Answer:
K = 0,00000135 = 1.35 * 10^-6
Explanation:
Step 1: Data given
The equilibrium constant, K, for any reaction is defined as the concentrations of the products raised by their coefficients divided by the concentrations of the reactants raised by their coefficients. In this case, the concentrations are given as partial pressures.
The partial pressures of H2O = 0.0500 atm
The partial pressures of H2 = 0.00150 atm
The partial pressures of O2 = 0.00150 atm
Step 2: The balanced equation
2H2O(g) ⇆ 2H2(g) + O2(g)
Step 3: Calculate K
K = [O2][H2]² / [H2O]²
K = 0.00150 * 0.00150² / 0.0500²
K = 0,00000135 = 1.35 * 10^-6
Consider the titration of 1L of 0.36 M NH3 (Kb=1.8x10−5) with 0.74 M HCl. What is the pH at the equivalence point of the titration? A) 4.69 B) 4.85 C) 4.94 D) 9.06 E) 9.31
Answer:
C
Explanation:
The question asks to calculate the pH at equivalence point of the titration between ammonia and hydrochloric acid
Firstly, we write the equation of reaction between ammonia and hydrochloric acid.
NH3(aq)+HCl(aq)→NH4Cl(aq)
Ionically:
HCl + NH3 ---> NH4 + Cl-
Firstly, we calculate the number of moles of the ammonia as follows:
from c = n/v and thus, n = cv = 0.36 × 1 = 0.36 moles
At the equivalence point, there is equal number of moles of ammonia and HCl.
Hence, volume of HCl = number of moles/molarity of HCl = 0.36/0.74 = 0.486L
Hence, the total volume of solution will be 1 + 0.486 = 1.486L
Now, we calculate the concentration of the ammonium ions = 0.36/1.486 = 0.242M
An ICE TABLE IS USED TO FIND THE CONCENTRATION OF THE HYDROXONIUM ION(H3O+). ICE STANDS FOR INITIAL, CHANGE AND EQUILIBRIUM.
NH4+ H2O ⇄ NH3 H3O+
I 0.242 0 0
C -X +x +X
E 0.242-X X X
Since the question provides us with the base dissociation constant value K b, we can calculate the acid dissociation constant value Ka
To find this, we use the mathematical equation below
K a ⋅ K b = K w
, where K w- the self-ionization constant of water, equal to
10 ^-14 at room temperature
This means that you have
K a = K w.K b = 10 ^− 14 /1.8 * 10^-5 = 5.56 * 10^-10
Ka = [NH3][H3O+]/[NH4+]
= x * x/(0.242-x)
Since the value of Ka is small, we can say that 0.242-x ≈ 0.242
Hence, K a = x^2/0.242 = 5.56 * 10^-10
x^2 = 0.242 * 5.56 * 10^-10 = 1.35 * 10^-10
x = 0.00001161895
[H3O+] = 0.00001161895
pH = -log[H3O+]
pH = -log[0.00001161895 ] = 4.94
The pH at the equivalence point of the titration is:
C) 4.94
Chemical equation:NH₃(aq)+HCl(aq)→NH₄Cl(aq)
Ionic chemical equation:
HCl + NH₃ ---> [tex]NH_4^+ + Cl^-[/tex]
Calculation for number of moles:c = n/v
n = cv = 0.36 × 1 = 0.36 moles
At the equivalence point, there is equal number of moles of ammonia and HCl.
Hence, volume of HCl = number of moles/molarity of HCl = 0.36/0.74 = 0.486L
Hence, the total volume of solution will be 1 + 0.486 = 1.486L
Calculation for concentration of the ammonium ions = 0.36/1.486 = 0.242M
[tex]NH_4^+[/tex] H₂O ⇄ NH₃ [tex]H_3O^+[/tex]
I 0.242 0 0
C -X +x +X
E 0.242-X X X
Since the question provides us with the base dissociation constant value K b, we can calculate the acid dissociation constant value :Ka
To find this, we use the mathematical equation below:
K a ⋅ K b = K w
Kw = [tex]10^{-14}[/tex]
[tex]K_a = K_w*K_b\\\\ 10^{− 14} /1.8 * 10^{-5} = 5.56 * 10^{-10}\\\\K_a = [NH_3][H_3O^+]/[NH_4^+]\\\\K_a = x * x/(0.242-x)[/tex]
Since, the value of Ka is small, we can say that 0.242-x ≈ 0.242
[tex]K_a = x^2/0.242 = 5.56 * 10^{-10}\\\\x^2 = 0.242 * 5.56 * 10^{-10} = 1.35 * 10^{-10}\\\\x = 0.00001161895\\\\H_3O^+= 0.00001161895\\\\pH = -log[H_3O^+]\\\\pH = -log[0.00001161895 ] = 4.94[/tex]
Thus, correct option is C. The pH at the equivalence point of the titration is: 4.94
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How would the calculated ratio of water molecules present in Epsom salt be affected if the procedure was changed so the analysis was preformed using 2.500g of hydrated salt instead of 1.500g? Assume you still heated the sample to a constant mass. (Too High, Too Low, Unchanged) Explain
Answer:
Unchanged
Explanation:
Epsom salt is also referred to as magnesium sulphate heptahydrate with the chemical formula [tex]MgSO_4.7H_2O.[/tex]
There are 7 moles of water and 1 mole of magnesium sulphate in every 1 mole of epsom salt. The water ratio remains the same irrespective of the amount or weight of epsom salt taken.
Hence, if 2.500 g of epsom salt is taken and heated to a constant mass instead of 1.500 g, the ratio of water molecules present will be the same.
Duncan knows that it takes 36400 calcal of energy to heat a pint of water from room temperature to boiling. However, Duncan has prepared ramen noodles so many times he does not need to measure the water carefully. If he happens to heat 0.900 pintpint of room-temperature water, how many kilojoules of heat energy will have been absorbed by the water at the moment it begins to boil?
Final answer:
To calculate the amount of heat energy that will be absorbed by the water, we need to use the specific heat capacity of water and convert the volume of water to liters. By using the formula Q = mcΔT, we can calculate that 143.32 kJ of heat energy will be absorbed by the water.
