Answer:
C. Gas
Explanation:
A quantity of liquid methanol, CH 3OH, is introduced into a rigid 3.00-L vessel, the vessel is sealed, and the temperature is raised to 500K. At this temperature, the methanol vaporizes and decomposes according to the reaction CH 3OH(g) CO(g) + 2 H 2(g), K c= 6.90×10 –2. If the concentration of H 2 in the equilibrium mixture is 0.426M, what mass of methanol was initially introduced into the vessel?
Answer:
74.3g of methanol were introduced into the vessel
Explanation:
In the equilibrium:
CH₃OH(g) ⇄ CO(g) + 2H₂(g)
Kc is defined as the ratio between concentrations in equilibrium of :
Kc = 6.90x10⁻² = [CO] [H₂]² / [CH₃OH]
Some methanol added to the vessel will react producing H₂ and CO. And equilibrium concentrations must be:
[CH₃OH] = ? - X
[CO] = X
[H₂] = 2X
Where ? is the initial concentration of methanol
As [H₂] = 2X = 0.426M; X = 0.213M
[CH₃OH] = ? - 0.213M
[CO] = 0.213M
[H₂] = 0.426M
Replacing in Kc to solve equilibrium concentration of methanol:
6.90x10⁻² = [0.213] [0.426]² / [CH₃OH]
[CH₃OH] = 0.560
As:
[CH₃OH] = ? - 0.213M = 0.560M
? = 0.773M
0.7733M was the initial concentration of methanol. As volume of vessel is 3.00L, moles of methanol are:
3.00L * (0.773 mol / L) = 2.319 moles methanol.
Using molar mass of methanol (32.04g/mol), initial mass of methanol added was:
2.319 moles * (32.04g / mol) =
74.3g of methanol were introduced into the vesselDetermine whether each chemical substance would remain the same color or turn pink in the presence of phenolphthalein.
Answer:
See the answer below
Explanation:
The complete question can be seen in the attached image.
Phenolphthalein is an indicator that is often utilized in an acid-base reaction to indicate the endpoints of such reactions due to its ability to change color from pink/colorless to colorless/pink depending on if the final solution is acidic or basic.
Phenolphthalein is usually colorless in acidic solutions and appears pink in basic solutions. The more basic or alkaline a solution is, the stronger the pink color of phenolphthalein. Hence;
1. Ammonia with a pH of 11 is basic, phenolphthalein will turn pink.
2. Battery acid with a pH of 1 is acidic, it will remain colorless.
3. Lime juice with a pH of 2 is acidic, it will remain colorless.
4. Mashed avocado with a pH of 6.5 is acidic, it will remain colorless.
5. Seawater with a pH of 8.5 is basic, it will turn pink.
6. Tap water with a pH of 7 is neutral, it will remain colorless
Phenolphthalein is a chemical compound with the formula[tex]C_{20}H_{14}O_4[/tex]. Phenolphthalein is often used as an indicator in acid-base titrations. For this application, it turns colorless in acidic solutions and pink in basic solutions
Phenolphthalein works as in:-
Colorless in acidPink in baseAccording to the question, There are 5 solutions having different ph and the indication only turns basic solution to pink
The indicator only turn the basic solution pink and these solutions are as follows,
AmmoniaSea waterTap water.Hence, these are the answer.
For more information, refer to the link:-
How to do q solution, qrxn, moles of Mg , and delta Hrxn?
Answer:
14, 508J/K
ΔHrxn =q/n
where q = heat absorbed and n = moles
Explanation:
m = mass of substance (g) = 0.1184g
1 mole of Mg - 24g
n moles - 0.1184g
n = 0.0049 moles.
Also, q = m × c × ΔT
Heat Capacity, C of MgCl2 = 71.09 J/(mol K)
∴ specific heat c of MgCL2 = 71.09/0.0049 (from the formula c = C/n)
= 14, 508 J/K/kg
ΔT= (final - initial) temp = 38.3 - 27.2
= 11.1 °C.
