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It is true that when a distance between gas particles increases, the volume of the gas will also increase.
Answer: Option A
<u>Explanation:</u>
The volume of any material can be the space occupied by any material. And the space required by any material to occupy a desired shape or size is determined by its bonding and distance between the particles making that material. If the particles which are soft and small can be compressed easily to occupy very small volume compared to compound made up of hard and large particles.
Along with this, the distance between the neighbouring particles also play a major role, as if we consider a laddo, we can compress it to small size by reducing the distance between neighbouring particles while we will not be able to compress the size or volume occupied by a table as we cannot reduce the distance of particle separation. Thus, it is true that if the distance between gas particles increased then the volume of gas particles will also increase.
You can solve this by using the equation (P1V1/T1) = (P2V2/T2). Plug in 0.50 atm for P1, leave V1 as the unknown, and plug in 325 K as T1. Then substitute 1.2 atm for P2, 48 L for V2, and 320 K for T2. Solve for V1, which is 117L, but since you round using two sig figs, your answer is C, 120 L. Hope this helps!
Answer: <span>Yes, a hydrocarbon is any compound that contains hydrogen and oxygen.
Justification:
First, I will try to rewrite the given structural formula since it is misrepresented.
This should be the right structural formula:
H H
| |
H - C - C - OH
| |
H H
And, yes the definition of hydrocarbon is that it is most simple organic compound constitured only by atoms of carbon (C) and hydrogen (H).
By the way, that hydrocarbon is also represented as CH3 - CH2 OH, it is an alcohol (since it has the functional group OH) and its name is ethanol.
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Move the decimal<span> so there is one non-zero digit to the left of the </span>decimal<span> point. The number of </span>decimal<span>places you move will be the exponent on the 10 . If the </span>decimal<span> is being moved to the right, the exponent will be negative. If the </span>decimal<span> is being moved to the left, the exponent will be positive.</span>
