Explanation:
Grey Goose vodka has an alcohol content of 40.0 % (v/v).
Volume of vodka = V = 100 mL
This means that 40.0 mL of alcohol is present 100 mL of vodka.
Volume of ethanol=V' = 40.0 mL
Mass of ethanol = m
Density of the ethanol = d = 0.789 g/mL
[tex]m=d\times V' = 0.789 g/ml\times 40.0 mL=31.56 g[/tex]
Volume of water = V''= 100 ml - 40.0 mL = 60.0 mL
Mass of water = m'
Density of the water = d' = 1.00 g/mL
[tex]m'=d'\times V'' = 1.00 g/ml\times 60.0 mL=60.0 g[/tex]
a.)
Moles of ethanol = n= [tex]\frac{31.56 g}{46g/mol}=0.6861 mol[/tex]
Volume of vodka = V = 100 mL = 0.100 L ( 1mL=0.001 L)
Molarity of the ethanol:
[tex]=\frac{0.6861 mol}{0.100 L}=6.861 M[/tex]
6.861 M the molarity of ethanol in this vodka.
b) Mass of ethanol = 31.56 g
Moles of ethanol = n= [tex]\frac{31.56 g}{46g/mol}=0.6861 mol[/tex]
Volume of vodka = V = 100 mL
Mass of vodka = m
Density of the water = D = 0.935 g/mL
[tex]M=D\times V=0.935 g/ml\times 100 ml=93.5 g[/tex]
The percent by mass of ethanol % (m/m):
[tex]\frac{31.56 g}{93.5 g}\times 100=33.75\%[/tex]
33.75% is the percent by mass of ethanol % (m/m) in this vodka.
c)
Moles of ethanol = n= [tex]\frac{31.56 g}{46g/mol}=0.6861 mol[/tex]
Mass of solvent that is water = 60.0 g = 0.060 kg ( 1g = 0.001 kg)
Molality of ethanol in vodka :
[tex]m=\frac{0.6861 mol}{0.060 kg}=11.435 m[/tex]
11.435 m is the molality of ethanol in this vodka.
d)
Moles of ethanol = [tex]n_1=\frac{31.56 g}{46g/mol}=0.6861 mol[/tex]
Moles of water = [tex]n_2=\frac{60.0 g}{18 g/mol}=3.333 mol[/tex]
Mole fraction of ethanol = [tex]\chi_1[/tex]
[tex]\chi_1=\frac{n_1}{n_1+n_2}=\frac{0.6861 mol}{0.6861 mol+3.333 mol}[/tex]
= 0.1707
Mole fraction of water = [tex]\chi_2[/tex]
[tex]\chi_2=\frac{n_2}{n_1+n_2}=\frac{3.3333 mol}{0.6861 mol+3.333 mol}[/tex]
= 0.8290
e)
The vapor pressure of vodka = P
Mole fraction of ethanol = [tex]\chi_1=0.1707[/tex]
Mole fraction of water = [tex]\chi_2=0.8290[/tex]
The vapor pressures of ethanol = [tex]p_1=45.0 Torr[/tex]
The vapor pressures of pure water = [tex]p_2=23.8Torr[/tex]
[tex]P=\chi_1\times p_1+\chi_2\times p_2[/tex]
[tex]P=0.1707\times 45.0torr+0.8290\times 23.8 Torr=27.41 torr[/tex]
The vapor pressure of vodka is 27.41 Torr.
A chemist prepares a solution of sodium nitrate (NaNO3) by measuring out 286. umol of sodium nitrate into a 450. mL volumetric flask and filling the flask to the mark with water.Calculate the concentration in mmol/L of the chemist's sodium nitrate solution.
Answer:
Concentration of sodium nitrate solution is 0.636 mmol/L
Explanation:
We know, [tex]1micromol=1\times 10^{-3}mmol[/tex]
So, [tex]286.micromol=(286.\times 10^{-3})mmol=0.286mmol[/tex]
Concentration in mmol/L is defined as number of mmol of solute dissolved in 1000 mL of solution
Here solute is [tex]NaNO_{3}[/tex]
Total volume of solution = Total volume of volumetric flask = 450. mL
Hence concentration of [tex]NaNO_{3}[/tex] solution = [tex]\frac{0.286mmol}{450.mL}\times 1000=0.636mmol/L[/tex]
So, concentration of sodium nitrate solution is 0.636 mmol/L
The gas phase reaction between NO2 and F2 is first order in [NO2] and first order in [F2l. What would happen to the reaction rate if the concentrations of both reactants were halved with everything else held constant?a.It would decrease by a factor of 2.b.It would increase by a factor of 4.c.It would increase by a factor of 2. d.It would remain unchanged. e.It would decrease by a factor of 4.
Answer:
e.It would decrease by a factor of 4.
Explanation:
first order in [NO2]
first order in [F2]
The rate is then given as;
Rate = k [NO2][F2]
where k = rate constant = And is constant for a reaction.
Let's insert some dummy values (Any values work, just be consistent);
[NO2] = 2
[F2] = 2
K = 3
Rate = 3*2*2
Rate = 12
What would happen to the reaction rate if the concentrations of both reactants were halved with everything else held constant?
[NO2] = 2 / 2 = 1
[F2] = 2 / 2 = 1
K = 3
Rate = 3*1*1
Rate = 3
Comparing both rates (12 and 3); the correct option is;
e.It would decrease by a factor of 4.
A metallurgical firm wishes to dispose of 1200 gallons of waste sulfuric acid whose molarity is 1.05 M. Before disposal, it will be reacted with calcium hydroxide (slaked lime), which costs $0.25 per pound. Write the balanced chemical equation for this process. (Use the lowest possible coefficients. Use the pull-down boxes to specify states such as (aq) or (s). If a box is not needed, leave it blank.) Determine the cost that the firm will incur from this use of slaked lime. Cost
Answer:
The balanced chemical equation for this process:
[tex]H_2SO_4(aq)+Ca(OH)_2(s)\rightarrow 2H_2O(l)+CaSO_4(aq)[/tex]
$194.51 is the cost that the firm will incur from this use of slaked lime.
Explanation:
The balanced chemical equation for this process
[tex]H_2SO_4(aq)+Ca(OH)_2(s)\rightarrow 2H_2O(l)+CaSO_4(aq)[/tex]
Moles of sulfuric acid = n
Volume of sulfuric acid disposed = V = 1200 gallons = 3.785 × 1200 L = 4,542 L
1 gallon = 3.785 Liter
Morality of the sulfuric acid = M = 1.05 m
[tex]Molarity(M)=\frac{n}{V(L)}[/tex]
[tex]n=m\times V=1.05 M\times 4,542 L=4,769.1 mol[/tex]
According to reaction, 1 mol of sulfuric acid reacts with 1 mole of calcium hydroxide.Then 4,769.1 moles of sulfuric acid will recat with ;
[tex]\frac{1}{1}\times 4,769.1 mol=4,769.1 mol[/tex] of calcium hydroxide
Mass of 4,769.1 moles of calcium hydroxide:
4,769.1 mol 74 g/mol = 352,913.4 g
= [tex]\frac{352,913.4 }{453.6} pounds =778.03 pounds[/tex]
(1 pound = 453.6 grams)
Cost of 1 pound of slaked lime = $0.25
Cost of 778.03 pounds of slaked lime = $0.25 × 778.03 = $194.51
$194.51 is the cost that the firm will incur from this use of slaked lime.
Final answer:
To neutralize 1200 gallons of 1.05 M sulfuric acid, a metallurgical firm will need 352.89 kg of calcium hydroxide (slaked lime) costing a total of $194.53.
