According to the Vaporization Heat table, the heat needed for 1 mol of H2O to evaporate at 100°C is 40.7KJ and 44.0KJ/mol is needed to evaporate H2O at 25°C. Thus 44.0-40.7=3.7KJ is the energy needed to heat H2O to 100°C from 25°C.
However, according to the heat capacity of H2O, 3.7KJ will only warm the water by ~+43°C, which is not enough to reach 100°C starting from 25°C!
Am I missing something?!
Answers
Suppose you have a material in it's liquid phase. As you give energy to that liquid, the temperature of the liquid will increase gradually, and the relation between the increase of temperature and the given energy is the specific heat.
Now, there is a point, a critical point, where the temperature stops to increase, which means that we are near a change of phase. So from this point on, the energy is not used to increase the kinetic energy of the particles (which would increase the temperature), the energy is used to break the bonds and allow a change of phase, for example, from liquid to gas.
So, we know that if you have a mol of water at 100°C, then you need to add 40.7 kJ of energy to change the phase of the water from liquid to gas phase.
This means that if you have a mol of water and you give that exact energy, the temperature will not change, instead, you now will have a mol of water at the temperature of 100°C.
Similarly with the case at 25°C (which happens for a particular pressure only)
So the heat of vaporization can not really be related to increases in temperature as you thought.
For changes in temperature, you need to use the specific heat.
We know that for water it is:
c = 4.184 J/g*°C = 76.15 J/mol*°C
So, if you want to increase the temperature from 25° to 100°
This means an increase of 75°C of one mol of water.
We just need to multiply the above number by:
1mol*(75°C)
Energy needed = (76.15 J/mol*°C)*1mol*(75°C) = 5,711.25 J
If you want to learn more, you can read:
https://brainly.com/question/11297584
Related Questions
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?
Answers
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%
As discussed in class, the Fischer esterification reactants and products are at equilibrium. How was the equilibrium of the reaction that you performed shifted towards the products
Answers
Answer:
See explanation
Explanation:
The particular reactants in the Fischer esterification reaction were not stated.
Generally, a Fischer esterification is a reaction that proceeds as follows;
RCOOH + R'OH ⇄RCOOR' + H2O
This reaction occurs in the presence of an acid catalyst.
We can shift the equilibrium of this reaction towards the products side in two ways;
I) use of a large excess of either of the reactants
ii) removal of one of the products as it is formed.
Any of these methods shifts the equilibrium of the Fischer esterification reaction towards the products side.
2. Write the chemical equation for the reaction NaOH Sodium Hydroxide AgNO3 Silver Nitrate
Answers
Answer:
AgNO3 + NaOH = AgOH + NaNO3.
Explanation:
Balancing Strategies: In this reaction, the products are initially NaNO3 + AgOH. However the AgOH would break down into Ag2O and H2O. This would give us NaNO3 + Ag2O + H2O as our products for the overall reaction.
Balancing Strategies: In this reaction, the products are initially NaNO3 + AgOH. However the AgOH would break down into Ag2O and H2O. This would give us NaNO3 + Ag2O + H2O as our products for the overall reaction.However, the equation balanced here is the initial reaction which produces AgOH and NaNO3.
Wet helium gas is placed into a balloon at 24.4 degrees Celsius and a pressure of 765.3 mm Hg. What volume (in L) does the dry gas occupy if the water vapor pressure is 24.3 torr and the mass of dried helium gas in the balloon is 0.498 g
Answers
Answer:
Hence the Volume of Gas = 3.04 L.
Explanation:
pressure of dry gas = 765.3 - 24.3 = 741 mmhg
Temperature of gas = 24.4+273.15 = 297.55 k
No of mol of gas = 0.498/4 = 0.1245 mol
R = gas constant = 0.0821 l.atm.k-1.mol-1
From ideal-gas equation
PV = nRT
(741/760) x v = 0.1245 x 0.0821 x 297.55
V = Volume of Gas = 3.04 L
How many Noble gases we have in Periodic Table???
Answers
Answer:
Six
Explanation:
The six naturally occurring noble gases are helium (He), neon (Ne), argon (Ar), krypton (Kr), xenon (Xe), and the radioactive radon (Rn).