mass of MgCl2 = 95.211 × 0.1184 = 11.27
⇒ q = 11.27g × 11.1 °C × 14, 508 j/K/kg
= 1,7117.7472 J °C-1 g-1
∴ ΔHrxn = q/n
=1,7117.7472 ÷ 0.1184
= 14, 508J/K
1. What's the concentration of hydronium ions if a water-base solution has a temperature of 25°C (Kw = 1.0x10-14), with a concentration of
hydroxide ions of 2.21x10-6 M? A. 3.1x10-6 M
B. 4.52 X10-9 M
C. 2.8x10-8 M
D. 1.6x10-9 M
Answer:
B. 4.52 X10-9 M
Explanation:
Our goal for this question is to calculate the concentration of hydronium ions [tex]H^+[/tex] produced by water in a vessel with a concentration of hydroxide ions of [tex]2.21X10^-^6~M[/tex]. So, our first approach can be the ionization reaction of water:
[tex]H_2O_(_l_)~->~H^+~_(_a_q_)~+~OH^-~_(_a_q_)[/tex]
If we write the Keq expression for this reaction we will have:
[tex]Keq=[H^+][OH^-][/tex]
Now, water is the universal solvent, so, Keq has a special name. In the equilibrium problems for water we have to use "Kw" instead of "Keq":
[tex]Kw=[H^+][OH^-][/tex]
From this equation, we know the Kw value () and the concentration of the hydroxide ions ([2.21X10^-^6~M]). If we replace these values into the equation we can solve for [tex][H^+][/tex]:
[tex]1.0X10^-^1^4=[H^+][2.21X10^-^6~M][/tex]
[tex][H^+]=\frac{1.0X10^-^1^4}{2.21X10^-^6}=4.52X^-^9[/tex]
I hope it helps!
Answer:
B. 4.52 * 10^-9M
Explanation:
did the test
Gaseous indium dihydride is formed from the elements at elevated temperature:
In(g)+H2(g)⇌InH2(g),Kp=1.48 at 973 K
The partial pressures measured in a reaction vessel are
PIn =0.0540atm
PH2= 0.0250atm
PInH2 =0.0780atm
Calculate Qp and give equal partial pressure for In, H2, and InH2.
Answer:
The reaction given is:
In (g) + H₂ (g) ⇔ InH₂ (g), the Kp is 1.48 at 973 K.
The partial pressures measured in the reaction vessel is Partial pressure of In is 0.0540 atm, partial pressure of H₂ is 0.0250 atm, and the partial pressure of InH₂ is 0.0780 atm. By using the table given in the attachment below, the value of PInH₂ is (0.078-x), PIn is (0.054 + x), and the value of PH2 is (0.025 + x).
Kp = PInH₂/PIn × PH₂ = (0.078 - x) / (0.054 +x) (0.025 + x)
1.48 = (0.078 - x) / (0.054 +x) (0.025 + x)
x = 0.06689
Now the partial pressures of In, H₂ and InH₂ will be,
PH₂ = 0.025 + x = 0.025 + 0.0668 = 0.0918 atm
PIn = 0.054 + 0.0668 = 0.1208 atm
PInH₂ = 0.078 - 0.0668 = 0.0112 atm
Now the Qp or the reaction quotient will be,
Qp = (0.078) / (0.054) (0.025) = 57.78.
How many grams of PtBr4 will dissolve in 250.0 mL of water that has 1.00 grams of KBr dissolved in it
Answer:
[tex]m_{PtBr_4}=0.306gPtBr_4[/tex]
Explanation:
Hello,
In this case, since the solubility product of platinum (IV) bromide is 8.21x10⁻⁹, and the dissociation is:
[tex]PtBr_4(s)\rightleftharpoons Pt^{4+}(aq)+4Br^-(aq)[/tex]
The equilibrium expression is:
[tex]Ksp=[Pt^{4+}][Br^-]^4[/tex]
Thus, since the salt is added to a solution initially containing 1.00 grams of potassium bromide, there is an initial concentration of bromide ions:
[tex][Br^-]_0=\frac{1.00gKBr*\frac{1molKBr}{119gKBr}*\frac{1molBr^-}{1molKBr} }{0.250L}=0.0336M[/tex]
Hence, in terms of the molar solubility [tex]x[/tex], we can write:
[tex]8.21x10^{-9}=(x)(0.0336+4x)^4[/tex]
In such a way, solving for [tex]x[/tex], we obtain:
[tex]x=0.00238M[/tex]
Which is the molar solubility of platinum (IV) bromide. Then, since its molar mass is 514.7 g/mol, we can compute the grams that get dissolved in the 250.0-mL solution:
[tex]m_{PtBr_4}=0.00238\frac{molPtBr_4}{1L}*0.250L *\frac{514.7gPtBr_4}{1molPtBr_4} \\\\m_{PtBr_4}=0.306gPtBr_4[/tex]
Best regards.