Explanation:
A metallurgical firm wishes to dispose of 1200 gallons of waste sulfuric acid whose molarity is 1.05 M by reacting it with calcium hydroxide (slaked lime). The balanced chemical equation for this process is:
H2SO4(aq) + Ca(OH)2(s) → CaSO4(aq) + 2H2O(l)
To find the mass of slaked lime needed, first, calculate the moles of sulfuric acid in 1200 gallons (4536 liters, assuming 1 gallon = 3.78541 liters). The total moles of H2SO4 are 1.05 moles/L × 4536 L = 4762.8 moles. According to the stoichiometry of the reaction, 1 mole of H2SO4 reacts with 1 mole of Ca(OH)2, so 4762.8 moles of Ca(OH)2 are required. The molar mass of Ca(OH)2 is 74.09 g/mol, which means (4762.8 moles × 74.09 g/mol) = 352889.352 grams or 352.89 kg of Ca(OH)2 needed. Considering the cost of slaked lime is $0.25 per pound, and 1 kg = 2.20462 pounds, the total cost would be (352.89 kg × 2.20462 pounds/kg × $0.25/pound) = $194.53.
100 mL of 1.00 M HCl solution is titrated with 1.00 M NaOH solution. You added the following quantities of 1.00 M NaOH to the reaction flask. Classify the following conditions based on whether they are before the equivalence point, at the equivalence point, or after the equivalence point/endpoint.
1.) 150 mL of 1 M NaOH
2.) 200 mL of 1 M NaOH
3.) 50 mL of 1 M NaOH
4.) 100 mL of 1 M NaOH
5.) 5.00 mL of 1 M NaOH
6.) 10.0 mL of 1 M NaOH
Answer:
1. After Equivalence Point
2. After Equivalence Point
3. Before Equivalence Point
4. At the Equivalence Point
5. Before Equivalence Point
6. Before Equivalence Point
Explanation:
First let's write down the balanced equation for this reaction:
[tex]HCL + NaOH -->H_2O + NaCL[/tex]
The above equation proves that we need one mole of NaOH to reach the equivalence point with one mole of HCL.
Thus for 100 ml of the 1 mole solution of HCL, we would need a corresponding 100 ml of 1 mole NaOH solution.
So anything less than 100 ml will be before the equivalence point, and anything larger than 100 ml will be after the equivalence point.
The equivalence point will be reached at exactly 100ml.
The answers reflect the above statements.
53. Consider an electrochemical cell made with zinc in zinc sulfate and copper in copper (II) sulfate. Identify items a through h, for h determine the standard cell potential given that the standard reduction potential for Zn2+ is - 0.763 V and for Cu2+ is + 0.337 V (16 points for answers a – h, and 9 points for i - k):
Answer: Ecell = -0.110volt
Explanation:
Zn--->Zn^+2 + 2e^-.........(1) oxidation
Cu^2+ 2e^- --->Cu........(2)reduction
Zn + Cu^2+ ----> Cu + Zn^+2 (overall
For an electrochemical cell, the reduction potential set up is given by
E(cell) = E(cathode) - E(anode)
E(cell) = E(oxidation) - E(reduction)
E(cathode) = E(oxidation)
E(anode) = E(reduction)
Given that
E(oxidation) = -0.763v
E(reduction) = +0.337v
E(cell) = -0.763 - (+0.337)
E(cell) = -0.763- 0.337
E(cell) = -0.110volt
Write out the BALANCED NET IONIC equation for the neutralization reaction that occurs between aqueous phosphoric acid and aqueous sodioum hydroxide. Include all states of matter.
Answer: Balanced net ionic equation is [tex]3H^{+}(aq)+3OH^{-}(aq)\rightarrow 3H_2O(l)[/tex]
Explanation:
Neutralization is a chemical reaction in which an acid and a base reacts to form salt and water.
Spectator ions are defined as the ions which does not get involved in a chemical equation or they are ions which are found on both the sides of the chemical reaction present in ionic form.
The given chemical equation is:
[tex]H_3PO_4(aq)+3NaOH(aq)\rightarrow Na_3PO_4(aq)+3H_2O(l)[/tex]
The ions which are present on both the sides of the equation are sodium and phosphate ions and hence are not involved in net ionic equation is:
[tex]3H^{+}(aq)+3OH^{-}(aq)\rightarrow 3H_2O(l)[/tex]
Final answer:
The net ionic equation for the neutralization reaction between phosphoric acid and sodium hydroxide is H₃PO₄(aq) + 3 NaOH(aq) → Na₃PO₄(aq) + 3 H₂O(l). It showcases the essential reactions involved in the process.
Explanation:
Net Ionic Equation: H₃PO₄(aq) + 3 NaOH(aq) → Na₃PO₄(aq) + 3 H₂O(l)
Overall Balanced Equation: H₃PO₄(aq) + 3 NaOH(aq) → Na₃PO₄(aq) + 3 H₂O(l)
The net ionic equation represents the reaction at the molecular level, focusing on the species involved in the chemical change. In this case, phosphoric acid (H₃PO₄) reacts with sodium hydroxide (NaOH) to form sodium phosphate (Na₃PO₄) and water (H₂O).
A 30 wt % solution of NaOH is diluted in a mixer to 5 wt%. If the streams entering and leaving the mixer are at 40 oC, find the heat removed from the mixer on a basis of 100 kg of feed solution.
Explanation:
It is given that 100 kg feed solution contains 30 wt% solution NaOH. Hence, feed contains 70 kg of water and 30 kg of NaOH.
Therefore, mixer outlet shows 5 wt% NaOH.
Let us assume that the mixture outlet be F kg so, by applying mass balance of NaOH we get the value of force as follows.
[tex]F \times 0.05[/tex] = 30
F = 600 Kg
In that 30 kg is NaOH and 570 kg is water which also means that initially water present is 70 kg. And, additional water added is 500 kg .
Thus, water feed rate is 500 kg/hr.
A 25.0 mL sample of a solution of a monoprotic acid is titrated with a 0.115 M NaOH solution. The end point was obtained at about 24.8 mL. The concentration of the monoprotic acid is about ........ mol/L.
A) 25.0
B) 0.0600
C) 0.240
D) 0.120
E) None of the abov
Answer:
The concentration of the monoprotic acid is about 0.114 mol/L.
The correct answer is E none of the above
Explanation:
Step 1: Data given
Volume of a monoprotic acid = 25.0 mL = 0.025 L
Molarity of the NaOH solution = 0.115 M
The end point was obtained at about 24.8 mL.
Step 2: Calculate the concentration of the monoprotic acid
b*Ca*Va = a*Cb*Vb
⇒with B = the coefficient of NaOH = 1
⇒with Ca = the concentration of the monoprotic acid = ?
⇒with Va = the volume of the monoprotic acid = 0.025 L
⇒with a = the coefficient of the monoprotic acid = 1
⇒with Cb = the concentration of NaOH = 0.115M
⇒with Vb = the volume of NaOH= 0.0248 L
1*Ca*0.025 = 1*0.115*0.0248
Ca = (0.115*0.0248)/0.025
Ca = 0.114 M
The concentration of the monoprotic acid is about 0.114 mol/L.
The correct answer is E none of the above
Consider a solution formed by mixing 50.0 mL of 0.100 M H2SO4, 30.0 mL of 0.1133 M HOCl, 25.0 mL of 0.200 M NaOH, 25.0 mL of 0.100 M Ba(OH)2, and 10.0 mL of 0.170 M KOH. Calculate the pH of this solution. Ka(HOCl) = 3.5×10-8 pH =
The pH of the given solution is approximately 12.1.
Let's go through the calculations step by step:
Step 1: Moles of Hydrogen Ions in Sulfuric Acid Solution (a)
a = Molarity × Volume
a = 0.100 M × 0.050 L
a = 0.005 mol
Step 2: Moles of Hydrogen Ions in HOCl Solution (b)
Dissociation: HOCl ↦ H+ + OCl-
Ka = [H+][OCl-]/[HOCl] = 3.5 × 10^-8
At equilibrium: (0.1133 - x) M (x) M
x^2/(0.1133 - x) = 3.5 × 10^-8
Solving for x gives x ≈ 1.87 × 10^-4 mol
b = moles of H+ in HOCl solution = x
Step 3: Total Moles of Hydrogen Ions
Total moles of H+ = a + b
Step 4: Moles of Hydroxide Ions in NaOH Solution (c)
c = Molarity × Volume
c = 0.200 M × 0.025 L
c = 0.005 mol
Step 5: Moles of Hydroxide Ions in Ba(OH)2 Solution (d)
d = Molarity × Volume
d = 0.100 M × 0.025 L
d = 0.0025 mol
Step 6: Moles of Hydroxide Ions in KOH Solution (e)
e = Molarity × Volume
e = 0.170 M × 0.010 L
e = 0.0017 mol
Step 7: Total Moles of Hydroxide Ions
Total moles of hydroxide ions = c + d + e
Step 8: Determine Excess Ions (Hydrogen or Hydroxide)
Since Total moles of H+ < Total moles of hydroxide ions, there is excess hydroxide ions.
