At standard pressure, a certain compound has a low boiling point and is insoluble in water. At STP, this compound most likely exists as(1) ionic crystals
(2) metallic crystals
(3) nonpolar molecules
(4) polar molecules
Answers
At standard pressure, a certain compound has a low boiling point and is insoluble in water. At STP, this compound most likely exists as nonpolar molecules. Hence, option C is correct.
What are nonpolar molecules?
A molecule in which the individual dipoles cancel each other & result in zero net dipole moment is called a non-polar molecule.
Nonpolar molecules make weak attractions between molecules called van der Waals forces and they are loosely bound with each other.
Therefore, nonpolar molecules can easily go to the atmosphere by breaking those weak bonds and having a boiling point low.
Hence, option C is correct.
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Answer is (3) - nonpolar molecules.
Nonpolar molecules make weak attractions between molecules called van der Waals forces. Hence, they are looesly bound with each other. Therefore, nonpolar molecules can easily go to the atmosphere by breaking those weak bonds. Hence, the boiling point is low.
Since water is a polar solvent, nonpolar molecules do not soluble in water.
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As the temperature of a sample of matter is increased, what happens to the average kinetic energy of the particles in the sample?A. It decreases. B. It increases. C. It does not change.
Answers
The element that has the electron configuration is Darmstadtium.
Electronic configuration:
Since An element with the electronic configuration 1s²2s²2p⁶3s²3p⁶4s²3d¹⁰4p⁶5s²4d¹⁰5p⁶6s²4f¹⁴5d¹⁰6p⁶7s²5f¹⁴6d⁸ contains 110 electrons in its electron shells.
Also, the element should be a neutral atom, this number should also equal to its atomic number. Therefore, its atomic number is 110.
The element in the periodic table that has an atomic number of 110 is Darmstadtium, a d-block element, thus a transition metal.
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Answer:
Darmstadtium
Explanation:
An element with the electronic configuration 1s²2s²2p⁶3s²3p⁶4s²3d¹⁰4p⁶5s²4d¹⁰5p⁶6s²4f¹⁴5d¹⁰6p⁶7s²5f¹⁴6d⁸ has 110 electrons in its electron shells.
Since the element is a neutral atom, this number is also equal to its atomic number. Therefore, its atomic number is 110.
The element in the period table that has an atomic number of 110 is Darmstadtium, a d-block element, thus a transittion metal. It also belong to period 7 in the Periodic table of elements.
Answers
Answer: Option (1) is the correct answer.
Explanation:
A redox reaction is a reaction in which both oxidation and reduction takes place.
For example,
Oxidation-half reaction :
Reduction-half reaction :
Hence, we can see that electrons are involved in a redox reaction.
Therefore, we can conclude that in a redox reaction, electrons are the particles which are lost and gained in equal numbers.
0.001011 g
0.000001101 kg
0.0001011 g
Answers
For this case, the first thing we are going to do is convert all the numbers to the same unit.
For this, we use the following conversion:
1 Kg = 1000 g
We then have the following numbers:
1) 0.00001101 kg = (0.00001101) * (1000) = 0.01101 g
2) 0.001011 g
3) 0.000001101 kg = (0.000001101) * (1000) = 0.001101 g
3) 0.0001011 g
Therefore, the largest amount in grams is:
0.01101 g
Answer:
0.00001101 kg = 0.01101 g
option 1
Answers
Answer:
A just A
Explanation:
Answer:
so is c they have the same mass.
Answers
Answer:
-1215.9J is the work done
Explanation:
It is possible to find work done in the change of volume of a gas at constant pressure using:
W = -P×ΔV
Where W is work, P is pressure and V is change in volume}
Replacing:
W = -6atm×(5L-3L)
W = -12atmL
As 1atmL = 101,325J, work done in joules is:
-12atmL ×(101.325J / atmL) = -1215.9J is the work done
The value of work done when a volume increases from 3 liters to 5 liters at 6 atm of pressure is -1215.9 Joules.
How do we calculate workdone?
Workdone on any boby or by the body will be calculated as:
W = -P×ΔV, where
W = workdone
P = applied or exerted pressure = 6 atm
ΔV = change in volume due to workdone = (5-3) L
Negative sign in the formula shows that work is done on the opposite side of the pressure or volume.
On putting all these values on the above equation, we get
W = -6atm × (5L-3L)
W = -12 atmL
We know that, 1 atmL = 101,325 J,
So, workdone in joules will be written as:
-12 atmL × (101.325J / atmL) = -1215.9 J
Hence, -1215.9 J is the workdone.
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