You have to prepare a pH 3.65 buffer, and you have the following 0.10M solutions available: HCOOH, CH3COOH, H3PO4, HCOONa, CH3COONa, and NaH2PO4. How many mL of HCOOH and HCOONa would you use to make approximately a liter of the buffer?
Answer:
550mL of HCOOH 0.1M and 450mL of HCOONa 0.1M
Explanation:
It is possible to find the pH of a buffer by using H-H equation:
pH = pKa + log [A⁻]/[HA]
For the formic buffer (HCOOH/HCOONa):
pH = 3.74 + log [HCOONa]/[HCOOH]
As you need a buffer of pH 3.65:
pH = 3.74 + log [HCOONa]/[HCOOH]
3.65 = 3.74 + log [HCOONa]/[HCOOH]
0.81283 = [HCOONa]/[HCOOH] (1)Where [HCOONa]/[HCOOH] can be taken as the moles of each specie.
As molarity of both solutions is 0.10M (0.10mol / L) and you need 1L of solution, total moles of the buffer are:
0.10 moles = [HCOONa] + [HCOOH] (2)Replacing (2) in (1):
0.81283 = 0.10 - [HCOOH] /[HCOOH]
0.81283[HCOOH] = 0.10 - [HCOOH]
1.81283[HCOOH] = 0.10
[HCOOH] = 0.055 molesAnd moles of HCOONa are:
[HCOONa] = 0.1 mol - 0.055mol =
[HCOONa] = 0.045 molesAs concentration of the solutions is 0.1M, the volume you need to add of both solutions is:
HCOOH = 0.055 mol ₓ (1L / 0.1mol) = 0.55L = 550mL of HCOOH 0.1M
HCOONa = 0.045 mol ₓ (1L / 0.1mol) = 0.45L = 450mL of HCOONa 0.1M
The number should be considered like 550mL of HCOOH 0.1M and 450mL of HCOONa 0.1M.
Calculation of mL:Here we used the H-H equation:
pH = pKa + log [A⁻]/[HA]
Now
For the formic buffer (HCOOH/HCOONa):
So,
pH = 3.74 + log [HCOONa]/[HCOOH]
Now
need a buffer of pH 3.65:
So,
pH = 3.74 + log [HCOONa]/[HCOOH]
3.65 = 3.74 + log [HCOONa]/[HCOOH]
0.81283 = [HCOONa]/[HCOOH] (1)
here [HCOONa]/[HCOOH] can be considered as the moles of each specie.
Now the total moles should be
0.10 moles = [HCOONa] + [HCOOH] (2)
Now
0.81283 = 0.10 - [HCOOH] /[HCOOH]
0.81283[HCOOH] = 0.10 - [HCOOH]
1.81283[HCOOH] = 0.10
[HCOOH] = 0.055 moles
And moles of HCOONa should be
[HCOONa] = 0.1 mol - 0.055mol =
[HCOONa] = 0.045 moles
Now
HCOOH = 0.055 mol ₓ (1L / 0.1mol) = 0.55L = 550mL of HCOOH 0.1M
HCOONa = 0.045 mol ₓ (1L / 0.1mol) = 0.45L = 450mL of HCOONa 0.1M
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The ceramic glaze on a red-orange Fiestaware plate is U2O3 and contains 50.1 grams of 238U, but very little 235U. (a) What is the activity of the plate (in Ci)
Answer:
The correct answer is 1.68 × 10⁻⁵ Ci
Explanation:
The activity of the uranium is determined by using the formula,
R = 0.693 N/t1/2 -------------- (i)
The number of atoms is, N = nNA
Here, NA is the Avogadro number and n is the number of moles. The value of n is m/M, that is, mass/molecular mass. Now the value of N becomes,
N = (m/M) NA
The m or mass of uranium given is 50.1 grams, and the molecular mass is 238 g/mol, now putting the values we get,
N = (50 g/238 g) (6.023 × 10²³) = 1.26 × 10²³
The half-life of 238U from year to second is,
t1/2 = (4.468 × 10⁸ year) (3.16 × 10⁷ s/ 1 year) = 1.412 × 10¹⁶ s
Substituting the values of t1/2 as 1.412 × 10¹⁶, and 1.26 × 10²³ for N in equation (i) we get,
R = 0.639 (1.26 × 10²³) / 1.412 × 10¹⁶ s
= 6.18 × 10⁶ Bq (2.7027 × 10⁻¹¹ Ci/1 Bq)
= 1.68 × 10⁻⁵ Ci
Hence, the activity of the plate is 1.68 × 10⁻⁵ Ci
Write the Arrhenius Base reaction for the following:
Sr(OH)2
Answer:
Explanation:
Sr(OH)2 (aq) ⇔ Sr+2 (aq) + 2OH- (aq)
Classify each of the following fatty acids as saturated, monounsaturated, or polyunsaturated. Drag the appropriate items to their respective bins.
1. myristic acid
2. oleic acid
3. linoleic acid
4. palmitic acid
A. Saturated acids
B. Monounsaturated acids
C. Polyunsaturated acids
Answer:
A. Saturated acids - 1. myristic acid, 4. palmitic acid
B. Monounsaturated acids - 2. oleic acid
C. Polyunsaturated acids - 3. linoleic acid
Explanation:
Saturated fatty acids are the fats combined together with a single bond and no double or triple bond while unsaturated fatty acids are the fats having a double or triple bond between them.
Monosaturated acids have only one double or triple bond while polyunsaturated acids have more than one double or triple bond.
So, an appropriate match of given acids are:
A. Saturated acids - Myristic acid (CH3(CH2)12COOH) and Palmitic acid (CH3(CH2)14COOH) dont have any double or triple bonds.
B. Monounsaturated acids - Oleic acid (CH3(CH2)7CH=CH(CH 2)7COOH) have only one double bond.
C. Polyunsaturated acids - Linoleic acid (HOC–7CH=CH–CH 2–CH=CH–3H) has two double bonds.
Hence, the correct answer is:
A. Saturated acids - 1. myristic acid, 4. palmitic acid
B. Monounsaturated acids - 2. oleic acid
C. Polyunsaturated acids - 3. linoleic acid
How many cups are in five gallons?
Answer:
In 5 US liquid gallons, there are 80 cups.
Explanation:
To get from gallons to cups, just multiply the amount of gallons you have by 16.
Combustion reactions are a notable source of carbon dioxide in the environment. Using the following balanced equation, how many grams of carbon dioxide are formed when 100.00 g of propane is burned? Express your answer to the correct number of significant figures. Be sure to show all steps completed to arrive at the answer. Equation: C3H8 + 5O2 ->>>>>>> 3CO2 + 4H2O
Answer:
Explanation:
Number of moles of propane:
=Mass in grams ÷ Relative molecular Mass
= 100/((12*3) + (1*8))
= 100 ÷ 44
= 2.2727
Mole ratio propane:carbon (IV) oxide = 1:3(from the equation)
Number of moles of CO2 = 3*2.2727 = 6.8181
Mass in grams = Relative molecular Mass * Number of moles
= 44 * 6.8181
= 299.9964 grams
A number of moles of propane:
Mass in grams ÷ Relative molecular Mass
= 100/((12*3) + (1*8))
= 100 ÷ 44
= 2.2727
Mole ratio propane:carbon (IV) oxide = 1:3(from the equation)
Number of moles of CO2 = 3*2.2727 = 6.8181
Mass in grams = Relative molecular Mass * Number of moles
=44 * 6.8181
= 299.9964 grams
What is carbon dioxide useful for?Carbon dioxide is used as a refrigerant, in fireplace extinguishers, for inflating lifestyles rafts and life jackets, blasting coal, foaming rubber and plastics, selling the increased vegetation in greenhouses, and immobilizing animals earlier than slaughter, and in carbonated liquids.
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What happens when an atom of sulfur combines with two atoms of chlorine to produce SCl2?
Answer:
Each chlorine atom shares a pair of electrons with the sulfur atom. ... Each chlorine atom shares all its valence electrons with the sulfur atom
What is the electron configuration for the transition metal ion in each of the following compounds?
[Ni(H2O)6]Br2
[Cr(H2O)4(NO2)2]I
Answer:
1)Ni=1s2, 2s2, 2p6, 3s2, 3p6, 4s0, 3d10 called full-filled
2)Cr=1s2, 2s2, 2p6, 3s2, 3p6, 4s1, 3d5 called half-filled
Calculate the number of hydrogen atoms in a 110.0 sample of tetraborane(B4H10) . Be sure your answer has a unit symbol if necessary, and round it to 4 significant digits.
Answer:
[tex]1.242 \times 10^{25}\text{ atoms H}[/tex]
Explanation:
You must convert the mass of B₄H₁₀ to moles of B₄H₁₀, then to molecules of B₄H₁₀, and finally to atoms of H.