Explanation:
To calculate the amount of heat energy that will be absorbed by the water, we need to use the specific heat capacity of water, which is 4184 J/kg/°C. We also need to convert the volume of water from pint to liters. There are approximately 0.473 liters in a pintpint.
First, we convert the volume of water to liters: 0.900 pint × (0.473 liters/pint) = 0.426 liters.
Next, we calculate the temperature change: boiling point (100°C) - room temperature (20°C) = 80°C.
Finally, we use the formula Q = mcΔT, where Q is the heat energy absorbed, m is the mass of water, c is the specific heat capacity of water, and ΔT is the temperature change: Q = (0.426 kg) × (4184 J/kg/°C) × (80°C) = 143,315.52 J. To convert this to kilojoules (kJ), divide by 1000: 143,315.52 J / 1000 = 143.32 kJ.
A generic metal thiocyanate, M(SCN)2, has a Ksp value of 2.00×10−5. Calculate the molar solubility of the metal thiocyanate in 0.421 M KSCN. Express your answer numerically in units of mM to 4 decimal places.
Answer:
The molar solubility of the metal thiocyanate is [tex]1.127\times 10^{-4} M[/tex].
Explanation:
Concentration of potassium thiocyanate = 0.421 M
Concentration of thiocyanate ion =[tex][SCN^-]= 0.421 M[/tex]
Concentration of metal ion =[tex][M^{2+}]= ?[/tex]
The solubility product of metal thiocyanate = [tex]K_{sp}=2.00\times 10^{-5}[/tex]
[tex]M(SCN)_2\rightleftharpoons M^{2+}+2SCN^-[/tex]
S 2S
At equilbrium
S (2S+0.421)
The expression of solubility product is given by :
[tex]K_{sp}=[M^{2+}]\times [SCN^-]^2[/tex]
[tex]2.00\times 10^{-5}=S\times (2S+0.421)^2[/tex]
Solving for S:
[tex]S = 1.128\times 10^{-4} M[/tex]
[tex][M^{2+}]=\frac{2.00\times 10^{-5}}{(0.421 M)^2}=1.127\times 10^{-4} M[/tex]
The molar solubility of the metal thiocyanate is [tex]1.127\times 10^{-4} M[/tex].
A student prepared an equilibrium solution by mixing the following solutions:A.2.00 mL of 0.00250 M Fe(NO3)3B.5.00 mL of 0.00250 M KSCNC.3.00 mL of 0.050 M HNO3Calculate the initial concentrations of all ions, after mixing, prior to the reaction occurring. The equilibrium concentration of Fe(NCS)2+ was determined using a spectrophotometer to be 3.6 x 10-5 M. Calculate the concentrations of all ions at equilibrium. Calculate the value for the equilibrium constant, K.
Explanation:
[tex]Molarity=\frac{\text{Moles of solute}}{\text{Volume of solution Liter}}[/tex]
A. 2.00 mL of 0.00250 M [tex]Fe(NO_3)_3[/tex]
Moles of ferric nitrate = n
Volume of ferric nitrate = 2.00 ml = 0.002 L ( 1 mL=0.001 L)
Molarity of ferric nitrate = 0.00250 M
[tex]n=0.00250 M\times 0.002 L=0.000005 mol[/tex]
B. 5.00 mL of 0.00250 M [tex]KSCN[/tex]
Moles of KSCN = n'
Volume of KSCN = 5.00 ml = 0.005 L ( 1 mL=0.001 L)
Molarity of KSCN = 0.00250 M
[tex]n'=0.00250 M\times 0.005 L=0.0000125 mol[/tex]
C. 3.00 mL of 0.050 M [tex]HNO_3[/tex]
Moles of nitric acid = n''
Volume of nitric acid = 3.00 ml = 0.003 L ( 1 mL=0.001 L)
Molarity of nitric acid = 0.050 M
[tex]n=0.050 M\times 0.003 L=0.00015 mol[/tex]
After mixing A, B and C together and their respective initial concentration before reaction.
After mixing A, B and C together the volume of the solution becomes = V
V = 0.002 L=0.005 L+0.003 L= 0.010 L
Concentration of ferric nitrate :
[tex][Fe(NO_3)_3]=\frac{0.000005 mol}{0.010 L}=0.0005 M[/tex]
Concentration of ferric ions :
[tex][Fe^{3+}]=1\times [Fe(NO_3)_3]=0.0005 M[/tex]
Concentration of nitrate ions from ferric nitrate:
[tex][NO_3^{-}]=3\times [Fe(NO_3)_3]=0.0015 M[/tex]
Concentration of KSCN :
[tex][KSCN]=\frac{0.0000125 mol}{0.010 L}=0.00125 M[/tex]
Concentration of [tex]SCN^-[/tex] ions:
[tex][SCN^-]=1\times [KSCN]=0.00125 M[/tex]
Concentration of potassium ions:
[tex][K^+]=1\times [KSCN]=0.00125 M[/tex]
Concentration of nitric acid :
[tex][HNO_3]=\frac{0.00015 mol}{0.010 L}=0.015 M[/tex]
Concentration of hydrogen ion :
[tex][H^+]=1\times [HNO_3]=0.015 M[/tex]
Concentration of nitrate ions from nitric acid :
[tex][NO_3^{-}]=1\times [HNO_3]=0.0015 M[/tex]
Concentration of nitrate ion in mixture = 0.0015 M + 0.0015 M = 0.0030 M
[tex]Fe^{3+}+SCN^-\rightleftharpoons Fe(NCS)^{2+}[/tex]
given concentration of [tex] Fe(NCS)^{2+}[/tex] at equilbrium = [tex]3.6\times 10^{-5} M = 0.000036 M[/tex]
initially :
0.0005 M 0.00125 M 0
At equilibrium
(0.0005-0.000036) M (0.00125-0.000036) M 0.000036 M
0.000464 M 0.001214 M 0.000036 M
The expression of an equilibrium constant will be given as;
[tex]K_c=\frac{[Fe(NCS)^{2+}]}{[Fe^{3+}][SCN^{-}]}[/tex]
[tex]=\frac{0.000036 M}{0.000464 M\times 0.001214 M}=63.91 [/tex]
The value for the equilibrium constant is 63.91.