Cesium- is radioactive and has a half life of years. Calculate the activity of a sample of cesium-. Give your answer in becquerels and in curies. Round your answer to significant digits.
Answers
The question is incomplete, the complete question is;
Cesium-137 is radioactive and has a half life of 30. years. Calculate the activity of a 6.8 mg sample of cesium-137. Give your answer in becquerels and in curies. Round your answer to 2 significant digits Bq Ci
Answer:
See explanation
Explanation:
The formula for activity is;
R= 0.693N/t1/2
N= 6.02 ×10^23 mol × 6.8 ×10^-3g/137 g/mol = 3 × 10^19
Substituting into the formula;
R= 0.693 × 3 × 10^19/30 years
R= 6.93 ×10^17 y^-1
In Bq;
6.93 ×10^17 y^-1 × 1.00y/3.16 ×10^7 seconds
= 2.19 ×10^10 Bq
In Ci;
2.19 ×10^10 Bq/3.7 ×10^10 Bq/Ci
= 0.59 Ci
90
1
39
is the
In the following decay equation,
90
Sy →
38
et
90
39
-1
A. alpha particle
B. parent element
C. daughter element
D. beta particle
Answers
Answer:
D. beta particle
Explanation:
Number of protons increases from 38 to 39 indicating beta decay (only one proton up from parent isotope to daughter isotope) Also atomic mass (on top of an isotope), 90 stays the same as beta particle is very small.
Predict the products (if any) that will be formed by the reaction below. If no reaction occurs, write NR after the reaction arrow.
2HClO4(aq) + Co(s) -->
Answers
Answer:
The product is aqueous [tex]CO(HCl)_2[/tex] and [tex]O_2(g)[/tex].
Explanation:
Given:
⇒ [tex]2HClO_4(aq) +CO(s)[/tex]
then,
The reaction will be:
⇒ [tex]2HClO_4(aq)+CO(s) \rightarrow CO(HCl)_2 +O_2 (g)[/tex]
In the above reaction, we can see that
The products is:
aqueous [tex]CO(HCl)_2[/tex] and [tex]O_2(g)[/tex]
Thus the above is the correct answer.
A quantity of ideal gas requires 800 kJ to raise the temperature of the gas by 10.0 K when the gas is maintained at constant volume. The same quantity of gas requires 900 kJ to raise the temperature of the gas by 10.0 K when the gas is maintained at constant pressure. What is the adiabatic gas constant of this gas
Answers
Answer:
[tex]\gamma=1.125[/tex]
Explanation:
From the question we are told that:
Initial Heat [tex]Q_1=800kJ[/tex]
initial Temperature [tex]T_1=10.0K[/tex]
Final Heat [tex]Q_2=800kJ[/tex]
Final Temperature [tex]T_2=10.0K[/tex]
Generally the equation for Adiabatic constant is mathematically given by
[tex]\gamma=\frac{Cp}{Cv}[/tex]
Since
Equation for Heat [tex]dQ=nCdT[/tex]
Where
[tex]n_1=n_2\\\\T_1=T_2[/tex]
Therefore
[tex]Q_1=Cv\\\\Cv=800[/tex]
And
[tex]Cp=900[/tex]
Therefore
[tex]\gamma=\frac{900}{800}\\\\\gamma=\frac{9}{8}[/tex]
[tex]\gamma=1.125[/tex]
A sample of neon gas occupies 105 L at 27°C under a pressure of
985 torr. What volume would it occupy at standard condition
Answers
Answer: Volume occupied by given neon sample at standard condition is 123.84 L.
Explanation:
Given: [tex]V_{1}[/tex] = 105 L, [tex]T_{1} = 27^{o}C = (27 + 273) K = 300 K[/tex], [tex]P_{1}[/tex] = 985 torr
At standard conditions,
[tex]T_{2}[/tex] = 273 K, [tex]P_{2}[/tex] = 760 K, [tex]V_{2}[/tex] = ?
Formula used to calculate the volume is as follows.
[tex]\frac{P_{1}V_{1}}{T_{1}} = \frac{P_{2}V_{2}}{T_{2}}[/tex]
Substitute the values into above formula as follows.
