Answer:
pH = 5.22
Explanation:
As you can see, your initial concentration of sodium acetate (NaCH₃COO) is 0.02M (0.02mol /L). When you add HCl, the reaction is:
NaCH₃COO + HCl → CH₃COOH + NaCl.
If you add HCl, and final concentration of NaCH₃COO is 0.015M, the concentration of CH₃COOH is 0.005M.
You can know the pH of this solution using H-H equation:
pH = pKa + log {NaCH₃COO} / {CH₃COOH}
pH = 4.74 + log {0.015M} / {0.005M}
pH = 5.22Air contains nitrogen, oxygen, argon, and trace gases. Ifthe partial pressure of nitrogen is 592 mm Hg, oxygen is160 mm Hg, argon is 7 mm Hg, and trace gas is 1 mm Hg,what is the atmospheric pressure
Answer:
760 mmHg
Explanation:
Step 1: Given data
Partial pressure of nitrogen (pN₂): 592 mmHgPartial pressure of oxygen (pO₂): 160 mmHgPartial pressure of argon (pAr): 7 mmHgPartial pressure of the trace gas (pt): 1 mmHgStep 2: Calculate the atmospheric pressure
Since air is a gaseous mixture, the atmospheric pressure is equal to the sum of the gases that compose it.
P = pN₂ + pO₂ + pAr + pt = 592 mmHg + 160 mmHg + 7 mmHg + 1 mmHg = 760 mmHg
Which resulted from the study of chemistry?
A) Alchemy to turn base metals into noble metals
B) The understanding of earth, air, fire, and water as the basic components of matter.
C) A supernatural, mystical view of the world.
D) Discovering the role of oxygen in combustion
Arrange the following kinds of electromagnetic radiation in order of increasing wavelength: infrared, green light, red light, radio waves, X rays, ultraviolet light.
Rank from shortest to longest. To rank items as equivalent, overlap them.
infrared
green light
red light
radio waves
X rays
ultraviolet
Answer:
In the other of increasing wavelength we arrange as
X-rays
ultraviolet
green light
red light
infrared
radio waves
Explanation:
In the electromagnetic spectrum, wavelength decreases with increase in the energy of the electromagnetic wave. Since the e-m wave spectrum is arranged in the order of increasing energy (decreasing wavelength) as: Radio wave; infrared; visible light; ultraviolet; x-rays; gamma rays. Within the visible light, the green light has more energy than the red light. Therefore, the arrangement should be in the reverse direction of their increasing energy.
Which one of the following reactions must be carried out in an electrolytic cell rather than in an electrochemical cell?
a. Zn2+ + Ca → Zn + Ca2+
b. Al3+ + 3Br– → Al + (3/2)Br2
c. 2Al + 3Fe2+ → 2Al3+ + 3Fe
d. H2 + I2(s) → 2H+ + 2I–
Answer:
b. Al3+ + 3Br– → Al + (3/2)Br2
Explanation:
If we look at this reaction closely, aluminum was reduced while bromine was oxidized. The reduction potential of aluminum is -1.66 V while that of bromine is + 1.087. Recall that the more negative the redox potential of a chemical specie, the greater its tendency to function as a reducing agent donating electrons in an electrochemical reaction.
However, in this reaction, aluminium is found to accept rather than donate electrons. Therefore, the process is non spontaneous and can only occur in an electrolytic cell, hence the answer.
b. [tex]Al^{3+} + 3 Br^{-}[/tex] → [tex]Al + \frac{3}{2} Br_{2}[/tex]
Electrolytic Cell v/s Electrochemical Cell:
Electrochemical cells convert chemical energy into electrical energy or vice versa while an electrolytic cell is a type of electrochemical cell in which electrical energy is converted into chemical energy.Electrochemical cells consist of a cathode (+) and an anode(-) while Electrolytic cells consist of a positively charged anode and a negatively charged cathode.Out of the given reactions, [tex]Al^{3+} + 3 Br^{-}[/tex] → [tex]Al + \frac{3}{2} Br_{2}[/tex] is carried out in an electrolytic cell rather than in an electrochemical cell.
As we know, In electrolytic cells, like galvanic cells, are composed of two half-cells
one is a reduction half-cell, the other is an oxidation half-cell.In the given reaction, aluminum is being reduced while bromine gets oxidized and the value of reduction potential of aluminum is -1.66 V while that of bromine is + 1.087. Therefore, the process is non spontaneous and can only occur in an electrolytic cell.
