 M [tex]Na_3PO_4(aq)[/tex] and 0.1 M [tex]HC_2H_3O_2(aq)[/tex]
Answer: 0.1 M [tex]Na_3PO_4[/tex]
Explanation:
As the relative lowering of vapor pressure is directly proportional to the amount of dissolved solute.
The formula for relative lowering of vapor pressure will be,
[tex]\frac{p^o-p_s}{p^o}=i\times x_2[/tex]
where,
[tex]\frac{p^o-p_s}{p^o}[/tex]= relative lowering in vapor pressure
i = Van'T Hoff factor
[tex]x_2[/tex] = mole fraction of solute
1. For 0.1 M [tex]NH_4NO_3[/tex]
[tex]NH_4NO_3\rightarrow NH_4^{+}+NO_3^{-}[/tex]
, i= 2 as it is a electrolyte and dissociate to give 2 ions. and concentration of ions will be [tex]1\times 0.1+1\times 0.1=0.2[/tex]
2. For 0.1 M [tex]NaF[/tex]
[tex]NaF\rightarrow Na^{+}+F^{-}[/tex]
, i= 2 as it is a electrolyte and dissociate to give 2 ions, concentration of ions will be [tex]1\times 0.1+1\times 0.1=0.2[/tex]
3. For 0.1 M [tex]LiNO_3[/tex]
[tex]LiNO_3\rightarrow Li^{+}+NO_3^{-}[/tex]
, i= 2 as it is a electrolyte and dissociate to give 2 ions. and concentration of ions will be [tex]1\times 0.1+1\times 0.1=0.2[/tex]
4. For 0.1 M [tex]Na_3PO_4[/tex]
[tex]Na_3PO_4\rightarrow 3Na^{+}+PO_4^{3-}[/tex]
, i= 4 as it is a electrolyte and dissociate to give 4 ions. and concentration of ions will be [tex]3\times 0.1+1\times 0.1=0.4[/tex]
5. For 0.1 M [tex]HC_2H_3O_2(aq)[/tex]
[tex]HC_2H_3O_2(aq)\rightarrow CH_3COO^{-}+H^{+}[/tex] [/tex]
, i= 2 as it is a electrolyte and dissociate to give two ions, concentration of ions will be [tex]2\times 0.1=0.2[/tex]
Thus as concentration of solute is highest for 0.1 M [tex]Na_3PO_4[/tex] , the vapor pressure will be lowest.
Sulfurous acid is a polyprotic acid. Write balanced chemical equations for the sequence of reactions that sulfurous acid can undergo when it's dissolved in water.
Answer:
The reactions are:
H₂SO₃ + H₂ O ⇄ HSO₃⁻ + H₃O⁺
HSO₃⁻ + H₂O ⇄ SO₃⁻² + H₃O⁺
Explanation:
This is the sulfurous acid → H₂SO₃.
It is a weak diprotic acid and it is diprotic because it can release 2 protons to increase H₃O⁺ from water. As every weak acid, the reactions are in equilibrium, that's why it has two Ka. As it can release protons, it is an acid from Bronsted Lowry and an acid from Arrhenius.
H₂SO₃ + H₂ O ⇄ HSO₃⁻ + H₃O⁺ Ka₁
HSO₃⁻ → this is called acid sulfite
HSO₃⁻ + H₂O ⇄ SO₃⁻² + H₃O⁺ Ka₂
2H₂O ⇄ H₃O⁺ + OH⁻ Kw
(Don't forget to put water equation, always)
Sulfurous acid, a diprotic acid, ionizes in two stages when dissolved in water. The first stage produces the hydrogen sulfite ion, HSO3¯, while the second stage further ionizes this ion to form the sulfite ion, SO3²-. The first ionization is more prominent than the second, reflecting the moderate strength of sulfurous acid.
