Answer:
The heat absorbed by the sample of water is 3,294.9 J
Explanation:
Calorimetry is the measurement and calculation of the amounts of heat exchanged by a body or a system.
The sensible heat of a body is the amount of heat received or transferred by a body when it undergoes a temperature variation (Δt) without there being a change of physical state (solid, liquid or gaseous). Its mathematical expression is:
Q = c * m * ΔT
Where Q is the heat exchanged by a body of mass m, made up of a specific heat substance c and where ΔT is the temperature variation.
In this case:
- Q=?
- m= 45 g
- c= 4.184
- ΔT= Tfinal - Tinitial= 38.5 C - 21 C= 17.5 C
Replacing:
Q= 4.184 * 45 g* 17.5 C
Solving:
Q=3,294.9 J
<u><em>The heat absorbed by the sample of water is 3,294.9 J</em></u>
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I believe it is A because the plant is offering protection while the bacteria is converting it in to nitrogen that the plant can use!
Answer:
Lysosome contains hydrolytic enzymes associated with the intracellular digestion of macromolecules.
Explanation:
Lysosome is found in all types of eukaryotic cells, and it is responsible for the digestion of old cell parts, micro-organism and macromolecules. Lysosome is surrounded by a membrane which maintains the interior acidic environment with the help of proton pump.
Lysosome contains unique membrane proteins and wide variety of hydrolytic enzymes such as acid hydrolyses that are helping in the breakdown of macromolecules (proteins, nucleic acid and polysaccharides). Lysosome acid dependent activity helping to protect the cell from self degradation in the situation of lysosomal rupture or leakage, while the pH of the cell is neutral to slightly alkaline.
Answer:
60 grams of ice will require 30.26 calories to raise the temperature 1°C.
Explanation:
The amount of heat (Q) to raise the temperature of 60.0 g of ice by 1°C can be calculated from:
<em>Q = m.c.ΔT,</em>
where, Q is the amount of heat released or absorbed by the system.
m is the mass of the ice (m = 60.0 g).
c is the specific heat capacity of ice (c = 2.108 J/g.°C).
ΔT is the temperature difference (ΔT = 1.0 °C).
∴ Q = m.c.ΔT = (60.0 g)(2.108 J/g.°C)(1.0 °C) = 126.48 J.
<em>It is known that 1.0 cal = 4.18 J.</em>
<em>∴ Q = (126.48 J)(1.0 cal / 4.18 J) = 30.26 cal.</em>
They are called precipitates.