1. Moles of B₄H₁₀
[tex]\text{Moles of B$_{4}$H}_{10} = \text{110.0 g B$_{4}$H}_{10} \times \dfrac{\text{1 mol B$_{4}$H}_{10}}{\text{53.32 g B$_{4}$H}_{10}} = \text{2.063 mol B$_{4}$H}_{10}[/tex]
2. Molecules of B₄H₁₀
[tex]\text{No. of molecules} = \text{2.063 mol B$_{4}$H}_{10} \times \dfrac{6.022 \times 10^{23}\text{ molecules B$_{4}$H}_{10}}{\text{1 mol B$_{4}$H}_{10}}\\\\=1.242 \times 10^{24}\text{ molecules B$_{4}$H}_{10}[/tex]
3. Atoms of H
[tex]\text{Atoms of H} = 1.242 \times 10^{24}\text{ molecules B$_{4}$H}_{10} \times \dfrac{\text{10 atoms H}}{\text{1 molecule B$_{4}$H}_{10}}\\\\= \mathbf{1.242 \times 10^{25}}\textbf{ atoms H}[/tex]
People take antacids, such as milk of magnesia, to reduce the discomfort of acid stomach or heartburn. The recommended dose of milk of magnesia is 1 teaspoon, which contains 500 mg of Mg(OH)2. What volume of HCl solution with a pH of 1.25 can be neutralized by 1 dose of milk of magnesia
Answer:
[tex]V_{HCl}=0.208L=208mL[/tex]
Explanation:
Hello,
In this case, since the chemical reaction is:
[tex]2HCl+Mg(OH)_2\rightarrow MgCl_2+2H_2O[/tex]
We can see that hydrochloric acid and magnesium hydroxide are in a 2:1 mole ratio, which means that the neutralization point, we can write:
[tex]n_{HCl}=2*n_{Mg(OH)_2}[/tex]
In such a way, the moles of magnesium hydroxide (molar mass 58.3 g/mol) in 500 mg are:
[tex]n_{Mg(OH)_2}=500mg*\frac{1g}{1000mg}*\frac{1mol}{58.3g} =0.00858mol[/tex]
Next, since the pH of hydrochloric acid is 1.25, the concentration of H⁺ as well as the acid (strong acid) is:
[tex][H^+]=[HCl]=10^{-pH}=10^{-1.25}=0.0562M[/tex]
Then, since the concentration and the volume define the moles, we can write:
[tex][HCl]*V_{HCl}=2*n_{Mg(OH)_2}[/tex]
Therefore, the neutralized volume turns out:
[tex]V_{HCl}=\frac{2*0.00858mol}{0.0562\frac{mol}{L} }\\ \\V_{HCl}=0.208L=208mL[/tex]
Best regards.
What are the conditions that are favorable for extensive solid solubility of one element in another (Hume-rothery rules)
Answer:
Atomic radius less than 15%, similar structure and same valency.
Explanation:
The conditions that are favorable for extensive solid solubility of one element in another are the following.
The atomic radius of the solute and solvent atoms must be less than 15%. The structure of both solute and solvent are similar. Solubility completes when both have same valency. Valency means number of electrons in the outermost shell. If both solute and solvent has same number of electrons so it will be completely soluble in each other.
The conditions that are favorable for extensive solid solubility of one element in another is the same size, electrongativity and valency.
What is Hume - Rothery rules?Hume - Rothery rules are the sets of some important rules which gives idea about the desired condition for the formation of solid solution.
Following main points are described in this rule:
Difference between the size of the solute and the solvent should be less than 15%.Electronegativity difference between the solute and solvents should be small.And they both should have same valency, means same no. of electrons in the outermost shell.Hence size, electronegativity and valency are the conditions.
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A chamber contains equal molar amounts of He, Ne, Ar, and Kr. If the total chamber pressure is 1 atm, then the partial pressure (in atm) of Kr is:
Answer:
The correct answer is 0.25 atm.
Explanation:
As mentioned in the given question that the chamber comprises equal molar concentrations of He, Ne, Ar and Kr gas. So, let us assume that the moles of all the gases will be x then the total number of moles will be 4x.
The formula for calculating mole fraction is,
Mole fraction = mole of the substance/total moles
The mole fraction of Kr = x/4x = 0.25
The total pressure given in the chamber is 1 atm. Therefore, the partial pressure will be,
Partial pressure = mole fraction * Total pressure
Partial pressure = 0.25 * 1 = 0.25 atm.
The partial pressure of Kr is 0.25 atm.