Explanation:Sulfurous acid, being a polyprotic acid, ionizes in two stages when dissolved in water. In the first step, sulfurous acid gives up one proton (H⁺) to produce the hydrogen sulfite ion, HSO3¯:
H2SO3(aq) ⟶ H⁺(aq) + HSO3¯(aq)
In the second stage, the hydrogen sulfite ion can further ionize to give up another proton to form the sulfite ion, SO3²-:
HSO3¯(aq) ⟶ H⁺(aq) + SO3²-(aq)
It's important to note that the ionization is much more significant in the first stage than the second, due to sulfurous acid being a moderately strong acid.
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A scientist adds a solution of HCl (aq) to a buffer comprised of CH3COOH (aq) and CH3COONa (aq). Which net ionic reaction would initially occur?
Explanation:
Net ionic between HCl and the buffer equation
CH3COOH --> acetic acid, CH3COO- acetate , H+ proton from HCl
there is neutralization of the acetate, and an increase in acid
the ionic equation
CH3COO-(aq) + H+(aq) + Cl-(aq) --> CH3COOH(aq) + Cl-(aq)
Net ionic:
CH3COO-(aq) + H+(aq) --> CH3COOH(aq).
Final answer:
H+ ions from added HCl react with the acetate ions in a buffer to form acetic acid, preventing a significant pH change.
Explanation:
When a strong acid like HCl is added to a buffer made of acetic acid (HC₂H₃O₂) and sodium acetate (CH₃COONa), the hydrogen ions (H+) from HCl will react with the acetate ions (C₂H₃O₂-) from sodium acetate in the buffer. The net ionic reaction for this process is:
H*(aq) + C₂H₃O₂-(aq) → HC₂H₃O₂(aq)
The reaction shows that H+ ions are intercepted by the buffer's acetate anions to form acetic acid (HC₂H₃O₂), preventing a significant change in pH and maintaining the buffer's purpose.
½ H₂(g) + ½ I₂(s) --> HI(g) ΔH= 26 kJ/mol
½ H₂(g) + ½ I₂(g) --> HI(g) ΔH= -5.0 kJ/mol
Based on the information above, what is the enthalpy change for the sublimation of iodine, represented by I₂(s) --> I₂(g)?
a) 15 kJ/mol
b) 21 kJ/mol
c) 31 kJ/mol
d) 42 kJ/mol
e) 62 kJ/mol
The enthalpy change for the sublimation of iodine, represented by I₂(s) → I₂(g), is 31 kJ/mol.
Explanation:The enthalpy change for the sublimation of iodine, represented by I₂(s) → I₂(g), can be determined by comparing the enthalpy changes of the two given reactions. Since the equation ½ H₂(g) + ½ I₂(g) → HI(g) has a lower enthalpy change (-5.0 kJ/mol) compared to the equation ½ H₂(g) + ½ I₂(s) → HI(g) (26 kJ/mol), it means that the phase change from solid iodine to gaseous iodine requires an additional amount of energy. Thus, the enthalpy change for the sublimation of iodine is the difference between the two reaction enthalpies, which is 31 kJ/mol.
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Which atom(s) in the molecule CH3OCH3 will display tetrahedral electron-pair geometry? (Select all that apply)a. Ob. Cc. Hd. none of the above
In a tetrahedral molecular geometry, a central atom is located at the center with four substituents that are located at the corners of a tetrahedron.
The number of bonds in the structure is as follows:-
5 bond10 electronsThe angle between the elements is 109.5.
The geometry is how the bonds and the lone pairs of electrons are distributed around the atom, and the angles formed by them occur to minimize the repulsive forces of the bonds and the lone pairs.
For example, the molecule of CO₂ is linear, because the central atom C has no lone pairs (the valence shell of it has 4 electrons, so it shares all of them to be stable), and does only two bonds. But the molecule of H₂O is angular, because the central atom O has two lone pairs (its valence shell has 6 electrons, so it only shares 2 electrons to be stable), and the repulsive forces are minimized with an angle of 104.45°.
