Proton mass = 1.672 x 10^-27 kilo. So the approximate mass of a proton would be 1 g or kg.
Answer:
The mass of 10 cm³of a 0.4 g/dm³ solution of sodium carbonate is 0.004 grams
Explanation:
The question is with regards to density calculations
The density of the given sodium carbonate solution, ρ = 0.4 g/dm³
The volume of the given solution of sodium carbonate, V = 10 cm³ = 0.01 dm³
Therefore, we have;
The mass, "m", of the sodium carbonate in = ρ×V = 0.4 g/dm³ × 0.01 dm³ = 0.004 g
The mass of 10 cm³ (10 cm³ = 0.01 dm³) of a 0.4 g/dm³ solution of sodium carbonate, m = 0.004 g.
Answer:
An amide may be produced by reacting an acid chloride with ammonia.
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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Ca(s)+2Hcl(aq) ------>CaCl2(s)+H2(g)
