Answer:
D
Explanation:
It would definitely affect size because breeding will change that trait
Answer/ explanation :
Protist can be multicellular or unicellular organisms
Plants are all multicellular and also exhibit cellular differentiation.
Protist can be autotroph, heterotrophic or decomposer
Plants are only autotrophs because they manufacture their own food through photosynthesis
Protists are microscopic, more diverse and abundant in nature
Plants are big and complex in nature
Nuclear DNA strands in plants are of higher complexity than those of protist
Plants require oxygen for cellular respiration process unlike protist which can be aerobic and some other species facultative anaerobic
Plants only can reproduce asexually through bulbs and tubers as in yam, potatoes while protists reproduce either sexually through meiosis or asexually through simple cell division.
Answer:
strong enough to hold molecules relatively close together but not strong enough to keep molecules from moving past each other.
Explanation:
In liquids, the attractive intermolecular forces are <u>strong enough to hold molecules relatively close together but not strong enough to keep molecules from moving past each other</u>.
Intermolecular forces are the forces of repulsion or attraction.
Intermolecular forces lie between atoms, molecules, or ions. Intramolecular forces are strong in comparison to these forces.
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Explanation:
Hydrogen does not obey the octet rule. Boron does not always
obey the octet rule and in fact forms Lewis acids such as BF3 which
only has 6 electrons.
Answer:
1. d[H₂O₂]/dt = -6.6 × 10⁻³ mol·L⁻¹s⁻¹; d[H₂O]/dt = 6.6 × 10⁻³ mol·L⁻¹s⁻¹
2. 0.58 mol
Explanation:
1.Given ΔO₂/Δt…
2H₂O₂ ⟶ 2H₂O + O₂
-½d[H₂O₂]/dt = +½d[H₂O]/dt = d[O₂]/dt
d[H₂O₂]/dt = -2d[O₂]/dt = -2 × 3.3 × 10⁻³ mol·L⁻¹s⁻¹ = -6.6 × 10⁻³mol·L⁻¹s⁻¹
d[H₂O]/dt = 2d[O₂]/dt = 2 × 3.3 × 10⁻³ mol·L⁻¹s⁻¹ = 6.6 × 10⁻³mol·L⁻¹s⁻¹
2. Moles of O₂
(a) Initial moles of H₂O₂
(b) Final moles of H₂O₂
The concentration of H₂O₂ has dropped to 0.22 mol·L⁻¹.
(c) Moles of H₂O₂ reacted
Moles reacted = 1.5 mol - 0.33 mol = 1.17 mol
(d) Moles of O₂ formed