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If the heat of combustion for a specific compound is −1380.0 kJ/mol and its molar mass is 44.53 g/mol, how many grams of this compound must you burn to release 822.00 kJ of heat?mass:g
Answer:
Mass = 26.53 g
Explanation:
Heat of combustion = −1380.0 kJ/mol
This means 1 mol of the compound releases 1380 kJ
Molar mass = 44.53 g/mol
This means 1 mol of the compound has a mass of 44.53 g
How many grams would release 822kJ..?
First, we have to obtain the number of moles
1 mol = 1380
x = 822
x = 0.5957 moles
Moles = Mass / Molar mass
Mass = Molar mass * moles
Mass = 44.53 * 0.5957
Mass = 26.53 g
Mrs. Wilson leaves her freshly-baked blueberry pie on the windowsill to cool. The delicious fragrance diffuses through the air with a diffusion coefficient of D = 0.2 cm2/s. How long does it take for Dennis to smell the pie in his treehouse 10 meters away? Give your answer in days, without entering the unit.
Answer:
Poop Buttt.
Explanation:
How many grams are in 5.87 x 10^21 molecules of sulfur?
Answer:
0.312g
Explanation:
From Avogadro's hypothesis, 1mole of any substance contains 6.02x10^23 molecules. This means that 1mole of sulphur also contains 6.02x10^23 molecules
1mole of sulphur = 32g
If 1 mole(i.e 32g) of sulphur contains 6.02x10^23 molecules
Then, Xg of sulphur will contain 5.87x10^21 molecules i.e
Xg of sulphur = (32x5.87x10^21)/6.02x10^23 = 0.312g
just saying this is not my work, thank Eduard22sly
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A chemistry student weighs out of formic acid into a volumetric flask and dilutes to the mark with distilled water. He plans to titrate the acid with solution. Calculate the volume of solution the student will need to add to reach the equivalence point. Round your answer to significant digits.
The given question is incomplete, the complete question is:
A chemistry student weighs out 0.0349g of formic acid HCHO2 into a 250.mL volumetric flask and dilutes to the mark with distilled water. He plans to titrate the acid with 0.1500M NaOH solution. Calculate the volume of NaOH solution the student will need to add to reach the equivalence point. Round your answer to 3 significant digits.
Answer:
The correct answer is 5.06 ml.
Explanation:
Based on the given information, the weight of formic acid given is 0.0349 grams. The volume of formic acid of V1 given is 250 ml. The molecular mass of formic acid is 46 grams per mole. Now the molarity of formic acid will be,
[HCOOH] = weight * 1000 / molecular mass * volume (ml)
= 0.0349 * 1000 / 46 * 250
= 0.003035 M or M1
The molarity of NaOH given is 0.1500 M or M2
Let us assume that the volume needed to attain equivalence point is V2 ml. The volume V2 can be determined by using the dilution equation,
M1V1 = M2V2
V2 = M1V1/M2
V2 = 0.003035 * 250 / 0.1500
V2 = 5.06 ml.
Hence, the volume of NaOH needed is 5.06 ml.
Predict the order of acid strengths in the following series of cationic
species: CH3CH2NH3
+, CH3CH=NH2
Answer:
CH3CH=NH2+>CH3CH2NH3+
Explanation:
If we look at the both species under review, we will realize that they are both amines hence they possess the polar N-H bond.
Electrons are ordinarily attracted towards the nitrogen atom hence making both compounds acidic. It is worthy of note that certain features of a compound may make it more acidic than another of close structural proximity. 'More acidic' simply means that the proton is more easily lost.
CH3CH=NH2+ contains an sp2 hybridized carbon atom which is highly electronegative and further withdraws electron density from the N-H bond thereby leading to a greater acidity of CH3CH=NH2+ compared to CH3CH2NH3+
1. The following thermochemical equation is for the reaction of water(l) to form hydrogen(g) and oxygen(g). 2H2O(l)2H2(g) + O2(g) H = 572 kJ How many grams of H2O(l) would be made to react if 110 kJ of energy were provided? _____ grams
2. The following thermochemical equation is for the reaction of carbon monoxide(g) with hydrogen(g) to form methane(g) and water(g). CO(g) + 3H2(g) CH4(g) + H2O(g) H = -206 kJ When 6.27 grams of carbon monoxide(g) react with excess hydrogen(g),_____ kJ of energy are ____ a.evolved b.absorbed
Answer:
1. 6.92 g of H2O
2i. - 46 KJ of energy.
ii. Option A. Evolved.
