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Answer:
The melting of the caps increases the concentration of CO2 in the water, making its absorption faster.
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Explanation:
<h2><u>CO2 absorption</u><u> </u></h2>
The polar caps of our planet have in their composition several elements, among them the CO2 that is absorbed by the atmosphere. Cold waters, which are present in the Arctic, have an easier time absorbing CO2 compared to other waters.
When glaciers melt, the CO2 that is present in the mixture is dissolved in the ocean, increasing its concentration. The cold waters that came from the ice caps increase the absorption of CO2.
Learn more about natural methods remove CO2 from the atmosphere in:
brainly.com/question/14323197
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Answer:
Because you can physically see the object melting when it comes to the melting point. The objects texture, color, temperature, shape, and state of matter (solid, liquid, gas) are possibly changing.
The answer for the following problem is described below.
<em><u> Therefore the standard enthalpy of combustion is -2800 kJ</u></em>
Explanation:
Given:
enthalpy of combustion of glucose(Δ of ) =-1275.0
enthalpy of combustion of oxygen(Δ of ) = zero
enthalpy of combustion of carbon dioxide(Δ of ) = -393.5
enthalpy of combustion of water(Δ of ) = -285.8
To solve :
standard enthalpy of combustion
We know;
Δ = ∈Δ (products) - ∈Δ (reactants)
(s) +6 (g) → 6 (g)+ 6 (l)
Δ = [6 (-393.5) + 6(-285.8)] - [6 (0) + (-1275)]
Δ = [6 (-393.5) + 6(-285.8)] - [0 - 1275]
Δ = 6 (-393.5) + 6(-285.8) - 0 + 1275
Δ = -2361 - 1714 - 0 + 1275
Δ =-2800 kJ
<em><u> Therefore the standard enthalpy of combustion is -2800 kJ</u></em>
Answer:
Between oxygen and selenium, oxygen is more electronegative because of more ionisation energy and small size
Explanation:
Answer:
320 g
Step-by-step explanation:
The half-life of Co-63 (5.3 yr) is the time it takes for half of it to decay.
After one half-life, half (50 %) of the original amount will remain.
After a second half-life, half of that amount (25 %) will remain, and so on.
We can construct a table as follows:
No. of Fraction Mass
half-lives t/yr Remaining Remaining/g
0 0 1
1 5.3 ½
2 10.6 ¼
3 15.9 ⅛ 40.0
4 21.2 ¹/₁₆
We see that 40.0 g remain after three half-lives.
This is one-eighth of the original mass.
The mass of the original sample was 8 × 40 g = 320 g