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sashaice [31]
2 years ago
8

In summary, a heritable characteristic that helps an animal or plant to have more offspring which survive to reproduce will tend

to become more common in a population as a result of evolution by natural selection. PROMPT: Explain why a characteristic which helps an animal to live longer will generally tend to become more common in the population as a result of evolution by natural selection.
Chemistry
1 answer:
aleksklad [387]2 years ago
4 0

A characteristic which helps an animal to live longer will generally tend to become more common in the population because as the animal produces more offspring those traits are passed down to them. Eventually it will be a common characteristic in the population.

Hope that helped :)

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Use the pictures below to answer the following question what type of bond is this
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At STP how many moles or helium would occupy a volume of 12 liters?
butalik [34]
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?? mole ---------- 12 L

12 x 1 / 22.4 => 0.5357 moles

hope this helps! 
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3 years ago
What makes the great pyramid unusual among the places on antipater’s list?
Brut [27]

Answer : It is only one of those ancient sights which still exist and gives ancient antiquity to the great pyramid in Egypt. It is believed that the pyramids are the definition of prayers that still remains in the heart of the local people.

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7 0
3 years ago
The following reactions can be used to prepare samples of metals. Determine the enthalpy change under standard state conditions
mamaluj [8]

Answer:

a) 62.1 kJ/mol

b) 2.82 kJ/mol

c) 270.91 kJ/mol

d) -851.5 kJ/mol

Explanation:

The enthalpy change for a reaction in standard conditions (ΔH°rxn) can be calculated by:

ΔH°rxn = ∑n*ΔH°f, products - ∑n*ΔH°f, reagents

Where n is the number of moles in the stoichiometry reaction, and ΔH°f is the enthalpy of formation at standard conditions. ΔH°f = 0 for substances formed by only a single element. The values can be found in thermodynamics tables.

a) 2Ag₂O(s) → 4Ag(s) + O₂(g)

ΔH°f, Ag₂O(s) = -31.05 kJ/mol

ΔH°rxn = 0 - (2*(-31.05)) = 62.1 kJ/mol

b) SnO(s) + CO(g) → Sn(s) + CO₂(g)

ΔH°f,SnO(s) = -285.8 kJ/mol

ΔH°f,CO(g) = -110.53 kJ/mol

ΔH°f,CO₂(g) = -393.51 kJ/mol

ΔH°rxn = [-393.51] - [-110.53 - 285.8] = 2.82 kJ/mol

c) Cr₂O₃(s) + 3H₂(g) → 2Cr(s) + 3H₂O(l)

ΔH°f,Cr₂O₃(s) = -1128.4 kJ/mol

ΔH°f,H₂O(l) = -285.83 kJ/mol

ΔH°rxn = [3*(-285.83)] - [( -1128.4)] = 270.91 kJ/mol

d) 2Al(s) + Fe₂O₃(s) → Al₂O₃(s) + 2Fe(s)

ΔH°f,Fe₂O₃(s) = -824.2 kJ/mol

ΔH°f,Al₂O₃(s) = -1675.7 kJ/mol

ΔH°rxn = [-1675.7] - [-824.2] = -851.5 kJ/mol

3 0
3 years ago
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