[tex]\frac{P_{1}V_{1}}{T_{1}} = \frac{P_{2}V_{2}}{T_{2}}\\\frac{985 torr \times 105 L}{300 K} = \frac{760 torr \times V_{2}}{273 K}\\V_{2} = \frac{94116.75}{760} L\\= 123.84 L[/tex]
Thus, we can conclude that volume occupied by given neon sample at standard condition is 123.84 L.
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.
Answers
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!
LION
If 3.0L of helium at 20°C is allowed to expand to 4.4L, with pressure remain the same
Answers
Answer:
This question is asking to find the new temperature
The answer for the final temperature is 429.73K
Explanation:
Using Charles law equation as follows:
V1/T1 = V2/T2
Where;
V1 = initial volume (L)
V2 = final volume (L)
T1 = initial temperature (K)
T2 = final temperature (K)
According to this question;
V1 = 3.0L
V2 = 4.4L
T1 = 20°C = 20 +273 = 293K
T2 = ?
Using V1/T1 = V2/T2
3/293 = 4.4/T2
Cross multiply
293 × 4.4 = 3 × T2
1289.2 = 3T2
T2 = 1289.2 ÷ 3
T2 = 429.73K
Write the balanced half-reactions and an overall cell reaction below. Be sure to include states of matter.
Answers
Answer: Hello your question is incomplete attached below is the missing image
answer:
Reduction half-reaction: Zn(s) ⇒ Zn⁺² (aq) + 2e⁻ ( occurs at anode )
Oxidation half-reaction; CO²⁺(aq) + 2e⁻ ⇒ CO (s) ( occurs at the cathode )
Overall cell reaction ; Zn(s) + CO²⁺(aq) ⇒ Zn⁺² (aq) + CO (s)
Explanation:
stating the standard reduction potentials
E° zn²⁺/zn = -0.76 v
E°Co²⁺ / Co = -0.28 v
since ; -0.76 v < -0.28 v. Zn will be oxidized while Co²⁺ will be reduced .
Reduction half-reaction: Zn(s) ⇒ Zn⁺² (aq) + 2e⁻ ( occurs at anode )
Oxidation half-reaction; CO²⁺(aq) + 2e⁻ ⇒ CO (s) ( occurs at the cathode )
hence the
Overall cell reaction ;
Zn(s) + CO²⁺(aq) ⇒ Zn⁺² (aq) + CO (s)
Given the equation: 2C6H10(l) 17 O2(g) ---> 12 CO2(g) 10 H2O(g) MM( g/mol): 82 32 44 18 If 115 g of C6H10 reacts with 199 g of O2 and 49 g of H2O are formed, what is the percent yield of the reaction
Answers
Answer:
74%
Explanation:
Step 1: Write the balanced equation
2 C₆H₁₀(l) + 17 O₂(g) ⇒ 12 CO₂(g) + 10 H₂O(g)
Step 2: Determine the limiting reactant
The theoretical mass ratio (TMR) of C₆H₁₀ to O₂ is 164:544 = 0.301:1.
The experimental mass ratio (EMR) of C₆H₁₀ to O₂ is 115:199 = 0.578:1.
Since EMR > TMR, the limiting reactant is O₂.
Step 3: Calculate the theoretical yield of H₂O
The theoretical mass ratio of O₂ to H₂O 544:180.
199 g O₂ × 180 g H₂O/544 g O₂ = 65.8 g H₂O
Step 4: Calculate the percent yield of H₂O
%yield = (experimental yield/theoretical yield) × 100%
%yield = (49 g/65.8 g) × 100% = 74%
Answer:
Percentage yield of H₂O = 74.24%
Explanation:
The balanced equation for the reaction is given below:
2C₆H₁₀ + 17O₂ —> 12CO₂ + 10H₂O
Next, we shall determine the masses of C₆H₁₀ and O₂ that reacted and the mass of H₂O produced from the balanced equation. This is can be obtained as follow:
Molar mass of C₆H₁₀ = 82 g/mol
Mass of C₆H₁₀ from the balanced equation = 2 × 82 = 164 g
Molar mass of O₂ = 32 g/mol
Mass of O₂ from the balanced equation = 17 × 32 = 544 g
Molar mass of H₂O = 18 g/mol
Mass of H₂O from the balanced equation = 10 × 18 = 180 g
SUMMARY:
From the balanced equation above,
164 g of C₆H₁₀ reacted with 544 g of O₂ to produce 180 g of H₂O.