We are told that there is an equal molar amounts of He, Ne, Ar, and Kr. If we decide to say, let the molar amount of each gas be x, the total number of moles of all the gases will now be; x + x+ x + x = 4x
Let us recall that partial pressure of a gas can be obtained form the formula;
Partial pressure = mole fraction × total pressure
Mole fraction of Kr = x/4x = 1/4
Total pressure = 1 atm
Partial pressure of Kr = 1/4 × 1 = 0.25 atm
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When the following molecular equation is balanced using the smallest possible integer coefficients, the values of these coefficients are: potassium hydrogen sulfate (aq) potassium hydroxide (aq) potassium sulfate (aq) water (l)\
Answer:
Explanation:
Answer:
1, 1, 1, 1
Explanation:
potassium hydrogen sulfate + potassium hydroxide ⟶ potassium sulfate + water(l)
KHSO₄ + KOH ⟶ K₂SO₄ + H₂O
1. Put a 1 in front of the most complicated-looking formula (K₂SO₄?):
KHSO₄ + KOH ⟶ 1K₂SO₄ + H₂O
2. Balance S:
We have fixed 1 S on the right. We need 1 S on the left. Put a 1 in front of KHSO₄ to fix it.
1KHSO₄ + KOH ⟶ 1K₂SO₄ + H₂O
3. Balance K:
We have fixed 2 K on the right and 1 K on the left. We need 1 more K on the left. Put a 1 in front of KOH.
1KHSO₄ + 1KOH ⟶ 1K₂SO₄ + H₂O
4. Balance O
We have fixed 4 O on the right and 5 O on the left. We need 1 more O on the right. Put a 1 in front of H₂O.
1KHSO₄ + 1KOH ⟶ 1K₂SO₄ + 1H₂O
Every formula has a coefficient. The equation should be balanced.
5. Check that atoms balance:
[tex]\begin{array}{ccc}\textbf{Atom} & \textbf{On the left} & \textbf{On the right}\\\text{K} & 2 &2\\\text{H} & 2 & 2\\\text{S} & 1 & 1\\\text{O}&5&5\\\end{array}[/tex]
It checks.
The coefficients are 1, 1, 1, 1.
Predict the order of acid strengths in the following series of cationic
species: CH3CH2NH3
+, CH3CH=NH2
+ and CH3C-NH+
Answer:
CH3C-NH+> CH3CH=NH2 >CH3CH2NH3+
Explanation:
The acid strength has to do with the ease of the loss of hydrogen ion from the cationic specie. Hydrogen ion will easily be lost from any specie which contains an atom, group of atoms or bond which withdraws electrons along the chain of the N-H bond.
The pi bond system is known to be highly electronegative and withdraws electrons along the chain hence a withdrawal of electron density along the chain which makes the hydrogen ion to be easily lost from a system which contains a pi bond along the chain. A triple bond is more electronegative than a double bond, hence the answer above.
For the following reaction, 3.76 grams of iron are mixed with excess oxygen gas . The reaction yields 4.29 grams of iron(II) oxide . iron ( s ) oxygen ( g ) iron(II) oxide ( s ) What is the theoretical yield of iron(II) oxide
Answer:
4.84g of FeO is the theoretical yield
Explanation:
The Iron, Fe(s), reacts with oxygen, O₂(g), producing Iron (II) oxide, as follows:
2Fe(s) + O₂(g) → 2FeO
Theoretical yield is the yield of a reaction in which you assume the 100% of reactants is converted in products.
To find theoretical yield we need to find moles of Iron, and, knowing 2 moles of Fe produce 2 moles of FeO (Ratio 1:1), we can find theoretical yield of FeO as follows:
Moles Fe (Molar mass: 55.845g/mol)
Using the molar mass of the compound we can convert grams to moles, thus:
3.76g Fe × (1mol / 55.845g) = 0.0673 moles of Fe
Moles and mass of FeO
As there are in reaction 0.0673 moles Fe, assuming a theoretical yield (And as ratio of the reaction is 1:1), you will obtain 0.0673 moles of FeO.
Theoretical yield is given in grams, As molar mass of FeO is 71.844g/mol, theoretical yield of the reaction is:
0.0673 moles FeO × (71.844g / mol) =
4.84g of FeO is the theoretical yieldacid-catalyzed hydration of 1-methylcyclohexene gives two alcohols. The major product does not undergo oxidation, while the minor product will undergo oxidation. Explain
Answer:
Major product does not undergo oxidation since it is a tertiary alcohol whereas minor product undergoes oxidation to ketone as it is secondary alcohol.
