Answer:
ΔT = Tfinal − Tinitial = 150°C − 35.0°C = 125°C
given the specific heat of iron as 0.108 cal/g·°C
heat=(100.0 g)(0.108 cal /g· °C )(125°C) =
100x 0.108x125= 1350 cal
<span>Jet streams are the major means of transport for weather systems. A jet stream is an area of strong winds ranging from 120-250 mph that can be thousands of miles long, a couple of hundred miles across and a few miles deep. Jet streams usually sit at the boundary between the troposphere and the stratosphere at a level called the tropopause. This means most jet streams are about 6-9 miles off the ground. Figure A is a cross section of a jet stream.
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The dynamics of jet streams are actually quite complicated, so this is a very simplified version of what creates jets. The basic idea that drives jet formation is this: a strong horizontal temperature contrast, like the one between the North Pole and the equator, causes a dramatic increase in horizontal wind speed with height. Therefore, a jet stream forms directly over the center of the strongest area of horizontal temperature difference, or the front. As a general rule, a strong front has a jet stream directly above it that is parallel to it. Figure B shows that jet streams are positioned just below the tropopause (the red lines) and above the fronts, in this case, the boundaries between two circulation cells carrying air of different temperatures.
Answer:
The calorimeter constant is = 447 J/°C
Explanation:
The heat absorbed or released (Q) by water can be calculated with the following expression:
Q = c × m × ΔT
where,
c is the specific heat
m is the mass
ΔT is the change in temperature
The water that is initially in the calorimeter (w₁) absorbs heat while the water that is added (w₂) later releases heat. The calorimeter also absorbs heat.
The heat absorbed by the calorimeter (Q) can be calculated with the following expression:
Q = C × ΔT
where,
C is the calorimeter constant
The density of water is 1.00 g/mL so 50.0 mL = 50.0 g. The sum of the heat absorbed and the heat released is equal to zero (conservation of energy).
Qabs + Qrel = 0
Qabs = - Qrel
Qcal + Qw₁ = - Qw₂
Qcal = - (Qw₂ + Qw₁)
Ccal . ΔTcal = - (cw . mw₁ . ΔTw₁ + cw . mw₂ . ΔTw₂)
Ccal . (30.31°C - 22.6°C) = - [(4.184 J/g.°C) × 50.0 g × (30.31°C - 22.6°C) + (4.184 J/g.°C) × 50.0 g × (30.31°C - 54.5°C)]
Ccal = 447 J/°C
True because molo means two so you would need more than 1 atom
Answer:
c. decarboxylation of an a-keto acid.
Explanation:
Decarboxylation refers to the removal of the carboxyl group from a carboxylic acid and thus releasing carbon dioxide. Decarboxylases are enzymes that speed up the removal of the carboxyl group from acids. These reactants could be amino acids, alpha-keto acids, and beta-keto acids. Biotin is known to catalyze the decarboxylation of malonyl CoA to acetyl CoA during fatty acid synthesis.
Malonyl CoA is converted to acetyl CoA after decarboxylation assisted by biotin also known as Vitamin H. Alpha keto acids are involved in fatty acids synthesis and Malonyl CoA is an alpha-keto acid because the keto group is located in the first carbon near the carboxylic acid group. Keto acids have both a carboxyl group and a ketone group.