Next, we shall determine the limiting reactant. This can be obtained as follow:
From the balanced equation above,
164 g of C₆H₁₀ reacted with 544 g of O₂.
Therefore, 115 g of C₆H₁₀ will react to produce = (115 × 544)/164 = 381 g of O₂.
From the calculations made above, we can see that a higher mass (i.e 381 g) of O₂ than what was given (i.e 199 g) is needed to react with 115 g of C₆H₁₀.
Therefore, O₂ is the limiting reactant and C₆H₁₀ is the excess reactant.
Next, we shall determine the theoretical yield of H₂O. This can be obtained by using the limiting reactant as shown below:
From the balanced equation above,
544 g of O₂ reacted to produce 180 g of H₂O.
Therefore, 199 g of O₂ will react to produce = (199 × 180)/544 = 66 g of H₂O.
Thus, the theoretical yield of H₂O is 66 g.
Finally, we shall determine the percentage yield. This can be obtained as follow:
Actual yield of H₂O = 49 g
Theoretical yield of H₂O = 66 g
Percentage yield of H₂O =?
Percentage yield = Actual yield /Theoretical yield × 100
Percentage yield of H₂O = 49/66 × 100
Percentage yield of H₂O = 74.24%
A sample of gas contains 0.1200 mol of H2(g) and 0.1200 mol of O2(g) and occupies a volume of 11.5 L. The following reaction
takes place:
H2(g) + O2(g)>H2O2(g)
Calculate the volume of the sample after the reaction takes place, assuming that the temperature and the pressure remain constant.
L
Answers
Answer:
5.75L is the volume of the sample after the reaction
Explanation:
Based on the reaction, 1 mole of H2 reacts with 1 mole of O2 to produce 1 mole of H2O2.
As in the reaction, 0.1200 moles of H2 and 0.1200 moles of O2 are added, 0.1200 moles of H2O2 are produced.
Before the reaction, the moles of gas are 0.2400 moles and after the reaction the moles are 0.1200 moles of gas.
Based on Avogadro's law, the moles of a gas are directly proportional to the volume under temperatura and pressure constant. The equation is:
V1/n1 = V2/n2
Where V is volume and n are moles of 1, initial state and 2, final state.
Replacing:
V1 = 11.5L
n1 = 0.2400 moles
V2 = ?
n2 = 0.1200 moles
11.5L*0.1200 moles / 0.2400 moles = V2
V2 = 5.75L is the volume of the sample after the reaction
Enough of a monoprotic weak acid is dissolved in water to produce a 0.0118 M solution. The pH of the resulting solution is 2.32 . Calculate the Ka for the acid.
Answers
Answer:
1.94 × 10⁻³
Explanation:
Step 1: Calculate the concentration of H⁺ ions
We will use the definition of pH.
pH = -log [H⁺]
[H⁺] = antilog -pH = antilog -2.32 = 4.79 × 10⁻³ M
Step 2: Calculate the acid dissociation constant (Ka) of the acid
For a monoprotic weak acid, whose concentration (Ca) is 0.0118 M, we can use the following expression.
Ka = [H⁺]²/Ca
Ka = (4.79 × 10⁻³)²/0.0118 = 1.94 × 10⁻³
Determine whether each of the examples represents a colligative property or a non-colligative property. boiling point elevation Choose... color Choose... freezing point depression Choose... vapor pressure lowering Choose... density Choose...
Answers
Answer:
boiling point elevation - colligative property
color - non-colligative property
freezing point depression - colligative property
vapor pressure lowering - colligative property
density - non-colligative property
Explanation:
A colligative property is a property that depends on the number of particles present in the system.
Freezing point depression, boiling point elevation and vapour pressure lowering are all colligative properties of solutions.
