Answer:
1.8 g
Explanation:
Step 1: Write the balanced equation
CH₃CH₃(g) + 3.5 O₂(g) ⇒ 2 CO₂(g) + 3 H₂O(g)
Step 2: Determine the limiting reactant
The theoretical mass ratio of CH₃CH₃ to O₂ is 30.06:112.0 = 0.2684:1.
The experimental mass ratio of CH₃CH₃ to O₂ is 0.60:3.52 = 0.17:1.
Thus, the limiting reactant is CH₃CH₃
Step 3: Calculate the mass of CO₂ produced
The theoretical mass ratio of CH₃CH₃ to O₂ is 30.06:88.02.
0.60 g CH₃CH₃ × 88.02 g CO₂/30.06 g CH₃CH₃ = 1.8 g
In what areas of the periodic table do you find the most highly reactive elements?
Answer:
The elements toward the bottom left corner of the periodic table are the metals that are the most active in the sense of being the most reactive.
The most highly reactive elements are typically found at the far left (Group 1) and far right (Group 17) of the periodic table.
Highly reactive elements in the periodic tableGroup 1 elements, also known as alkali metals, are located on the far left of the periodic table. They have one electron in their outermost energy level and are highly reactive due to their tendency to lose that electron to achieve a stable electron configuration. This makes them very reactive with water and other substances.
Group 17 elements, known as halogens, are located on the far right of the periodic table. They have seven electrons in their outermost energy level and are highly reactive due to their strong tendency to gain one electron to achieve a stable electron configuration. This makes them reactive with metals and other elements.
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What would be the name of this compound?
Answer:
2,3 Dimethyl hexane
Explanation:
First, start the count from which side is given the CH3 smallest number
first; the longest carbon chain in this compound is 6
and you don't have any double and triple bonds or functional groups so it is Hexane
you start to count from the right side to give the branch molecules the smallest number ..
CH3 = methyl
and you have 2 methyl in this compound ..
and 2 mean you must write ( Di )
you write the name in this way
2,3 Dimethyl hexane
hope this helps you.
stay safe ...
An ordinary gasoline can measuring 30.0 cm by 20.0 cm by 15.0 cm is evacuated with a vacuum pump.
1a. Assuming that virtually all of the air can be removed from inside the can, and that atmospheric pressure is 14.7 psi, what is the total force (in pounds) on the surface of the can?
1b. Do you think that the can could withstand the force?
Answer:
Explanation:
From the given information:
The surface area of the can = (30 × 20 × 2) +(20× 15 × 2) +(30 × 15 × 2)
= 1200 + 600 + 900
= 2700 cm²
Since 1 inch² = 0.155 inch²
The surface area in inches² = 2700 × 0.155 inch²
= 418.5 inches²
The total force can be determined by using the expression:
Force = Pressure ×Area
Force = 14.7 psi × 418.5 inches²
Force = 6151.95 lbs
Yes, the gasoline can will be able to withstand the force.
using the balanced equation below how many grams of lead(||) sulfate would be produced from the complete reaction of 23.6 g lead (|V) oxide
Answer:
59.8 g of PbSO₄.
Explanation:
The balanced equation for the reaction is given below:
Pb + PbO₂ + 2H₂SO₄ —> 2PbSO₄ + 2H₂O
Next, we shall determine the mass of PbO₂ that reacted and the mass of PbSO₄ produced from the balanced equation. This can be obtained as follow:
Molar mass of PbO₂ = 207 + (16×2)
= 207 + 32
= 239 g/mol
Mass of PbO₂ from the balanced equation = 1 × 239 = 239 g
Molar mass of PbSO₄ = 207 + 32 + (16×4)
= 207 + 32 + 64
= 303 g/mol
Mass of PbSO₄ from the balanced equation = 2 × 303 = 606 g
SUMMARY:
From the balanced equation above,
239 g of PbO₂ reacted to produce 606 g of PbSO₄.
Finally, we shall determine the mass of PbSO₄ that will be produced by the reaction of 23.6 g of PbO₂. This can be obtained as follow:
From the balanced equation above,
239 g of PbO₂ reacted to produce 606 g of PbSO₄.
Therefore, 23.6 g of PbO₂ will react to produce = (23.6 × 606) / 239 = 59.8 g of PbSO₄.
Thus, 59.8 g of PbSO₄ were obtained from the reaction.
You perform a distillation to separate a mixture of propylbenzene and cyclohexane, and you obtain 2.9949 grams of cyclohexane (density -0.779 g/mL, MW - 84.16 g/mol) and 1.6575 grams of propylbenzene (density = 0.862 g/mL, MW = 120.2 g/mol). What is the volume percent composition of cyclohexane in the mixture?