Explanation:
1. Determination of the mass of H2O that would be made to react if 110 kJ of energy were provided.
This can be obtained as follow:
The equation for the reaction is given below
2H2O(l) —> 2H2(g) + O2(g) H = 572 kJ
Next, we shall determine the mass of H2O required to produce 572 kJ from the balanced equation.
Molar mass of H2O = (2x1) + 16 = 18 g/mol
Mass of H2O from the balanced equation = 2 x 18 = 36 g
From the balanced equation above, 36 g of H2O reacted to produce 572 kJ of energy.
Finally, we shall determine the mass of water (H2O) needed to produce 110 kJ of energy.
This is illustrated below:
From the balanced equation above, 36 g of H2O reacted to produce 572 kJ of energy.
Therefore, Xg of H2O will react to 110 kJ of energy i.e
Xg of H2O = (36 x 110)/572
Xg of H2O = 6.92 g
Therefore, 6.92 g of H2O is needed to react in order to produce 110 KJ of energy.
2i. Determination of the energy.
The balanced equation for the reaction is given below:
CO(g) + 3H2(g) —> CH4(g) + H2O(g) H = -206 kJ
Next, we shall determine the mass of CO that reacted to produce -206 kJ of energy from the balanced equation.
This is illustrated below:
Molar mass of CO = 12 + 16 = 28 g/mol
Mass of CO from the balanced equation = 1 x 28 = 28 g
From the balanced equation above,
28 g of CO reacted to produce -206 kJ of energy.
Finally, we shall determine the amount of energy produced by reacting 6.27 g of CO. This is illustrated below:
From the balanced equation above,
28 g of CO reacted to produce -206 kJ of energy.
Therefore, 6.27 g of CO will react to produce = (6.27 x -206)/28 = - 46 KJ of energy.
Therefore, - 46 KJ of energy were produced from the reaction.
2ii. Since the energy obtained is negative, it means heat has been given off to the surroundings.
Therefore, the heat is evolved.
The decomposition of hydrogen peroxide to form water and oxygen gas releases 196.6 kJ per mole of hydrogen peroxide. This reaction occurs when hydrogen peroxide is placed on a cut to sterilize it. How much heat is released when 5.70 mol H2O2 decompose
Answer:
1120.62 kJ
Explanation:
In order to find how much heat is released for 7.70 mol, we have to compare it with the heat released from one mole.
So from the question, we have;
196.6 kJ = 1 mol
x = 5.70
x = 5.70 * 196.6 / 1
x = 1120.62 kJ
Help please.☺️☺️☺️☺️☺️
Answer:
A.∆s>0contribute to spontaneity.
An aqueous solution containing 5.06 g of lead(II) nitrate is added to an aqueous solution containing 6.03 g of potassium chloride.Enter the balanced chemical equation for this reaction. Be sure to include all physical states.balanced chemical equation:What is the limiting reactant?lead(II) nitratepotassium chlorideThe percent yield for the reaction is 82.9% . How many grams of the precipitate are formed?precipitate formed:gHow many grams of the excess reactant remain?excess reactant remaining:
Answer:
Pb(NO3)2(aq) + 2KCl(aq) ------> 2KNO3(aq) + PbCl2(s)
3.52 g of PbCl2
3.76 g of KCl
Explanation:
The equation of the reaction is;
Pb(NO3)2(aq) + 2KCl(aq) ------> 2KNO3(aq) + PbCl2(s)
Number of moles of Pb(NO3)2 =mass/molar mass 5.06g/331.2 g/mol = 0.0153 moles
Number of moles of KCl= mass/ molar mass= 6.03g/74.5513 g/mol= 0.081 moles
Next we obtain the limiting reactant; the limiting reactant yields the least number of moles of products.
For Pb(NO3)2;
1 mole of Pb(NO3)2 yields 1 mole of PbCl2
Therefore 0.0153 moles of Pb(NO3)2 yields 0.0153 moles of PbCl2
For KCl;
2 moles of KCl yields 1 mole of PbCl2
0.081 moles of KCl yields 0.081 moles ×1/2 = 0.041 moles of PbCl2
Therefore Pb(NO3)2 is the limiting reactant.
