<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:
partial pressure of gas D Pd = 15.5 kPa
Explanation:
As per the Dalton's law of partial pressure, in a mixture, pressure exerted by each gas when summed gives the total partial pressure exerted by mixture.
P(Total) = P1+P2+P3.....
Given P(Total) = 35.7 kPa
Partial pressure of gas A Pa = 7.8 kPa
Partial pressure of gas B Pb = 3.7 kPa
Partial pressure of gas C Pc = 8.7 kPa
There, Partial pressure of gas D Pd = P(Total) -(Pa+Pb+Pc)
Pd = 35.7-(7.8+3.7+8.7) = 35.7-20.2 kPa = 15.5 kPa
Therefore, partial pressure of gas D Pd = 15.5 kPa
Chemical reaction: Ba(NO₃)₂ + H₂SO₄ → BaSO₄ + 2HNO₃.
V(H₂SO₄) = 250 mL ÷ 1000 mL/L = 0,25 L.
m(BaSO₄) = 0,55 g.
n(BaSO₄) = m(BaSO₄) ÷ M(BaSO₄).
n(BaSO₄) = 0,55 g ÷ 233,38 g/mol.
n(BaSO₄) = 0,00235 mol.
From chemical reaction: n(BaSO₄) : n(Ba(NO₃)₂) = 1 : 1.
n(Ba(NO₃)₂) = 0,00235 mol.
c(Ba(NO₃)₂) = n(Ba(NO₃)₂) ÷ V.
c(Ba(NO₃)₂) = 0,00235 mol ÷ 0,25 L.
c(Ba(NO₃)₂) = 0,0095 mol/L.
Doubling the amplitude of a wave means that you increase the energy of a wave factor of quadrupling
