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8090 [49]
3 years ago
9

HELP ASAP PLEASE!!! WILL MARK BRAINLIEST, ONLY ANSWER IF YOU'RE SURE

Chemistry
2 answers:
krok68 [10]3 years ago
7 0

1. dropping the ball converts potential energy PE to kinetic energy KE. when ball hits ground, it bounces back up. KE is converted back to PE at dat pt. cuz' ball bounces back almost up 2 ur hand, the PE is almost conserved.

2. energy changes with ball as it is accelerated by gravity. the increase in velocity converts ball's PE to its KE.

3. two changes in speed n position: first during drop of ball. speed from zero at hand postion to its max at ground position. second during the bounce back. speed from max at ground poisiton to zero at almost hand position.

GalinKa [24]3 years ago
5 0

1. As you drops a ball, its potential energy is converted into kinetic energy during its fall. It moves faster and faster as more PE is converted into KE until the ball hits the ground.

2. The conversion from PE to KE is due to the work done by gravity acceleration. As the ball is accelerated by gravity towards ground, its PE is converted to KE.

3. Total energy of the ball is conserved. Its position changes as it gets closer to the ground. Its PE decreases as its speed and hence KE increases until it hits the ground.

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What is the mass of H2O produced when 14.0 grams of H2 reacts completely with 2.0 grams of O2?
Vladimir79 [104]
Balanced chemical equation:

2 H2 + 1 O2 = 2 H2O

4 g H2  -------> 32 g O2 -----------> 36 g H2O
   ↓                       ↓                             ↓
14.0 g ---------> 2.0 g O2 ----------> mass H2O ?

32 * mass H2O = 2.0 * 36

32 * mass H2O = 72

mass of H2O = 72 / 32

mass of H2O = 2.25 g

hope this helps!.


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3 years ago
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cupoosta [38]

Answer:

c.boron-11

Explanation:

The atomic mass of boron is 10.81 u.

And 10.81 u is a lot closer to 11u than it is to 10u, so there must be more of boron-11.

To convince you fully, we can also do a simple calculation to find the exact proportion of boron-11 using the following formula:

(10u)(x)+(11u)(1−x)100%=10.81u

Where u is the unit for atomic mass and x is the proportion of boron-10 out of the total boron abundance which is 100%.

Solving for x we get:

11u−ux=10.81u

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x=0.19

1−x=0.81

And thus the abundance of boron-11 is roughly 81%.

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