Theoretical Mass of precipitate obtained = 0.0153 moles of PbCl2 × 278.1 g/mol = 4.25 g of PbCl2
% yield = actual yield/theoretical yield ×100
Actual yield = % yield × theoretical yield /100
Actual yield= 82.9 ×4.25/100
Actual yield = 3.52 g of PbCl2
If 1 mole of Pb(NO3) reacts with 2 moles of KCl
0.0153 moles of Pb(NO3)2 reacts with 0.0153 moles × 2 = 0.0306 moles of KCl
Amount of excess KCl= 0.081 moles - 0.0306 moles = 0.0504 moles of KCl
Mass of excess KCl = 0.0504 moles of KCl × 74.5513 g/mol = 3.76 g of KCl
The final overall chemical equation is Upper Ca upper O (s) plus upper C upper O subscript 2 (g) right arrow upper C a upper C upper O subscript 3 (s).. When the enthalpy of this overall chemical equation is calculated, the enthalpy of the second intermediate equation is halved and has its sign changed. is halved. has its sign changed. is unchanged.
Answer:
the enthalpy of the second intermediate equation is halved and has its sign changed.
Explanation:
Let us take a look at the first and second intermediate reactions as well as the overall reaction equation for the process under review;
First reaction;
Ca (s) + CO₂ (g) + ½O₂ (g) → CaCO₃ (s) ΔH₁ = -812.8 kJ
Second reaction;
2Ca (s) + O₂ (g) → 2CaO (s) ΔH₂ = -1269 kJ
Hence the overall equation is now;
CaO (s) + CO₂ (g) → CaCO₃ (s) ΔH = ?
According to the Hess law of constant heat summation, the enthalpy of the overall reaction is supposed to be obtained as a sum of the enthalpy of both reactions but this will not give the enthalpy of the overall reaction in this case. The enthalpy of the overall reaction is rather obtained by halving the enthalpy of the second intermediate reaction and reversing its sign before taking the sum as shown below;
Enthalpy of Intermediate reaction 1 + ½(- Enthalpy of Intermediate reaction 2) = Enthalpy of Overall reaction
Answer:
A.
Explanation:
Did it on Edge.
Carbon-14 has a half-life of 5720 years and this is a fast-order reaction. If a piece of wood has converted 75 % of the carbon-14, then how old is it?
Answer:
11445.8years
Explanation:
Half-life of carbon-14 = 5720 years
First we have to calculate the rate constant, we use the formula :
What is the name of the molecule below?
A) 2-pentene
B) pentane
C) 2-pentyne
D) 2-pentane
The name of the molecule which is given below is 2-pentene.
What are alkene?Alkenes are the organic compounds which are composed of carbon and hydrogen atoms, in which double bond is present.
In the given diagram:
Each corner and joints shows the carbon atoms and number of carbon atoms in it is 5.One double bond is present in the 2nd position.So the compound is 2 pentene.
Hence, 2 pentene is the name of the compound.
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A compound X has a molecular ion peak in its mass spectrum at m/z 136. What information does this tell us about X
Explanation:
The mass to charge ratio =136
Calculate the pH of a buffer solution obtained by dissolving 18.0 g of KH2PO4(s) and 35.0 g of Na2HPO4(s) in water and then diluting to 1.00 L.
Answer:
pH of the buffer is 7.48
Explanation:
The H₂PO₄⁻/HPO₄²⁻ buffer has a pKa of 7.21. You can find pH of this buffer following H-H equation:
pH = pKa + log [A⁻] / [HA]
pH = 7.21 + log [HPO₄²⁻] / [ H₂PO₄⁻]
Where [] represents molarity of each specie of the buffer and, as volume is 1.00L, also represents its moles.
Thus, to find pH of the buffer we need to calculate moles of each specie, thus
Moles of 18.0g of KH₂PO₄(Molar mass: 136.086g/mol) = moles of H₂PO₄⁻ are:
18.0g KH₂PO₄ ₓ (1mol / 136.086g) = 0.132 moles of KH₂PO₄= H₂PO₄⁻
Moles of 35.0g of Na₂HPO₄(Molar mass: 141.96g/mol) = moles of HPO₄²⁻ are:
35.0g Na₂HPO₄ ₓ (1mol / 141.96g) = 0.2465 moles of Na₂HPO₄= HPO₄²⁻
Replacing in H-H equation:
pH = 7.21 + log [HPO₄²⁻] / [ H₂PO₄⁻]
pH = 7.21 + log [0.2465] / [0.132]
pH = 7.48
pH of the buffer is 7.48
Amanda is doing a report for her Earth Science class about the four seasons. Which of the following would be an effective scientific model to incorporate in her project? a. A calendar indicating the first days of autumn, winter, spring, and summer b. Four earth models, each with a different tilt to represent the earth’s position relative to the sun, lined up in order of season c. A poster board with pictures of weather characteristic of the four seasons d. A bar graph indicating average high and low temperatures for Amanda’s area in the autumn, winter, spring, and summer Please select the best answer from the choices provided
Answer:
(B)
Explanation:
edg 2020
The four seasons in earth is originating from the earth's revolution around the sun. Therefore, the most suitable model for Amanda is Four earth models, each with a different tilt to represent the earth’s position relative to the sun, lined up in order of season.