To the geometry be tetrahedral, the atom must have 4 bonds and no lone pairs, and it only happens at the two C in the molecule given.
H only does one bond, and O does 2 bonds and has 2 lone pairs.
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In the molecule CH3OCH3, the carbon (C) atoms will display a tetrahedral electron-pair geometry due to the sp³ hybridization commonly found in organic compounds.
Explanation:In the molecule CH3OCH3, the atoms that will display tetrahedral electron-pair geometry are the carbon (C) atoms. This is because carbon atoms, particularly when involved in organic compounds like this one, often form bonds using sp³ hybridization, which results in a tetrahedral shape. The tetrahedral geometry of carbon occurs when it forms four bonds, often with hydrogen and other carbon atoms.
This conformation is one of the most common in organic chemistry and is noteworthy for the fact that all the bonds are 109.5° apart - the ideal bond angle in a true tetrahedral geometry.
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Identify the functional group(s) that appear in the molecule cortisone (a major hormone synthesized by the human adrenal cortex).
Answer:
Ketone, alcohol and unsaturation
Explanation:
The functional groups are the groups that identify the organic function of the molecule. For example, the hydroxyl (-OH) represents alcohol, the carbonyl (represented in the figure below) can represent a ketone or an aldehyde, and so each function has its representation.
The cortisone molecule is represented below, and each group is marked in blue. They are:
a, c, and d = carbonyl of ketone;
b = unsaturation (double bond) of alkene;
e, and f = hydroxyl of alcohol.
What are the coefficients for the following reaction when it is properly balanced? ?potassium iodide +?lead (II) acetate lead (II) iodide +-potassium acetate 3,2,2,1 1,1,2,2 none of these
Answer: none of the above
Explanation:
It should be 2,1,1,2 to give a balanced chemical reaction
To balance the chemical equation between potassium iodide and lead (II) acetate, producing lead (II) iodide and potassium acetate, the coefficients must be 2 (for potassium iodide), 1 (for lead (II) acetate), 1 (for lead (II) iodide), and 2 (for potassium acetate).
Explanation:The question is asking for the correct coefficients that balance the chemical equation between potassium iodide and lead (II) acetate to produce lead (II) iodide and potassium acetate. The balanced chemical equation is:
2KI + Pb(C2H3O2)2 → PbI2 + 2KC2H3O2
The resultant balanced equation shows that two molecules of potassium iodide react with one molecule of lead (II) acetate to yield one molecule of lead (II) iodide and two molecules of potassium acetate. Therefore, the coefficients for the balanced equation are 2, 1, 1, 2 respectively.
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Salts are always ________. Select one: a. single covalent compounds b. hydrogen bonded c. ionic compounds d. double covalent compounds
Answer:
C
Explanation:
The chemical formula of salt is NaCl (sodium chloride).
Na is a metal and Cl is a non-metal.
This means that NaCl is an ionic compound.
Covalent compound is a compound made up of non-metal atoms only.
Salts are always ionic compounds. Therefore, option C is correct.
An ionic compound is a type of chemical compound that is composed of ions held together by electrostatic forces of attraction.
Ionic compounds play vital in various applications, such as in the formation of salts, the electrolysis of compounds, and the construction of solid-state electronic devices.
Salts are a specific type of ionic compound. They are compounds formed when an acid reacts with a base through a chemical reaction called neutralization.
Salts consist of cations (positively charged ions) and anions (negatively charged ions) held together by ionic bonds.
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Using molecular orbital theory, the bond order of the C-C bond in the C2 molecule is _____. g
Answer:
The bond order for C2 molecule is 2.
Explanation:
Bond order can be defined as the half of the difference between the number of electrons in the bonding orbital and the number of electrons in the antibonding orbitals. It can be represented mathematically by; .