Molecular orbitals formation involved in the combination of same type atomic orbitals which also have same symmetry for diatomic molecules.So the correct combinations to the formation of molecular orbitals are,
1s + 1s -----> Ï1s + Ï*1s
2s + 2s -----> Ï2s + Ï*2s
2pz + 2pz -----> Ï2pz + Ï*2pz
2py + 2py -----> Ï2py + Ï*2py
2px + 2px -----> Ï2px + Ï*2px
Answer:2py + 2py -----> Ï2py + Ï*2py
2px + 2px -----> Ï2px + Ï*2px
Explanation:
Molecular orbitals are constructed from atomic orbitals by linear combination of atomic orbitals. The 1s, 2s and 2pz orbitals overlap in an end to end manner hence they only form sigma bonding and anti bonding orbitals. The 2px and 2py orbitals overlap side by side and form pi bonding and anti bonding orbitals. Hence the answer.
Rank the following substances in order from most soluble in water to least soluble in water: methane, CH4; 2-pentanol, C5H11OH; copper(II) sulfate, CuSO4; and propane, C3H8.
The ranking from most soluble to least soluble in water is:
1. Copper(II) sulfate, CuSO4
2. 2-Pentanol, C5H11OH
3. Methane, CH4
4. Propane, C3H8
To rank the substances in order of solubility in water, we can consider the types of intermolecular forces involved. Water is a polar molecule, and substances with polar or ionic characteristics tend to be more soluble in water.
1. **Copper(II) sulfate, CuSO4:**
- This substance is an ionic compound containing copper ions (Cu^2+) and sulfate ions (SO4^2-). Ionic compounds often dissolve well in water through ion-dipole interactions. Thus, copper(II) sulfate is likely to be the most soluble.
2. **2-Pentanol, C5H11OH:**
- 2-Pentanol is a polar molecule due to the presence of the hydroxyl (OH) functional group. It can form hydrogen bonds with water molecules, making it moderately soluble in water.
3. **Methane, CH4:**
- Methane is a nonpolar molecule. It lacks a permanent dipole moment and cannot form strong interactions with water molecules. Thus, it is expected to be less soluble in water than polar substances.
4. **Propane, C3H8:**
- Similar to methane, propane is a nonpolar molecule. It also lacks a permanent dipole moment and is expected to be the least soluble in water among the given substances.
So, the ranking from most soluble to least soluble in water is:
1. Copper(II) sulfate, CuSO4
2. 2-Pentanol, C5H11OH
3. Methane, CH4
4. Propane, C3H8
A sculptor has asked you to help electroplate gold onto a brass statue. You know that the charge carriers in the ionic solution are monovalent (charge e) gold ions, and you've calculated that you must deposit 0.60 g of gold to reach the necessary thickness.
How much current do you need, in mA, to plate the statue in 3.5 hr?
Answer: The current needed, in mA, to plate the statue in 3.5 hr is 20 mA
Explanation:
Moles of electron = 1 mole
According to mole concept:
1 mole of an atom contains [tex]6.022\times 10^{23}[/tex] number of particles.
We know that:
Charge on 1 electron = [tex]1.6\times 10^{-19}C[/tex]
Charge on 1 mole of electrons = [tex]1.6\times 10^{-19}\times 6.022\times 10^{23}=9.6352\times 10^4C[/tex]
[tex]Au^++e^-\rightarrow Au[/tex]
197 g of gold is deposited by = 96500 C of electricity
Thus 0.60 g of gold is deposited by =[tex]\frac{96500}{197}\times 0.60=294 C[/tex] of electricity
To calculate the current required, we use the equation:
[tex]I=\frac{q}{t}[/tex]
where,
I = current passed = ?
q = total charge = [tex]294C[/tex]
t = time required = 3.5 hrs =[tex]3.5\times 3600=12600s[/tex]
Putting values in above equation, we get:
[tex]I=\frac{294C}{12600}\\\\I=0.02A=20mA[/tex]
Hence, the current needed, in mA, to plate the statue in 3.5 hr is 20 mA
) Any gas exerts a pressure against its surroundings. For example, if you put more gas in a balloon, it exerts more pressure and expands the balloon. Describe what causes the pressure in a gas, in terms of the fact that molecules of gas are moving.
Explanation:
As we know that molecules of a gas move far away from each other because they are held by weak Vander waal forces. So, when a balloon is filled by a gas then molecules of a gas strike at its walls leading to more number of collisions.
This will also lead to more expansion of the balloon. As we know that pressure is the force exerted n per unit area of a substance or object.
Thus, more is the kinetic energy of the molecules of the gas more will be the force exerted by them. Hence, more will be the pressure exerted.
A solution is prepared by dissolving 27.0 g of urea [(NH2)2CO], in 150.0 g of water. Calculate the boiling point of the solution. Urea is a nonelectrolyte.
Answer: The boiling point of solution is 101.56°C
Explanation:
Elevation in boiling point is defined as the difference in the boiling point of solution and boiling point of pure solution.
The equation used to calculate elevation in boiling point follows:
[tex]\Delta T_b=\text{Boiling point of solution}-\text{Boiling point of pure solution}[/tex]
To calculate the elevation in boiling point, we use the equation:
[tex]\Delta T_b=iK_bm[/tex]
Or,
[tex]\text{Boiling point of solution}-\text{Boiling point of pure solution}=i\times K_b\times \frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams)}}[/tex]
where,
Boiling point of pure water = 100°C
i = Vant hoff factor = 1 (For non-electrolytes)
[tex]K_b[/tex] = molal boiling point elevation constant = 0.52°C/m.g
[tex]m_{solute}[/tex] = Given mass of solute (urea) = 27.0 g
[tex]M_{solute}[/tex] = Molar mass of solute (urea) = 60 g/mol
[tex]W_{solvent}[/tex] = Mass of solvent (water) = 150.0 g
Putting values in above equation, we get:
[tex]\text{Boiling point of solution}-100=1\times 0.52^oC/m\times \frac{27\times 1000}{60\times 150}\\\\\text{Boiling point of solution}=101.56^oC[/tex]
Hence, the boiling point of solution is 101.56°C
A solution is prepared by dissolving 27.0 g of urea in 150.0 g of water has a boiling point of 101.54 °C.
