Answer:
Q = 81.59kJ
Explanation:
Hello,
The heat of condensation is the energy required to to convert the steam into water.
Mass = 195g
Specific heat capacity of water = 4.184J/g°C
Initial temperature(T1) = 100°C
Final temperature(T2) = 0°C
Heat energy (Q) = ?
Heat energy (Q) = mc∇T
M = mass of the substance
C = specific heat capacity of the substance
∇T = T2 - T1 = change in temperature of the substance
Q = 195 × 4.184 × (0 - 100)
Q = -81588J
Q = -81.588kJ
The heat required for the condensation of 195g of steam is 81.59kJ
Spell out the full name of the compound.
Answer:
4–octene.
Explanation:
To name the compound given in the question, we must determine the following:
1. Determine the functional group of the compound and locate its position by giving it the lowest possible count.
2. Locate the longest continuous carbon. This gives the parent name of the compound.
3. Combine the above to obtain the name of the compound.
Now, let us determine the name of the compound bearing in mind the information given above. This is illustrated below:
1. The functional group of the compound is double bond i.e alkene and it located at carbon 4.
2. The longest continuous carbon chain is 8. Since the compound is an alkene, the name becomes octene.
3. Therefore, the name of the compound is:
4–octene.
under the same conditions carbon (iv) oxide,propane and nitrogen (i) oxide diffuse at the same rate.Explain
Answer:
Rate of diffusion is same .
Explanation:
As we know that Rate of the diffusion is directly proportional to the [tex]\frac{1}{\sqrt{M} }[/tex] .They have same mass if there is same rate and similar condition therefore the mass of carbon (iv) oxide,propane and nitrogen (i) oxide will be similar.
The mass is directly proportional to the Rate of the diffusion.Therefore the rate of diffusion is similar in all carbon (iv) oxide,propane and nitrogen (i) oxide .Using appropriate chemical equation distinguish between cation and anion hydrolysis
Answer:
HCO3- (aq) + H2O (I) <--> H2CO3 (aq) + OH- (aq)
Explanation:
The equation to distinguish between cation and anion hydrolysis is given below :
HCO3- (aq) + H2O (I) <--> H2CO3 (aq) + OH- (aq)
The important thing to remember is their origin. The anions can react with water and can produce hydroxide ions while hydroxide ions make a solution basic.
Cathodic protection of iron involves using another more reactivemetal as a sacrificial anode. Classify each of thefollowing metals by whether they would or would not act as asacrificial anode to iron.
a. Sn
b. Cu
c. Zn
d. Au
e. Pb
f. Ag
g. Mg
An old iron beam was coated with an unknown metal. There is a crackon the coating and it is observed that the iron is rusting at thefracture. The beam is in a structure that experiences high stress,resulting in frequent fractures to the coating.
What was the old metal coating likely made of and what metal youwould use to repair the fractures to avoid further corrosion?
Choices: tin, aluminum, gold
1. The old coating was made of __________________.
2. __________________would be a good choice for repairing thefracture.
Answer:
1.) zinc and aluminum
2.)
a.) The old coating was made of tin.
b.) Aluminum would be a good choice for repairing the fracture.
Vanadium (V) and oxygen (O) form a series of compounds with the following compositions: Mass % V 76.10 67.98 61.42 56.02 Mass % O 23.90 32.02 38.58 43.98 Compound Mass % N 1 33.28 2 39.94 Mass % Si 66.72 60.06 10. What are the relative numbers of atoms of oxygen in the compounds for a given mass of vanadium
Answer:
For every given mass of Vanadium, the relative number of oxygen atoms present or the mole ratio of Oxygen to Vanadium is:
A. 1:1
B. 3:2
C. 2:1
D. 5:2
Note: The question is stated more clearly below:
Vanadium (V) and oxygen (O) form a series of compounds with the following compositions: Mass % V 76.10 67.98 61.42 56.02 Mass % O 23.90 32.02 38.58 43.98 Compound Mass % N 1 33.28 2 39.94.
