Body systems often interact with each other. The image below shows parts of body systems that interact to fulfill a function for
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Considering the definition of STP conditions, 3.35 moles of helium are contained within the balloon.
<h3>Definition of STP condition</h3>The STP conditions refer to the standard temperature and pressure. Pressure values at 1 atmosphere and temperature at 0 ° C are used and are reference values for gases. And in these conditions 1 mole of any gas occupies an approximate volume of 22.4 liters.
<h3>Amount of moles of helium within the balloon</h3>In this case, you know that scientist fills a large, tightly sealed balloon with 75,000 mL of helium at STP.
So, you can apply the following rule of three: if by definition of STP conditions 22.4 liters are occupied by 1 mole of helium, 75 L (75 L= 75000 mL, being 1 L= 1000 L) are occupied by how many moles of helium?
<u><em>amount of moles of helium= 3.35 moles</em></u>
Finally, 3.35 moles of helium are contained within the balloon.
Learn more about STP conditions:
The highest energy occupied molecular orbital in the C-C bond of the C₂ molecule is 2pπ orbitals.
<h3>What is Molecular Orbital Theory?</h3>According to this theory,
- Molecular orbitals are formed by intermixing of atomic orbitals of two or more atoms having comparable energies
- The number of molecular orbitals formed is equal to the number of atomic orbitals combined.
- The shape of molecular orbitals formed depends on the type of atomic orbitals combined
- Only atomic orbitals having comparable energies and the same orientation can intermix
- Bonding M.O. is formed by the additive effect of atomic orbitals and thus, has lower energy and high stability.
- Antibonding M.O. is formed by the subtractive effect of atomic orbitals and thus, has higher energy and low stability.
- Bonding M.O. is represented by
while Antibonding M.O. is represented by
Molecular Orbital Diagram of C₂
Learn more about Molecular Orbital Theory:
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<u>Answer:</u> The mass percent of lead in lead (IV) carbonate is 63.32 %
<u>Explanation:</u>
The given chemical formula of lead (IV) carbonate is
To calculate the mass percentage of lead in lead (IV) carbonate, we use the equation:
Mass of lead = (1 × 207.2) = 207.2 g
Mass of lead (IV) carbonate = [(1 × 207.2) + (2 × 12) + (6 × 16)] = 327.2 g
Putting values in above equation, we get:
Hence, the mass percent of lead in lead (IV) carbonate is 63.32 %