The normal boiling point of water is 100 °C. However, when 27.0 g of urea is dissolved in 150.0 g of water, we expect the boiling point of the solution to be higher.
What is the boiling point elevation?Boiling point elevation is the phenomenon that occurs when the boiling point of a liquid is increased when another compound is added. It is a colligative property. We can calculate the increase in the boiling point using the following expression.
ΔT = i × Kb × m
where,
i is the Van't Hoff factor (i = 1 for nonelectrolytes).Kb is the molal boiling point constant (Kb = 0.513 °C/m for water).m is the molality.What is molality?Molality (m) is defined as the total moles of a solute contained in a kilogram of a solvent. We can calculate it using the following expression.
m = mass solute / molar mass solute × kg solvent
m = 27.0 g / (60.06 g/mol) × 0.1500 kg = 3.00 m
The boiling point elevation is:
ΔT = i × Kb × m = 1 × (0.513 °C/m) × 3.00 m = 1.54 °C
Then, the boiling point of the solution is:
T = 100 °C + 1.54 °C = 101.54 °C
A solution is prepared by dissolving 27.0 g of urea in 150.0 g of water has a boiling point of 101.54 °C.
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Germanium forms a substitutional solid solution with silicon. Compute the weight percent of germanium that must be added to silicon to yield an alloy that contains 3.43 × 1021 Ge atoms per cubic centimeter. The densities of pure Ge and Si are 5.32 and 2.33 g/cm3, respectively. The atomic weights for germanium and silicon are 72.64 and 28.09 g/mol, respectively.
Answer:
The answer to tnhe question is;
The weight percent of germanium to be added is 16.146 %.
Explanation:
To solve the question, we note that the formula to calculate the weight percent of an element in terms of the number of atoms per cm³ in a 2 element alloy is given by,
[tex]C_1=\frac{100}{1+\frac{N_A\rho_2}{N_1A_1} -\frac{\rho_2}{\rho1} }[/tex]
Where
N[tex]_A[/tex] = Avogadro's Number
ρ₁ = Density of alloy whose weight percent is sought
ρ₂ = density of the other alloy
N₁ = Number of atoms per cubic centimeter
A₁ = Atomic weight of the element whose weight percent is sought
Therefore
[tex]C_{Ge}=\frac{100}{1+\frac{(6.022*10^{23})*(2.33)}{(3.43*10^{21})*(72.64)} -(\frac{2.33}{5.32}) } = \frac{100}{1+5.632 -0.43797 } = 16.146 %[/tex]
[tex]C_{Ge}[/tex] = 16.146 %.
Changes in pressure have a measurable effect: (select all that apply) Select all that apply:
in any system only in systems in which gases are involved
when there are equal numbers of moles of gas on the reactant and product sides of the equilibrium only
when the chemical reaction produces a change in the total number of gas molecules in the system
Answer:
1. only in systems in which gases are involved
2. only when the chemical reaction produces a change in the total number of gas molecules in the system
Explanation:
According to Le Chatelier's principle, pressure will only affect a system in equilibrium containing gaseous reactants and products. However, change in the pressure will only affect the gaseous system in which the total number of moles of the reactants are different from the total number of moles of the products.
Pressure changes affect equilibrium systems involving gases, particularly when the reaction results in different numbers of gas molecules on each side. Thus option A is correct.
Pressure changes in a system at equilibrium have a measurable effect primarily in gaseous systems. Here's how they work:
Pressure changes impact systems in which gases are involved.The effect is significant when there is a change in the total number of gas molecules in the system.If the number of gas moles is equal on both sides of the reaction, there is no effect on equilibrium.This occurs because increasing the pressure (by decreasing the volume) will shift the equilibrium toward the side with fewer moles of gas, while decreasing the pressure (by increasing the volume) will shift it toward the side with more moles of gas.
Complete question:
Changes in pressure have a measurable effect: (select all that apply) Select all that apply:
A. in any system only in systems in which gases are involved
B. when there are equal numbers of moles of gas on the reactant and product sides of the equilibrium only
C. when the chemical reaction produces a change in the total number of gas molecules in the system
D. None of these
A 0.9679-g sample containing dimethylphthalate, (194.19 g/mol), and unreactive species was refluxed with 50.00 mL of 0.1215 M to hydrolyze the ester groups (this process is called saponification).
C6H4(COOCH3)2 + 2OH----->> C6H4(COO)-2 + H2O
After the reaction was complete, the excess NaOH was back titrated with 32.25mL of 0.1251M HCl. Calculate the percentage of dimethylphthalate in the sample.
Answer:
20.44% is the percentage of dimethylphthalate in the sample.
Explanation:
Molarity of NaOH = [tex]M_1=0.1215 M[/tex]
Volume of NaOH consumed in back titration = [tex]V_1=?[/tex]
Molarity of HCl =[tex]M_2=0.1251 M[/tex]
Volume of HCl = [tex]V_2=32.25 ml[/tex]
[tex]M_1V_1=M_2V_2[/tex]
[tex]V_1=\frac{M_2V_2}{M_1}=\frac{0.1251 M \times 50.0 mL}{0.1215 M}[/tex]
[tex]V_1=33.21 mL[/tex]
Volume of NaOH used in hydrolysis of ester = 50.00 mL - 33.21 mL = 16.79 mL[/tex]
Moles of NaOH in 16.79 ml of 0.1215 M solution : n
Volume of solution = 16.79 mL = 0.01679 L ( 1mL=0.001 L)
[tex]n=0.1215 M\times 0.01679 L =0.002040 mol[/tex]
1 mole of NaOH has 1 mole of hydroxide ions than 0.002040 moles of NaOH will have :
1 × 0.002040 mol = 0.002040 mol of hydroxide ions
Moles of hydroxide ions = 0.002040 mol
[tex]C_6H_4(COOCH_3)_2 + 2OH^-\rightarrow C_6H_4(COO)^{2-} + 2H_2O[/tex]
According to reaction, 2 moles of hydroxide ion reacts with 1 mole of dimethylphthalate , then 0.002040 moles of hydroxide ion swill react with ;
[tex]\frac{1}{2}\times 0.002040 mol=0.001020 mol[/tex] of dimethylphthalate
Mass of 0.001020 moles of dimethylphthalate :
194 g/mol × 0.001020 mol = 0.1979 g
Mass of sample = 0.9679 g
Mass of dimethylphthalate = 0.1979 g
Percentage of dimethylphthalate in sample;
[tex]=\frac{0.1979 g}{0.9679 g}\times 100=20.44\%[/tex]
20.44% is the percentage of dimethylphthalate in the sample.