Bond order,n= [number of electrons in the bonding molecular orbitals(BMO) - the number or electrons in the anti-bonding molecular orbitals(AMO) ] / 2.
The electronic configuration of the C2 molecule is given below;
C2 = (1sσ)^2 (1s^*σ)^2 (2sσ)^2 (2s^*σ)^2 (2pπ)^4.
The ones with the (*) are known as the Anti-bonding molecular orbitals while the ones without (*) are known as the bonding molecular orbitals. Hence, we have 8 Electrons from the bonding molecular orbitals and 4 Electrons from the anti-bonding molecular orbitals.
So, from the formula given above, the bond order of C2 molecule is;
===> 8-4/2= 4/2.
===> 2.
The bond order of the C-C bond in the C₂ molecule is 1. This is calculated using molecular orbital theory by identifying bonding and antibonding electrons and applying the bond order formula.
The bond order of the C-C bond in the C₂ molecule can be calculated using molecular orbital theory. First, we identify the number of bonding and antibonding electrons in the molecule. The molecular orbital configuration for C₂ is (σ2s)² (σ*2s)² (π2p)⁴. This gives us 4 bonding electrons (from π2p) and 2 antibonding electrons (from σ*2s).
To calculate the bond order, we use the formula:
Bond order = (Number of bonding electrons - Number of antibonding electrons) / 2
Substituting the numbers, we get Bond order = (4 - 2) / 2 = 1.
Therefore, the bond order of the C-C bond in the C₂ molecule is 1.
Carbon dioxide and an unknown gas start to effuse from a container and the carbon dioxide takes 4.69 times as long to escape as the unknown gas. What is the identity of the unknown gas?
Answer: the gas is lighter than CO2 and it is hydrogen
Explanation:Please see attachment for explanation
A block of metal has a width of 3.2 cmcm, a length of 17.1 cmcm, and height of 5.0 cmcm . Its mass is 1.2 kgkg . Calculate the density of the metal. Express your answer to two significant figures and include the appropriate units.
Answer:
≅ 4.39 gm/cm^3
Explanation:
Given data:
width of block =3.2 cm
length = 17.1 cm
height = 5.0 cm
Mass = 1.2 kg
Therefore volume= width×length×height = 3.2×17.1×5= 273.6 cm^3
now, density = mass/volume
calculating density in gm/cm^3
= 1200/273.6 = 4.38596
≅ 4.39 gm/cm^3
A 0.1014 g sample of a purified CHO compound was burned in a combustion apparatus and produced 0.1486 g CO2 and 0.0609 g of H2O. What is the empirical formula of this CHO compound? Enter as C#H#O#, e.g. C2H3O2
Answer: the empirical formula is CH2O
Explanation:Please see attachment for explanation
Determine the pH of (a) a 0.40 M CH3CO2H solution, (b) a solution that is 0.40 M CH3CO2H and 0.20 M NaCH3CO2
Answer:
a) pH = 2.573
b) pH = 4.347
Explanation:
a) weak acid: CH3COOH
CH3COOH + H2O ↔ CH3COO- + H3O+∴ Ka = 1.8 E-5 = [H3O+][CH3COO-] / [CH3COOH]
∴ C CH3COOH = 0.40 M
mass balance:
⇒ 0.40 M = [CH3COO-] + [CH3COOH].........(1)
charge balance:
⇒ [H3O+] = [CH3COO-].........(2)
(2) in (1):
⇒ [CH3COOH] = 0.40 - [H3O+]
replacing in Ka:
⇒ Ka = 1.8 E-5 = [H3O+]² / ( 0.40 - [H3O+] )
⇒ [H3O+]² = 7.2 E-6 - 1.8 E-5[H3O+]
⇒ [H3O+]² + 1.8 E-5[H3O+] - 7.2 E-6 = 0
⇒ [H3O+] = 2.6743 E-3 M
∴ pH = - Log [H3O+]
⇒ pH = 2.573
b) balanced reations:
CH3COONa + H2O → Na+ + CH3COO-CH3COOH + H2O ↔ CH3COO- + H3O+∴ C CH3COOH = 0.40 M
∴ C CH3COONa = 0.20 M
mass balanced:
⇒ C CH3COOH + C CH3COONa = [CH3COO-] + [CH3COOH]
⇒ 0.60 = [CH3COO-] + [CH3COOH]......(1)
charge balanced:
⇒ [H3O+] + [Na+] = [CH3COO-]
∴ [Na+] = 0.20 M
⇒ [H3O+] + 0.20 M = [CH3COO-]........(2)
(2) in (1):
⇒ 0.60 M = ( [H3O+] + 0.20 ) + [CH3COOH]
⇒ [CH3COOH] = 0.40 - [H3O+]
replacing in Ka:
⇒ 1.8 E-5 = ([H3O+])([H3O+] + 0.20) / (0.40 - [H3O+])
⇒ 7.2 E-6 - 1.8 E-5[H3O+] = [H3O+]² + 0.20[H3O+]
⇒ [H3O+]² + 0.20[H3O+] - 7.2 E-6 = 0
⇒ [H3O+] = 4.499 E-5 M
⇒ pH = 4.347
The pH of the solution in (a) is 2.57 The pH of the solution in (b) is 4.4.
We have to set up the ICE table for the reaction;
CH3CO2H + H2O ⇄ H3O^+ + CH3CO2^-
I 0.40 0 0
C -x +x +x
E 0.40 - x x x
The pKa of CH3CO2H is 1.8 x 10-5
Hence,
1.8 x 10-5 = x^2/0.40 - x
1.8 x 10-5 (0.40 - x ) = x^2
7.2 x 10-6 - 1.8 x 10-5x = x^2
x^2 + 1.8 x 10^-5x - 7.2 x 10^-6 = 0
x = 0.00267 M
Hence;
pH = -log [0.00267 M] = 2.57
Using the Henderson Hasselbaclch equation;
pH = pKa + log [A-]/[HA]
pKa = - log Ka = -log[1.8 x 10^-5] = 4.7
Hence;
pH = 4.7 + log [0.20 M]/[0.40 M ]
pH = 4.4
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How long will it take 10.0 mL of Ne gas to effuse through a porous barrier if it has been observed that 125 minutes are required for 10.0 mL of Ar gas to effuse through the same barrier
Answer:
88.8 minutes
Explanation:
Graham's law of diffusion relates rate of difusion by the following formula
Rate1 / rate 2 = √( Mass of argon / Mass of Neon)
Where rate = volume divided by time
Rate 1 = 10 ml / t1
Rate 2 = 10 ml / t2
Rate 1/ rate 2 = 10 ml / t1 ÷ 10 ml/ t2 = t2/ t1
t2/t1 = √(Mass of argon / mass of Neon) = √( 39.984/20.179)
125 / t1 = 1.4026
t1 = 125 / 1.4026 = 88.8 minutes
It will take approximately 95.3 seconds for 10.0 mL of Ne gas to effuse through the porous barrier.
Explanation:To calculate the time it will take for 10.0 mL of Ne gas to effuse through a porous barrier, we can use the effusion rate ratio. In Example 9.21, it is stated that it takes 243 seconds for 4.46 × 10-5 mol Xe to effuse through a tiny hole. Using the effusion rate ratio, we can calculate that it will take 95.3 seconds for 4.46 × 10-5 mol Ne to effuse. Since Ne is lighter than Xe, its effusion rate will be larger, resulting in a smaller effusion time.
Consider the five balanced chemical reactions listed below, all using O2 as a reactant. Normally, O2 is an excess reagent for reactions because there is a large amount of it available in the air. For this question, suppose that you have5.73 moles of O2 and1.70 moles of any of the other reactants in every equation. Select those equations below in which O2 would be the limiting reactant.