What are the relative numbers of atoms of oxygen in the compounds for a given mass of vanadium?
Explanation:
Number of moles in 100 g mass = % mass / molar mass
Molar mass of Vanadium, V = 51 g/mol
Molar mass of oxygen atom, O = 16 g/mol
1. Percentage mass of V and O is 76.10% and 23.90% respectively.
Number of moles of each atom;
V = 76.10/51.0 = 1.5 moles
O = 23.9/16 = 1.5 moles
Mole ratio of oxygen to vanadium = 1.5/1.5 = 1 : 1
2. Percentage mass of V and O is 67.98% and 32.02% respectively
Number of moles of each atom:
V = 67.98/51 = 1.33
O = 32.02/16 = 2
Mole ratio of oxygen to vanadium = 2/1.33 = 1.5 : 1 = 3 : 2
3. Percentage mass of V and O is 61.42% and 38.58% respectively
Number of moles of each atom:
V = 61.42/51 = 1.2
O = 38.58/16 = 2.4
Mole ratio of oxygen to vanadium = 2.4/1.2 = 2 : 1
4. Percentage mass of V and O is 56.02% and 43.98% respectively
Number of moles of each atom:
V = 56.02/51 = 1.10
O = 43.98/16 = 2.75
Mole ratio of oxygen to vanadium = 2.75/1.10 = 2.5 : 1 = 5 : 2
Mass of the Vanadium, number of O2 atoms present, or the mole ratio of 1:1 , 3:2 , 2:1 , 5:2 . As Vanadium (V) and oxygen (O) form a series of compounds is given with masses of 76.10 67.98, 23.90 32.02, 33.28 2 39.94, etc.
As per No of moles in 100 g mass = % mass / molar mass Mass of Vanadium, V = 51 g/ mol e, Mass of oxygen atom, O = 16 g/mole O = 23.9/16 = 1.5 moles for oxygen to vanadium = 1.5/1.5 = 1 : 1 2. Percentage mass of V and O is 67.98% and 32.02%. Mole ratio of oxygen to vanadium = 2/1.33 = 1.5 : 1 = 3 : 2 3. Percentage mass of V and O is 61.42% and 38.58% Mole ratio of oxygen to vanadium = 2.4/1.2 = 2 : 1 4. Percentage mass of V and O is 56.02% and 43.98%. Mole ratio of oxygen to vanadium = 2.75/1.10 = 2.5 : 1 = 5 : 2Learn more about the Vanadium (V) and oxygen (O).
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Most modern medications are given in doses of milligrams. Thyroid medications, however, are typically given in doses of micrograms. How many milligrams are in a dose labeled 125 µg? View Available Hint(s) Most modern medications are given in doses of milligrams. Thyroid medications, however, are typically given in doses of micrograms. How many milligrams are in a dose labeled 125 µg? 1.25 x 105 mg 0.125 mg 1.25 x 10?4 mg 1.25 x 102 mg
Answer:
0.125 mg
Explanation:
The correct answer would be 0.125 mg
According to the conversion factor, one milligram of a sample is equivalent to one thousand micrograms of the same sample.
milligram = [tex]10^{-3}[/tex]
microgram = [tex]10^{-6}[/tex]
Hence,
1 milligram = 1000 micrograms or 1 microgram = [tex]10^{-3}[/tex] milligram
Therefore, 125 micrograms will be:
125/1000 = 0.125 milligram
What does a complete ionic equation look like?
A. All substances are written as ionic compounds bonded together.
B. All substances are labeled with the oxidation states of the atoms.
C. All ionic substances are written as separate ions in solution.
D. All ionic substances are written with the state symbol (eo) after it.
Answer:
All ionic substances are written as separate ions in solution
All ionic substances are written as separate ions in solution in a complete ionic equation. Therefore, option (C) is correct.