Step 9: Moles of Excess Hydroxide Ions
Moles of hydroxide left after neutralization = Total moles of hydroxide ions - Total moles of H+
Step 10: Concentration of Hydroxide Ions
Concentration of hydroxide ions left in the solution = Moles of hydroxide left/Total volume of solution
Step 11: Calculate pOH
pOH = -log(OH- concentration)
Step 12: Calculate pH
pH = 14 - pOH
Let's substitute the values and calculate:
pOH = -log(0.001698/0.140) ≈ 1.9
pH = 14 - 1.9 ≈ 12.1
So, the pH of the solution is approximately 12.1.
The calculated pH of the solution is approximately 12.09.
Let's start with the calculations:
1. Calculate Moles of Each Component
H₂SO₄: Moles = 0.050 L * 0.100 M = 0.005 moles of H₂SO₄.HOCl: Moles = 0.030 L * 0.1133 M = 0.003399 moles of HOCl.NaOH: Moles = 0.025 L * 0.200 M = 0.005 moles of NaOH.Ba(OH)₂: Moles = 0.025 L * 0.100 M * 2 = 0.005 moles of OH⁻ (since Ba(OH)₂ dissociates to give 2 OH⁻).KOH: Moles = 0.010 L * 0.170 M = 0.0017 moles of KOH.2. Determine Neutralization
Strong acids (H₂SO₄) and bases (NaOH, KOH, Ba(OH)₂) will neutralize each other:
Total moles of OH⁻ from NaOH, Ba(OH)₂, and KOH = 0.005 + 0.005 + 0.0017 = 0.0117 moles of OH⁻.Total moles of H⁺ from H₂SO₄ = 0.005 * 2 = 0.01 moles (since each H₂SO₄ dissociates to provide 2 H⁺).Neutralization reaction: 0.0117 moles OH⁻ neutralizes 0.01 moles H⁺, leaving 0.0017 moles of OH⁻.3. Calculate Final pH
Since 0.0017 moles of OH- remain:
Total volume of the solution = 50.0 + 30.0 + 25.0 + 25.0 + 10.0 = 140.0 mL = 0.140 L.Concentration of OH⁻ = 0.0017 moles / 0.140 L = 0.01214 M.[H+] = Kw / [OH⁻] = 1.0 × 10-14 / 0.01214 M ≈ 8.24 × 10⁻¹³ M.pH = -log[H⁺] ≈ -log(8.24 × 10⁻¹³) ≈ 12.09.The resulting pH of the mixed solution is approximately 12.09.
Water freezes at 0∘C and CO freezes at −205∘C. Which type of intermolecular force accounts for this difference in freezing point between the two compounds?
The difference in freezing point between water and carbon monoxide is due to the different types of intermolecular forces they possess. Water has stronger hydrogen bonds whereas carbon monoxide has weaker dipole-dipole forces.
Explanation:The difference in freezing point between water (H2O) and carbon monoxide (CO) can be explained by the type of intermolecular forces present in these molecules. Water molecules have stronger hydrogen bonds which require more energy to break, hence it freezes at a higher temperature (0°C). On the other hand, carbon monoxide molecules primarily exhibit dipole-dipole forces which are not as strong as hydrogen bonds, resulting in a much lower freezing point (-205°C).
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You are in space and running out of water. You do have a great deal of magnesium carbonate pentahydrate. It is possible to extract the water from this. Determine the percent of water by mass in the hydrate magnesium carbonate pentahydrate (MgCO3·5H2O).
Answer:
= 51.72%
Explanation: Use the formula
percentage by mass of water
= (molar mass of water / molar mass of MgCO3.5H2O) x 100
Molar mass of MgCO3.5H2O = 24+12+(16X3)+[5(2+16)
=36+48+90
=174g/mol
Molar mass of water 5H2O =5(2 + 16)
=5X18
=90
Therefore percentage of water = (90/174) x 100
= 51.72%
Note the molar masses: Mg = 24, C = 12, O = 16, H = 1
The percent of the water by mass extracted from magnesium carbonate pentahydrate in space is 51.72%.
What is percent mass?The percent mass of water in the compound is given by:
[tex]\rm percent \;mass=\dfrac{mass\;of\;water}{mass\;of\;compound}\;\times\;100[/tex]
The mass of magnesium carbonate pentahydrate is 174 grams.
The compound is composed of 5 water units. The mass of 1 water unit is 18 grams. The mass of 5 water units is:
[tex]\rm 1\;H_2O=18\;g\\5\;H_2O=18\;\times\;5\;g\\5\;H_2O=90\;g[/tex]
The percent mass of water in magnesium carbonate pentahydrate is given as:
[tex]\rm percent\;mass=\dfrac{90}{174}\;\times\;100\\ percent\;mass=0.5172\;\times\;100\\percent\;mass=51.72\;\%[/tex]
The percent mass of water in magnesium carbonate pentahydrate is 51.72%.
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When a 26.3 mL sample of a 0.465 M aqueous nitrous acid solution is titrated with a 0.461 M aqueous barium hydroxide solution, what is the pH after 19.9 mL of barium hydroxide have been added
Answer: The pH of the solution after addition of barium hydroxide is 3.78
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}\times 1000}{\text{Volume of solution (in mL)}}[/tex] .......(1)
For nitrous acid:Molarity of nitrous acid = 0.465 M
Volume of solution = 26.3 mL
Putting values in equation 1, we get:
[tex]0.465M=\frac{\text{Moles of nitrous acid}\times 1000}{26.3mL}\\\\\text{Moles of nitrous acid}=\frac{0.465\times 26.3}{1000}=0.0122mol[/tex]
For barium hydroxide:Molarity of barium hydroxide = 0.461 M
Volume of solution = 19.9 mL
Putting values in equation 1, we get:
[tex]0.461M=\frac{\text{Moles of barium hydroxide}\times 1000}{19.9mL}\\\\\text{Moles of barium hydroxide}=\frac{0.461\times 19.9}{1000}=0.0092mol[/tex]
The chemical reaction for nitrous acid and barium hydroxide follows the equation:
[tex]2HNO_2+Ba(OH)_2\rightarrow Ba(NO_2)_2+2H_2O[/tex]
Initial: 0.0122 0.0092
Final: 0.003 - 0.0092
Volume of solution = 26.3+ 19.9 = 46.2 mL = 0.0462 L (Conversion factor: 1 L = 1000 mL)
To calculate the pH of acidic buffer, we use the equation given by Henderson Hasselbalch:
[tex]pH=pK_a+\log(\frac{[salt]}{[acid]})[/tex]
[tex]pH=pK_a+\log(\frac{[NO_2^-]}{[HNO_2]}[/tex]
We are given:
[tex]pK_a[/tex] = negative logarithm of acid dissociation constant of nitrous acid = 3.29
[tex][NO_2^-]=\frac{0.0092}{0.0462}[/tex]
[tex][HNO_2]=\frac{0.003}{0.0462}[/tex]
pH = ?
