Answer: The distance is slightly less than 3.5 m
Explanation: assuming wall and target are the same thing, and the bullet has constant velocity, the bullet will travel 7 m in half a second, so half that distance is 3.5 m.
In reality, the bullet is decelerating (at an unknown rate) so the distance is slightly less than 3.5 m.
There is also a vertical velocity component, which means it hits the target/wall at an angle. The trajectory is such that it hits the wall above the shooter because the ricochet hits at ~the level at which it left the firearm.
If the wall was absent, the bullet would have described a parabola which brough it back to the initial level after 7m. This could be calculated, but it means that the actual distance between the shooter and the wall is slightly less than 3.5 m
In addition, the collision with the wall is not 100% elastic, so the velocity aftercthe ricochetvis further reduced.
A calculation would be complex because these confounding factors are not completely independent of each other, but all reduce the average velocity and therefore the distance.
Therefore it is only possible to say that the distance was somewhat less than 3.5 m
We have to first write a balanced equation.
so2 + o2 -> so3
this is not balanced though. we have 3 oxygen on right and 4 on left
2so2 + o2 -> 2so3
now it is same on both sides. we have to figure out which is limiting reagent with the given amounts of reagents. we do this by comparing the ratio between them in terms of moles. we see that so2 has a coefficient of 2 and o2 has none which implies 1 and so3 has 2. this means that for every 2 moles of so2 reacting with 1 mole of o2, we get 2 moles of so3.
lets convert the given values to moles. to do this we know that molecular weight is measured in grams per mole. we are given grams and need to cancel out the grams to get moles. so the molecular weight:
so2 =32.1 + 2 * 16 = 64.1 g/mol
o2 = 2 * 16 = 32 g/mol
so3 = 32.1 + 3 * 16 = 80.1 g/mol
now to convert 90 g of 2so2 under ideal conditions.
90g / 64.1g/mol = 1.404 moles
convert this amount of moles of so2 to moles of o2. we have 2 moles of so2 to 1 of o2
1.404moles so2 / 2 moles so2 * 1 mole o2= 0.702 moles o2
so we see under ideal conditions that 90g of so2 would react with .702g of o2. lets see how many we actually have with 100g of o2
100g / 32g/mol =3.16 mol.
so we have a lot more o2 than needed. we are looking for how much is left in grams. we have to figure out how much was used. to do this convert our ideal moles of o2 into grams.
.702 moles o2 * 32g/mol = 22.5g o2
so what we startrd with (100g) minus what we needed (22.5g) is what we have left
100 - 22.5 = 77.5g o2
Answer: Very small fragments of solid materials or liquid droplets suspended in air are called particulates. ... For example, solid particulates between roughly 1 and 100 μm in diameter are called dust particles, whereas airborne solids less than 1 μm in diameter are called fumes.
Brainliest would be nice!
Explanation:
