He Sprite and Banana Challenge consists of eating two bananas and then drinking a liter of Sprite<span>. This causes you to projectile vomit and throw up all over the place!</span>
Answer:

Explanation:
Given
485 L
Required
Determine the measurement not equal to 485L

<em>From standard unit of conversion;</em>
1 KL = 1000 L
<em>Multiply both sides by 485</em>


<em>Divide both sides by 1000</em>

---- This is equivalent
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<em>From standard unit of conversion;</em>
1000 mL = 1 L
<em>Multiply both sides by 485</em>


Convert to standard form

Hence;
is not equivalent to 485L
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<em>From standard unit of conversion;</em>
100 cL = 1 L
<em>Multiply both sides by 485</em>


Convert to standard form
---- This is equivalent
---------------------------------------------------------------------------------------------
μL
<em>From standard unit of conversion;</em>
1000000 μL = 1 L
<em>Multiply both sides by 485</em>


Convert to standard form
---- This is equivalent
From the list of given options;
is not equivalent to 485L
Bikes that are kept outside are not used as often, speeding up the chemical reaction of metal rusting.
Heat energy from the sun slows down the chemical reaction of the metal rusting.
Moisture and oxygen cause oxidation, which speeds up the chemical reaction of the metal rusting.
Wind energy outside speeds up the chemical reaction of the metal rusting.
Answer:
The molar mass of the gas is 44 g/mol
Explanation:
It is possible to solve this problem using Graham's law that says: Rates of effusion are inversely dependent on the square of the mass of each gas. That is:

If rate of effusion of nitrogen is Xdistance / 48s and for the unknown gas is X distance / 60s and mass of nitrogen gas is 28g/mol (N₂):

6,61 = √M₂
44g/mol = M₂
<em>The molar mass of the gas is 44 g/mol</em>
<em></em>
I hope it helps!
Answer:
Lewis structure in attachment.
Explanation:
Atoms of elements in and beyond the third period of the periodic table form some compounds in which more than eight electrons surround the central atom. In addition to the 3s and 3p orbitals, elements in the third period also have 3d orbitals that can be used in bonding. These orbitals enable an atom to form an <u>expanded octet</u>.
The central Xe atom in the XeF₄ molecule has <u>two</u> unbonded electron pairs and <u>four</u> bonded electron pairs in its valence shell.