Answer:
66.67%
Explanation:
From the given information:
mass of cyclohexane = 2.9949 grams
density of cyclohexane = 0.779 g/mL
Recall that:
Density = mass/volume
∴
Volume = mass/density
So, the volume of cyclohexane = 2.9949 g/ 0.779 g/mL
= 3.8445 mL
Also,
mass of propylbenzene = 1.6575 grams
density of propylbenzene = 0.862 g/mL
Volume of propylbenzene = 1.6575 g/ 0.862 g/mL
= 1.9229 mL
The volume % composition of cyclohexane from the mixture is:
[tex]= (\dfrac{v_{cyclohexane}}{v_{cyclohexane}+v_{propylbenzene}})\times 100[/tex]
[tex]= (\dfrac{3.8445}{3.8445+1.9229})\times 100[/tex]
[tex]= (\dfrac{3.8445}{5.7674})\times 100[/tex]
= 66.67%
Suppose you are studying the Ksp of CaCl2, which has a molar mass of 110.98 g/mol, at multiple temperatures. You dissolve 4.99 g of CaCl2 in 10.0 mL of water at 100 oC and cool the solution. At 90 oC, a solid begins to appear. What is the Ksp of CaCl2 at 90 oC
Answer:
Hence the Solubility product,
Ksp = [Ca2+] [Cl-]2
or, Ksp = (4.5) (9)2
or, Ksp = 364.5
Explanation:
Mass of CaCl2 = 4.99 g
Molar mass of CaCl2 = 110.98 g/mol
Moles of CaCl2
= given mass/ molar mass
= 4.99/ 110.98
= 0.045
Volume = 10.0 mL = 0.01 L
CaCl2 dissociates into its ion as:
CaCl2 (s) \rightleftharpoons Ca2+ (aq) + 2 Cl- (aq)
At 90°C, the solution is saturated with Ca2+ and Cl- ions.
Moles of Ca2+ = Moles of CaCl2 dissolved = 0.045
Moles of Cl- = 2 x ( Moles of CaCl2 dissolved) = 2 x 0.045 = 0.09
[Ca2+] = Moles/ Volume = 0.045/ 0.01 = 4.5 M
[Cl-] = 0.09/ 0.01 = 9 M
Solubility product,
Ksp = [Ca2+] [Cl-]2
or, Ksp = (4.5) (9)2
or, Ksp = 364.5
ort
Which is a primary alcohol?
0 3-pentanol
2-propanol
1-ethanol
4-octanol
urvey
Lig A Moving to another question will save this response.
Answer:
1 ethanol is right answer
Explanation:
CH3- CH2-OH
Which of the following ligands is not capable of exhibiting linkage isomerism?
a. NCO-
b. -OH
c. -CN
d. -SCN
Answer:
a
...
........
...........
Calculate the mass percent of each component in the following solution.
159 g NiCl2 in 500 g water
% Nicla
% water
Answer:
% NiCl2 = 24.13%
% water = 78.57%
Explanation:
Mass percentage = mass of solute/mass of solution × 100
According to this question, a solution contains 159 g of NiCl2 in 500 g of water. Hence, mass of the solution is calculated as follows:
Mass of solution = 159g + 500g
Mass of solution = 659g
Therefore;
A) % Mass of NiCl2 in solution = mass of NiCl2/mass of solution × 100
% Mass of NiCl2 in solution = 159/659 × 100
% Mass of NiCl2 in solution = 0.2413 × 100
= 24.13%
B) % Mass of water in solution = mass of water/mass of solution × 100
% Mass of water in solution = 500/659 × 100
% Mass of water in solution = 0.7587 × 100
% Mass of water in solution = 75.87%
Lead of mass 0.75kg is heated from 21°c to its melting point and continues to be heated unit it has all melted. Calculate how much energy is supplied to the lead. [Melting point of lead 372.5°c specific latent heat of fusion of lead = 23000 Jkg 'k ']
Answer:
65.5J
Explanation:
ML=Q
ML=MC(change in temperature)
0.75 X 23000 =0.75 X 351 X C
C= 65.5J
The energy supplied to the lead to melt from 21°c to its melting point is 51521 Joules.
What is the specific heat capacity?Specific heat is the amount of heat energy supplied to change the temperature of one unit mass of a substance by 1 °C. The SI unit of the specific heat capacity of a substance is J/Kg.
The mathematical expression for the specific heat capacity can be written as:
Q = mCΔT Where C is the specific heat of the substance.
The specific heat capacity depends upon the starting temperature and is an intensive characteristic of the material.
Given, the melting point of the lead T₂ = 327.5° C
The initial temperature of the lead, T₁ = 21° C
The latent heat of the lead given, L = 230000 J/Kg K
The specific heat of the lead, C = 130 J/Kg K
The heat required to melt the lead from 12°C to 327.5 °C is :
Q = m× [C × (T₂ - T₁) + L ]
Q = 0.75 × [0.130 (327.5 - 21) + 23000]
Q = 51521 J
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What is the molarity of an HCl solution if 25.0 mL of this solution required 17.80 mL of 0.108 M NaOH to reach the end point in a titration?
Answer:
[tex]\boxed {\boxed {\sf 0.0769 \ M}}[/tex]
Explanation:
We are asked to find the molarity of an acid given the details of a titration experiment. The formula for titration is as follows:
[tex]M_AV_A= M_B V_B[/tex]
In this formula, M is the molarity of the acid or base and V is the volume of the acid or base. The molarity of the hydrochloric acid (HCl) is unknown and the volume is 25.0 milliliters.
[tex]M_A * 25.0 \ mL = M_BV_B[/tex]
The molarity of the sodium hydroxide (NaOH) is 0.108 molar and the volume is 17.80 milliliters.
[tex]M_A * 25.0 \ mL = 0.108 \ M * 17.80 \ mL[/tex]
We are solving for the molarity of the acid and we must isolate the variable [tex]M_A[/tex]. It is being multiplied by 25.0 milliliters. The inverse operation of multiplication is division, so we divide both sides of the equation by 25.0 mL.
[tex]\frac {M_A * 25.0 \ mL }{25.0 \ mL}= \frac{0.108 \ M * 17.80 \ mL }{25.0 \ mL}[/tex]
[tex]M_A= \frac{0.108 \ M * 17.80 \ mL }{25.0 \ mL}[/tex]
The units of milliliters cancel.
[tex]M_A= \frac{0.108 \ M * 17.80 }{25.0 }[/tex]
[tex]M_A= \frac{1.9224}{25.0 } \ M[/tex]
[tex]M_A= 0.076896 \ M[/tex]
The original measurements have 3 and 4 significant figures. We must round our answer to the least number of sig figs, which is 3. For the number we calculated, that is the ten-thousandth place. The 9 to the right of this place tells us to round the 8 up to a 9.
[tex]M_A \approx 0.0769 \ M[/tex]
The molarity of the hydrochloric acid is 0.0769 Molar.
What process occurs during the corrosion of iron?
Answers
A.
Iron is oxidized.
B.
Iron is reduced.
C.
Iron (III) is oxidized.
D.
Iron (III) is reduced.
Answer:
A
Explanation:
The iron corrodes so it oxidized
How does the radiation from radioisotopes cause damage to human tissue?
Answers
A.
by ionization knocking electrons away from atoms
B.
by breaking the bonds between atoms in molecules
C.
by causing nuclear chain reactions inside cells
D.
by causing transmutations of atoms within cells
Answer:
Explanation:
B.
by breaking the bonds between atoms in molecules
Which of the following is classified as paramagnetic?
1.NO(+)
2.Be2
3.B2(+)
4.+N2(+)
Which of the following behaviors would best describe someone who is listening and paying attention? a) Leaning toward the speaker O b) Interrupting the speaker to share their opinion c) Avoiding eye contact d) Asking questions to make sure they understand what's being said
Answer:
a Leaning towards the speaker
Adding more than one equivalent of HCl to pent-1-yne will lead to which product:______.
a. 1,2-dichloro-1-butene.
b. 1,1-dichloropentane.
c. 2,2-dichloropentane.
d. 2,2-dichlorobutane.
Answer:
c. 2,2-dichloropentane.
Explanation:
Hello there!
In this case, according to the given information, it turns out possible for us to firstly draw the structure of the reactant, pent-1-yne:
[tex]CH\equiv C-CH_2-CH_2-CH_2[/tex]
Now, we infer the halogen is added to the carbon atom with the most carbon atoms next to it, in this case, carbon #2, in order to write the following product:
[tex]CH\equiv C-CH_2-CH_2-CH_2+2HCl\rightarrow CH_3- CCl_2-CH_2-CH_2-CH_2[/tex]
Whose name is 2,2-dichloropentane.
Regards!
GM 2 all ,What is an atom define it .Good Day
Answer:
An atom is the smallest particle of an element that can take part in chemical reaction.