A.C6H12O6(s) + 6 O2(g) → 6 CO2(g) + 6 H2O(g)B.C3H8(g) + 5 O2(g) → 3 CO2(g) + 4 H2O(g)C.H2S(g) + 2 O2(g) → SO3(g) + H2O(g)D.4 NH3(g) + 5 O2(g) → 4 NO(g) + 6 H2O(g)E.None because in all the reactions O2 is in excess
Answer: option E. None because in all the reactions O2 is in excess
Explanation:
In the above reactions, only Reaction A and Reaction B have oxygen (O2) as the limiting reactant. For the rest of the reactions, the other compound will be used up before all the O2 is spent.
Explanation:In every reaction, you have 5.73 moles of O2 and 1.70 moles of the other reactants. You can determine which of the two reactants are limiting (which will be consumed first) by comparing the number of moles of that reactant divided by its stoichiometric coefficient (the numbers in front of the chemicals in the reaction) to the same calculation for the other reactant.
For reaction A, C6H12O6(s) + 6 O2(g) → 6 CO2(g) + 6 H2O(g), the ratios are 5.73 moles O2/6 and 1.70 moles C6H12O6/1. So O2 is limiting. For reaction B, C3H8(g) + 5 O2(g) → 3 CO2(g) + 4 H2O(g), the ratios are 5.73 moles O2/5 and 1.70 moles C3H8/1. So O2 is limiting. For reaction C, H2S(g) + 2 O2(g) → SO3(g) + H2O(g), the ratios are 5.73 moles O2/2 and 1.70 moles H2S/1. In this case, H2S is limiting, not O2. For reaction D, 4 NH3(g) + 5 O2(g) → 4 NO(g) + 6 H2O(g), the ratios are 5.73 moles O2/5 and 1.70 moles NH3/4. In this case, NH3 is limiting, not O2. Learn more about Limiting Reactant here:
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A 5.000 gram sample of lenthionine contains 0.6357 g carbon and 0.1070 g hydrogen, and the remainder is sulfur. The molar mass of lenthionine is 188.4 g/mol. Determine the chemical formula for lenthionine.
Answer: C2H4S5
Explanation:
Since the total mass is 5.000g
Mass of sulphur = 5.000-(0.6357+0.1070)
Mass of sulphur = 4.2573g
Using Empirical relation
C= 0.6357 H= 0.1070 S= 4.2573
Divide through by their molar mass to obtain the smallest ratio
C= 0.6357/12 H=0.1070/1 S=4.2573/32
C= 0.053 H= 0.1070 S= 0.133
Divide through by the smallest ratio (0.053)
C=0.053/0.053 H=0.1070/0.053 S=0.133/0.053
C=1 H=2 S=2.5
1:2:2.5 ,multiply through by 2 ,to obtain whole numbera
2:4:5
Therefore the empirical formula is C2H4S5. Thus only gives the ratio
Molecular formula is the chemical formula .
(Empirical formula) n = molecular formula
(C2H4S5)n = molar mass
[(12×2) + ( 1×4) +(32×5)]n = 188.4
188n=188.4
n= 1
Molecular formula = (C2H4S5)×1
Therefore the chemical formula of
lenthionine is C2H4S5
Three atoms of iron (Fe) are to be combined with four molecules of oxygen (O2). How many total iron atoms and oxygen molecules will be combined during the formation of Fe3O4
Final answer:
In the formation of Fe3O4, three atoms of iron (Fe) combine with four molecules of oxygen (O2), involving a total of three iron atoms and eight oxygen atoms to maintain the stoichiometric ratio as per the compound's chemical formula.
Explanation:
The question relates to a chemical reaction where three atoms of iron (Fe) combine with four molecules of oxygen (O2) to form the compound Fe3O4. This compound, known as magnetite, is an example of an iron oxide where the ratio of iron to oxygen atoms is maintained as per the stoichiometry of the compound's chemical formula. In this case, three atoms of iron will combine with oxygen to utilize four oxygen molecules (which equates to eight oxygen atoms), resulting in the formula Fe3O4. This shows the direct application of stoichiometry in understanding chemical formulas and the ratios in which elements combine to form compounds.
