Molecular compounds can not conduct electricity because they do not break up into ions. Therefore, option (A) is correct.
Molecular compounds can be characterized as those compounds where the atoms are connected by covalent bonds. The covalent bond is generally created by the mutual sharing of electrons by the atoms. In molecular compounds, intermolecular forces are responsible for bond formation.
The melting point of the molecular compound is low as compared to the ionic compound as the energy required to break bonds is less compared to the ionic bond between ions in the crystalline ionic compound.
Molecular compounds are generally by the reaction between non-metals. Molecular compounds are also known as covalent compounds. The ionic and molecular compound differ from each in their physical properties,
The molecular compounds do not conduct electricity as they do not bark into ions. Therefore they can behave as insulators while ionic compounds conduct electricity.
Learn more about Molecular compounds, here:
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B. A sample of 2.00 g of O2 is removed.
C. A sample of 4.00 g of O2 is added to the 4.80 g of O2 gas in the container.
By using the ideal gas law and molar mass calculations, the final volumes are found to be A. 65.0 L, B. 8.75 L, and C. 27.5 L.
To calculate the final volume when additional O2 is added or when some O2 is removed, we can use the concept of the molar mass and the ideal gas law that states that volume is directly proportional to the amount of gas, assuming pressure and temperature is constant.
The molar mass of O2 is approximately 32.00 g/mol.
A. 0.500 moles of O2 is added. This equals 0.500 * 32 g = 16 g. The total mass in the system is now 20.8 g. If the original 15.0 L represented 4.80 g, now 20.8 g would represent 15.0 L * 20.8/4.80 = 65.0 L.
B. 2.00 g of O2 is removed. So, the total mass in the system is now 2.80 g. If the original 15.0 L represented 4.80 g, now 2.80 g would represent 15.0 L * 2.80/4.80 = 8.75 L.
C. 4.00 g of O2 is added. So, the total mass in the system is now 8.80 g. If the original 15.0 L represented 4.80 g, now 8.80 g would represent 15.0 L * 8.80/4.80 = 27.5 L.
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True
False
Answer:
It's true I got it wrong.
Explanation:
(2) H2O (4) Cu
per second after allowing for background radiation.
Calculate how long it would be before the activity fell from 800 counts per second to
200 counts per second.
Answer: The time for activity to fell from 800 counts per second to 200 counts per second is 20 minutes
Explanation:
Expression for rate law for first order kinetics is given by:
where,
k = rate constant
t = age of sample
a = let initial amount of the reactant
a - x = amount left after decay process
a) for finding the rate constant:
Half life is the amount of time taken by a radioactive material to decay to half of its original value.
b) for activity to fell from 800 counts per second to 200 counts per second.
The time for activity to fell from 800 counts per second to 200 counts per second is 20 minutes
The half-life of the given radioactive isotope is 10 minutes. The activity of the isotope falls to half its initial value every 10 minutes. Hence, it would take 20 minutes for the activity to fall from 800 counts per second to 200 counts per second.
The process you're describing relates to the concept of radioactive decay, which is a fundamental principle in nuclear physics. In this case, we're dealing with what's known as a half-life. The half-life of a radioactive isotope is the time it takes for half of the isotope in a sample to decay.
In this scenario, the radioactive isotope has a half-life of 10 minutes and it starts with a sample activity of 800 counts per second. According to the nature of half-life, it will reduce to half of its initial counts, 400 counts per seconds in 10 minutes. Importantly, another 10 minutes would be required for the number of counts to halve again to 200 counts per second.
Therefore, it would take 20 minutes for the activity to fall from 800 counts per second to 200 counts per second.
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b. decomposition
c. double replacement
d. substitution
B. Decomposition, I remember reading this in my student workbook. Hope this helped!