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Akimi4 [234]
3 years ago
6

**Help!! ** How do I do this question

Chemistry
1 answer:
swat323 years ago
5 0
The answer is D. Okay l hope this helps
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What is the potential energy of a 2 kg plant that is on a windowsill 1.3 m high
gulaghasi [49]

Answer:

P.E = 25.48 J

Explanation:

Given data:

Mass = 2 Kg

Height = 1.3 m

Potential energy = ?

Solution:

Formula:

P.E = m . g . h

P. E = potential energy

m = mass in kilogram

g = acceleration due to gravity

h = height

Now we will put the values in formula.

P.E = m . g . h

P.E = 2 Kg . 9.8 m /s² . 1.3 m

P.E = 25.48 Kg. m² / s²

 Kg. m² / s² = J

P.E = 25.48 J

8 0
3 years ago
Read 2 more answers
When a car engine burns gasoline, the results of the reaction are similar to when cells burn glucose. both reactions release car
-Dominant- [34]

the answer is heat. while a car is in idol, the tailpipe gets very hot, (motorcycle, car, etc.) this also produces h20 which you can see dripping out of the tailpipe.

3 0
3 years ago
Hi! Which of the following decreases kinetic energy of solvent molecules? ( I need this handed in quickly!)
Black_prince [1.1K]

Answer:

c

Explanation:

4 0
3 years ago
What is the change in boiling point for a 0.615m solution of Mgl2 in water?
ivanzaharov [21]

Answer :  The change in boiling point is, 0.94^oC

Explanation :

Formula used :

\Delta T_b=i\times K_f\times m

where,

\Delta T_b = change in boiling point = ?

i = Van't Hoff factor = 3 (for MgI₂ electrolyte)

K_f = boiling point constant for water = 0.51^oC/m

m = molality  = 0.615 m

Now put all the given values in this formula, we get

\Delta T_b=3\times (0.51^oC/m)\times 0.615m

\Delta T_b=0.94^oC

Therefore, the change in boiling point is, 0.94^oC

4 0
3 years ago
A sample of hydrogen occupies a volume of 351 mL at a temperature of 20 degrees Celsius. What is the new volume if the temperatu
xeze [42]
Volume of Hydrogen V1 = 351mL 
Temperature T1 = 20 = 20 + 273 = 293 K 
Temperature T2 = 38 = 38 + 273 = 311 K 
We have V1 x T2 = V2 x T1 
So V2 = (V1 x T2) / T1 = (351 x 311) / 293 = 372.56 
Volume at 38 C = 373 ml
6 0
3 years ago
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