What is the ionic equation?A complete ionic equation can be described as a particular chemical equation where charged atoms such as ions are expressed in a given solution. The complete ionic equations always contain all ions that are formed or act during a particular chemical reaction.
The net ionic equation can be described as an equation that provides information about ions that exists in an aqueous medium. Salts get dissolved in polar solvents such as water which are present as cations and anions in their dissolved state.
The ionic equation shows the chemical species that undergo a chemical change. The ions which are present on both sides of the equation are considered to be spectator ions. Therefore, in order to obtain the net ionic equation we can eliminate them.
Learn more about complete ionic equations, here:
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Activity: A Ferris wheel with a diameter of 60.0 m is moving at a speed of 2.09 m/s. What is the centripetal
acceleration?
Answer:
Centripetal acceleration of the wheel is [tex]0.145\ m/s^2[/tex].
Explanation:
We have,
Diameter of a Ferris wheel is 60 m
Radius of the wheel is 30 m
Speed of the wheel is 2.09 m/s
It is required to find the centripetal acceleration of the wheel. The formula of centripetal acceleration is given by :
[tex]a=\dfrac{v^2}{r}\\\\a=\dfrac{(2.09)^2}{30}\\\\a=0.145\ m/s^2[/tex]
So, the centripetal acceleration of the wheel is [tex]0.145\ m/s^2[/tex].
PdPd has an anomalous electron configuration. Write the observed electron configuration of PdPd. Express your answer in complete form in order of orbital filling. For example, 1s22s21s22s2 should be entered as 1s^22s^2. View Available Hint(s)
Answer:
1s²,2s²,2p⁶,3s²,3p⁶,4s²,3d¹⁰,4p⁶,5s⁰,4d¹⁰.
Explanation:
Palladium is a chemical element with the symbol Pd and atomic number 46.
The electronic configuration is;
[Kr] 4d¹⁰
The full electronic configuration observed for palladium is given as;
1s²,2s²,2p⁶,3s²,3p⁶,4s²,3d¹⁰,4p⁶,5s⁰,4d¹⁰.
The reason for for the anomlaous electron configuration is beacuse;
1. Full d orbitals are more stable than partially filled ones.
2. At higher energy levels, the levels are said to be degenerated which means that they have very close energies and then electrons can jump from one orbital to another easily.
How many milliliters of a 1.5 m h2so4 are needed to neutralize 35ml sample of a 1.5 m solution?
1) 17.5ml
2) 35ml
3) 52.5ml
4) 3.0ml
Answer:
1) 17.5 mL
Explanation:
Hello,
In this case, the reaction between sulfuric acid and potassium hydroxide is:
[tex]H_2SO_4+2KOH\rightarrow K_2SO_4+2H_2O[/tex]
In such a way, we notice a 1:2 molar ratio between the acid and the base, therefore, at the equivalence point we have:
[tex]2*n_{acid}=n_{base}[/tex]
And in terms of concentrations and volumes:
[tex]2*M_{acid}V_{acid}=M_{base}V_{base}[/tex]
Thus, we solve for the volume of acid:
[tex]V_{acid}=\frac{M_{base}V_{base}}{2*M_{acid}} =\frac{35mL*1.5M}{2*1.5M} \\\\V_{acid}=17.5mL[/tex]
Best regards.
1-hexanol was prepared by reacting an alkene with either hydroboration-oxidation or oxymercuration-reduction. Draw the structure of the alkene that was used to prepare the alcohol in highest yield. You do not have to consider stereochemistry. Indicate the method of preparation by drawing either BH3 (for hydroboration-oxidation), or Hg (for oxymercuration-reduction), in a separate sketcher. If there is more than one alkene that can be used for a given method, draw all of them. If either hydroboration-oxidation or oxymercuration-reduction can be used, just give the structures for one method. Separate structures with signs from the drop-down menu.
