Answer:
1. A/Thermal Energy
2. B/The particles within the system will have greater motion, and the temperature will increase.
3. C/Attractions occur due to electrostatic forces. When particles move fast enough, these forces can no longer keep particles together.
4. D/ For a phase change from solid to liquid, the bonds do not break completely and particles can still slide past each other.
5. D/ a sample of liquid benzene at 80ºC and a sample of gaseous benzene at 80ºC
Explanation:
I took the Test.
1. Internal energy associated with systems and particles is thermal energy. Thus, option A is correct.
The internal energy has been described as the energy that has been assessed by the molecules and bonds at the STP. Thermal energy has been the energy of particular molecules and the system comprising all the molecules have the summation of the thermal energy.
Thus, internal energy associated with systems and particles is thermal energy. Thus, option A is correct.
2. The addition of energy results in greater motion with an increase in temperature has been representation no phase change. Thus, option B is correct.
The addition of energy to the substance results in the increased motion of the particles. The energy higher than the threshold energy has been responsible for the phase change of the substance.
The addition of energy results in greater motion with an increase in temperature has been representation no phase change. Thus, option B is correct.
3. The movement of the particles under the gravitational attraction is not enough to hold the particles together at a longer distance. Thus, option B is correct.
The attraction has been the force that has been responsible for holding the particles in a molecule. Different attractions have a different levels of stability and strength.
The movement of the particles under the gravitational attraction is not enough to hold the particles together at a longer distance. Thus, option B is correct.
4. The phase change of the molecules has been mediated by overcoming the force of attraction that allows the molecules to move freely. The solid to liquid phase change has been mediated with the breaking of the attraction between the bonds allowing free movement.
Thus, statement 4 is incorrect, and option D is correct.
5. The kinetic energy has been the energy possessed by the moving molecules in the medium. The molecules in liquid and gas have been possessing the same energy at the same temperature, Thus, option D is correct.
For more information about energy and electrostatic interactions, refer to the link:
https://brainly.com/question/6170401
radium-223 decays with a half-life of 11.4 days, how long will it take for a 0.240-mol sample of radiuim to decay to 1.88 x 10-3 mol
The time taken for 0.240 mole sample of radiuim to decay to 1.88×10¯³ mole is 79.8 days
We'll begin by calculating the number of half-lives that has elapsed.
Original amount (N₀) = 0.240 mole
Amount remaining (N) = 1.88×10¯³ mole
Number of half-lives (n) =?N = 1/2ⁿ × N₀
1.88×10¯³ = 1/2ⁿ × 0.240
Cross multiply
1.88×10¯³ × 2ⁿ = 0.240
Divide both side by 1.88×10¯³
2ⁿ = 0.240 / 1.88×10¯³
2ⁿ = 128
2ⁿ = 2⁷
n = 7Thus, 7 half-lives has elapsed
Finally, we shall determine the time.Number of half-lives (n) = 7
Half-life (t½) = 11.4 days
Time (t) =?t = n × t½
t = 7 × 11.4
t = 79.8 daysTherefore, the time taken for 0.240 mole sample of radiuim to decay to 1.88×10¯³ mole is 79.8 days
Learn more: https://brainly.com/question/13266270
Why we use two different methods for detection of cogulase enzyme ? Or what other reason or what basic different between them?
Problem 07.020 - Heat transferred from cold to hot reservoir Assume that heat in the amount of 100 kJ is transferred from a cold reservoir at 600 K to a hot reservoir at 1050 K contrary to the Clausius statement of the second law. What is the total entropy change
This problem is describing a system in which 100 kJ of energy is transferred from a cold reservior at 600 K to a hot one at 1050 K. This situation is opposed to the Clausius statement of the second law of thermodynamics, because it is widely known that heat is transferred from hot to cold systems.
However, we can calculate the total entropy change by using the following formula:
[tex]\Delta S =\Delta S _{Low}+\Delta S _{High}[/tex]
Which is broken down as follows:
[tex]\Delta S =\frac{-Q}{T_{Low}} + \frac{Q}{T_{High}}\\\\\\\Delta S =Q(\frac{-1}{T_{Low}} + \frac{1}{T_{High}})[/tex]
Since the heat is assumed to be equal for the both of them. Then, we plug in the given heat and temperatures to obtain:
[tex]\Delta S =100kJ(\frac{-1}{600K} + \frac{1}{1050K})\\\\\Delta S =-0.0714\frac{kJ}{K}[/tex]
Learn more:
https://brainly.com/question/17145614https://brainly.com/question/13999732Tristan wraps some gifts and then brings them to the post office where they are delivered to people in different parts of the country. Which organelle is Tristan most like?
Answer:
Tristan wraps some gifts and then brings them to the post office where they are delivered to people in different parts of the country. Which organelle is Tristan most like?
Answer: Tristan is most like the Golgi Body
Explanation:
Atomic mass is an average of the elements_____.
Specific chemical element
Answer:
The atomic mass of an element is the average mass of the atoms of an element measured in atomic mass unit (amu, also known as daltons, D). The atomic mass is a weighted average of all of the isotopes of that element, in which the mass of each isotope is multiplied by the abundance of that particular isotope.