What are seasons?Seasons in earth is originating from the difference in the distance from the sun over each time period. Hence, revolution of earth around sun make these seasons.
The time period at which earth comes closer to sun more hot will be earth's surface and we experience summer season. When we far from sun winter season occurs.
Therefore, different season are coming based on the distance of earth from sun at each revolution point. This is also affected by the tilt of earth's in its own axis.
Hence, different earth models with different distance from sun is most suitable model here for Amanda. Thus option B is correct.
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Arrange the following oxides in order of increasing acidity.
Rank from least acidic to most acidic. To rank items as equivalent,overlap them.
CaO
P2O5
SO3
SiO2
Al2O3
CO2
Answer:
Based on the Modern Periodic table, there is an increase in the electropositivity of the atom down the group as well as increases across a period. On comparing the electropositivities of the mentioned oxides central atom, it is seen that Ca is most electropositive followed by Al, Si, C, P, and S is the least electropositive.
With the decrease in the electropositivity, there is an increase in the acidity of the oxides. Thus, the increasing order of the oxides from the least acidic to the most acidic is:
CaO > Al2O3 > SiO2 > CO2 > P2O5 > SO3. Hence, CaO is the least acidic and SO3 is the most acidic.
Since acidity increases across a period from left to right, and decreases down a group, the oxides can be ranked from the least acidic to the most acidic as follows:
[tex]\mathbf{CaO < Al_2O_3 < SiO_2 < CO_2 < P_2O_5 <SO_3}[/tex]
The least acidic is CaOThe most acidic is [tex]SO_3[/tex]Note the following:
Acidity of an oxide depends on its electronegativity.Non-metals are more electronegative, while metals are less electronegative.Acidity of oxides increases across a period as you move from left to the right side of a periodic table.Acidity of oxides decreases down a group (column) in a periodic table.Using the periodic table diagram given in the attachment below, we can rank the given oxides according to their increasing acidity.
CaO, is the least, because it is an oxide of the metal, Calcium, which is at the far left in group 2 in the periodic table.The next is, [tex]Al_2O_3[/tex]. Aluminum is a metal from group 3.[tex]SiO_2[/tex] is an oxide of Silicon, also in group 4 but below Carbon.[tex]CO_2[/tex] is an oxide of Carbon, from group 4.
[tex]P_2O_5[/tex] is an oxide of the non-metal, Phosphorus, a group 5 element
[tex]SO_3[/tex] is an oxide of the non-metal, Sulphur, a group 6 element.
Therefore, since acidity increases across a period from left to right, and decreases down a group, the oxides can be ranked from the least acidic to the most acidic as follows:
[tex]\mathbf{CaO < Al_2O_3 < SiO_2 < CO_2 < P_2O_5 <SO_3}[/tex]
The least acidic is CaOThe most acidic is [tex]SO_3[/tex]Learn more here:
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what is 1 +1 (a) 11 (b) 3 (c) 6 (d) 2
Answer:
(d) 2
Explanation:
Lets say that there are 2 apples or 1+1 if you count them you would do 1,2 so 2 would be the final answer
Why do you think sodium bicarbonate is included to neutralize an acidic spill rather than sodium hydroxide?
Imagine a hypothetical situation in which 250 mL of diethyl ether (SDS) has spilled inside of a chemical fume hood onto a stir plate that is plugged in and stirring. Discuss the risks associated with this situation (location, size, compound spilled, and external hazards), and then explain how this spill should be managed.
Answer:
Acid spills should be neutralized with sodium bicarbonate and then cleaned up with a paper towel or sponge.
Explanation:
Determine the cell notation for the redox reaction given below.