Colour and density do not depend on the number of particles present hence they are not colligative properties.
The boiling point elevation, freezing point depression, and vapor pressure lowering are colligative properties. And color and density are non-colligative properties.
Explanation:
The colligative properties are the properties depending upon the number of particles of solute not on the nature of the solute.Example of colligative properties:Vapor pressure loweringElevation boiling pointDepression in freezing pointOsmotic pressureThe non-colligative properties are the properties depending upon the nature of solute and solvent.Example of non-colligative properties :ViscositySurface tensionDensitySolubilitySo, from this, we can conclude that boiling point elevation, freezing point depression, and vapor pressure lowering are colligative properties. And color and density are non-colligative properties.
Learn more about colligative properties here:
brainly.com/question/10323760?referrer=searchResults
brainly.com/question/4227527?referrer=searchResults
World leaders met in Copenhagen, Denmark in December 2009, to try to come up with an agreement that
would lead to reducing greenhouse gas emissions. They agreed that we need to reduce carbon dioxide
emissions 80% by the year 2050. How does Lester Brown feel about that?
Select one:
O a. He thinks it is a great achievement.
b. He thinks it is not fast enough
O c. He thinks that we do not need international cooperation
d. He thinks that greenhouse gas emissions are not the most important factor.
Clear my choice
Answers
Lester Brown thinks reducing carbon dioxide emissions 80% by 2050 is not fast enough.
Lester Brown is an American environmentalist who has focused on studying the environment and its protection. In recent years, he has made alerts for world leaders and large industries to strive to stop CO2 emissions because this greenhouse gas has a massive influence on global warming.
Therefore, Lester Brown considers that the projections of reduction of greenhouse gases (especially CO2) made by the world powers for the year 2050, ignore the reality because he considers that CO2 emissions must decrease by at least one 80% in 2020 to avoid drastic consequences in current living conditions. Therefore, the answer is B.
Learn more in: https://brainly.com/question/1577730?referrer=searchResults
What do scientist use to form a hypothesis
Answers
Answer:
an if/then statement
Explanation:
Which of the following statements is correct concerning the class of reactions to be expected for benzene and cyclooctatetraene?
A) Both substances undergo addition reactions.
B) Both substances undergo substitution reactions.
C) Benzene undergoes addition; cyclooctatetraene undergoes substitution.
D) Benzene undergoes substitution; cyclooctatetraene undergoes addition.
Answers
Answer:
Both substances undergo substitution reactions.
Explanation:
Let us go back to the idea of aromaticity. Aromatic substances are said to possess (4n + 2) π electrons according to Huckel rule.
Aromatic substances are unusually stable and the aromatic ring can not be destroyed by addition reactions.
Since both benzene and cyclooctatetraene are both aromatic, they do not undergo addition reactions whereby the aromatic ring is destroyed. They both undergo substitution reaction in which the aromatic ring is maintained.
GM 2 all ,What is an atom define it .Good Day
Answers
Answer:
An atom is the smallest particle of an element that can take part in chemical reaction.
Explanation:
hope it will help u Amri
Evaluate the validity of the following statement: Spontaneous processes are ones that occur quickly and have a low activation energy. A) False. Spontaneous processes always require an input of energy to overcome the activation energy, but always react quickly. B) False. Spontaneous processes can occur slowly, but always have a low activation energy. C) False. Spontaneous reactions can react slowly and can have a high activation energy. D) False. Spontaneous processes always react slowly and always have a high activation energy. E) True.
Answers
Answer:
Spontaneous processes are ones that occur quickly and have a low activation energy. - False -
Spontaneous processes always require an input of energy to overcome the activation energy, but always react quickly - False
Spontaneous processes can occur slowly, but always have a low activation- false
Spontaneous reactions can react slowly and can have a high activation energy - True
Spontaneous processes always react slowly and always have a high activation energy- False
Explanation:
A spontaneous reaction is reaction that proceeds on its own without us having to do a thing at all!
A spontaneous reaction may be fast or slow depending on the activation energy of the reaction. A spontaneous reaction having a high activation energy will be slow. However, if the spontaneous reaction has a low activation energy then it will be fast.