Explanation:
hope it will help u Amri
Nitric acid can be formed in two steps from the atmospheric gases nitrogen and oxygen, plus hydrogen prepared by reforming natural gas. In the first step, nitrogen and hydrogen react to form ammonia: N2 (g) + 3H2 (g) â 2NH3 (g) =ÎHâ92.kJ In the second step, ammonia and oxygen react to form nitric acid and water:
NH3 (g) + 2O2 (g) â HNO3 (g) + H2O (g) =ÎHâ330.kJ
Required:
Calculate the net change in enthalpy for the formation of one mole of nitric acid from nitrogen, hydrogen and oxygen from these reactions.
Answer:
-376 kJ
Explanation:
The first step equation:
[tex]\mathsf{N_{2(g)} + 3H_2{(g)} \to 2NH_3{(g)} \ \ \ \Delta H = -92\ kJ}[/tex] ---- (1)
The second step equation:
[tex]\mathsf{NH_{3(g)} + 2O_2{(g)} \to HNO_3{(g)} +H_2O_{(g)} \ \ \ \Delta H = -330\ kJ}[/tex] ---- (2)
To determine the enthalpy of formation for 1 mole of HNO₃ (nitric acid), we have the following.
From the above equations; let multiply equation (1) by 1 and equation (2) by 2.
[tex]\mathsf{N_{2(g)} + 3H_2{(g)} \to 2NH_3{(g)} \ \ \ \Delta H = -92\ kJ}[/tex] ---- (3)
[tex]\mathsf{2NH_{3(g)} + 4O_2{(g)} \to 2HNO_3{(g)} +2H_2O_{(g)} \ \ \ \Delta H = 2(-330)\ kJ}[/tex] ----- (4)
adding the above two equations, we have:
[tex]\mathsf{N_{2(g)} + 3H_2{(g)}+ 2NH_{3(g)} + 4O_{2(g)} \to 2HNO_{3(g)} + 2NH_3{(g)} +2H_2O_{(g)} \ \ \ \Delta H = (-660 \ kJ -92\ kJ)}[/tex][tex]\mathsf{N_{2(g)} + 3H_2{(g)} + 4O_{2(g)} \to 2HNO_{3(g)} +2H_2O_{(g)} \ \ \ \Delta H = (-752 \ kJ)}[/tex]
Now, from the recent equation, we have:
2 moles of nitric acid = -752 kJ
∴
1 mole of nitric acid will be: = (1 mole × (-752 kJ)) ÷ 2 moles
1 mole of nitric acid will be: = -376 kJ
công thức phân tử của glucozo
C₆H₁₂O₆ is the molecular formula of gulcozo.
The sample concentration was measured at 50mg/ml. The loading concentration needs to be 10mg/ml. The final volume needs to be 25ul. What is the volume of sample needed and the amount of buffer needed to reach 25ul
Answer:
a) [tex]V_1=5ul[/tex]
b) [tex]v=20ul[/tex]
Explanation:
From the question we are told that:
initial Concentration [tex]C_1=50mg/ml[/tex]
Final Concentration [tex]C_2=10mg/ml[/tex]
Final volume needs [tex]V_2 =25ul[/tex]
Generally the equation for Volume is mathematically given by
[tex]C_1V_1=C_2V_2[/tex]
[tex]V_1=\frac{C_1V_1}{C_2}[/tex]
[tex]V_1=\frac{10*25}{50}[/tex]
[tex]V_1=5ul[/tex]
Therefore
The volume of buffer needed is
[tex]v=V_2-V_1\\\\v=25-5[/tex]
[tex]v=20ul[/tex]
Choose the correct statement
a) The maximum value of principal quantum number (n) is 7
b) The angular quantum number (l) can receive value from 1 to (n-1)
c) The magnetic quantum number (ml) shows the energy of electron
d) The magnetic quantum number (ml) show how many orbitals in each subshell
Answer:
maybe number b is correct. ...thank you
In the reoxidation of QH2 by purified ubiquinone-cytochrome c reductase (Complex III) from heart muscle, the overall stoichiometry of the reaction requires 2 mol of cytochrome c per mole of QH2 because:
Answer: Options related to your question is missing below are the missing options
a. cytochrome c is a one-electron acceptor, whereas QH2 is a two-electron donor.
b. cytochrome c is a two-electron acceptor, whereas QH2 is a one-electron donor.
c. cytochrome c is water soluble and operates between the inner and outer mitochondrial membranes
d. heart muscle has a high rate of oxidative metabolism, and therefore requires twice as much cytochrome c as QH2 for electron transfer to proceed normally.
e. two molecules of cytochrome c must first combine physically before they are catalytically active.
answer:
cytochrome c is a one-electron acceptor, whereas QH2 is a two-electron donor. ( A )
Explanation:
The overall stoichiometry of the reaction requires 2 mol of cytochrome per mole of QH2 because a cytochrome is simply a one-electron acceptor while QH2 is not a one-electron donor ( i.e. it is a two-electron donor )
An electron donor in a reaction is considered a reducing agent because it donates its electrons to another compound thereby self oxidizing itself in the process.