Answer:
Alkene form hexan-1-ol with oxidation in presence of NaOH with highest yield
Explanation:
AgNO3 is added to a solution containing Cl- and CrO42- in order to separate the ions. If the Cl- and CrO42- concentrations are 0.020 and 0.010 M, respectively, what are the minimum Ag+ concentrations required to precipitate out the anions?
Answer: The minimum [tex][Ag^{+}][/tex] concentrations required to precipitate out the anions is [tex]9 \times 10^{-9}[/tex] M.
Explanation:
We know that,
[tex]K_{sp}[/tex] for AgCl is [tex]1.8 \times 10^{-10}[/tex]
and, [tex]K_{sp}[/tex] for [tex]Ag_{2}CrO_{4}[/tex] is [tex]9 \times 10^{-12}[/tex]
Now, we will calculate the concentration of at which these ions precipitate out are as follows.
For AgCl :
[tex][Ag^{+}] = \frac{K_{sp}}{[Cl^{-}]}[/tex]
= [tex]\frac{1.8 \times 10^{-10}}{0.02}[/tex]
= [tex]9 \times 10^{-9}[/tex] M
For [tex]Ag_{2}CrO_{4}[/tex] :
[tex][Ag^{+}]^{2} = \frac{K_{sp}}{CrO^{2-}_{4}}[/tex]
= [tex]\frac{9 \times 10^{-12}}{0.01}[/tex]
= [tex]9 \times 10^{-10}[/tex]
[tex][Ag^{+}] = \sqrt{(9 \times 10^{-9})}[/tex]
= [tex]3 \times 10^{-5}[/tex] M
This shows that concentration of ions in AgCl is less than the concentration of AgCl will precipitate first.
glucose 6‑phosphate+H2O⟶glucose+Pi glucose 6‑phosphate+H2O⟶glucose+Pi K′eq1=270 K′eq1=270 ATP+glucose⟶ADP+glucose 6‑phosphate ATP+glucose⟶ADP+glucose 6‑phosphate K′eq2=890 K′eq2=890 Using this information for equilibrium constants determined at 25∘C,25∘C, calculate the standard free energy of hydrolysis of ATP. standard free energy:
Answer:
-30.7 kj/mol
Explanation:
The standard free energy for the given reaction that is the hydrolysis of ATP is calculated using the formula: ∆Go ’= -RTln K’eq
where,
R = -8.315 J / mo
T = 298 K
For reaction,
1. K′eq1=270,
∆Go ’= -RTln K’eq
= - 8.315 x 298 x ln 270
= - 8.315 x 298 x 5.59
= - 13,851.293 J / mo
= - 13.85 kj/mol
2. K′eq2=890
∆Go ’= -RTln K’eq
= - 8.315 x 298 x ln 890
= - 8.315 x 298 x 6.79
= - 16.82 kj/mol
therefore, total standard free energy
= - 13.85 + (-16.82)
= -30.7 kj/mol
Thus, -30.7 kj/mol is the correct answer.
The simplest carboxylic acid is called *
O Formaldehye
O formic acid
acetic acid
O
acetone
The following reaction: NO2 (g) --> NO (g) 1/2 O2 (g) is second-order in the reactant. The rate constant for this reaction is 3.40 L/mol*min. Determine the time needed for the concentration of NO2 to decrease from 2.00 M to 1.50 M.
Answer:
t = 0.049 mins or 2.94 secs
Explanation:
For a simple second order reaction, the integrated law which describes the concentration of reactants at a given time t, is as follows: 1/[A] = 1/[A]o + Kt;
Where [A] is concentration of reactant at time, t, [A]o is initial concentration of A; K is rate constant; t is time at a given instant.
Using the integrated rate law:
I/[NO2]t - 1/[NO2]o = Kt
Where K = 3.40 L/mol/min
[NO2]t = 1.5 mol/L
[N02]o = 2.0 mol/L
t = ?