Sn(s) + 2H+(aq) ⟶ Sn2+(aq) + H2(g)
a. H+(aq) | H2(g) | Pt ∥ Sn(s) | Sn2+(aq)
b. H2(g) | H+(aq) | Pt ∥ Sn2+(aq) | Sn(s)
c. Sn2+(aq) | Sn(s) ∥ H2(g) | H+(aq) | Pt
d. Sn(s) | Sn2+(aq) ∥ H+(aq) | H2(g) | Pt
e. Sn(s) | H2(g) ∥ Sn2+(aq) | H+(aq) | Pt
Answer:
The correct answer is d. Sn(s) | Sn²⁺(aq) ∥ H⁺(aq) | H₂(g) | Pt
Explanation:
The half reactions are:
2H⁺(aq) + 2 e- ⟶ H₂(g) (reduction)
Sn(s) ⟶ Sn²⁺(aq) + 2 e- (oxidation)
In the cell notation, there are two electrodes in which are separated the reduction reaction from the oxidation reaction. In the left electrode occurs the oxidation reaction (anode) while in the right electrode occurs the reduction reaction (cathode). The general form of the cell notation is the following:
anode reaction∥ cathode reaction
where the two bars ( ∥ ) represent the physical barrier between the electrodes. A single bar ( | ) is used to represent a phase separation.
In this redox reaction, the half reaction of the anode is Sn(s) ⟶ Sn²⁺(aq) + 2 e-; whereas the half reaction of the cathode is 2H⁺(aq) + 2 e- ⟶ H₂(g).
The componens are written in order according to the half reaction. Since Sn²⁺ and H⁺ ions are in solution, a platinum electrode is used and represented as Pt. Thus, the cell notation is:
Sn(s) | Sn²⁺(aq) ∥ H⁺(aq) | H₂(g) | Pt
some students believe that teachers are full of hot air. If I inhale 3.5 liters of gas at a temperature of 19 degrees Celsius and it heats to a temperature of 58 degrees celsius in my lungs. what is the new volume of the gas?
Answer:
3.97 L
Explanation:
Data obtained from the question include the following:
Initial volume (V1) = 3.5 L
Initial temperature (T1) = 19 °C
Final temperature (T2) = 58 °C
Final volume (V2) =..?
Next, we shall convert celsius temperature to Kelvin temperature. This can be done as shown below:
Temperature (K) = temperature (°C) + 273
T (K) = T (°C) + 273
Initial temperature (T1) = 19 °C
Initial temperature (T1) = 19 °C + 273 = 292 K
Final temperature (T2) = 58 °C
Final temperature (T2) = 58 °C + 273 = 331 K
Finally, we shall determine the new volume of the gas by using Charles' law equation as shown below:
Initial volume (V1) = 3.5 L
Initial temperature (T1) = 292 K
Final temperature (T2) = 331 K
Final volume (V2) =..?
V1 /T1 = V2 /T2
3.5 /292 = V2 /331
Cross multiply
292 x V2 = 3.5 x 331
Divide both side by 292
V2 = (3.5 x 331) / 292
V2 = 3.97 L
Therefore, the new volume of the gas is 3.97 L.
Drag each image to the correct location on the model. Each image can be used more than once. Apply the rules and principles of electron configuration to draw the orbital diagram of aluminum. Use the periodic table to help you.
Answer:
The answer to your question is given below.
Explanation:
Aluminium has atomic number of 13. Thus, the electronic configuration of aluminium can be written as:
Al (13) —› 1s² 2s²2p⁶ 3s²3p¹
The orbital diagram is shown on the attached photo.
Answer: screen shot
Explanation:
Find the molecular formula of each compound CCl , 189.83 g/mol C3H2N , 156.23 g/mol
Answer:
Explanation:
The given formula is empirical formula
Let the molecular formula of first be
[tex]( CCl )_n[/tex]
molecular weight = n x ( 12 + 35.5 )
= 47.5 n
Given molecular weight = 189.83 so
47.5 n = 189.83
n = 3.99 or 4 approx
Molecular formula =
[tex]( CCl )_4[/tex]
= C₄ Cl₄
Let the molecular formula of second compound be
[tex]( C_3H_2N)_n[/tex]
molecular weight = n x ( 3 x 12 +2+14 )
= 52 n
Given molecular weight = 156.23 so
52 n = 156.23
n = 3.0044 or 3 approx
Molecular formula =
[tex]( C_3H_2N )_3[/tex]
= C₉H₆ N₃
Automotive antifreeze is typically a 50:50 mixture (by volume) of water and ethylene glycol. Discuss why this solution is useful for protecting automobile engines from both summer and winter temperature extremes.
Answer:
A 50:50 mixture of ethylene glycol and water is effective both summer and winter extremes in temperature because of high boiling point of 106°C and low freezing point of about -37°C. In summer when the average daily temperature rises to about 22°c, the mixture will be effective in keeping the automobile engine cool. Also in winter, when the average temperature falls below 0°C, the mixture will be effective as an antifreeze as it remains a liquid well below 0°C.
