The dispersion force and the dipole force has been the attractive forces that result in the formation of bond within the molecules and result in the change in the properties of the compounds.
The dipole force is a strong force and results in a higher boiling point.
The statements regarding the forces are:
(a) For molecules with similar molecular weights, the dispersion forces become stronger as the molecules become more polarizable.
The statement is true.
(b) For the noble gases the dispersion forces decrease while the boiling points increase as you go down the column in the periodic table.
The statement is true.
(c) In terms of the total attractive forces for a given substance, dipole-dipole interactions, when present, are always greater than dispersion forces.
The statement is false.
(d) All other factors being the same, dispersion forces between linear molecules are greater than those between molecules whose shapes are nearly spherical.
The statement is true.
For more information about the dispersion force, refer to the link:
Answer:
A ,B- false
C,D- true
Explanation:
Dipole forces always lead to stronger attraction and boiling points than dispersion forces. When linear molecules are involved, they often posses greater dipole forces and higher boiling points. Linear alkanes posses higher boiling points than branched alkanes.
Answer:
The solution in the buret, during a titration is called titrant.
Explanation:
A titration is a useful process, that makes you know the concentration of a solution. A titrant solution (burette) is evaluated against a titrand to control the pH changes against the volume aggregate. Only a strong acid with a strong base, a strong base with a strong acid, a weak acid with a strong base and a weak base with strong acid are valued.
When the pH reaches the equivalence point, it is said that the normality of the acid by the milliequivalents, is equal to the basic normality by its milliequivalents. In conclusion, the entire base / acid became its conjugate pair.
To check this sudden change in pH, a substance is used, called Indicator that changes the color of the titrand (analyte).
In a titration analysis, the substance in the buret is called the 'titrant'. It is used to react with the analyte, the sample solution whose concentration we're measuring. The goal is to reach the endpoint, the point when a distinct visual change indicates that the titrant has completely reacted with the analyte.
In a titration analysis, the solution in the buret is called the titrant. This solution contains a known concentration of a substance. During a titration, this titrant is added incrementally to a sample solution, called the analyte, which contains the substance whose concentration is to be measured. The titrant and analyte undergo a chemical reaction of known stoichiometry.
By measuring the volume of the titrant solution needed to completely react with the analyte, scientists can calculate the concentration of the analyte. This point where the titrant has completely reacted with the analyte is termed the equivalence point of the titration. The process of adding the titrant is halted when a distinct change is visually detected in the solution - this could be a color change, for example. This is known as the end point.
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For the whole set of problems, always remember the Avogadro’s number is 6.023*10^23 units per mole of a substance. Units could be atoms, molecules or formula units.
The first question asks for the number of molecules of NaNO3. The molar mass of NaNO3 is 85 grams per mole. So,
150g NaNO3(1mole NaNO3/85 grams NaNO3)(6.023*10^23 molecules/1mole NaNO3)=1.063*10^24 molecules of NaNO3
5.7*10^46 molecules of NaNO3(1mole NaNO3/6.023*10^23 molecules)(85 grams NaNO3/1mole NaNO3) = 8.044*10^24 grams NaNO3
For the molar mass of water, we have 18.02grams per mole.
301 moles H2O(18.02 grams H2O/1 mole H2O) = 5424.02 grams H2O
For the molar mass of sulfuric acid, we have 98.08 grams per mole.
25g H2SO4(I mole H2SO4/98.08g H2SO4) = 0.2549 mole H2SO4
For the molar mass of Ca(OH)2, we have 74.1 grams per mole.
252gCa(OH)2(1mol/74.1g)(6.023*10^23/1mol) = 2.048*10^24 molecules of Ca(OH)2
For the molar mass of calcium, we have 40 grams of Ca per mole.
6.7*10^35 atoms Ca(1 mole Ca/6.023*10^23 atoms)(40g Ca/1mol Ca) = 4.45*10^13 grams Ca
B) Cs
C) O
D) F
b. buffer
c. acid
d. catalyst
Answer:
1. D. cal......
2.A. iron
3. D
4.2.44j/g°C A
5,Lf=334J/g B
Explanation:
1: Which of the following is the abbreviation for a unit of energy? A. K / B. °C/ C. W / D. cal...............
calorie is the unit of energy
#2: A 200 g block of a substance requires 1.84 kJ of heat to raise its temperature from 25°C to 45°C. Use the table attached to identify the substance. A. iron/ B. aluminum/ C. gold/ D. copper.....................
Q=mcdt
1840=0.2*C*(45-25)
C=460J/KgK
if the specific heat capacity is the above then he substance is iron
#3: In a calorimeter, the temperature of 100 g of water decreased by 10°C when 10 g of ice melted. How much heat was absorbed by the ice? A. 418 kJ / B. 100 kJ / C. 10 J / D. 4.18 kJ .................
Q=mcdT
Q=0.1*10*4180
Q=4180j. answer D
.#4: The amount of heat needed to raise the temperature of 50 g of a substance by 15°C is 1.83 kJ. What is the specific heat of the substance? A. 2.44 J/g-°C / B. 2.22 J/g-°C / C. 2.13 J/g-°C / D. 2.05 J/g-°C ................
Q=mcdT
1830=50/1000*C*15
C=2440j/kg/k
change it to j/g°C
2.44j/g°C A
#5: In a calorimeter, 3.34 kJ of heat was absorbed when 10 g of ice melted. What is the enthalpy of fusion of the ice? A. 6.68 J/g / B. 334 J/g / C. 6.68 kJ/g/ D. 334 kJ/g
Q=mLf
Lf=enthalpy of fusion
3340/10=Lf
Lf=334J/g B
Enthalpy of fusion quantity of heat to convert 1 unit mass of a solid to liquid without any noticeable change in temperature.