Making t subject of formula;
t = (1/[NO2]t - 1/[NO2]o) / K
t = (1/1.5 - 1/2.0)/3.40
t = 0.049 mins or 2.94 secs
Copper sulfate is a blue solid that is used to control algae growth. Solutions of copper sulfate that come in contact with the surface of galvanized ( Zinc-plated) steel pails undergo the following reaction that forms copper metal on the zinc surface. How many grams of Zinc would react with 454g (1lb) of copper sulfate (160g/mol)?
CuSO4(aq)+ Zn(s)>>>>Cu(s) + ZnSO4(aq)
Answer:
185.49 grams of Zinc would react with 454g (1lb) of copper sulfate
Explanation:
Yo know the following balanced reaction:
CuSO₄(aq)+ Zn(s) →Cu(s) + ZnSO₄(aq)
You can see that by stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of reagents and products are part of the reaction:
CuSO₄: 1 moleZn: 1 moleCu: 1 moleZnSO₄: 1 moleBeing:
Cu: 63.54 g/moleS: 32 g/moleO: 16 g/moleZn: 65.37 g/molethe molar mass of the compounds participating in the reaction is:
CuSO₄:63.54 g/mole + 32 g/mole + 4*16 g/mole= 159.54 g/mole ≅ 160 g/moleZn: 65.37 g/moleCu: 63.54 g/moleZnSO₄: 65.37 g/mole + 32 g/mole + 4*16 g/mole= 161.37 g/moleThen, by stoichiometry of the reaction, the following amounts of mass of reagent and product participate in the reaction:
CuSO₄: 1 moles* 160 g/mole= 160 gZn: 1 mole* 65.37 g/mole= 65.37 gCu: 1 mole* 63.54 g/mole= 63.54 gZnSO₄: 1 mole* 161.37 g/mole= 161.37 gNow you can apply the following rule of three: if 160 grams of CuSO₄ react with 65.37 grams of Zn by this reaction stoichiometry, 454 grams of CuSO₄ with how much mass of Zn will it react?
[tex]mass of Zn=\frac{454 grams of CuSO_{4} *65.37 grams of Zn}{160 grams of CuSO_{4}}[/tex]
mass of Zn= 185.49 grams
185.49 grams of Zinc would react with 454g (1lb) of copper sulfate
An 8.5 mL sample of gasoline has a mass of .75 g. What is the density of the gasoline?
Answer:
density = 8.824g/mL
Explanation:
given
mass = 75g
volume = 8.5mL
density = mass/volume
density = 75g/8.5mL
density = 8.824g/mL
Answer:0.088g/ml
Explanation:
Density=mass/volume
d=0.75g/8.5ml
d=0.088g/ml
a) What substances are present in an aqueous buffer composed of HC2H3O2 and C2H3O2 - ?b) What happens when LiOH is added to a buffer composed of HC2H3O2 and C2H3O2 - ? Write a chemical equation for that reaction.c) What happens when HBr is added to this buffer? Write a chemical equation for that reaction.
Answer:
a) HC₂H₃O₂, C₂H₃O₂⁻, H₃O⁺, H₂O, OH⁻
b) HC₂H₃O₂ + LiOH ⇄ H₂O + LiC₂H₃O₂
c) C₂H₃O₂⁻ + HBr ⇄ HC₂H₃O₂ + Br⁻
Explanation:
a) In a HC₂H₃O₂/C₂H₃O₂⁻ buffer system, the following reactions take place:
HC₂H₃O₂ + H₂O ⇄ C₂H₃O₂⁻ + H₃O⁺
C₂H₃O₂⁻ + H₂O ⇄ HC₂H₃O₂ + OH⁻
Thus, the species present are: HC₂H₃O₂, C₂H₃O₂⁻, H₃O⁺, H₂O, OH⁻.
b) When LiOH is added to the buffer system, it is partially neutralized according to the following equation.
HC₂H₃O₂ + LiOH ⇄ H₂O + LiC₂H₃O₂
c) When HBr is added to the buffer system, it is partially neutralized according to the following equation.
C₂H₃O₂⁻ + HBr ⇄ HC₂H₃O₂ + Br⁻
A student mixes baking soda and vinegar in a glass. Are there any new substances created from this mixture?
Answer:
Explanation:
1. A student mixes baking soda and vinegar in a glass. The results are shown at left. ... Yes I do belive that new substances are being formed because there is a chemical reaction between the baking soda and vinegar turning it into a bubbly substances instead of a powder and liquid.
Yes, there are new substances created from this mixture.
The heat capacity of air is much smaller than that of water, and relatively modest amounts of heat are needed to change its temperature. This is one of the reasons why desert region, although very hot during the day, are bitterly cold at night. The heat capacity of air at room temperature and pressure is appoximately 21 J/K*mol. How much energy is required to raise the temperature of a room of dimensions 5.5m x 6.5m x 3.0m by 10 degrees Celsius? If losses are neglected, how long will it take a heater rated at 1.5 kW to achieve that increase given that 1 W = 1 J/s?
Answer:
[tex]Q=9.2x10^5J[/tex]
[tex]t=614s=10.2min[/tex]
Explanation:
Hello,
In this case, we can compute the energy by using the following formula for air:
[tex]Q=nCp\Delta T[/tex]
Whereas the moles of air are computed via the ideal gas equation at room temperature inside the 5.5m x 6.5m x 3.0m-room:
[tex]n=\frac{PV}{RT}\\\\V=5.5m*6.5m*3.0m=107.25m^3*\frac{1000L}{1m^3}=107250L\\ \\n=\frac{1atm*107250L}{0.082\frac{atm*L}{mol*K}*298.15K}\\ \\n=4386.8mol[/tex]
Now, we are able to compute heat, by considering that the temperature raise is given in degree Celsius or Kelvins as well:
[tex]Q=4386.8mol*21\frac{K}{mol*K}*10K \\\\Q=9.2x10^5J[/tex]
Finally, we compute the time required for the heating by considering the heating rate and the required heat, shown below:
[tex]t=\frac{9.2x10^5J}{1.5\frac{kJ}{s}*\frac{1000J}{1kJ} } \\\\t=614s=10.2min[/tex]
Regards.
When copper is heated with an excess of sulfur, copper(I) sulfide is formed. In a given experiment, 0.0970 moles of copper was heated with excess sulfur to yield 1.76 g copper(I) sulfide. What is the percent yield?
Answer:
Percent yield = 22.8 %
Explanation:
Step 1: Data given
Numbers of moles copper = 0.0970 moles
Mass of copper(I) sulfide = 1.76 grams
Step 2: The balanced equation
2Cu + S ⇒ Cu2S
Step 3: Calculate moles of Cu2S
For 2 moles Cu we need 1 mol S to produce 1 mol Cu2S
For 0.0970 moles Cu we'll hace 0.0970 / 2 = 0.0485 moles
Step 4: Calculate mass of Cu2S
Mass Cu2s = moles Cu2S * molar mass Cu2S
Mass Cu2S = 0.0485 moles * 159.16 g/mol
Mass Cu2S = 7.72 grams
Step 5: Calculate percent yield
Percent yield = (actual yield/ theoretical mass) * 100%
Percent yield = (1.76 grams / 7.72 grams)*100%
Percent yield = 22.8 %
The percentage yield of the experiment obtained by the reaction of 0.0970 mole of copper with excess sulfurs is 22.8%
We'll begin by calculating the number of mole of Cu₂S produced from the reaction. This can be obtained as follow:
2Cu + S —> Cu₂S
From the balanced equation above,
2 moles of Cu reacted to produce 1 mole of Cu₂S.
Therefore,
0.0970 mole of Cu will react to produce = [tex]\frac{0.0970}{2}[/tex] = 0.0485 mole of Cu₂S.