Explanation:
Hello,
In this case, given the attached picture, the hydration of the 1 methylcyclohexene yields to alcohols; 1-methylcyclohexan-1-ol and 1-methylcyclohexan-2-ol. Thus, since the OH in the 1-methylcyclohexan-1-ol (major product) is bonded to a tertiary carbon (bonded with other three carbon atoms) it is not able to increase the number of oxygen bonds (oxidation) as it already attained the octet whereas the 1-methylcyclohexan-2-ol (minor product) is able to undergo oxidation to ketone as the carbon bonded to it is secondary (bonded with other two carbon atoms), so one extra bond the oxygen is allowed to be formed to carbonyl.
Best regards.
In the experiment, you heated a sample of metal in a water bath and quickly removed and dried the sample prior to inserting it into the cold water "system". Why did you need to dry the sample prior to placing it into the cold water "system"
a. The extra water might react with the metal which would ruin the sample.
b. Any residual hot water would have added to the heat transferred from the metal to the cold water "system".
c. The wet metal would not transfer any heat thus causing inaccuracy in you measurements.
d. The metal would oxidize in the presence of water thus ruining the sample
Answer:
b
Explanation:
Provided the experiment is about the transfer of heat from one medium to another, any residual hot water on the metal would have added to the heat transferred from the metal to the cold water system.
The residual hot water on the metal posses its own heat and when such is transferred along with the metal into the cold water, the heat from the residual hot water will interfere with the measurement of the actual heat transferred to the cold water by the metal. Hence, the accuracy of the result would be impacted.
The correct option is b.
The two common properties of all solids are fixed _____ and _____.
Answer:
shape
volume
Hope this helps! （づ￣3￣）づ╭❤～
Explanation:
A 1.0 L buffer solution is 0.250 M HC2H3O2 and 0.050 M LiC2H3O2. Which of the following actions will destroy the buffer?
A. adding 0.050 moles of NaOH
B. adding 0.050 moles of LiC2H3O2
C. adding 0.050 moles of HC2H3O2
D. adding 0.050 moles of HCl
E. None of the above will destroy the buffer.
Answer:
D
Explanation:
Addition of 0.05 M HCl, will react with all of the C2H3O2- from LiAc which will give 0.05 M more HAc. So there will be no Acetate ion left to make the solution buffer. Hence, the correct option for the this question is d, which is adding 0.050 moles of HCl.
The action that destroys the buffer is option c. adding 0.050 moles of HCl.
What is acid buffer?It is a solution of a weak acid and salt.
Here, The buffer will destroy at the time when either HC2H3O2 or NaC2H3O2 should not be present in the solution.
The addition of equal moles of HCl finishly reacts with equal moles of NaC2H3O2. Due to this, there will be only acid in the solution.
Since
moles of HC2H3O2 = 1*0.250 = 0.250
moles of NaC2H3O2 = 1*0.050 = 0.050.
moles of HCl is added = 0.050
Now
The reaction between HCl and NaC2H3O2
[tex]HCl + NaC_2H_3O_2 \rightarrow HC_2H_3O_2 + NaCl[/tex]
Now
BCA table is
NaC2H3O2 HCl HC2H3O2
Before 0.050 0.050 0.250
Change -0.050 -0.050 +0.050
After 0 0 0.300
Now, the solution contains the acid (HC2H3O2 ) only.
Therefore addition of 0.050 moles of HCl will destroy the buffer.
Learn more about moles here: https://brainly.com/question/24817060
A buffer is prepared such that [H2PO4-] = 0.095M and [HPO42-] = 0.125M? What is the pH of this buffer solution? (pKa = 7.21 for H2PO4-)
Answer:
pH of the buffer is 7.33
Explanation:
The mixture of the ions H₂PO₄⁻ and HPO₄²⁻ produce a buffer (The mixture of a weak acid, H₂PO₄⁻, with its conjugate base, HPO₄²⁻).
To find pH of a buffer we use H-H equation:
pH = pka + log [A⁻] / [HA]
Where A⁻ is conjugate base and HA weak acid.
For the H₂PO₄⁻ and HPO₄²⁻ buffer:
pH = pka + log [HPO₄²⁻] / [H₂PO₄⁻]
Computing values of the problem:
pH =7.21 + log [0.125M] / [0.095M]
pH = 7.33
pH of the buffer is 7.33
11mg of cyanide per kilogram of body weight is lethal for 50% of domestic chickens. How many grams per kilogram of body weight is a lethal dose for 50% of domestic chickens?
