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What is the mass of CO2 gas inside a 7.949 L glass bulb at a pressure of 2.906 x 10-1 ATM at 52 degrees C?
Wiki User
∙ 11y ago
Use the Ideal Gas law ... PV = nRT
n = number of moles. Temp must be kelvin not celsius. R is the ideal gas constant
so ... n = PV/(RT)
Substitute in appropriately and get an answer for # of moles.
CO2 has a molar mass of 44
multiply moles by 44 to find mass in grams.
Wiki User
∙ 11y ago
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Why is maths and physics so important in space exploration?
The problem with space is the distance between things.If the Sun was 1 meter in diameter, at their closestapproach to Earth the planets would be far away. Mercury would be about 51 meters from Earth. Venus would be about 28 meters away. Mars would be about 52 meters away. Those distances would be far enough, but when you look at the distances to the other places we might want to explore, the numbers are far worse. Jupiter would be 400 meters away. Saturn would be 953 meters away. Uranus would be 1819 meters away. Neptune would be 2906 meters away, and Pluto would be 3900 meters away. Those distances are as close as they ever get. If they are on the other side of the Sun, add another 99 meters.Also, these objects are moving. Even if you can see them, you are seeing them only at the speed of light. For example when you look at Mars, you are seeing it where it was 14 minutes ago. So, to explore a place, you have to calculate where it will be at the moment our explorer gets out there to it. Exactly. That takes very good math. To move an object requires energy. Now you are dealing with Physics. If you are going to send a 100 kilogram object to Mars, you need less fuel than if you are going to send a 1,000 kilogram object. How much more? Ask Physics.One example: to do space exploration you have to get from A to B. Traveling in the solar system you don't just point at the target and step on the gas. A transfer orbit is needed, and calculating orbits is a job for Physics and Math.
Assuming no heat losst by the system what will be the final temp. when 11.2 g of water at 10.0 deg. C are mixed with 50.8 g of water at 55.0 deg C?
you can't. You can calculate the heat released, and then convert that. Hmmm...how about: 11.2*10=112 gC 50.8*55=2794 gC 112+2794 = 2906gC 2906/(11.2+50.8)= 46.87C Quick check: 10C < 46.87C < 55C the answer is also closer to 55C because of the greater mass (almost 5x) of the hotter water. Anyone see that as correct? :)
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