If the balance were not tared prior to weighing out the KHP.... how would you expect this to affect the molarity of NaOH calculated? What type of error is this?
Answer:
Following are the response to the given question:
Explanation:
In the given scenario, When the balance has never been tainted before the KHP is weighted, which can affect the molar concentration of NaOH because its molarity is directly proportional to the weight including its substance. In this question it is the mistake is systemic because it may be corrected by modifying balancing parameters.
A sample of a compound is found to consist of 0.44g H and 6.92g O what’s its formula
Answer:
Explanation:
H = 0.44/1.01 = 0.4356
O = 6.92/16 = 0.4319
This gives a 1:1 ratio. So the closest thing you could say is the formula is 0H
Going to your chemical storage room, you could justify that it is H2O2 or hydrogen peroxide. The question needs one more fact to make the answer certainty.
calculate the hydrogen ion concentration of a solution who's pH is 2.4
Answer:
I don't know sorry yyyyyyy6yyyyyyyyyyyyyyyyyyyyyyyyyyy
Please help answering 11)
Answer:
the answer is C
Explanation:
does anyone know how to solve this and what the answer would be?
Dynamic equilibrium is showed at the point at which solid liquid and gas intersect.
At the point at which solid liquid and gas intersect represents a system that shows dynamic equilibrium. There is equal amount of reactants and products at the point of dynamic equilibrium because the transition of substances occur between the reactants and products at equal rates, means that there is no net change. Reactants and products are formed at the rate that no change occur in their concentration.
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Given the equation representing a nuclear reaction in
which X represents a nuclide:
232Th → He + x
Which nuclide is represented by X?
A) 236
B) 228
Ra
SS
C) 236
Ra
92
U
92
D) 228
.
Ss U
The nuclide represented as X is thorium and this is an alpha decay.
The equation shown represents an alpha decay. In an alpha decay, an alpha particle is given off.
The atomic number of the parent nuclide is greater than that of the daughter nuclide by two units while the mass number of the parent is greater than that of the daughter nuclide by four units.
Hence the equation occurs as follows;
[tex]\frac{232}{92} Th ------> \frac{228}{88} Ra + He[/tex]
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To prepare a standard (calibration) curve for a spectroscopy experiment, start by preparing ___________ with ______________ Then, measure the ______________ of each solution at _____________ and create a plot of ____________ for the measured values. Finally, find the best-fit line of the data set.
Answer: See explanation
Explanation:
The calibration curve is the method used for the determination of the concentration of a substance such that the unknown sample will be compared to some standard samples of the known concentration.
To prepare a standard (calibration) curve for a spectroscopy experiment, start by preparing (multiple solutions) with (different known concentrations). Then, measure the (absorbance) of each solution at (thesame wavelength) and create a plot of (absorbance vs. concentration) for the measured values. Finally, find the best-fit line of the data set.
If a buffer is composed of 23.34 mL of 0.147 M acetic acid and 33.66 mL of 0.185 M sodium acetate, how many mL of 0.100 M NaOH can be added before the buffer capacity is reached
Answer:
25.5mL of 0.100M NaOH are needed to reach buffer capacity.
Explanation:
The buffer capacity is reached when the ratio between moles of conjugate base (Sodium acetate) and moles of weak acid (Acetic acid) is 10:
Moles sodium acetate / Moles Acetic acid = 10
The reaction of acetic acid, HA, with NaOH, to produce sodium acetate, NaA is:
HA + NaOH → H2O + NaA
That means the moles of NaOH added = Moles of HA that are being subtracted and moles of NaA that are been produced.
The initial moles of each species is:
Acetic acid:
23.34mL = 0.02334L * (0.147mol / L) = 0.00343 moles Acetic Acid
Sodium Acetate:
33.66mL = 0.03366L * (0.185mol / L) = 0.00623 moles Sodium Acetate
We can write the moles of each species when NaOH is added as:
Moles sodium acetate / Moles Acetic acid = 10
0.00623 moles + X / 0.00343 moles - X = 10
Where X are moles of NaOH added
Solving for X:
0.00623 moles + X = 0.0343 moles - 10X
11X = 0.0281
X = 0.00255 moles of NaOH are needed
In Liters:
0.0255mol NaOH * (1L / 0.100mol) = 0.0255L of 0.100M NaOH are needed =
25.5mL of 0.100M NaOH are needed to reach buffer capacity
