Answer: Rubber source, temperature, thinkness, thread design, driving pattenrs, weather, etc.
Explanation: There are many variables. Here are a few I would include in a tire lifetime study:
1. Type of rubber, including source
2. Thickness of tire
3. Design of tire thread
4. Life as a function of average speed and road surface
5. Expected outside temperature and wet conditions
6. Driving conditions of speed and both acceleration and deceleration parameters (e.g., tire life when slamming on the brakes or accelerating quickly)
Answer:
Subbituminous coal can form at temperatures as low as 35 to 80 °C (95 to 176 °F) while anthracite requires a temperature of at least 180 to 245 °C (356 to 473 °F).
Sub-types: Cannel coal
Child material class: Lignite
Explanation:
Answer:
All offspring are tall when a homozygous tall parent with homozygous short parent.
Explanation:
When we crossed homozygous tall parent with homozygous short parent, we conclude that all offspring are tall, because homozygous short parent are supressed under the homozygous tall parent, due to law of dominance.
Law of dominance states that, recessive alleles are suppressed by dominant alleles but they can appear in F2 generation.
Using a punett square, we can predict the cross between homozygous tall and homozygous short parent.
The phenotypes are: All are tall plants (4:0).
Answer:
Explanation:
Ph is the major of acidity or basicity of a solution. On a scale of 0 to 14
PH = -log 10 ^(H+)
Therefore PH= -log 10^(6.5) =0.81
Therefore PH of the solution is 0.81
This solution is therefore an acidic solution
Answer:
Kc = 8.05x10⁻³
Explanation:
This is the equilibrium:
2NH₃(g) ⇄ N₂(g) + 3H₂(g)
Initially 0.0733
React 0.0733α α/2 3/2α
Eq 0.0733 - 0.0733α α/2 0.103
We introduced 0.0733 moles of ammonia, initially. So in the reaction "α" amount react, as the ratio is 2:1, and 2:3, we can know the moles that formed products.
Now we were told that in equilibrum we have a [H₂] of 0.103, so this data can help us to calculate α.
3/2α = 0.103
α = 0.103 . 2/3 ⇒ 0.0686
So, concentration in equilibrium are
NH₃ = 0.0733 - 0.0733 . 0.0686 = 0.0682
N₂ = 0.0686/2 = 0.0343
So this moles, are in a volume of 1L, so they are molar concentrations.
Let's make Kc expression:
Kc= [N₂] . [H₂]³ / [NH₃]²
Kc = 0.0343 . 0.103³ / 0.0682² = 8.05x10⁻³