Final answer:
The percentage of dimethylphthalate in the sample is calculated by first determining the moles of NaOH that reacted with it, then converting those moles to grams of dimethylphthalate, and finally dividing this by the original sample mass. The calculated percentage is approximately 20.48%.
Explanation:
The question pertains to the calculation of the percentage of dimethylphthalate in a sample after a saponification reaction followed by a back titration. Firstly, we need to calculate the amount of NaOH that reacted with dimethylphthalate. This is done by determining the amount of NaOH remaining after the reaction, which is found through the back titration with HCl. The total amount of NaOH initially added to the reaction is 50.00 mL of 0.1215 M, which equals 6.075 mmol. The back titration used 32.25 mL of 0.1251 M HCl, which equals 4.033 mmol of NaOH remaining.
Thus, the amount of NaOH that reacted with dimethylphthalate is 6.075 mmol - 4.033 mmol = 2.042 mmol. Since the saponification of dimethylphthalate requires two moles of NaOH for each mole of ester, the moles of dimethylphthalate are half this amount, which is 1.021 mmol. Multiplying by the molar mass of dimethylphthalate (194.19 g/mol) gives 0.1983 g of dimethylphthalate in the sample.
Finally, the percentage of dimethylphthalate in the sample is calculated by dividing the mass of dimethylphthalate by the original sample mass and multiplying by 100. Therefore, it is (0.1983 g / 0.9679 g) × 100 = 20.48%.
NaCl is purified by adding HCl to a saturated solution of NaCl (317 g/L). Will pure NaCl precipitate when 30 mL of 4.5 M HCl is added to 0.15 L of saturated solution? Group of answer choices NaCl will precipitate from solution No NaCl will precipitate
The thing which will happen when NaCl is purified by adding HCl to a saturated solution of NaCl (317 g/L) is
NaCl will not precipitate.What is a Saturated Solution?This refers to the type of solution which has a lot of solute that enables the solution to dissolve.
Hence, if NaCl is purified by the addition of HCl to a saturated solution of NaCl (317 g/L) and 30 mL of 4.5 M HCl is added to 0.15 L of saturated solution, then the NaCl will not precipitate because it cannot be transformed the dissolved substance to an insoluble solid as a result of the extra HCL added.
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Assuming each solution to be 0.10 M , rank the following aqueous solutions in order of decreasing pH.
Rank the solutions from the highest to lowest pH. To rank items as equivalent, overlap them
a. N2H2
b. Ba(OH)2
c. HOCL
d. NAOH
e. HCL
Answer: Highest to lowest pH: Ba(OH)2, NaOH, N2H2, HOCL HCl
Explanation:
Stronger the acid, lower the pH, stronger the base, higher the pH.
N2H2 weak base, Ba(OH)2 strong base, HOCL weak acid, NaOH strong base, HCl strong acid.
Ba(OH)2 has produces more H ions, so it has a higher pH than NaOH.
Identify the statement that is FALSE. Identify the statement that is FALSE. Entropy increases with dissolution. Entropy generally increases with increasing molecular complexity. For noble gasses, entropy increases with size. The entropy of a gas is greater than the entropy of a liquid. Free atoms have greater entropy than molecules.
Answer:
The statement that is FALSE.
Entropy increases with dissolution.
Entropy generally increases with increasing molecular complexity.
For noble gasses, entropy increases with size.
The entropy of a gas is greater than the entropy of a liquid.
Free atoms have greater entropy than molecules.
Explanation:
Free atoms have greater entropy than molecules.
The energy is increased in faster moving particles, the opposite happens in slower ones. The entropy increases if there is randomized distribution of the energy, the system becomes then more stable as energy is released in a cluster.
The false statement is: 'Free atoms have greater entropy than molecules'. Because complexity brings more possibilities for microstates, actual molecules usually have greater entropy than single atoms. All of the other statements provided in the original query are correct.
Explanation:In the given context, the statement that is false is: "Free atoms have greater entropy than molecules". Entropy, which is a measure of randomness or disorder in a system, is influenced by the structure of the particles (atoms or molecules) that comprise a substance.
More complex molecules, with greater number of atoms, typically have a higher entropy than solitary atoms because they have more ways to vibrate and thus more potential microstates, increasing the entropy of the system. This goes contrary to the false statement made.All the other statements are true. For instance, entropy does increase with dissolution, reflecting the increase in molecular disorder and the number of potential microstates.
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The following Lewis diagram represents the valence electron configuration of a main-group element. This element is in group 2A According to the octet rule, this element would be expected to form a(n) with a charge of cation anion If X is in period 4, the ion formed has the same electron configuration as the noble gas The symbol for the ion is________
Answer: The symbol of the ion formed is [tex]Ca^{2+}[/tex]
Explanation:
An ion is formed when a neutral atom looses or gains electrons.
When an atom looses electrons, it results in the formation of positive ion known as cation.When an atom gains electrons, it results in the formation of negative ion known as anion.Electronic configuration is defined as the representation of electrons around the nucleus of an atom.
Number of electrons in an atom is determined by the atomic number of that atom.
The element present in Group 2-A and in period 4 is Calcium (Ca)
Electronic configuration of Ca atom: [tex]1s^22s^22p^63s^23p^64s^2[/tex]
This atom will loose 2 electrons to attain stable electronic configuration similar to Argon element (noble gas)
The electronic configration of [tex]Ca^{2+}\text{ ion}=1s^22s^22p^63s^23p^6[/tex]
Hence, the symbol of the ion formed is [tex]Ca^{2+}[/tex]
Answer: The octet rule says that in forming ions, main-group atoms gain or lose electrons in order to attain a noble gas electron configuration. Except for the He configuration, this means that atoms gain or lose electrons to have an octet of electrons in the outermost shell.