Explanation:
Ethylene glycol or antifreeze is an organic compound which is used in automobile engines as a coolant and also as an anti-freezing agent, however it does not conduct heat effectively as water due to its lower heat capacity. It has a freezing point of -12.9°C and boiling point of 197.3°C.
Water is also used as a coolant in automobile engine but it has a limited range due to its boiling point of 100°C. It is also not a good anti-freezing agent due to it high freezing point of 0.°C
However, when ethylene glycol is mixed with water in a ratio of 50:50, the property of the mixture is enhanced to both serve as a coolant and as an antifreeze. The boiling point is elevated to about 106°C while its freezing point is lowered to about -37°C.
This temperature range is effective for both summer and winter temperatures. In summer when the average daily temperature rises to about 22°c, the mixture will be effective in keeping the the automobile engine cool. Also in winter, when the average temperature falls below 0°C, the mixture will be effective as an antifreeze as it remains a liquid well below 0°C.
how salt solution can be determined by using hydrometer
Answer:
Salt solution may be calculated by measuring the specific gravity of a sample of water using a hydrometer.
Hope this answer correct (^^)....
Suppose you titrate 25.00 mL of 0.200 M KOBr with 0.200M H2SO4. The pH at half-equivalence point is 7.75 a). What is the initial pH of the 25.00mL of 0.200M KOBr mentioned above
Answer:
Approximately [tex]10.88[/tex].
Explanation:
Equilibrium constant[tex]\rm OBr^{-}[/tex] can act as a weak Bronsted-Lowry base:
[tex]\rm OBr^{-}\; (aq) + H_2O\; (l) \rightleftharpoons HOBr\; (l) + OH^{-}\; (aq)[/tex].
(Side note: the state symbol of [tex]\rm HOBr[/tex] in this equation is [tex]\rm (l)[/tex] (meaning liquid) because [tex]\rm HOBr[/tex] is a weak acid.)
However, the equilibrium constant of this reaction, [tex]K_\text{eq}[/tex], isn't directly given. The idea is to find [tex]K_\text{eq}[/tex] using the [tex]\rm pH[/tex] value at the half-equivalence point. Keep in mind that this system is at equilibrium all the time during the titration. If temperature stays the same, then the same [tex]K_\text{eq}[/tex] value could also be used to find the [tex]\rm pH[/tex] of the solution before the acid was added.
At equilibrium:
[tex]\displaystyle K_\text{eq} = \frac{[\rm HOBr\; (l)]\cdot [\rm OH^{-}\; (aq)]}{[\rm OBr^{-}\; (aq)]}[/tex].
At the half-equivalence point of this titration, exactly half of the base, [tex]\rm OBr^{-}[/tex], has been converted to its conjugate acid, [tex]\rm HOBr[/tex]. Therefore, the half-equivalence concentration of [tex]\rm OBr^{-}[/tex] and [tex]\rm HOBr[/tex] should both be equal to one-half the initial concentration of [tex]\rm OBr^{-}[/tex].
As a result, the half-equivalence concentration of [tex]\rm OBr^{-}[/tex] and [tex]\rm HOBr[/tex] should be the same. The expression for [tex]K_\text{eq}[/tex] can thus be simplified:
[tex]\begin{aligned}& K_\text{eq} \\&= \frac{\left(\text{half-equivalence $[\rm HOBr\; (l)]$}\right)\cdot \left(\text{half-equivalence $[\rm OH^{-}\; (aq)]$}\right)}{\text{half-equivalence $[\rm OBr^{-}\; (l)]$}}\\ &=\text{half-equivalence $[\rm OH^{-}\; (aq)]$}\end{aligned}[/tex].
In other words, the [tex]K_\text{eq}[/tex] of this system is equal to the [tex]\rm OH^{-}[/tex] concentration at the half-equivalence point. Assume that [tex]\rm p\mathnormal{K}_\text{w}[/tex] the self-ionization constant of water, is [tex]14[/tex]. The concentration of [tex]\rm OH^{-}[/tex] can be found from the [tex]\rm pH[/tex] value:
[tex]\begin{aligned}& \text{half-equivalence $[\rm OH^{-}\; (aq)]$} \\ &= 10^{\rm pH - p\mathnormal{K}_\text{w}}\;\rm mol \cdot L^{-1} \\ &= 10^{7.75 - 14}\; \rm mol \cdot L^{-1}\\ &= 10^{-6.25}\; \rm mol \cdot L^{-1}\end{aligned}[/tex].