Next, we shall determine the theoretical yield by calculating the mass of 0.0485 mole of Cu₂S.
Molar mass of Cu₂S = (63.5×2) + 32 = 159 g/mol
Mole of Cu₂S = 0.0485 mole
Mass of Cu₂S =?Mass = mole × molar mass
Mass of Cu₂S = 0.0485 × 159
Mass of Cu₂S = 7.7115 gThus, the theoretical yield of Cu₂S is 7.7115 g
Finally, we shall determine the percentage yield of Cu₂S.
Actual yield = 1.76 g
Theoretical yield = 7.7115 g
Percentage yield =?[tex]Percentage yield = \frac{Actual}{Theoretical} * 100\\\\= \frac{1.76}{7.7115} * 100\\\\[/tex]
Percentage yield = 22.8%Therefore, the percentage yield of the experiment is 22.8%
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True/False: ________ To study the effect of sunlight on different plants, I expose the plants to the same amount of sunlight. The independent variable is the sunlight.
Answer: True
Explanation:
The independent variable is the one which can be changed or manipulated in an experiment. The independent variable exerts its influence on the dependent variable. The dependent variable is the result of the experiment.
The amount of sunlight, can be regulated or changed in an experiment, thus it is an independent variable. The effect of sunlight on different plants is the dependent variable.
Calculate the number of grams in sodium in 8.4g of Na2C6H6O7 (sodium hydrogen citrate) express your answer using two significant figures
please help!
Answer:
2.0 g Na
Explanation:
Stoichiometry.
8.4g sodium hydrogen citrate x (1 mol sodium hydrogen citrate / 192 g sodium hydrogen citrate) x (2 mol Na/1 mol sodium hydrogen citrate) x (23g Na/1 mol Na)
^write it out it makes more sense that way
Consider the three isomeric alkanes n-hexane,2,3-dimethylbutane, and 2-methylpentane. Which of the following correctly lists these compounds in order of increasing boiling point
a. 2,3-dimethylbutane < 2-methylpentane < n-hexane
b. 2-methylpentane
c. 2-methylpentane < 2,3-dimethylbutane
d. n-hexane < 2-methylpentane < 2,3-dimethylbutane
e. n-hexane < 2,3-dimethylbutane < 2-methylpentane
Answer:
a. 2,3-dimethylbutane < 2-methylpentane < n-hexane
Explanation:
The boiling point of alkanes is highly affected by the degree of branching in the molecule. Branched alkanes generally have a lower boiling point than unbranched alkanes.
The reason for the higher boiling point of unbranched alkanes is because they have greater vanderwaals forces acting between their molecules due to their larger surface area. Recall that branched alkanes have a lesser surface area compared to unbranched alkanes.
n-hexane is an unbranched alkane hence it will have the highest boiling point followed by 2-methyl pentane and lastly 2,3-dimethyl butane. The boiling point continues to decrease as the extent of branching increases.
4. Which of the following statements explains the cause of lanthanide contraction?
A. All lanthanides and actinides are radioactive
B. Protons exhibit a stronger pull on outer f orbitals
C. The d orbitals in lanthanides have unpair electrons
D. The d orbitals in actinides have paired electrons
Answer:
B. PROTONS EXHIBIT STRONGER PULL ON OUTER f ORBITALS
Explanation:
Lanthanide contraction is the greater than normal decrease in the ionic radius of the lanthanide series from atomic number 57 to atomic number 71. This decrease is rather not expected of the ionic radii of these elements and they result in the greater decrease in the subsequent series of the lanthanides from the atomic number 72. The cause of which is as a result of the poor shielding effects of the nuclear charge around the electrons of the f orbitals. So therefore, protons are strongly pulled out of the 4f orbital and as a result of the poor shielding effect which causes the electrons of the 6s orbitals to be drawn more closer to the nucleus and hence resulting in a smaller atomic radii. It is worthy to note that the shielding effects of the inner electrons decreasing from s orbital to the f orbital; that is s > p > d > f. So from the decrease in the shielding effects from s to the f orbitals, lanthanide contraction results from the inability of the orbitals far away from s like the 4f orbiatls to shield the outermost shells of the lanthanide elements. So the cause of lanthanide contraction is the action of the protons which strongly pull the electrons of the f orbitals because of the poor shielding effects due to the distance of this orbital from the nucleus.