Putting values in above equation, we get:
[tex]pH=3.29+\log(\frac{0.0092/0.0462}{0.003/0.0462})\\\\pH=3.78[/tex]
Hence, the pH of the solution after addition of barium hydroxide is 3.78
Ionizing radiation causes cancer by Select one: a. forming ions, causing reactions that mutate DNA b. forming ions, causing reactions that mutate bacteria c. forming ions, causing reactions that form products that cancer cells eat d. forming ions, causing reactions that result in the breakdown of mitochondrial phospholipid bilayers Previous page
Answer:
a. forming ions, causing reactions that mutate DNA
Explanation:
Ionizing radiation is a type of energy released by atoms in the form of electromagnetic waves or particles. The spontaneous decay of atoms is called radioactivity, and the surplus energy emitted is a form of ionizing radiation. The unstable elements that decay and emit ionizing radiation are called radionuclides.
Exposure of a human being to certain doses of ionizing radiation can cause irreparable changes in the structure of their DNA, that is, in the genes that control the function of cells, and therefore cause a series of changes that can lead to a cancer.
Ionizing radiation contributes to cancer formation by mutating DNA through the formation of ions and hydroxyl radicals that can damage the DNA structure.
Explanation:Ionizing radiation causes cancer by forming ions, causing reactions that mutate DNA. This occurs because ionizing radiation, such as X-rays and gamma rays, can create hydroxyl radicals upon exposure which can cause single- and double-stranded breaks in the DNA backbone or modify the bases within the DNA. The damage to DNA can lead to mutations which, if not properly repaired, can result in the uncontrolled cell division characteristic of cancer.
The following two compounds are constitutional isomers. Identify which of these is expected to be more acidic, and explain your choice. a. The compound below is more acidic because its conjugate base is more resonance stabilized.b. The conjugate base of the other compound is not as much resonance stabilized.c. The compound below is more acidic because its conjugate base is more resonance stabilized.d. The conjugate base of the other compound is not as much resonance stabilized.e. The compound below is more acidic because its conjugate base is resonance stabilized.f. The conjugate base of the other compound is not resonance stabilized.Tg. he compound below is more acidic because its conjugate base is resonance stabilized.h. The conjugate base of the other compound is not resonance stabilized.
Answer:
c. Compound 2 is more acidic because its conjugate base is more resonance stabilized
Explanation:
You haven't told us what the compounds are, so let's assume that the formula of Compound 1 is HCOCH₂OH and that of Compound 2 is CH₃COOH.
The conjugate base of 2 is CH₃COO⁻. It has two important resonance contributors, and the negative charge is evenly distributed between the two oxygen atoms.
CH₃COOH + H₂O ⇌ CH₃COO⁻ + H₃O⁺
The stabilization of the conjugate base pulls the position of equilibrium to the right, so the compound is more acidic than 1.
The compound with the more resonance stabilized conjugate base is the more acidic compound. This is because the stability of an acid's conjugate base directly influences acidity, and resonance stabilization enhances the stability of the conjugate base.
Explanation:In comparing the acidity of two constitutional isomers, the strength of an acid is directly related to the stability of its conjugate base. An acid is considered strong if its conjugate base is stable. One important factor contributing to the stability of the conjugate base is its ability to distribute the negative charge throughout the ion (resonance).
In this situation, if the conjugate base of compound 'a' (for instance) is more resonance stabilized than that of the conjugate base of compound 'b', compound 'a' would be considered more acidic (assuming the compounds are constitutional isomers). On the other hand, if the conjugate base of compound 'b' is not as resonance stabilized as compound 'a's, compound 'b' would be less acidic. This is because resonance stabilization allows the negative charge to be distributed across a larger volume, thus reducing the concentration of charge, which in turn increases stability.
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Use molecular orbital theory to predict whether or not each of the following molecules or ions should exist in a relatively stable form. Drag the appropriate items to their respective bins. C2 2+ Be2 2+ Li2 Li2 2
Answer:
C2 2+ stable and should exist
Be2 2+ stable and should exit
Li2 stable and should exit
Li2 2- unstable and doesn't exist
Explanation:
The first step in predicting the stability of a specie is knowing its molecular orbital configuration and bond order. If the Specie has a bond order of one or more, it is expected to exist in a stable form. The image attached shows the bond order and molecular orbital configuration of all the species mentioned in the question. This will aid you in understanding the stability of each specie.
The ions or molecule C2 2+ Be2 2+ Li2 exists in stable form but Li2 2 does not exist in stable form when using molecular orbital theory to predict their stability.
The production of molecular orbitals arises by combining atomic orbitals in a linear array. Based on the electronic configuration of molecular orbitals, the stability of molecules or ions can be determined by calculating the bond order for each molecule.
The bond order can be expressed as: [tex]\mathbf{= \dfrac{1}{2} \Big[no \ of \ electrons \ in \ B.O - No \ of \ electrons \ in \ Anti \ B.O \Big]}[/tex]
where:
B.O = Bonding OrbitalFor C₂²⁺:
The carbon atom has an electronic configuration of 1s²2s²2p². It has 12 electrons. For the formation of C₂ from C₂²⁺, there is the removal of 2 electrons.
As such, C₂²⁺ has 10 electrons.
Now, the electronic configuration for the molecular orbital for C₂²⁺ can be written as:
[tex]\mathbf{ \sigma_{1s}^2 < \sigma _{1s}^{*2} < \sigma _{2s}^{2}< \sigma _{2s}^{*2} < \pi _{2py}^{1}=\pi _{2pz}^{1} < \pi _{2px}^{0} < \pi _{2py}^{*0}= \pi _{2pz}^{0} < \pi _{2px}^{0} }[/tex]
where;
the orbitals with (*) = antibonding orbitals the orbitals without (*) = bonding orbitalsBond Order [tex]= \mathbf{\dfrac{1}{2} (6-4)}[/tex]
[tex]= \mathbf{\dfrac{1}{2} (2)}[/tex]
= 1
Thus, since Bond order = 1, the C₂²⁺ ion exist in a stable form
For Be₂²⁺ is a beryllium atom with a configuration is 1s² 2s². It has 8 electrons, for the formation of Be₂ from Be₂²⁺, there is the removal of 2 electrons. As such, Be₂²⁺ has 6 electrons, the electronic configuration for the molecular orbital can be expressed as:
[tex]\mathbf{ \sigma_{1s}^2 < \sigma _{1s}^{*2} < \sigma _{2s}^{2}< \sigma _{2s}^{*0} < \pi _{2py}^{0}=\pi _{2pz}^{0} < \pi _{2px}^{0} < \pi _{2py}^{*0}= \pi _{2pz}^{0} < \pi _{2px}^{0} }[/tex]
Bond Order [tex]= \mathbf{\dfrac{1}{2} (4-2)}[/tex]
[tex]= \mathbf{\dfrac{1}{2} (2)}[/tex]
= 1
Thus, since Bond order = 1, the Be₂²⁺ ion exist in a stable form
Lithium Li₂ has an atomic number 3 with an electronic configuration 1s² 2s¹. Thus, it comprises 6 electrons. The electronic configuration of its molecular orbital is expressed as;
[tex]\mathbf{ \sigma_{1s}^2 < \sigma _{1s}^{*2} < \sigma _{2s}^{2}< \sigma _{2s}^{*0} < \pi _{2py}^{0}=\pi _{2pz}^{0} < \pi _{2px}^{0} < \pi _{2py}^{*0}= \pi _{2pz}^{*0} < \pi _{2px}^{*0} }[/tex]
Bond Order[tex]= \mathbf{\dfrac{1}{2} (4-2)}[/tex]
[tex]= \mathbf{\dfrac{1}{2} (2)}[/tex]
= 1
Thus, since Bond order = 1, the Li₂ ion exists in a stable form
In Li₂²⁻, the number of electrons for its formation is 8; Since it needs 2 more electrons to its initial 6 electrons in Li₂. so, the electronic configuration of its molecular orbital can be expressed as;
[tex]\mathbf{ \sigma_{1s}^2 < \sigma _{1s}^{*2} < \sigma _{2s}^{2}< \sigma _{2s}^{*2} < \pi _{2py}^{0}=\pi _{2pz}^{0} < \pi _{2px}^{0} < \pi _{2py}^{*0}= \pi _{2pz}^{*0} < \pi _{2px}^{*0} }[/tex]
Bond Order [tex]= \mathbf{\dfrac{1}{2} (4-4)}[/tex]
[tex]= \mathbf{\dfrac{1}{2} (0)}[/tex]
= 0
Thus, since Bond order = 0, the Li₂²⁻ ion does not exist in a stable form
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Dimethylhydrazine is a carbon-hydrogen-nitrogen compound used in rocket fuels. When burned in an excess of oxygen, a 0.312 gg sample yields 0.458 gg CO2CO2 and 0.374 gg H2OH2O. The nitrogen content of a 0.486 gg sample is converted to 0.226 gg N2N2.What is the empirical formula of dimethylhydrazine?Express your answer as a chemical formula.
Answer:
CH₄N
Explanation:
Given that;
mass of the sample = 0.312 g
mass of CO2 = 0.458 g
mass of H2O = 0.374 g
nitrogen content of a 0.486 gg sample is converted to 0.226 gg N2N2.
Let start with calculating the respective numbers of moles of Carbon Hydrogen and Nitrogen from the given data.
numbers of moles of Carbon from CO2 = [tex]\frac{mass of CO_2}{molarmass}*\frac{1 mole of C}{1 mole of CO_2}[/tex]
= [tex]\frac{0.458}{44}*\frac{1mole of C}{1 mole of CO_2}[/tex]
= 0.0104 mole
numbers of moles of hydrogen from H2O = [tex]\frac{mass of H_2O}{molarmass}*\frac{2 mole of H}{1 mole of H_2O}[/tex]
= [tex]\frac{0.374}{18.02}*\frac{2 mole of H}{1 mole of H_2O}[/tex]
= 0.02077 × 2
= 0.0415 mole
The nitrogen content of a 0.486 g sample is converted to 0.226 g N2
Now, in 1 g of the sample; The nitrogen content = [tex]\frac{0.226}{0.486}*1[/tex]
in 0.312 g of the sample, the nitrogen content will be; [tex]\frac{0.226}{0.486}*0.312[/tex]
= 0.1450 g of N2
number of moles of N2 = [tex]\frac{mass}{molar mass}* \frac{2 mole}{1 mole}[/tex]
= [tex]\frac{0.1450}{28.0134} *\frac{2 mole}{1 mole}[/tex]
= 0.0103 mole
Finally to determine the empirical formula of Carbon Hydrogen and Nitrogen; we have:
Carbon Hydrogen Nitrogen
number of moles 0.0104 0.0415 0.0103
divided by the
smallest number [tex]\frac{0.0104}{0.0103}[/tex] [tex]\frac{0.0415}{0.0103}[/tex] [tex]\frac{0.0103}{0.0103}[/tex]
of moles
1 : 4 : 1
∴ The empirical formula = CH₄N
To find the empirical formula of dimethylhydrazine, we calculate the mass of carbon, hydrogen, and nitrogen from the combustion products. We then convert these masses to moles and obtain their simplest whole number ratio to derive the empirical formula.
Explanation:To determine the empirical formula of dimethylhydrazine, we will analyze the combustion products and use them to find the amount of carbon, hydrogen, and nitrogen in the original compound.
Step 1: Determine the amount of Carbon
From the given 0.458 g of CO2, we can calculate the mass of carbon in the CO2 using the molar mass of CO2 (44.01 g/mol). The molar mass of carbon is 12.01 g/mol.
Step 2: Determine the amount of Hydrogen
The 0.374 g of H2O contains a certain mass of hydrogen, which can be calculated by using the molar mass of H2O (18.015 g/mol) and the fact that there are 2 moles of hydrogen atoms for every mole of H2O.
Step 3: Determine the amount of Nitrogen
A 0.486 g sample of dimethylhydrazine produced 0.226 g of N2. Using the molar mass of N2 (28.02 g/mol), we can find the mass of nitrogen in the sample.
Step 4: Calculate the Empirical Formula
We then convert the masses of carbon, hydrogen, and nitrogen to moles and find their simplest whole number ratio to get the empirical formula of dimethylhydrazine.
How many significant figures are there in the following numbers: 10.78, 6.78, 0.78? If these were pH values, to how many significant figures can you express the [H]? Explain any discrepancies between your answers to the two questions.
Final answer:
The number 10.78 has four significant figures, 6.78 has three, and 0.78 has two. However, when expressed as pH values, significant figures are determined by the number of decimal places, so the significant figures for [H+] would match the decimal places of the pH.
Explanation:
The number of significant figures in a given number indicates how many digits are meaningful in expressing the precision of a measurement. For the numbers provided, 10.78 has four significant figures, 6.78 has three significant figures, and 0.78 has two significant figures. When expressing these numbers as pH values, however, the approach to significant figures changes slightly.
In the context of pH calculations, the significant figures are indicated by the number of digits after the decimal point. This is because the whole number part of a pH value indicates the power of 10 and is related to the magnitude of the hydrogen or hydronium ion concentration ([H+]), while the decimal portion represents the precision of the measurement. So, if 10.78, 6.78, and 0.78 were pH values, the respective [H+] concentrations would be expressed with the same number of significant figures as the decimal places in the pH value.
For example, if the given number is 0.010 M which has two significant figures, the corresponding pH 12.00 also reflects two significant figures. Similarly, a pH of 7.56 would result in a [H+] concentration rounded to two significant figures. This difference arises because the number of significant figures in the pH value is determined by the number of decimal places, unlike normal numerical significant figure rules.
The numbers 10.78, 6.78, and 0.78 have four, three, and two significant figures, respectively. If these were pH values, the corresponding hydrogen ion concentrations, [H], can be expressed with one, one, and two significant figures, respectively.
To determine the number of significant figures in a given number, one must count all the digits starting from the first non-zero digit and ending with the last digit, whether it is zero or not. This rule includes all digits except:
1. Leading zeros, which are zeros before the first non-zero digit.
2. Trailing zeros in a number without a decimal point.
3. Trailing zeros in a number with a decimal point that are to the right of the last non-zero digit.
Applying these rules:
- For 10.78, there are four significant figures: 1, 0, 7, and 8. The zero is significant because it is between two non-zero digits.
- For 6.78, there are three significant figures: 6, 7, and 8.
- For 0.78, there are two significant figures: 7 and 8. The zero is not significant because it is a leading zero.
When dealing with pH values and converting them to hydrogen ion concentrations, [H], the relationship is given by the equation pH = -log[H]. To find the [H] from a pH value, one would use the equation [H] = [tex]10^{(-pH)[/tex].
The number of significant figures in the pH value does not directly translate to the number of significant figures in [H] because the logarithm and antilogarithm [tex](10^x)[/tex] transformations can affect the number of significant figures:
- For a pH of 10.78, the [H] would be [tex]10^{(-10.78)[/tex]. The result of this calculation would typically be rounded to one significant figure because the pH value has uncertainty in the hundredths place.
- For a pH of 6.78, the [H] would be [tex]10^{(-6.78)[/tex]. Similarly, this would also be rounded to one significant figure.
- For a pH of 0.78, the [H] would be [tex]10^{(-0.78)[/tex]. This calculation would yield two significant figures because the pH value itself has two significant figures.
The discrepancy arises because the mathematical operations involved in converting pH to [H] introduce uncertainty that limits the number of significant figures that can be reliably reported for [H]. The pH values given are precise to the hundredths place, but when converted to [H], the precision is reduced due to the nature of the logarithmic scale.
the determination of a chemical formula n this experiment, you will use the law of definite proportions to find the chemical formula for ahydrated compound containing copper, chlorine, and water molecules locked in the crystal structure of the solid compound. The general formula for the compound is CuxCly•zH2O, and its name is copper chloride hydrate. The letters x, y, and z represent integers that will establish the proper chemical formula for this substance. First, you will gently heat a sample
Answer: Strictly a laboratory analysis and can only be done using the data obtained during analysis
Explanation:
To find a solution to this problem, you need to use the data collected during the lab work. A guide could be finding the possible forms of hydrated copper chlorides in reference books. Since it's also a lab work, you can definitely compare your data with lab mates.