Explanation: If x is in period 4, the ion formed has the same electron configuration as the noble ga
The boiling point of chloroform is 61.7 C. The enthalapy of vaporization is 31.4 kj/mol. Caculate the entropy of vaporization. Does the sign for entropy of vaporizatiou make sense
Answer:
Δ S = 93.8 J/mol-K
Explanation:
Given,
Boiling point of chloroform = 61.7 °C
= 273 + 61.7 = 334.7 K.
Enthalapy of vapourization = 31.4 kJ/mol.
Using Gibbs free energy equation
Δ G = Δ H - T (ΔS)
at equilibrium (when the liquid is boiling), Δ G = 0
so, 0 = ΔH - T (Δ S)
T (Δ S) = Δ H
and ΔS = ΔH / T
Δ S = (31400 J/mol.) / 334.7 K
Δ S = 93.8 J/mol-K
Final answer:
The molar entropy of vaporization (ΔvapS) of acetone can be calculated using the formula ΔvapS = ΔvapH / T. By converting the boiling point to Kelvin and substituting the values, we find ΔvapS to be 88.4 J/mol·K. The positive sign of the entropy change is expected as vaporization increases disorder.
Explanation:
The student's question pertains to the calculation of the molar entropy of vaporization (ΔvapS) of acetone at its normal boiling point. Given the normal boiling point of acetone is 56°C, and its molar enthalpy of vaporization (ΔvapH) is 29.1 kJ/mol, we can calculate the molar entropy of vaporization using the equation:
ΔvapS = ΔvapH / T
Where T is the absolute temperature in Kelvin (K). To obtain T, convert the boiling point from Celsius to Kelvin:
T = 56°C + 273.15 = 329.15 K
Now substituting the values into the equation:
ΔvapS = 29.1 kJ/mol / 329.15 K
ΔvapS = 0.0884 kJ/mol·K or 88.4 J/mol·K
The sign for entropy of vaporization is positive, which makes sense since the entropy of the system is expected to increase when a liquid turns into a gas due to the increase in disorder.
Fission reactions are induced in nuclear power plants because they produce great amounts of energy. Often, a nucleus must be forced to undergo fission by being hit by another particle. A U-235 nucleus being hit with some other particle causes it to undergo fission and emit more particles. Those particles eventually hit other U-235 nuclei, which propagates the reaction. What is this type of reaction called
Answer:
Chain reaction
Explanation:
A chain reaction is a reaction that sustains itself. It has the ability to continue for a very long time without adding any more materials to the reaction system. It may be succinctly described as a self propagating reaction.
In a nuclear fission, uranium-235 is bombarded with neutrons to produce unstable uranium-236 which disintegrates to form daughter nuclei and produce more neutrons that bombard more uranium-235 and the reaction continues indefinitely.
Initial alcohol is converted to compound A. Compound A undergoes three different transformations. Name compound A and each the final product in three transformations.
Explanation:
Initial alcohol that is formed from methane is methanol that can be converted to acetic acid.Methanol in presence of [tex]CO_2[/tex] and hydrogen gas gets oxidized to acetic acid with the release of water.Hence, compound A will be Acetic acid.
[tex]CH_3OH + CO_2 +H_2[/tex] → [tex]CH_3COOH + H_2O[/tex]
The acetic acid formed can be transformed into -[tex]CH_3COOH[/tex] → [tex]CH_4 + CO[/tex]The product formed is methane and carbon monoxide.
2. [tex]CH_3COOH[/tex] → [tex]CH_2CHO + H_2O[/tex]
The product formed is formaldehyde and water.
3. [tex]CH_3COOH + NaHCO_3[/tex] → [tex]CH_3COONa +CO_2 +H_2O[/tex]
The product formed is sodium acetate, carbon dioxide, and water.
When aqueous solutions of magnesium iodide and sodium carbonate are combined, solid magnesium carbonate and a solution of sodium iodide are formed. The net ionic equation for this reaction is:______
Answer:
Mg+2 (aq) + CO3-2 (aq) ---> MgCO3 (s)
Explanation:
The net ionic equation for the reaction of magnesium iodide and sodium carbonate is Mg⁺² (aq) + CO₃⁻²(aq) ---> MgCO₃ (s)
What is an ionic equation?The ionic equations are those in which the electrolytes are dissociated into the ion of a solution.
The equations in which the written in dissociated ions of an electrolyte solution.
Thus, the ionic equation is Mg⁺² (aq) + CO₃⁻²(aq) ---> MgCO₃ (s).
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Some CH2Cl2 is placed in a sealed flask and heated to 517 K. When equilibrium is reached, the flask is found to contain CH2Cl2 (3.42×10-2 M), CH4 (3.69×10-2 M), and CCl4 (4.12×10-2 M). What is the value of the equilibrium constant for this reaction at 517 K?
Answer: The value of equilibrium constant for the given reaction at 517 K is 1.30
Explanation:
The chemical equation for the dissociation of [tex]CH_2Cl_2[/tex] follows:
[tex]2CH_2Cl_2(g)\rightleftharpoons CH_4(g)+CCl_4(g)[/tex]
The expression of [tex]K_{eq}[/tex] for above equation follows:
[tex]K_{eq}=\frac{[CH_4][CCl_4]}{[CH_2Cl_2]^2}[/tex]
We are given:
[tex][CH_4]_{eq}=3.69\times 10^{-2}M[/tex]
[tex][CCl_4]_{eq}=4.12\times 10^{-2}M[/tex]
[tex][CH_2Cl_2]_{eq}=3.42\times 10^{-2}M[/tex]
Putting values in above expression, we get:
[tex]K_{eq}=\frac{(3.69\times 10^{-2})\times (4.12\times 10^{-2})}{(3.42\times 10^{-2})^2}\\\\K_{eq}=1.30[/tex]
Hence, the value of equilibrium constant for the given reaction at 517 K is 1.30
A volume of 40.0 mLmL of aqueous potassium hydroxide (KOHKOH) was titrated against a standard solution of sulfuric acid (H2SO4H2SO4). What was the molarity of the KOHKOH solution if 16.2 mLmL of 1.50 MM H2SO4H2SO4 was needed? The equation is 2KOH(aq)+H2SO4(aq)→K2SO4(aq)+2H2O(l)
Answer: The concentration of KOH solution is 1.215 M
Explanation:
For the given chemical equation:
[tex]2KOH(aq.)+H_2SO_4(aq.)\rightarrow K_2SO_4(aq.)+2H_2O(l)[/tex]
To calculate the concentration of base, we use the equation given by neutralization reaction:
[tex]n_1M_1V_1=n_2M_2V_2[/tex]
where,
[tex]n_1,M_1\text{ and }V_1[/tex] are the n-factor, molarity and volume of acid which is [tex]H_2SO_4[/tex]
[tex]n_2,M_2\text{ and }V_2[/tex] are the n-factor, molarity and volume of base which is KOH.
