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laiz [17]
3 years ago
7

If a hockey puck is made to slide along a straight line on a completely frictionless surface, which statement about the motion o

f the hockey puck is true? The hockey puck will come to a halt immediately. The hockey puck will continue moving with constant velocity along a straight line. The hockey puck will accelerate and continue moving in straight line. The hockey puck will continue moving in random direction. NextReset
Chemistry
2 answers:
JulsSmile [24]3 years ago
3 0
The hockey puck will continue in the same speed in a continues straight line
Art [367]3 years ago
3 0

Answer:

The hockey puck will continue moving with constant velocity along a straight line.

Explanation:

Friction decelerates the motion of a moving body. In the absence of friction on smooth surface an object in moving with uniform velocity will continue to move in a straight line until acted upon by some force. A rough surface has a higher friction and it will decelerate a moving object such that the object will not be able to travel very far in a uniform velocity but quickly come to halt.

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Would u die eating a full spoon of salt whe you are dehydrated?
zysi [14]
Yes, but u wouldn't die if u weren't dehyrated
3 0
3 years ago
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50 POINTS! PLEASE HELP!A gas in a balloon at constant pressure has a volume of 120.0mL at -12.30C. What is its volume at 197.00C
Cerrena [4.2K]

Answer:

Final volume=V₂ = 216.3 mL

Explanation:

Given data:

Initial volume = 120.0 mL

Initial temperature = -12.3 °C (-12.3 +273 = 260.7 K)

Final volume = ?

Final temperature = 197.0 °C (197+273 = 470 K)

Solution:

We will apply Charles Law to solve the problem.

According to this law, The volume of given amount of a gas is directly proportional to its temperature at constant number of moles and pressure.

Mathematical expression:

V₁/T₁ = V₂/T₂

V₁ = Initial volume

T₁ = Initial temperature

V₂ = Final volume  

T₂ = Final temperature

Now we will put the values in formula.

V₁/T₁ = V₂/T₂

V₂ = V₁T₂/T₁

V₂ = 120 mL × 470 K /260.7K

V₂ = 56400 mL.K /260.7K

V₂ = 216.3 mL

4 0
3 years ago
2H2 + O2 → 2H2O
pantera1 [17]
Okay
Mr (H2O)= 18g
therefore moles of H2O
is 720.8/18= 40.04mol
the ratio of H2 to O2 to H2O is
2 : 1 : 2
so moles of H2 is same as H2O here
H2= 40.04moles

moles of O2 is half
so 40.04 x 0.5
20.02moles

grams of O2 is
its moles into Mr of O2
that's 20.02 x 32 = 640.64g

6 0
2 years ago
Write the balanced chemical equation for the reaction of glucose (C6,H12,O6) with oxygen gas to produce carbon dioxide gas
Fittoniya [83]

Answer:

C6H12O6+6O2=6Co2+6H2o

4 0
2 years ago
Consider a general reaction
choli [55]

Answer:

a) K = 5.3175

b) ΔG = 3.2694

Explanation:

a) ΔG° = - RT Ln K

∴ T = 25°C ≅ 298 K

∴ R = 8.314 E-3 KJ/K.mol

∴ ΔG° = - 4.140 KJ/mol

⇒ Ln K = - ( ΔG° ) / RT

⇒ Ln K = - ( -4.140 KJ/mol ) / (( 8.314 E-3 KJ/K.mol )( 298 K ))

⇒ Ln K = 1.671

⇒ K = 5.3175

b) A → B

∴ T = 37°C = 310 K

∴ [A] = 1.6 M

∴ [B] = 0.45 M

∴ K = [B] / [A]

⇒ K = (0.45 M)/(1.6 M)

⇒ K = 0.28125

⇒ Ln K = - 1.2685

∴ ΔG = - RT Ln K

⇒ ΔG = - ( 8.314 E-3 KJ/K.mol )( 310 K )( - 1.2685 )

⇒ ΔG = 3.2694

7 0
3 years ago
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