Answer:
B) Protons exhibit a stronger pull on outer f orbitals than on d orbitals.
Explanation:
In the process of making soap, I poured some of the cooked mixture through some muslin fabric, in order to separate the solid particles from liquid. What am I doing to this mixture?
A) Serrating it
B) Decanting it
C) Mixing it
D) Filtering it
Answer:
filtering
Explanation:
you're pouring the mixture through muslin cloth to keep the particles and bigger peaces out of the soap.
25.00 mL of a H2SO4 solution with an unknown concentration was titrated to a phenolphthalein endpoint with 28.11 mL of a 0.1311 M NaOH solution. What is the concentration of the H2SO4 solution
Answer:
Concentration of the H₂SO₄ solution is 0.0737 M
Explanation:
Equation of the neutralization reaction between the acid, H₂SO₄, and the base, NaOH, is given below:
H₂SO₄ + 2NaOH -----> Na₂SO₄ + 2H₂O
From the above equation, one mole of acid requires 2 moles of base for complete neutralization which occurs at phenolphthalein endpoint.
mole ratio of acid to base, nA/nB = 1:2
Concentration of the base, Cb = 0.1311 M
Volume of base, Vb, = 28.11 mL
Concentration of acid, Ca = ?
Volume of acid, Va + 25.0 mL
Using the formula, CaVa/CbVb = nA/nB
making Ca subject of the formula, Ca = Cb*Vb*nA/Va*nB
substituting the values into the equation
Ca = (0.1311 * 28.11 * 1) / 25.0 * 2 = 0.0737 M
Therefore, concentration of the H₂SO₄ solution is 0.0737 M
Which of the following would be more reactive than magnesium (Mg)?
A. Calcium (Ca)
B. Potassium (K)
C. Argon (Ar)
D. Beryllium (Be)
Answer:potassium is more reactive than Mg because both lie in the same group and the element potassium has more electropositivity than magnesium
Explanation:
I hope it will help you
Answer: B. Potassium(K)
Explanation:
Analyze: The metallic character of an element is determined by how readily it loses electrons. Elements that lose electrons most easily have the greatest metallic character
A. Which group has the greatest metallic character?
B. Which group has the lowest metallic character?
C. What is the relationship between metallic character and ionization energy?
Answer:
Group 1 or akali metals have the greatest metallic property.
Group 17 has the lowest metallic character.
C. As you move from right to lefton the periodic table, metallic character increases which is the ability to lose electrons. Ionization energy decrease as we move from right to left on the periodic table.
Explanation:
Akali metals in group 1 have the greatest metallic property and they are the most reactive metals. Francium metal on the group has the most metallic characteristics. It is rare and very radioactive. Group 17 has the lowest metallic character. This is because while moving across the period, the number of electrons in the outermost shell increases. This make it difficult for atoms to leave see electrons and become electropositive . Group 17 has the highest tendency of accepting electrons.
Ionization energy is the energy use to remove electron from an atom in gaseous stage. Ionization energy decrease as we move from right to left on the periodic table and metallic character increases as we move from right to left on the periodic table.
what are the differences between strong and weak acids?
Strong acids are completely ionised and weak acids are partly ionised
Answer:
Como forman los iones en soluciión
Explanation:
Los ácidos fuertes y las bases fuertes se refieren a especies que se disocian completamente para formar los iones en solución. Por el contrario, los ácidos y bases débiles se ionizan solo parcialmente y la reacción de ionización es reversible.