We have to note here that a spontaneous reaction proceeds without a prolonged input energy. Sometimes energy may be supplied to the reaction at the beginning for instance in the case of the combustion of hydrocarbons.
So, spontaneous processes are not necessarily fast. Some of them may have a very high activation energy such as in the rusting of iron hence they are slow.
An aqueous solution contains 0.374 M ammonia (NH3). How many mL of 0.276 M nitric acid would have to be added to 125 mL of this solution in order to prepare a buffer with a pH of 8.970.
Answers
Answer:
111.95mL of HNO3 are needed to prepare the buffer
Explanation:
We can solve this equation using H-H equation for bases:
pOH = pKb + log [HA+] / [A]
Where pOH is the pOH of the solution
pOH = 14 - pH = 14 - 8.970 = 5.03
pKb is the pKb of NH3 = 4.74
[HA+] could be taken as moles of NH4+
[A] as moles of NH3
The NH3 reacts with nitric acid, HNO3, as follows:
NH3 + HNO3 → NH4+ + NO3-
That means the moles of HNO3 added = X = Moles of NH4+ produced
And moles of NH3 are initial moles NH3 - X
Initial moles of NH3 are:
0.125L * (0.374mol/L) = 0.04675 moles NH3
Replacing in H-H equation:
pOH = pKb + log [HA+] / [A]
5.03 = 4.74 + log [X] / [0.04675-X]
0.29 = log [X] / [0.04675-X]
1.95 = [X] / [0.04675-X]
0.0912 - 1.95X = X
0.0912 = 2.95X
X = 0.0309 moles
We need to add 0.0309 moles of HNO3. From a solution that is 0.276M:
0.0309 moles of HNO3 * (1L / 0.276moles) = 0.11195L of HNO3 are needed
In mL:
111.95mL of HNO3 are needed to prepare the buffer
By process of incineration, a mystery substance is empirically determined to contain 40.00% carbon by weight, 6.67% hydrogen, and 53.33% oxygen. Its molecular weight ranges between 55 and 62 g/mole. a. (6 points) Determine the chemical formula of this substance
Answers
Answer:
C₂H₄O₂
Explanation:
Step 1: Divide each percentage by the atomic mass of the element
C: 40.00/12.01 = 3.331
H: 6.67/1.01 = 6.60
O: 53.33/16.00 = 3.333
Step 2: Divide all the numbers by the smallest one
C: 3.331/3.331 = 1
H: 6.60/3.331 ≈ 2
O: 3.333/3.331 ≈ 1
The empirical formula is CH₂O, with a molecular weight of 12 g/mol + 2 × 1 g/mol + 16 g/mol = 30 g/mol. The molecular weight of the compound must be a product of 30, such as 60 (between 55 and 62 g/mol). Since we have to multiply by 2 (30 to 60) to get to the molecular weight of the compound, we also have to multiply the empirical formula by 2 to get the chemical formula of the compound.
CH₂O × 2 = C₂H₄O₂
A complex ion that forms in solution has a structure that:____.
a. can be determined simply by stoichiometry.
b. can be predicted on the basis of electrical charge.
c. can only be determined experimentally.
d. cannot be determined.
Answers
Answer:
can only be determined experimentally.
Explanation:
In the early days of inorganic chemistry, the structure of complex ions remained a mystery hence the name ''complex''.
These ions appear to have structures that defied accurate elucidation. However, by diligent laboratory investigation, Alfred Werner was able to accurately determine the structure of cobalt complexes. As a result of this, he is regarded as a pathfinder in coordination chemistry.
Hence, the structure of complex ions can only be determined experimentally.
Answer:
c. can only be determined experimentally
Explanation:
It is not possible to know for certain the structure of a complex ion on the basis of stoichiometry or by the electrical charges on the components. The structure of the resulting complex ion can only be known by experiment.
When naming organic compounds, there are strict rules regarding punctuation.
1. A comma is used to separate two numbers.
2. A hyphen is used to separate a number from a letter.
Rewrite the name of this compound using hyphens and commas as appropriate.
Answers
The question is incomplete, the complete question is shown in the image attached to this answer
Answer:
2,3,3-trimethylhexane
Explanation:
IUPAC nomenclature provides a universally acceptable method of naming organic compounds from its structure.