We are given:
[tex]n_1=2\\M_1=1.50M\\V_1=16.2mL\\n_2=1\\M_2=?M\\V_2=40.0mL[/tex]
Putting values in above equation, we get:
[tex]2\times 1.50\times 16.2=1\times M_2\times 40.0\\\\M_2=\frac{2\times 1.50\times 16.2}{1\times 40.00}=1.215M[/tex]
Hence, the concentration of KOH solution is 1.215 M
Calculate the equilibrium concentration of H 3 O H3O in a 0.20 M M solution of oxalic acid. Express your answer to two significant figures and include the appropriate units.
Answer: The equilibrium concentration of [tex]H_3O^+[/tex] ion is [tex]8.3064\times 10^{-2}M[/tex]
Explanation:
We are given:
Molarity of oxalic acid solution = 0.20 M
Oxalic acid [tex](H_2C_2O_4)[/tex] is a weak acid and will dissociate 2 hydrogen ions.
The chemical equation for the first dissociation of oxalic acid follows:[tex]H_2C_2O_4(aq.)+H_2O\rightleftharpoons H_3O^+(aq.)+HC_2O_4^-(aq.)[/tex]
Initial: 0.20
At eqllm: 0.20-x x x
The expression of first equilibrium constant equation follows:
[tex]Ka_1=\frac{[H_3O^+][HC_2O_4^{-}]}{[H_2C_2O_4]}[/tex]
We know that:
[tex]Ka_1\text{ for }H_2C_2O_4=0.059[/tex]
Putting values in above equation, we get:
[tex]0.059=\frac{x\times x}{(0.20-x)}\\\\x=-0.142,0.083[/tex]
Neglecting the negative value of 'x', because concentration cannot be negative.
So, equilibrium concentration of hydronium ion = x = 0.083 M
The chemical equation for the second dissociation of oxalic acid:[tex]HC_2O_4^-(aq.)+H_2O\rightarrow H_3O^+(aq.)+C_2O_4^{2-}(aq.)[/tex]
Initial: 0.083
At eqllm: 0.083-y 0.083+y y
The expression of second equilibrium constant equation follows:
[tex]Ka_2=\frac{[H_3O^+][C_2O_4^{2-}]}{[HC_2O_4^-]}[/tex]
We know that:
[tex]Ka_2\text{ for }H_2C_2O_4=6.4\times 10^{-5}[/tex]
Putting values in above equation, we get:
[tex]6.4\times 10^{-5}=\frac{(0.083+y)\times y}{(0.083-y)}\\\\y=-0.083,0.0000639[/tex]
Neglecting the negative value of 'x', because concentration cannot be negative.
So, equilibrium concentration of hydronium ion = y = 0.0000639 M
Total concentration of hydronium ion = [x + y] = [0.083 + 0.0000639] = 0.0830639 M
Hence, the equilibrium concentration of [tex]H_3O^+[/tex] ion is [tex]8.3064\times 10^{-2}M[/tex]
The equilibrium concentration of H3O+ in a solution of oxalic acid is linked to the acid dissociation constant Ka. Calculation of this requires specific values related to the nature of oxalic acid, a weak acid, and its behaviour in solution.
Explanation:In order to find the equilibrium concentration of H3O+ in a solution of oxalic acid, we must first understand that oxalic acid is a weak acid, and thus does not fully ionize in solution. In the ionization process of oxalic acid H2C2O4, it would donate a proton (H+) to water (H2O), forming a hydronium ion (H3O+) and a mono hydrogen oxalate ion. The equilibrium concentration of H3O+ (hydronium ions) thus corresponds to the acid dissociation constant Ka, with the equation Ka=[H3O+][HC2O4-]/[H2C2O4]. Considering the initial concentration of the oxalic acid, specific calculations will be required to find an accurate equilibrium concentration which is not possible without the value of Ka. So, the specific answer depends on the Ka value of oxalic acid.
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Group the elements into pairs that would most likely exhibit similar chemical properties. It does not matter which pair of elements is pair 1, pair 2, or pair 3, so long as the correct elements are paired. Pair 1 Pair 2 Pair 3
a. F K
b. P
c. As
d. Br Li
Explanation:
A property that leads to changes in chemical composition of a substance is known as chemical property. So, when we move down a group in periodic table then chemical properties of the elements remain the same.
This is because elements of the same group tend to have same number of valence electrons. Therefore, they have similar reactivity which leads to change in their chemical composition upon reaction with another substance.
For example, lithium (Li) and potassium (K) are both group 1 elements. And, phosphorous (P) and arsenic (As) are both group 15 elements.
Thus, we can conclude that pair of Li and K will show similar chemical properties. Also, pair of P and As will show similar chemical properties.
A cylindrical specimen of some metal alloy having an elastic modulus of 129 GPa and an original cross-sectional diameter of 4.4 mm will experience only elastic deformation when a tensile load of 1570 N is applied. Calculate the maximum length of the specimen before deformation if the maximum allowable elongation is 0.48 mm.
Explanation:
The cross-sectional area of the specimen is calculated as follows.
[tex]A_{o} = \frac{pi}{4} d^{2}[/tex]
= [tex]\frac{3.14}{4} \times (\frac{4.4}{1000})^{2}[/tex]
= [tex]1.5197 \times 10^{-5} m^{2}[/tex]
Equation of stress is as follows.
[tex]\sigma = \frac{F}{A_{o}}[/tex]
And, the equation of strain is as follows.
[tex]\epsilon = \frac{\Delta l}{l_{o}}[/tex]
Hence, the Hook's law is as follows.
