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Nesterboy [21]
3 years ago
12

a 15.0 kg chunk of ice falls off the top of an iceberg if the chunk of ice Falls 8.00 to the surface of the water what is the ki

netic energy of the chunk of ice when it hits the water and what is its velocity how do you work that out
Chemistry
1 answer:
kvv77 [185]3 years ago
8 0
When we can get the Kinetic energy from this formula KE= 1/2 M V^2and we can get the potential energy from this formula PE =  M g H
we can set that the kinetic energy at the bottom of the fall equals the potential energy at the top so, KE = PE
1/2 MV^2 = M g H
1/2 V^2 = g H
when V is the velocity, g is an acceleration of gravitational force (9.8 m^2/s) and H is the height of the fall (8 m).
∴ v^2 = 2 * 9.8 * 8 = 156.8
∴ v= √156.8 = 12.5 m/s
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what would be the final volume when 2.20 M solution is made from 25.0 mL of a 12.0 M solution? plzz show work
cupoosta [38]

Answer:

136.36 mL

Explanation:

Here we have to use the dilution formula

From C1V1= C2V2

Where;

C1= initial concentration of the solution= 12.0 M

C2= final concentration of the solution= 2.20 M

V1 = initial volume of the solution= 25.0 ml

V2= final volume of the solution= ?????

Then recall;

C1V1=C2V2

V2 = C1V1/C2

Substituting values from the parameters given;

V2= 12.0 × 25.0 / 2.20

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7 0
3 years ago
What amount (moles) of compound is present in 1.00 g of each of the compounds in Exercise 54?
Anika [276]

Answer:

1. Caffeine, C₈H₁₀N₄O₂

Amount = 1.00/194 = 0.00515 moles

2. Ethanol, C₂H₅OH

Amount = 0.0217 moles

3. Dry Ice, CO₂

amount = 0.0227 moles

<em>Note: The question is incomplete. The compound are as follows:</em>

<em> 1. Caffeine, C₈H₁₀N₄O₂;</em>

<em>2. Ethanol, C₂H₅OH;</em>

<em>3. Dry Ice, CO₂</em>

Explanation:

Amount (moles) = mass in grams /molar mass in grams per mole

1. Caffeine, C₈H₁₀N₄O₂

molar mass of caffeine = 194 g/mol

Amount = 1.00 g/194 g/mol = 0.00515 moles

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molar mass of ethanol = 46 g/mol

Amount = 1.00 g/46 g/mol = 0.0217 moles

3. Dry Ice, CO₂

molar mass of dry ice = 44 g/mol

amount = 1.00 g/44 g/mol = 0.0227 moles

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