The formula CuxCly.zH₂O and its name chloride hydrate already gives you an idea of the possibilities of the value of the integers, hence you can take a good guess for the identity of the unknown salt and calculate the theoretical formular weight for it. From the that you can proceed to also find the mass of water and copper from your lab analysis.
To determine the chemical formula of a hydrated compound like copper chloride hydrate, you heat the compound to find the mass of water lost and compare this to the mass of the anhydrous copper chloride. By using the molar masses, these can be turned into amounts in moles and thus, the integers x, y, z in the formula (CuxCly•zH2O) can be determined.
Explanation:In chemistry, when you have a hydrated compound like copper chloride hydrate, the x, y, and z in the formula (CuxCly•zH2O) represent the number of moles of each component in the compound. To find out these numbers, you first need to dry the compound by gently heating it. This will drive off the water molecules and leave you with the anhydrous copper chloride.
Then you can determine the mass of the copper chloride and the mass of water lost. Now, you will divide these by their respective molar masses, which will give you the number of moles of each. The ratio of these numbers will give you the values of x, y, and z in the formula. For instance, if you have one mole of copper, two moles of chlorine, and five moles of water, the formula would be Cu1Cl2•5H2O, normally written as CuCl2•5H2O.
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A procedure calls for the bromination of 12 mmol of trans‑cinnamic acid with 12 mmol of Br 2 in glacial acetic acid solvent. Select the likely consequence of accidentally using 6 mmol of Br 2 instead. The yield of dibromide product will be approximately one‑half of the expected yield. The yield of dibromide will not change because 2 mol of Br 2 react with 1 mol of trans‑cinnamic acid. The yield will not change because acetic acid is the limiting reactant. The yield of dibromide product will not change, as the equilibrium strongly favors the product.
Answer: The yield of dibromide product will be approximately one‑half of the expected yield.
Explanation:
Using half the required amount of Br2 in the bromination of trans-cinnamic acid would likely result in approximately half the yield of the expected dibromide product due to Br2 being the limiting reactant.
Explanation:The likely consequence of using 6 mmol of Br2 instead of the required 12 mmol in the bromination of trans-cinnamic acid is that the yield of the dibromide product will be approximately one-half of the expected yield. This is because the amount of Br2 would become the limiting reactant in this reaction, as not enough bromine is present to fully react with the 12 mmol of trans-cinnamic acid. In stoichiometric reactions, the yield is directly proportional to the amount of the limiting reagent. Therefore, if half the required amount of a reagent is used, it is logical to expect around half the yield, assuming that the reaction goes to completion under the given conditions.
A solution contains 2.0 ⨯ 10−4 M Ag+(aq) and 1.5 ⨯ 10−3 M Pb2+(aq). If NaI is added, will AgI (Ksp = 8.3 ⨯ 10−17) or PbI2 (Ksp = 7.9 ⨯ 10−9) precipitate first? Specify the concentration of I−(aq) needed to begin precipitation.
Answer:
The silver iodide will precipitate first.
At concentration more than [tex]4.15\times 10^{-13} M[/tex] of iodide ion will result in precipitation of silver iodide.
At concentration more than [tex]0.0023 M[/tex] of iodide ion will result in precipitation of lead iodide.
Explanation:
Solubility of silver iodide = [tex]K_{sp}=8.3\times 10^{-17}[/tex]
[tex]AgI\rightleftharpoons Ag^++I^-[/tex]
Concentration of silver ions = [tex][Ag^+]=2.0\times 10^{-4} M[/tex]
Suppose concentration of iodide ion due NaI added in solution be = [tex][I^-][/tex]
The expression of solubility product will be given as :
[tex]K_{sp}=[Ag^+][I^-][/tex]
[tex][I^-]=\frac{8.3\times 10^{-17}}{2.0\times 10^{-4} M}=4.15\times 10^{-13} M[/tex]
At concentration more than [tex]4.15\times 10^{-13} M[/tex] of iodide ion will result in precipitation of silver iodide.
Solubility of lead iodide = [tex]K_{sp}'=7.9\times 10^{-9}[/tex]
[tex]PbI_2\rightleftharpoons Pb^{2+}+2I^-[/tex]
Concentration of silver ions = [tex][Pb^{2+}]=1.5\times 10^{-3} M[/tex]
Suppose concentration of iodide ion due NaI added in solution be = [tex][I^-][/tex]
The expression of solubility product will be given as :
[tex]K_{sp}'=[Pb^{2+}][I^-]^2[/tex]
[tex][I^-]^2=\frac{7.9\times 10^{-9}}{1.5\times 10^{-3} M}=4.15\times 10^{-13}[/tex]
[tex][I^-]=0.0023 M[/tex]
At concentration more than [tex]0.0023 M[/tex] of iodide ion will result in precipitation of lead iodide.
For silver iodide , we need concentration of iodide more than [tex]4.15\times 10^{-13} M[/tex] and for lead iodide , we need concentration of iodide more than 0.0023 M. So , this indicates that silver iodide will precipitate out first.
Silver iodide is the first precipitate. silver iodide or lead iodide precipitates at concentrations greater than 4.15× 10⁻¹³ M or 2.3 ×10⁻³ M of iodide ions.
Concentration calculation:Calculating the value for the first question that is the AgI ppts first:
⇒ AgI
⇒ K = [Ag] [I]
⇒ 8.3 × 10⁻¹⁷= [2e⁻⁴][I]
⇒ [I] = 4.15× 10⁻¹³ Molar
Calculating the value for the second question, which is AgI to ppt, [I] > 4.15× 10⁻¹³ Molar
⇒ PbI₂
⇒ K = [Pb] [I]²
⇒ 7.9× 10⁻⁹ = [1.5 × 10⁻³] [I]²
⇒ I² = 5.26 × 10⁻⁶
⇒ I = 2.3 ×10⁻³ Molar
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While ethanol is produced naturally by fermentation, e.g. in beer- and wine-making, industrially it is synthesized by reacting ethylene with water vapor at elevated temperatures. A chemical engineer studying this reaction fills a flask with of ethylene gas and of water vapor. When the mixture has come to equilibrium she determines that it contains of ethylene gas and of water vapor. The engineer then adds another of ethylene, and allows the mixture to come to equilibrium again. Calculate the pressure of ethanol after equilibrium is reached the second time. Round your answer to significant digits.
The question is incomplete, here is the complete question:
While ethanol is produced naturally by fermentation, e.g. in beer- and wine-making, industrially it is synthesized by reacting ethylene with water vapor at elevated temperatures.
A chemical engineer studying this reaction fills a 1.5 L flask at 12°C with 1.8 atm of ethylene gas and 4.7 atm of water vapor. When the mixture has come to equilibrium she determines that it contains 1.16 atm of ethylene gas and 4.06 atm of water vapor.
The engineer then adds another 1.2 atm of ethylene, and allows the mixture to come to equilibrium again. Calculate the pressure of ethanol after equilibrium is reached the second time. Round your answer to 2 significant digits.
Answer: The partial pressure of ethanol after equilibrium is reached the second time is 1.0 atm
Explanation:
We are given:
Initial partial pressure of ethylene gas = 1.8 atm
Initial partial pressure of water vapor = 4.7 atm
Equilibrium partial pressure of ethylene gas = 1.16 atm
Equilibrium partial pressure of water vapor = 4.06 atm
The chemical equation for the reaction of ethylene gas and water vapor follows:
[tex]CH_2CH_2(g)+H_2O(g)\rightleftharpoons CH_3CH_2OH(g)[/tex]
Initial: 1.8 4.7
At eqllm: 1.8-x 4.7-x
Evaluating the value of 'x'
[tex]\Rightarrow (1.8-x)=1.16\\\\x=0.64[/tex]
The expression of [tex]K_p[/tex] for above equation follows:
[tex]K_p=\frac{p_{CH_3CH_2OH}}{p_{CH_2CH_2}\times p_{H_2O}}[/tex]
[tex]p_{CH_2CH_2}=1.16atm\\p_{H_2O}=4.06atm\\p_{CH_3CH_2OH}=0.64atm[/tex]
Putting values in above expression, we get:
[tex]K_p=\frac{0.64}{1.16\times 4.06}\\\\K_p=0.136[/tex]
When more ethylene is added, the equilibrium gets re-established.