According to this system of nomenclature;
A comma is used to separate two numbers.
A hyphen is used to separate a number from a letter.
Applying these rules, the name of the compound shown in the question should be written as 2,3,3-trimethylhexane.
Which statement describes the 3d, 4s, and 4p orbitals of Arsenic (As) based on its electronic configuration and position in the periodic table?
The 3d and 4s orbitals are completely filled, and the 4p orbital is partially filled.
The 3d orbital is completely filled, and the 4s and 4p orbitals are partially filled.
The 3d, 4s, and 4p orbitals are completely filled.
The 3d, 4s, and 4p orbitals are partially filled.
Answers
Answer:
The 3d and 4s orbitals are completely filled, and the 4p orbital is partially filled.
Explanation:
The correct answer is: The 3d and 4s orbitals are completely filled, and the 4p orbital is partially filled.
The d orbital contains 10 electrons, the s orbital takes 2 electrons and the p orbital takes six electrons.
The orbital in chemistry is defined as a region in space where there is a high probability of finding an electron. There are s, p, d, f orbitals in chemistry which correspond to sharp, principal, diffuse and fundamental.
The electronic configuration of arsenic is 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p3.
From this electronic configuration, we can see that the 4s and 3d orbitals are half filled while the 4p orbital is half filled.
For more about electronic configuration, see:
https://brainly.com/question/4949433
Hydrogengasand oxygengas react to form water vapor. Suppose you have of and of in a reactor. Calculate the largest amount of that could be produced. Round your answer to the nearest .
Answers
The question is incomplete. The complete question is :
Hydrogen [tex](H_2)[/tex] gas and oxygen [tex](O_2)[/tex] gas react to form water vapor [tex](H_2O)[/tex]. Suppose you have 11.0 mol of [tex]H_2[/tex] and 13.0 mol of [tex]O_2[/tex] in a reactor. Calculate the largest amount of [tex]H_2O[/tex] that could be produced. Round your answer to the nearest 0.1 mol .
Solution :
The balanced reaction for reaction is :
[tex]$2H_2(g) \ \ \ \ + \ \ \ \ \ O_2(g)\ \ \ \rightarrow \ \ \ \ 2H_2O(g)$[/tex]
11.0 13.0
11/2 13/1 (dividing by the co-efficient)
6.5 mol 13 mol (minimum is limiting reagent as it is completely consumed during the reaction)
Therefore, [tex]H_2[/tex] is limiting reagent. It's stoichiometry decides the product formation amount from equation above it is clear that number of moles for [tex]H_2O[/tex] will be produced = number of moles of [tex]H_2[/tex]
= 11.0 mol
At 25 oC the solubility of chromium(III) iodate is 2.07 x 10-2 mol/L. Calculate the value of Ksp at this temperature. Give your answer in scientific notation to 2 SIGNIFICANT FIGURES (even though this is strictly incorrect). [a]
Answers
Answer:
5.0 × 10⁻⁶
Explanation:
Step 1: Write the balanced equation for the solution of chromium(III) iodate
Cr(IO₃)₃(s) ⇄ Cr³⁺(aq) + 3 IO₃⁻(aq)
Step 2: Calculate the solubility product constant (Ksp)
To relate Ksp and the solubility (S), we will make an ICE chart.
Cr(IO₃)₃(s) ⇄ Cr³⁺(aq) + 3 IO₃⁻(aq)
I 0 0
C +S +3S
E S 3S
The solubility product constant is:
Ksp = [Cr³⁺] × [IO₃⁻]³ = S × (3S)³ = 27 S⁴ = 27 × (2.07 × 10⁻²)⁴ = 5.0 × 10⁻⁶
When 3-methyl-1-pentene is treated with in dichloromethane, the major product is 1-bromo-3-methyl-2-pentene.
a. True
b. False
Answers
Answer:
True
Explanation:
When Methyl Pentene is introduced in a chemical reaction with dichloromethane then the major product will be bromomethylpentene. There can be small amount of bromo methyl pentene than the amount of methyl pentene introduced for reaction.
