Study the images of two different fronts. Item A: Arrows for a cold front and warm front meet with high level, grey clouds. Item B: Arrows for a cold front and warm front are moving in the same direction. The warm front is above the cold front.

Based on arrow movement, why is the front shown in item A different from the front shown in item B?

Warm air rises above cold air in item A, while two air masses collide in item B.
Cold air mixes with warm air in item A, while two air masses barely move in item B.
Warm air is pushed upward near cold air in item A, while a boundary forms in item B.
Two air masses form a boundary in item A, while warm air rises above cold air in item B.

Problem #3 (ch. 1, problem 15)the ideal gas law provides one way to estimate the pressure exerted by a gas on a container. the law isí‘ťí‘ť=푛푛푛푛푛푛푉푉more accurate estimates can be made with the van der waals equationí‘ťí‘ť=푛푛푛푛푛푛푉푉â’푛푛푟푟â’푞푞푛푛2푉푉2where the term nb is a correction for the volume of the molecules and the term an2/v2is a correction for molecular attractions. the values of a and b depend on the type of gas. the gas constant is r, the absolutetemperature is t, the gas volume is v, and the number of moles of gas molecules is indicated by n. if n = 1 mol of an ideal gas were confined to a volume of v = 22.41 l at a temperature of 0â°c (273.2k), it would exert a pressure of 1 atm. in these units, r = 0.0826.for chlorine gas (cl2), a = 6.49 and b = 0.0562. compare the pressure estimates given by the ideal gas law and the van der waals equation for 1 mol of cl2 in 22.41 l at 273.2 k. what is the main cause of the difference in the two pressure estimates, the molecular volume or the molecular attractions?

You have 125g of a certain seasoning and are told that it contains 76.0 g of salt what is the percentage of salt by mass in this seasoning

Non renewable resources like petroleum eventually