Answer:
Parenchyma is the most simple and specialized tissue which is concerned mainly with the vegetative activities of the plant. The cells are isodiametric with well-developed intercellular spaces, vacuolated cytoplasm and cellulosic cell wall.
Collenchyma is the tissue of the primary body. The cells of the tissue contain protoplasm and are living without intercellular spaces. The cell wall articulate at the corners and are made up of cellulose, hemicellulose, and pectin.
Sclerenchyma is the thick-walled cell tissue. In the beginning, the cell is living and have protoplasm, but due to deposition of impermeable secondary board lignin, they become dead thick and hard.
Carbon dioxide dissolves faster
The answers include:
- A loaf of risen but unbaked bread - chemical change.
- Photo by Elinor D - chemical change.
- Bread dough rising - chemical change.
<h3>What is a Chemical change?</h3>
This involves the formation of a new products from substances. In this scenario, a rising bread contains alcohol which evaporates.
Photographs also fall under this category and is therefore an irreversible chemical change.
Read more about Chemical change here
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The change in internal energy of the combustion of biphenyl in Kj is calculated as follows
=heat capacity of bomb calorimeter x delta T where delta T is change in temperature
delta T = 29.4 -25.8= 3.6 c
= 5.86 kj/c x 3.6 c = 21.096 kj
Answer:
ΔH for formation of 197g Fe⁰ = 1.503 x 10³ Kj => Answer choice 'B'
Explanation:
Given Fe₂O₃(s) + 2Al⁰(s) => Al₂O₃(s) + 2Fe⁰(s) + 852Kj
197g Fe⁰ = (197g/55.85g/mol) = 3.527 mol Fe⁰(s)
From balanced standard equation 2 moles Fe⁰(s) => 852Kj, then ...
3.527 mole yield (a higher mole value) => (3.527/2) x 852Kj = 1,503Kj (a higher enthalpy value).
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NOTE => If 2 moles Fe gives 852Kj (exo) as specified in equation, then a <u>higher energy value</u> would result if the moles of Fe⁰(s) is <u>higher than 2 moles</u>. The ratio of 3.638/2 will increase the listed equation heat value to a larger number because 197g Fe⁰(s) contains more than 2 moles of Fe⁰(s) => 3.527 mole Fe(s) in 197g. Had the problem asked for the heat loss from <u>less than two moles Fe⁰(s)</u> - say 100g Fe⁰(s) (=1.79mole Fe⁰(s)) - then one would use the fractional ratio (1.79/2) to reduce the enthalpy value less than 852Kj.
