<em>Answer:</em>
<em>Chemical properties:</em>
Those properties which change the chemical nature of matter.
<em>Example:</em>
- Heat of combustion
- Enthalpy of formation
<em>Physical properties:</em>
Those properties which do not change the chemical nature of matter.
<em>Example</em>
<em>Differences between chemical and physical properties:</em>
Chemical properties Physical properties
1. Observed after the change bringing 1. Observed with out being
the change change
2. These changes the molecules 2. only change physical state
3. Chemical identity changes 3.Chemical identity not changes
4. Structure of material changes 4.Structure of material not change
5. Chemical reaction is needed 5. No need of Chemical reaction
6. depend on composition 6. Does not depend on composition
Answer:
You can't ice skate on a liquid, when it is frozen it is a solid, when it's unfrozen it is a liquid
Answer:
1.4 mols
4th answer
Explanation:
22. 5 g of O2 in moles = (22.5/32) mols = 0.703 mol
The stoichiometry between O2 and H2O =1: 2
Therefore H2O produced = 2 * 0.703 mols=1.406 mols
Answer:
64.52 mg.
Explanation:
The following data were obtained from the question:
Half life (t½) = 1590 years
Initial amount (N₀) = 100 mg
Time (t) = 1000 years.
Final amount (N) =.?
Next, we shall determine the rate constant (K).
This is illustrated below:
Half life (t½) = 1590 years
Rate/decay constant (K) =?
K = 0.693 / t½
K = 0.693/1590
K = 4.36×10¯⁴ / year.
Finally, we shall determine the amount that will remain after 1000 years as follow:
Half life (t½) = 1590 years
Initial amount (N₀) = 100 mg
Time (t) = 1000 years.
Rate constant = 4.36×10¯⁴ / year.
Final amount (N) =.?
Log (N₀/N) = kt/2.3
Log (100/N) = 4.36×10¯⁴ × 1000/2.3
Log (100/N) = 0.436/2.3
Log (100/N) = 0.1896
Take the antilog
100/N = antilog (0.1896)
100/N = 1.55
Cross multiply
N x 1.55 = 100
Divide both side by 1.55
N = 100/1.55
N = 64.52 mg
Therefore, the amount that remained after 1000 years is 64.52 mg