Partial pressure of ethylene added = 1.2 atm
[tex]CH_2CH_2(g)+H_2O(g)\rightleftharpoons CH_3CH_2OH(g)[/tex]
Initial: 2.36 4.06 0.64
At eqllm: 2.36-x 4.06-x 0.64+x
Putting value in the equilibrium constant expression, we get:
[tex]0.136=\frac{(0.64+x)}{(2.36-x)\times (4.06-x)}\\\\x=0.363,13.41[/tex]
Neglecting the value of x = 13.41 because equilibrium partial pressure of ethylene and water vapor will become negative, which is not possible.
So, equilibrium partial pressure of ethanol = (0.64 + x) = (0.64 + 0.363) = 1.003 atm
Hence, the partial pressure of ethanol after equilibrium is reached the second time is 1.0 atm
Convert .4076grams into moles
Element is copper
Answer:
0.00642mole
Explanation:
Molar Mass of Cu = 63.5g/mol
Mass of Cu from the question = 0.4076g
Number of mole =?
Number of mole = Mass /Molar Mass
Number of mole of Cu = 0.4076/63.5 = 0.00642mole
Excess aqueous copper(II) nitrate reacts with aqueous sodium sulfide to produce aqueous sodium nitrate and copper(II) sulfide as a precipitate. In this reaction 469 grams of copper(II) nitrate were combined with 156 grams of sodium sulfide to produce 272 grams of sodium nitrate.
The question in incomplete, complete question is;
Determine the theoretical yield:
Excess aqueous copper(II) nitrate reacts with aqueous sodium sulfide to produce aqueous sodium nitrate and copper(II) sulfide as a precipitate. In this reaction 469 grams of copper(II) nitrate were combined with 156 grams of sodium sulfide to produce 272 grams of sodium nitrate.
Answer:
The theoretical yield of sodium nitrate is 340 grams.
Explanation:
[tex]Cu(NO_3)_2(aq)+Na_2S(aq)\rightarrow 2NaNO_3(aq)+CuS(s)[/tex]
Moles of copper(II) nitrate = [tex]\frac{469 g}{187.5 g/mol}=2.5013 mol[/tex]
Moles of sodium sulfide = [tex]\frac{156 g}{78 g/mol}=2 mol[/tex]
According to reaction, 1 mole of copper (II) nitrate reacts with 1 mole of sodium sulfide.
Then 2 moles of sodium sulfide will react with:
[tex]\frac{1}{1}\times 2mol= 2 mol[/tex] of copper (II) nitrate
As we can see from this sodium sulfide is present in limiting amount, so the amount of sodium nitrate will depend upon moles of sodium sulfide.
According to reaction, 1 mole of sodium sulfide gives 2 mole of sodium nitrate, then 2 mole of sodium sulfide will give:
[tex]\frac{2}{1}\times 2mol=4 mol[/tex] sodium nitrate
Mass of 4 moles of sodium nitrate :
85 g/mol × 4 mol = 340 g
Theoretical yield of sodium nitrate = 340 g
The theoretical yield of sodium nitrate is 340 grams.
Explanation:
Below is an attachment containing the solution.
An aqueous solution contains 3.2 mM of total ions. Part A If the solution is NaCl(aq), what is the concentration of chloride ion?
Answer:
1.6mM
Explanation:
NaCl(aq) -----------> Na+ (aq) + Cl- (aq)
There are two ions present and the total concentration of both ions is 3.2mM. Since the total concentration of both sodium ions and chloride ions is 3.2mM, then one of the ions will have a concentration which is half of the given value. That is, concentration of chloride ion will be 1/2× 3.2 = 1.6mM. Since the two ions are present in a ratio of one is to one, when you add the concentrations of both ions, you obtain the total concentration of ions in solution.
Write an overall equation for the acid-base reaction that would be required to produce the following salt. (NH4 )2 SO4
Answer:
2NH₃(g) + H₂SO₄(g) → (NH₄)₂SO₄(s)
Explanation:
(NH₄)₂SO₄ → Ammonium sulfate
This salt comes from a weak base and a strong acid
Base → Ammonia
Acid → Sulfuric acid
The reaction is:
2NH₃(aq) + H₂SO₄(aq) → (NH₄)₂SO₄(aq)
This is an acid salt; the protons from the acid are gained by the weak base.
As sulfuric is a strong acid, the sulfate is the conjugate weak base (it has no reaction)
As ammonia is a weak base, the ammonium is the conjugate strong acid so in water, it can react as this: NH₄⁺ + H₂O ⇄ NH₃ + H₃O⁺ Ka
Hydroniums are released, that's why it is an acid salt
Answer: 2NH3 (aq) + H2SO4 (aq) ----> (NH4)2 SO4 (aq)
Explanation:
The salt in question is Ammonium Salt (NH4)2 SO4.
It is formed by the reaction between Tetraoxosulphate (VI) acid and Ammonia.
2NH3(aq) + H2SO4 (aq) ----> (NH4)2 SO4 (aq)
Base Acid Salt
Ammonium sulphate is mainly used as fertilizer.
The concentration from analysis question 5 represents the concentration in the 10.00 mL sample that was prepared in the volumetric flask using an aliquot of the solution in the 100.00 mL volumetric flask. Calculate the concentration of acetylsalicylic acid in the 100.00 mL volumetric flask. This is a simple dilution – as long as you use the correct volumes
Answer:
0.048 M
Explanation:
Note that the Concentration from Question 5= .00096 M
To Prepere of commercial aspirin solution, take the following steps:
Mix 1 aspirin tablet and 10 mL of 1 M NaOH in a 125 mL Erlenheyemer flask, heat to boil.Transfer solution to 100 mL volumetric flask and fill to 100 mL mark with deonized water. Cover and mix solution thoroughly.Using pipette transfer a 0.200 mL of solution to a 10 mL volumetric flask and dilute it with the 0.02 M buffered iron (III) chloride solution. Transfer it to test tube.
Final answer:
To find the concentration of acetylsalicylic acid in the original 100.00 mL solution, one would use the dilution formula C1V1 = C2V2, which relates concentrations and volumes before and after dilution. Exact calculations require the concentration in the 10.00 mL sample, which is not provided.
Explanation:
The calculation of the concentration of acetylsalicylic acid in the 100.00 mL volumetric flask, given its concentration in a 10.00 mL sample, is a classic example of a dilution problem in chemistry. To calculate the concentration in the original volumetric flask, you need to use the formula: C1V1 = C2V2, where C1 and V1 represent the concentration and volume of the starting solution, and C2 and V2 represent the concentration and volume after dilution. Unfortunately, without the specific concentration found in the 10.00 mL sample, the exact concentration in the 100.00 mL cannot be calculated here. However, this relationship allows one to understand how dilutions work and provides a method to find the concentration in the original solution if the concentration in the diluted solution is known.
Calculate the value of the equilibrium constant, K c , for the reaction Q ( g ) + X ( g ) − ⇀ ↽ − 2 M ( g ) + N ( g ) given that M ( g ) − ⇀ ↽ − Z ( g ) K c 1 = 3.15 6 R ( g ) − ⇀ ↽ − 2 N ( g ) + 4 Z ( g ) K c 2 = 0.509 3 X ( g ) + 3 Q ( g ) − ⇀ ↽ − 9 R ( g ) K c 3 = 12.5
Answer:
[tex]\large \boxed{0.0106}[/tex]
Explanation:
We have three equations:
1. M(g) ⇌ Z(g); Kc₁ = 3.15
2. 6R(g) ⇌ 2N(g) + 4Z(g); Kc₂ = 0.509
3. 3X(g) + 3Q(g) ⇌ 9R(g); Kc₃ = 12.5
From these, we must devise the target equation:
4. Q(g) + X(g) ⇌ 2M(g) + N(g); Kc = ?
The target equation has Q(g) on the left, so you divide Equation 1 by 3.
