Answer:
(1)=(A), (2)=(B), (3)=D, (4)=C, (5)=E, (6)=F
Explanation:
(1) Glassware used to accurately transfer small volumes = (A) Graduated pipette, that is basically a glass tube with graduation of different volumes to be dispensed.
(2) Glassware used to accurately transfer a small, single volume = (B) Volumetric pipette, that is a glass tube with a central glass bulb and is used to dispense accurately an unique volume of liquid everytime.
(3) Glassware to deliver a volume not known in advance = (D) Buret (or burette), that is used to dispense slowly a volume of liquid when a titration process is needed
(4) Glassware best used when greater access to the contents is needed = (C) Beaker, that is basically a very open glass cylinder with a spout
(5) Glassware used to prevent splashing or evaporation = (E) Erlenmeyer flask, that has a small open at the top and is useful when the liquid needs to be swirled as, for example, during a titration.
(6) Glassware used to make accurate solutions = (F) Volumetric flask, that has a long slim neck that provides a higher accuracy when a exact volume of liquid needs to be used for preparation of a solution.
Less friction to stop the wheel from turning
The question provides the data in an incorrect way, but what the question is asking is for the entropy change when combining 3 moles of water at 0 °C (273.15 K) with 1 mole of water at 100 °C (373.15 K). We are told the molar heat capacity is 75.3 J/Kmol. We will be using the following formula to calculate the entropy change of each portion of water:
ΔS = nCln(T₂/T₁)
n = number of moles
C = molar heat capacity
T₂ = final temperature
T₁ = initial temperature
We can first find the equilibrium temperature of the mixture which will be the value of T₂ in each case:
[(3 moles)(273.15 K) + (1 mole)(373.15 K)]/(4 moles) = 298.15 K
Now we can find the change in entropy for the 3 moles of water heating up to 298.15 K and the 1 mole of water cooling down to 298.15 K:
ΔS = (3 moles)(75.3 J/Kmol)ln(298.15/273.15)
ΔS = 19.8 J/K
ΔS = (1 mole)(75.3 J/Kmol)ln(298.15/373.15)
ΔS = -16.9 J/K
Now we combine the entropy change of each portion of water to get the total entropy change for the system:
ΔS = 19.8 J/K + (-16.9 J/K)
ΔS = 2.9 J/K
The entropy change for combining the two temperatures of water is 2.9 J/K.
Answer:
by adding water into the mix
Explanation:
this will dissolve the salt
Answer:
its height relative to some reference point, its mass, and the strength of the gravitational field
Explanation:
Gravitational energy is the potential energy associated with gravitational force, such as elevating objects against the Earth’s gravity. The potential energy due to elevated positions is called gravitational potential energy.
The factors that affect an object’s gravitational potential energy are the following; its height relative to some reference point, its mass, and the strength of the gravitational field it is in. For instance, consider a wallet lying on a table, it has less gravitational potential energy than the same wallet lying on top of a taller cupboard, and yet lesser gravitational potential energy than a heavier wallet lying on the same table.
If an object lies at a certain height above the Moon’s surface, it has less gravitational potential energy than the same object lying at the same height above the Earth’s surface because the Moon’s gravitational force is weaker.