Therefore, [tex]\begin{aligned} K_\text{eq} &= 10^{-6.25}\end{aligned}[/tex].
Initial pH of the solutionAgain, since [tex]\rm KOBr[/tex] is a soluble salt, all that [tex]0.200\; \rm M[/tex] of [tex]\rm KOBr[/tex] in this solution will be in the form of [tex]\rm K^{+}[/tex] and [tex]\rm OBr^{-}[/tex] ions. Before any hydrolysis takes place, the concentration of [tex]\rm OBr^{-}[/tex] should be equal to that of [tex]\rm KOBr[/tex]. Therefore:
[tex]\text{$[\rm OBr^{-}\; (aq)]$ before hydrolysis} = 0.200\; \rm M[/tex].
Let the equilibrium concentration of [tex][\rm OH^{-}\; (aq)][/tex] be [tex]x\; \rm M[/tex]. Create a RICE table for this reversible reaction:
[tex]\begin{array}{c|ccccccc} & \rm OBr^{-}\; (aq) &+&\rm H_2O\; (l)& \rightleftharpoons & \rm HOBr\; (l)& + & \rm OH^{-}\; (aq) \\ \textbf{I}& 0.200\; \rm M & & & & 0 \; \rm M & & 0\; \rm M \\ \textbf{C} & -x\; \rm M & & & & +x \; \rm M & & +x\; \rm M \\ \textbf{E}& (0.200 + x)\; \rm M & & & & x \; \rm M & & x\; \rm M \end{array}[/tex].
Assume that external factors (such as temperature) stays the same. The [tex]K_\text{eq}[/tex] found at the half-equivalence point should apply here, as well.
[tex]\displaystyle K_\text{eq} = \frac{[\rm HOBr\; (l)]\cdot [\rm OH^{-}\; (aq)]}{[\rm OBr^{-}\; (aq)]}[/tex].
At equilibrium:
[tex]\displaystyle \frac{[\rm HOBr\; (l)]\cdot [\rm OH^{-}\; (aq)]}{[\rm OBr^{-}\; (aq)]} = \frac{x^2}{0.200 + x}[/tex].
Assume that [tex]x[/tex] is much smaller than [tex]0.200[/tex], such that the denominator is approximately the same as [tex]0.200[/tex]:
[tex]\displaystyle \frac{[\rm HOBr\; (l)]\cdot [\rm OH^{-}\; (aq)]}{[\rm OBr^{-}\; (aq)]} = \frac{x^2}{0.200 + x} \approx \frac{x^2}{0.200}[/tex].
That should be equal to the equilibrium constant, [tex]K_\text{eq}[/tex]. In other words:
[tex]\displaystyle \frac{x^2}{0.200} \approx K_\text{eq} = 10^{-6.25}[/tex].
Solve for [tex]x[/tex]:
[tex]x \approx 3.35\times 10^{-4}[/tex].
In other words, the [tex]\rm OH^{-}[/tex] before acid was added was approximately [tex]3.35\times 10^{-4}\; \rm M[/tex], which is the same as [tex]3.35\times 10^{-4}\; \rm mol \cdot L^{-1}[/tex]. Again, assume that [tex]\rm p\mathnormal{K}_\text{w} = 14[/tex]. Calculate the [tex]\rm pH[/tex] of that solution:
[tex]\begin{aligned}\rm pH &= \rm p\mathnormal{K}_\text{w} + \log [\mathrm{OH^{-}}] \approx 10.88\end{aligned}[/tex].
(Rounded to two decimal places.)
Cyclohexane (C6H12) undergoes a molecular rearrangement in the presence of AlCl3 to form methylcyclopentane (CH3C5H9) according to the equation: C6H12 ⇌ CH3C5H9 If Kc = 0.143 at 25°C for this reaction predict the direction in which the system
Answer:
The reaction will shift leftwards, towards the formation of more cyclohexane at 25 °C
Explanation:
Hello,
In this case, for the given chemical reaction, we can write the law of mass action (equilibrium expression) as shown below:
[tex]Kc=\frac{[CH_3C_5H_9]}{[C_6H_{12} ]}[/tex]
Thus, since Kc < 1, we can conclude there are more moles of cyclohexane at equilibrium (denominator is greater than numerator), therefore, the reaction will shift leftwards, towards the formation of more cyclohexane at 25 °C.
Best regards.