When you divide an equation by 3, you take the cube root of its Kc.
5. X(g) + Q(g) ⇌ 3R(g): K₅ = ∛(Kc₃)
Equation 5 has 3R on the right, and that is not in the target equation.
You need an equation with 3R on the left, so you divide Equation 2 by 2.
When you divide an equation by 2, you take the square root of its Kc.
6. 3R(g) ⇌ N(g) + 2Z(g); K₆ = √ (Kc₂)
Equation 6 has 2Z on the right, and that is not in the target equation.
You need an equation with 2Z on the left, so you reverse Equation 2 by and double it.
When you reverse an equation, you take the reciprocal of its K.
When you double an equation, you square its K.
7. 2Z(g) ⇌ 2M(g); K₇ = (1/Kc₁)²
Now, you add equations 5, 6, and 7, cancelling species that appear on opposite sides of the reaction arrows.
When you add equations, you multiply their K values.
You get the target equation 4:
5. X(g) + Q(g) ⇌ 3R(g); K₅ = ∛(Kc₃)
6. 3R(g) ⇌ N(g) + 2Z(g); K₆ = √(Kc₂)
7. 2Z(g) ⇌ 2M(g); K₇ = (1/Kc₁)²
4. Q(g) + X(g) ⇌ 2M(g) + N(g); Kc = K₅K₆K₇ = [∛(Kc₃)√(Kc₂)]/(Kc₁)²
Kc = [∛(12.5)√(0.509)]/(12.5)² = (2.321 × 0.7120)/156.2 = 0.0106
[tex]K_{c} \text{ for the reaction is $\large \boxed{\mathbf{0.0106}}$}[/tex]
Answer:
The value of the equilibrium constant is 0.167
Explanation:
Step 1: The target equation
Q(g) + X(g) ⇔ 2M(g) + N(g)
Given is:
(1) M(g)⇔Z(g) c1=3.15
(2) 6R(g) ⇔ 2N(g) + 4Z(g) c2=0.509
(3) 3X(g) +3Q(g) ⇔ 9R(g) c3=12.5
Step 2: Rearange the equation
We have to rearange the equation to come to the final result
This is Hess' Law
In the target equation we have Q(g) + X(g)
In (3) we have 3X(g) +3Q(g) ⇔ 9R(g)
To get the target of Q(g) + X(g) we have to divide (3) by 3. This will give us:
X(g) +Q(g) ⇔ 3R(g) Kc = ∛12.5 = 2.32 (Note: to get Kc of the target equation we use cube root)
The target equation has as product 2M(g) + N(g)
To get M(g) we will use the (1) equation
Since M(g) is a product and not a reactant, we have to reverse the equation. Next to that we also have to double the equation because we need 2M(g) and not M(g)
2Z(g) ⇔ 2M(g) Kc = 1/(3.15)² = 0.101 (Note: to get Kc' after reversing the equation we calculate 1/Kc. To get Kc'' after doubling and reversing the equation we calculate 1/(Kc²)
To get N(g) we will use (2) 6R(g) ⇔ 2N(g) + 4Z(g)
Since we only need N(g) we will divide this equation by 2. This will get us:
3R(g) ⇔ N(g) + 2Z(g) Kc = √0.509 = 0.713 (Note: if we divide the equation by 2, to calculate Kc' we use square root)
Now we have all the components we will add the 3 equations:
X(g) +Q(g) + 2Z(g) + 3R(g)⇔ 3R(g) + 2M(g) + N(g) + 2Z(g)
We will simplify this equation:
X(g) +Q(g) ⇔ 2M(g) + N(g) this is our target equation
The value of the equilibrium constant, Kc is:
Kc = 2.32 * 0.101*0.713
Kc = 0.167
Note: to calculate Kc after adding several equations,we'll multiply Kc1* Kc2 * Kc3 etc...
The value of the equilibrium constant is 0.167
An unknown object is placed on a balance, which then reads 6.118 gg. Its volume is measured to be 3.04 cm3cm3. Find the density of this object by dividing mass by volume.
Answer:
2.013 g/cm3
Explanation:
Density is the ratio of mass to volume. It is measured in kg/m^3 or g/cm^3.
The density of an object whose mass is 6.118 g and a volume of 3.04 cm3
The density of the object is
= 6.118 g/3.04 cm3
= 2.013 g/cm3
Ron and Hermione begin with 1.50 g of the hydrate copper(II)sulfate·x-hydrate (CuSO4·xH2O), where x is an integer. Part of their practical exam is to determine this integer x. They are working in pairs, though Hermione is doing most of the work. This should be discouraged! After dehydration they find that they are left with 0.96 g of the an-hydrate CuSO4. What is the unknown integer x. Round the answer to the nearest integer.
Answer:
5
Explanation:
We can obtain the value of x by doing the following:
Mass of hydrated salt (CuSO4.xH2O) = 1.50g
Mass of anhydrous salt (CuSO4) = 0.96g
Mass of water molecule(xH2O) = 1.50 — 0.96 = 0.54g
Molar Mass of CuSO4.xH2O = 63.5 + 32 + (16x4) + x(2 +16) = 63.5 + 32 + 64 + 18x = 159.5 + 18x
Mass of water(xH2O) molecules in the hydrate salt is given by:
xH2O/CuSO4.xH2O = 0.54/1.5
18x/(159.5 + 18x) = 0.36
Cross multiply to express in linear form
18x = 0.36 (159.5 + 18x)
18x = 57.42 + 6.48x
Collect like terms
18x — 6.48x = 57.42
11.52x = 57.42
Divide both side by 11.52
x = 57.42/11.52
x = 5
Therefore, the unknown integer x is 5 and the formula for the hydrated salt is CuSO4.5H2O
Calculate the magnitude and direction (i.e., the angle with respect to the positive H-axis, measured positive as counter-clockwise) of the total force acting on M. Notice that the arrows representing the forces end on grid intersections.
The graphics in the attachment is part of the question, which was incomplete.
Answer: Fr = 102N and angle of approximately 11°.
Explanation: From the attachment, it is observed that from the three forces acting on M, two are perpendicular. So to find them, we have to show their x- and y- axis components. From the graph:
Fx = 70+40-10 = 100
Fy = 40-20 = 20
Now, as the forces form a triangle, the totalforce is:
Fr = [tex]\sqrt{Fx^{2} +Fy^{2} }[/tex]
Fr = [tex]\sqrt{10400}[/tex]
Fr = ≈ 102N
To determine the angle requested, we use:
arctg H = [tex]\frac{Fy}{Fx}[/tex]
arctg H = [tex]\frac{20}{100}[/tex]
H = tg 0.2 ≈ 11°.
Final answer:
Using trigonometry or graphical methods, the magnitudes and directions of forces such as F1 and F2 can be calculated. Applying principles of electromagnetism and mechanics, including the right-hand rule and cross-product notation, enables the determination of force magnitude and direction in a magnetic field.
Explanation:
The question involves calculating the magnitude and direction of forces acting on an object, which encompasses principles fundamental to physics, specifically in the area of mechanics and electromagnetism. Using trigonometry or graphical methods for force vector addition, one can find the magnitudes and directions of forces F1 and F2. Additionally, understanding the role of magnetic fields and applying the right-hand rule can assist in determining the direction of magnetic forces acting on a current-carrying wire, which is perpendicular to both the current and the magnetic field, leading to a simplified calculation of force magnitude and direction using cross-product notation.
For example, if given the values for the magnetic field, the charge, and the velocity of a particle moving through this field, one could apply the formula F = q(v x B) to determine the force's magnitude and direction. The understanding and application of Newton's Second Law are also crucial when analyzing forces that contribute to the work done on an object, taking into account the gravitational force component along the direction of displacement and the friction force.