E = [tex]\frac{\sigma}{\epsilon}[/tex]
E = [tex]\frac{\frac{F}{A_{o}}}{\frac{\Delta l}{l_{o}}}[/tex]
= [tex]\frac{F \times l_{o}}{A_{o} \times \Delta l}[/tex]
or, [tex]l_{o} = \frac{E \times \Delta l \times A_{o}}{F}[/tex]
= [tex]\frac{129 \times 10^{9} \times \frac{0.48}{1000} \times 1.662 \times 10^{-5}}{1570}[/tex]
= 0.6554 m
or, [tex]l_{o}[/tex] = 655.4 mm
Thus, we can conclude that the maximum length of the specimen before deformation if the maximum allowable elongation is 0.48 mm is 655.4 mm.
A moist soil sample weighing 5 lb has a moisture content of 2%. How many mL of water must be added to the sample to increase the moisture content to 10%? (This calculation is needed for your Compaction lab)
Answer:
The answer is 181.437 ml of water to increase the soil's moisture content to 10%.
Explanation:
We have been given:
Soil weight = 5 lb = 2.26796 kg
2% of moisture ⇒ [tex]\frac{2}{100}[/tex] × 2.26796 = 0.045359 kg of water in the soil
10% of moisture ⇒ [tex]\frac{10}{100}[/tex] × 2.26796 = 0.226796 kg of water in the soil
The difference in moisture content = (0.226796 - 0.045359) kg
=0.181432 kg
But 1kg = 1000 ml
∴ 0.181437 kg = 181.437 ml of water required to increase the moisture content to 10%
Carbon dioxide and water react to form methanol and oxygen, like this: CO2(g) + H2O (g) ----> CH3OH (l) + O2 (g) At a certain temperature, a chemist finds that 8.6 L a reaction vessel containing a mixture of carbon dioxide, water, methanol, and oxygen at equilibrium has the following composition: compound amount CO2 2.25 gH2O 2.72 gCH3OH 3.82 gO2 1.98 gCalculate the value of the equilibrium constant for this reaction. Round your answer to significant digits.
Final answer:
The concentration equilibrium constant (Kc) will not include CH₃OH because it is a pure liquid. Then, Kc = 68.
Explanation:
First, we will calculate the molar concentration (M) of each substance using the following expression.
M = mass of the substance / molar mass of the substance × volume of solution
CO₂
M = 2.25 g / 44.01 g/mol × 8.6 L = 0.0059 M
H₂O
M = 2.72 g / 18.02 g/mol × 8.6 L = 0.018 M
CH₃OH
M = 3.82 g / 32.04 g/mol × 8.6 L = 0.014 M
O₂
M = 1.98 g / 32.00 g/mol × 8.6 L = 0.0072 M
Let's consider the following reaction at equilibrium.
CO₂(g) + H₂O(g) ⇄ CH₃OH(l) + O₂(g)
The concentration equilibrium constant (Kc) will not include CH₃OH because it is a pure liquid. Then,
Kc = [O₂] / [CO₂] × [H₂O]
Kc = 0.0072 / 0.0059 × 0.018
Kc = 68
Gold has always been a highly prized metal, and it has been widely used from the beginning of history as a store of value. It does not rust like iron and does not become tarnished like silver. It is so chemically inert that it will not react with even the strongest concentrated acids. But it can be dissolved in aqua regia – a fresh-prepared mixture of concentrated HNO3 and HCl (1:3).
When Germany invaded Denmark in World War II, the Hungarian chemist George de Hevesy dissolved the gold Nobel Prizes of Max von Laue and James Franck in aqua regia to prevent the Nazis from stealing them. He placed the jar with the solution on a shelf in his laboratory, and after the war, precipitated the gold out of the acid and returned it to the Royal Swedish Academy of Sciences and the Nobel Foundation who recast the medals and again presented them to Laue and Franck.
The unbalanced equation for the reaction of gold with aqua regia is given below.
Add the stoichiometric coefficients to the equation to balance it.
Au(s) + HNO3(aq) + HCl(aq) → HAUCl4(aq) + NO2(g) + H20(l)
What's the function of HCL?
Answer: The balanced equation is
Au(s) + 3HNO3(aq) + 4HCl(aq) ---> HAuCl4(aq) + 3NO2(g) +3H2O(l)
The function of HCl in a solution of Aqua regia, that is used to dissolve gold is to dissolve other metals like quartz or iron stone that surround the gold.
Explanation: To balance the equation, we check the ratio of each element in the reacting side and the product side ( left and right hand side). Let their ratio be equal be adding moles to the compound of the element or the element it's self in either side of the equation.
HCl which is called hydrochloric acid, is an acid that does not react with gold, but it react with every other substance, like your skin, metals etc. it is used to clean a gold, by dipping the gold inside it, all the metals on the surface of the gold will dissolve.
When dissolving a gold in aqua regia solution, HCL is added to prepare this solution because it will help to dissolve all other substance on the surface of the gold.
Thermal energy is Question 2 options: A) solar energy, i.e. energy that comes from the sun. B) the energy stored within the structural units of chemical substances. C) energy available by virtue of an object's position. D) the energy associated with the random motion of atoms and molecules.
Answer:
D) the energy associated with the random motion of atoms and molecules.
Explanation:
Thermal energy is the manifestation of energy in the form of heat. In all materials, the atoms that make up their molecules are in continuous movement, either moving or vibrating, which implies that the atoms have a certain kinetic energy that we call heat or thermal energy. In a way, thermal energy is the internal energy of a body.
Final answer:
Thermal energy is the energy associated with the random motion of atoms and molecules, a type of kinetic energy that increases with object's temperature, making Option D the correct choice.
Explanation:
The question poses multiple options for defining thermal energy. Option A suggests solar energy, which isn't correct as solar energy is a form of radiant energy. Option B refers to the energy within chemical substances, known as chemical energy. Option C discusses energy by virtue of an object's position, commonly known as potential energy. The correct answer is Option D, which states that thermal energy is the energy associated with the random motion of atoms and molecules.
More specifically, thermal energy is related to the kinetic energy resulting from the random translational motions of particles and also includes vibrational and rotational energies within molecules. An increase in this motion, as measured by temperature, corresponds to an increase in thermal energy. Hence, thermal energy is a form of internal energy of an object, manifesting as the microscopic motion of its constituent particles.