Answer:
[tex]0.033g[/tex]
Explanation:
Hello,
In this case, since 11 mg per kilogram of body weight has the given lethality, the mg that turn out lethal for a chicken weighting 3 kg is computed by using a rule of three:
[tex]11mg\longrightarrow 1kg\\\\x\ \ \ \ \ \ \longrightarrow 3kg[/tex]
Thus, we obtain:
[tex]x=\frac{3kg*11mg}{1kg}\\ \\x=33mg[/tex]
That in grams is:
[tex]=33mg*\frac{1g}{1000mg} \\\\=0.033g[/tex]
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When 1604 J of heat energy is added to 48.9 g of hexane, C6H14, the temperature increases by 14.5 ∘C. Calculate the molar heat capacity of C6H14.
Answer:
THE MOLAR HEAT CAPACITY OF HEXANE IS 290.027 J/ C
Explanation:
1604 J of heat is added to 48.9 g of hexane
To calculate the molar heat capacity of hexane, it is important to note that the molar heat capacity of a substance is the measure of the amount of heat needed to raise 1 mole of a substance by 1 K.
Since 1604 J of heat = 48.9 g of hexane
Molar mass of hexane = 86 g/mol = 1 mole
then;
1604 J = 48.9 g
x = 86 g
x = 1604 * 86 / 48.9
x = 4205.4 J
Hence, 4205.4 J of heat will be added to 1 mole or 86 g of hexane to raise the temperature by 14.5 C.
In other words,
heat = molar heat capacity * temperature change
molar heat capacity = heat/ temperature change
Molar heat capacity = 4205.4 J / 14.5 C
Molar heat capacity = 290.027 J/C
The molar heat capacity of hexane is 290.027 J/ C
What are the correct half reactions for the following reaction: Cu2+ + Mg -> Cu + Mg2+
Answer:
Cu2 + 2Mg-> 2Cu+ Mg2
Explanation:
Balance the equation and make sure both the reactant and the products are the same
Hope it will be helpful
[tex]Cu^{+2} + 2Mg[/tex] -> [tex]2Cu + Mg^+2[/tex] is the correct half-reactions.
What is a balanced equation?A balanced equation is an equation for a chemical reaction in which the number of atoms for each element in the reaction and the total oxidation numbers is the same for both the reactants and the products.
[tex]Cu^{+2} + 2Mg[/tex] -> [tex]2Cu + Mg^+2[/tex] is the correct half-reactions.
Magnesium is oxidized because its oxidation state increased from 0 to +2 while Cu is reduced because its oxidation state decreased from +2 to 0.
Learn more about balanced equations here:
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Draw a structure for an alcohol that exhibits a molecular ion at M+ = 88 and that produces fragments at m/z = 73, m/z = 70 and m/z = 59.
Answer:
3-pentanol
Explanation:
In this case, we have alcohol as the main functional group (OH) with a molecular ion at 88. If the molecular ion is 88 the molar mass is also 88 g/mol therefore the formula for the unknown molecule is [tex]C_5H_1_2O[/tex].
Additionally, if the mass spectrum shows the molecular ion peak we can not have tertiary alcohols (tertiary alcohols often do not show M+ at all). So, the structures only can be primary and secondary structures.
With this in mind, our options are:
-) 1-pentanol
-) 2-pentanol
-) 3-pentanol
Now we can analyze each structure:
-) 1-pentanol
The structure must explain all the fragments produced (73, 70, and 59). In this primary alcohol, we will have an alpha cleavage (the red bond would be broken). If this has to happen, we will have fragments at 31 and 57. These fragments dont fit with the reported ones, therefore this is not a possible structure (See figure 1).
-) 2-pentanol
On this structure, we will have also an alpha cleavage (red bond). In this rupture we will have fragments at 45 and 43, these m/z values dont fit with the reported ones, therefore this is not a possible structure (See figure 1).
-) 3-pentanol
In this structure, we have the "OH" bonded to carbon three. So, we can analyze each fragment:
-) m/z 59
This fragment, can be explained as an alpha cleavage. But, in this case we have two ruptures that can produce the same ion. The carbons on both sides of the C-OH bond.
-) m/z 71
This fragment, can be explained as a loss of water (M-18) in which we have the production of a carbocation in the carbon where we previously have the C-OH bond.
-) m/z 73
This fragment, can be explained as a beta cleavage. But, in this case, also we have two ruptures that can produce the same ion. The methyl groups on each end molecule.
See figure 2
I hope it helps!