Explanation:
What is the freezing point, in Celsius, of a sucrose-water solution containing 2.23g sucrose per 100g water. The molal freezing point depression constant for water is 1.86oC/m.
The freezing point of the solution is - 0.12 oC.
We know that;
ΔT = K m i
ΔT = Freezing point depression
K = Freezing constant
m = molality
i = Van't Hoff factor
Number of moles of sucrose= 2.23g/342 g/mol = 0.0065 moles
Mass of solvent in Kg = 0.1 Kg
Molality of the solution = 0.0065 moles/ 0.1 Kg = 0.065 m
Now;
ΔT = 1.86 oC/m × 0.065 m × 1
ΔT = 0.12 oC
Freezing point of pure water = 0 oC
Freezing point of solution = 0 oC - 0.12 oC = - 0.12 oC
Learn more: https://brainly.com/question/2292439
How much does a cloud weigh in pounds?
Answer:
1.1 million pounds
Explanation:
Answer:
1.1 million pounds
Explanation:
that's a lot of water above us!!
Explain why there might be a change in the density of a forged product as compared to that of the cast blank.
Answer:
Forged parts are often tougher than cast parts. This can be determined by performing tensile tests on various areas on the parts. Additionally, the microstructures of forged and cast parts can be used to determine if a part was forged or cast. The microstructure of a cast part will have a more uniform grain structure.
Explanation:
An element with 5 valence electrons would be a
0 - 3 anion
O +5 anion
0-5 cation
O + 3 cation
Answer:
ask you subject teacher
Explanation:
What is caused by temperature differences in the mantle resulting in plate movement?
Crust
Convection Currents
Lithosphere
Answer:
Convection Currents
Explanation:
A pan containing 40 grams of water was allowed to cool from a temperature of 91.0 °C. If the amount of heat released is 1,300 joules, what is the approximate final temperature of the water? 74 °C 78 °C 81 °C 83 °C
Answer:
d. 83 °c
Explanation:
Answer:
D. 83 c
Explanation:
took the test
A mixture of solids containing a ketone, a carboxylic acid, and an amine, are dissolved in DCM. What is the best way to begin an extraction to separate the amine from the mixture
There are different ways of extraction. The best way to begin an extraction to separate the amine from the mixture is to extract with dilute NaOH.
An acid-base extraction is often used in the extraction of carboxylic acids from the organic layer and thereafter into the aqueous layer.NaOH is known to be the most common compound that is used to convert a carboxylic acid into its more water-soluble ionic carboxylate form.
But if the mixture has a compound that you want, and that can react with NaOH, another milder base such as sodium bicarbonate is preferably used.
See full question below
A mixture of solids containing a ketone, a carboxylic acid, and an amine, are dissolved in DCM. What is the best way to begin an extraction in order to separate the carboxylic acid from the mixture?
A) Extract with dilute NaOH
B) Extract with dilute HCl
C) Extract with dichloromethane
D) Extract with water
Learn more from
https://brainly.com/question/2646724
Doing Labs at home
I’m a junior and I’m staying home for this semester and I have to take chemistry and a lot of my work is Labs but I don’t know how to do them since I don’t have the materials at home to do the labs. Someone please help!!!
Answer:
go get the stuff.
Explanation:
HELP!! what are the usual products of combustion reactions?
Explanation:
Carbon dioxide and water
I hope it helps
Answer:
The usual products of combustion reactions are carbon dioxide and water.
Explanation:
Combustion reaction is when a substance reacts with oxygen gas, resulting in a release of energy in the form of light and heat. Combustion reactions must have oxygen (O2) as one of the reactants.
Lithium carbonate, Li2CO3, is commonly used to treat bipolar disorder. What is the percent by mass of carbon in this compound (to the nearest whole percent)
Answer:
18.78%
Hence, the mass percent of lithium in lithium carbonate is 18.78%.
Explanation:
The percent by mass of carbon in this compound, lithium carbonates is
16.2%
The compound is lithium carbonates.
The molar mass of the compound will be calculated and the molar mass of carbon in the compound will also be calculated to find the percentage of carbon in it.
Molar mass of Li₂CO₃ = (7 ×2) + 12 + 48 = 74 g
Molar mass of carbon in the compound = 12g
percentage mass of carbon = 12 / 74 × 100
percentage mass of carbon = 1200 / 74
percentage mass of carbon = 16.2162162162
percentage mass of carbon = 16.2 %
learn more: https://brainly.com/question/5596184?referrer=searchResults
Which statement best describes mechanical energy?
A.the energy of heat and magnetism
B.the energy of electricity and motion
C.the energy of moving parts and motion
D.the energy of electricity and heat
Answer:
the answer is B.
Explanation:
your welcome
Answer:
the answer is B
Explanation:
i have the same exact questionn
PLEASE HELP ME!!! When two atoms combine to form a compound, one atom pulls electrons from the other atom towards itself. The atom that pulls electrons is (reduced or oxidized). The atom whose electrons are being pulled is (reduced or oxidized)?
Answer:
The pulling atom is oxidized while the pulled atom is reduced. Grade 9 Chemistry
Explanation:
How many moles of c3h6will be consumed when 4.11 mol of co2 are produced in the following equation 2c3h6 +9o2 - 6co2 + 6H2o
Answer: a.1.37 b.4.11 c.6.00 d.12.3
Explanation:
3.00 L of a gas is collected at 35.0 C and 0.93 atm. What is the volume at STP
10. Select the more polar bond in each of the following pairs: a) C and N or C and o b) N and F or N and O.
Chemical properties of mercury hg
Answer:
Mercury has a relatively high vapour pressure and the highest volatility of any metal, vapourizing to become a colourless, odourless gas. The metal is a fair conductor of electricity, but a poor conductor of heat. Mercury's atomic number is 8.
A radioactive sample has a half life of 1 hour. If you start with 1.000 gram of it at noon, how much of it remains at 4pm
The amount of the sample remaining at 4pm is 0.0625 g
We'll begin by calculating the number of half-lives that has elapsed
Half-life (t½) = 1 hour
Time (t) = 4 hour
Number of half-lives (n) =?n = t / t½
n = 4 / 1
n = 4Finally, we shall determine the amount the sample remaining at 4pmNumber of half-lives (n) = 4
Initial amount (N₀) = 1 g
Amount remaining (N) =?[tex]N = \frac{N_0}{ {2}^{n}} \\ \\ N = \frac{1}{ {2}^{4}} \\ \\ N = \frac{1}{16} \\ \\ N = 0.0625 \: g[/tex]
Thus, the amount remaining at 4pm is 0.0625 g
Learn more: https://brainly.com/question/25750315
A solution that contains less than the amount of solute that would be dissolved at equilibrium is considered to be
Answer:
dilute solution
A solution containing less solute than the equilibrium amount is called a dilute solution. The solvent has a limited capacity to dissolve a solute.
Explanation:
2. A solution NaF is add dropwise to a solution that is .0122 M in Ba . When the concentration of F exceeds ______M, BaF2 will precipitate. Neglect volume changes. BaF2 K
Explanation:
BaF2(s) <------> Ba2+(aq) + 2F-(aq)
Ksp BaF2 = 1.0 x 10^-6.
Ksp BaF2 = [Ba2+(aq)]×[F-(aq)]^2 at equilibrium
When Qsp >Ksp, BaF2 will precipitate
Qsp = [Ba2+(aq)]×[F-(aq)]^2
[Ba2+(aq)]×[F-(aq)]^2 > 1.0 x 10^-6.
0.0122 moldm-3 × [F-(aq)]^2 > 1.0 x 10^-6
[F-(aq)]^2 > 1×10^-6 / 0.0122 mol2dm-6
[F-(aq)]^2 > 81.96 × 10^-6 mol2dm-6
[F- (aq)] > 9.05 × 10^-3 moldm-3
So F- concentration should be more than 9.05 × 10^-3 moldm-3
Identify the strongest intermolecular force that is likely to affect each of the samples described below.
A mixture of chlorine gas (Cl) and fluorine gas (F): V London dispersion forces
COMPLETE
Tweaks
Menu
A mixture of hydrogen sulfide (H2S) and hydrogen chloride (HCI): V dipole-dipole interactions
Search
Selection
COMPLETE
Guess
this
hydrogen bonding
A mixture of water (H2O) and ammonia (NH3):
Answer:
A mixture of chlorine gas (Cl2) and fluorine gas (F2):
✔ London dispersion forces
Explanation:
The group of elements that make up the smallest percentage of the periodic table are the
Answer:
Explanation:wwwwwwwwwwwwwwwwwwww
What are the steps to Photosynthesis
Answer:
Step 1-Light Dependent. CO2 and H2O enter the leaf.
Step 2- Light Dependent. Light hits the pigment in the membrane of a thylakoid, splitting the H2O into O2.
Step 3- Light Dependent. ...
Step 4-Light Dependent.
Step 5-Light independent.
Step 6-Light independent.
Kinetic energy is described as - The answers are stored energy - destroyed energy movement or created energy if you answer the question I will give you the brainiest.
1Touch and hold a clip to pin it. Unpinned clips will be deleted after 1 hour.
Suppose you have samples of three unknown solids. Explain how you could use their properties to
determine whether or not they are ionic solids.
Using melting and boiling temperature, hardness and electric current passing testing.
Ionic solidsIonic solids are materials that have a strong bond between their ions, thus producing well-defined shapes.
In addition, due to this strong attraction, the boiling and melting temperatures of these materials are very high, in addition to the resistance to breakage presented by them.
Finally, ionic solids are also excellent conductors of electricity.
So, their properties used to determine whether or not they are ionic solids are melting and boiling temperature, hardness and electric current passing testing.
Learn more about ionic solids: brainly.com/question/8236583
Which is true of protons and neutrons?
1. They have approximately the same mass and the same charge.
2) They have approximately the same mass but different charge.
by The have different mass and different charge.
O sette
4) They have different mass but the same charge.
Answer:
[tex]\blue{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}[/tex]
[tex]\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}[/tex]
