Answer:
40.4 kJ
Explanation:
Step 1: Given data
- Heat of sublimation of CO₂ (ΔH°sub): 32.3 kJ/mol
Step 2: Calculate the moles corresponding to 55.0 g of CO₂
The molar mass of CO₂ is 44.01 g/mol.
n = 55.0 g × 1 mol/44.01 g = 1.25 mol
Step 3: Calculate the heat (Q) required to sublimate 1.25 moles of CO₂
We will use the following expression.
Q = n × ΔH°sub
Q = 1.25 mol × 32.3 kJ/mol = 40.4 kJ
Answer:
D.
Explanation:
D is the correct answer because, in aqueous solution, solvent is water and solute (in this example carbon dioxide CO₂) is a substance dissolved in water. The amount of solute that can be dissolved in a solvent depends of chemical composition, temperature and pressure
Answer:
Atoms must have similar electronegativities in order to share electrons in a covalent bond.
Explanation:
Covalent bonding is one of the bondings that occurs between the atoms of elements. It is the bonding in which atoms share their valence electrons with one another. However, the ELECTRONEGATIVITY, which is the ability of an atom to be attracted to electrons play a major role in the formation of covalent bonds.
When atoms of different electronegativities combine, the more electronegative atom pulls more electrons towards itself, hence, an IONIC bond is formed. However, when the electronegativities of the atoms are similar, the sharing of their electrons becomes stronger. Hence, ATOMS MUST HAVE SIMILAR ELECTRONEGATIVITIES in order to share electrons in a covalent bond.
It is in period 3
It is in group 17
It is a chlorine atom because it has 17 electrons which means the atomic number is 17
The isotope that is more abundant, given the data is isotope Li7
<h3>Assumption</h3>
- Let Li6 be isotope A
- Let Li7 be isotope B
<h3>How to determine whiche isotope is more abundant</h3>
- Molar mass of isotope A (Li6) = 6.02 u
- Molar mass of isotope B (Li7) = 7.02 u
- Atomic mass of lithium = 6.94 u
- Abundance of A = A%
- Abundance of B = (100 - A)%
Atomic mass = [(mass of A × A%) / 100] + [(mass of B × B%) / 100]
6.94 = [(6.02 × A%) / 100] + [(7.02 × (100 - A)) / 100]
6.94 = [6.02A% / 100] + [702 - 7.02A% / 100]
6.94 = [6.02A% + 702 - 7.02A%] / 100
Cross multiply
6.02A% + 702 - 7.02A% = 6.94 × 100
6.02A% + 702 - 7.02A% = 694
Collect like terms
6.02A% - 7.02A% = 694 - 702
-A% = -8
A% = 8%
Thus,
Abundance of B = (100 - A)%
Abundance of B = (100 - 8)%
Abundance of B = 92%
SUMMARY
- Abundance of A (Li6) = 8%
- Abundance of B (Li7) = 92%
From the above, isotope Li7 is more abundant.
Learn more about isotope:
brainly.com/question/24311846
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