If you have 16 g of manganese (II) nitrate tetrahydrate, how much water is required to prepare 0.16 M solution from this amount of salt?
Answer: The amount of water required to prepare given amount of salt is 398.4 mL
Explanation:
To calculate the volume of solution, we use the equation used to calculate the molarity of solution:
[tex]\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}[/tex]
We are given:
Molarity of solution = 0.16 M
Given mass of manganese (II) nitrate tetrahydrate = 16 g
Molar mass of manganese (II) nitrate tetrahydrate = 251 g/mol
Putting values in above equation, we get:
[tex]0.16M=\frac{16\times 1000}{251\times \text{Volume of solution}}\\\\\text{Volume of solution}=\frac{16\times 1000}{251\times 0.16}=398.4mL[/tex]
Volume of water = Volume of solution = 398.4 mL
Hence, the amount of water required to prepare given amount of salt is 398.4 mL
What is the final pH of a solution when 0.1 moles of acetic acid is added to water to a final volume of 1L?
Answer:
The pH of solution is 2.88 .
Explanation:
The reaction is :
[tex]CH_3COOH-->CH_3COO^-+H^+[/tex]
We know, [tex]K_a[/tex] for this reaction is = [tex]1.76\times 10^{-5}.[/tex]
Also, since volume of water is 1 L.
Therefore, molarity of solution is equal to number of moles.
Also, [tex]K_a=\dfrac{[CH_3COO^-][H^+]}{[CH_3COOH]}[/tex]
Let, amount of [tex]CH_3COO^- and\ H^+[/tex] produce is x.
So,
[tex]K_a=\dfrac{[x][x]}{[0.1]}\\1.76\times 10^{-5}=\dfrac{[x][x]}{[0.1]}[/tex]
[tex]x=0.0013\ mol.[/tex]
We know, [tex]pH=-log{x}=-log(0.0013)=2.88[/tex]
Therefore, pH of solution is 2.88 .
Hence, this is the required solution.
Final answer:
The final pH of a 1L solution containing 0.1 moles of acetic acid is determined by using the hydrogen ion concentration resulting from its dissociation. The pH is calculated to be approximately 2.879 following the calculation procedures outlined for weak acids.
Explanation:
Calculating pH of an Acetic Acid Solution
The final pH of a solution can be determined by calculating the hydrogen ion concentration ([H+]) resulting from the dissociation of acetic acid in water. Given that we have 0.1 moles of acetic acid added to water to make 1L of solution, the molarity of acetic acid is 0.100 M. Acetic acid (CH₃COOH) is a weak acid, and its dissociation in water is represented by the following equation:
CH₃COOH <--> CH₃COO⁻ + H⁺
We use the acid dissociation constant (Ka) to solve for [H+]. For acetic acid, Ka = 1.8 x 10⁻⁵ M. The calculation involves setting up an ICE table and using the quadratic formula, as outlined in the subject's textbook or reference. The calculation typically results in a hydrogen ion concentration of 1.32 x 10⁻³ M.
The pH is found by taking the negative logarithm of the hydrogen ion concentration:
pH = -log(1.32 x 10⁻³) = 2.879
Therefore, the pH of a 0.100 M solution of acetic acid is approximately 2.879.
Gas is $2.50/gal at Station A and 85 cents/L at Station B. A) Convert the cost of gas at Station B to $/gal. B) At which station would you buy gas?
Answer:
85 cents/L is equal to 3.2176$/gallon.
gas wiuld be heaper to buy from the station A.
Explanation:
As 85 cents/L is more than $2.5/gallon therefore buying gas from station A would be cheaper than the other one.