Answer:
1. 0.178 moles ; 2. 8x10²³ atoms ; 3. 7.22x10²³ molecules ; 4. 89.6 g ; 5. 1.34x10²² atoms ; 6. 1.67x10²⁵ molecules
Explanation:
1. Mass / Molar mass = Mol
5g / 28 g/m = 0.178 moles
2. 1 molecule of N₂ has 2 atoms, it is a dyatomic molecule.
4x10²³ x2 = 8x10²³ atoms
3. 1 mol of anything, has 6.02x10²³ particles
6.02x10²³ molecules . 1.2 mol = 7.22x10²³
4. 1 atom of C weighs 12 amu.
4.5x10²⁴ weigh ( 4.5x10²⁴ . 12) = 5.24x10²⁵ amu
1 amu = 1.66054x10⁻²⁴g
5.24x10²⁵ amu = (5.24x10²⁵ . 1.66054x10⁻²⁴) = 89.6 g
5. Molar mass NaCl = 58.45 g/m
1.3 g / 58.45 g/m = 0.0222 moles
1 mol has 6.02x10²³ atoms
0.0222 moles → ( 0.0222 . 6.02x10²³) = 1.34x10²²
6. Density of water is 1 g/mL, so 500 mL are contained in 500 g of water
Molar mass H₂O = 18 g/m
500 g / 18 g/m = 27.8 moles
6.02x10²³ molecules . 27.8 moles = 1.67x10²⁵
-58 °C
The melting point is the same as the freezing point.
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:
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Answer:
Heat is the transfer of energy that results from the difference in temperature between a system and its surroundings. At a molecular level, heat is the transfer of energy that makes use of or stimulates disorderly molecular motion in the surroundings.
Explanation:
