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Amiraneli [1.4K]
3 years ago
10

What is the kinetic energy of a 1.0-kg billiard ball that moves at 5.0 m/s?

Chemistry
1 answer:
bekas [8.4K]3 years ago
8 0
Given,
          m = 1 kg
          v = 5 m/s 
We know that 
              K.E.= mv²/2
                     =1×5²/2
                     =25/2
                     =12.5 Joules
The answer is 12.5 Joules
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0.90 g of sodium hydroxide ( NaOH ) pellets are dissolved in water to make 3.0 L of solution. What is the pH of this solution
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Answer:

Explanation:

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2 years ago
Explain obser made when flourine is bubbled in bromine water for sometime
lisabon 2012 [21]

Explanation:

Alkenes react in the cold with pure liquid bromine, or with a solution of bromine in an organic solvent like tetrachloromethane. The double bond breaks, and a bromine atom becomes attached to each carbon. The bromine loses its original red-brown color to give a colorless liquid. In the case of the reaction with ethene, 1,2-dibromoethane is formed.

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3 years ago
A student constructs a coffee cup calorimeter and places 50.0 mL of water into it. After a brief period of stabilization, the te
Grace [21]

Answer:

The calorimeter constant is  = 447 J/°C

Explanation:

The heat absorbed or released (Q) by water can be calculated with the following expression:

Q = c × m × ΔT

where,

c is the specific heat

m is the mass

ΔT is the change in temperature

The water that is initially in the calorimeter (w₁) absorbs heat while the water that is added (w₂) later releases heat. The calorimeter also absorbs heat.

The heat absorbed by the calorimeter (Q) can be calculated with the following expression:

Q = C × ΔT

where,

C is the calorimeter constant

The density of water is 1.00 g/mL so 50.0 mL = 50.0 g. The sum of the heat absorbed and the heat released is equal to zero (conservation of energy).

Qabs + Qrel = 0

Qabs = - Qrel

Qcal + Qw₁ = - Qw₂

Qcal = - (Qw₂ + Qw₁)

Ccal . ΔTcal = - (cw . mw₁ . ΔTw₁ + cw . mw₂ . ΔTw₂)

Ccal . (30.31°C - 22.6°C) = - [(4.184 J/g.°C) × 50.0 g × (30.31°C - 22.6°C) +  (4.184 J/g.°C) × 50.0 g × (30.31°C - 54.5°C)]

Ccal  = 447 J/°C

5 0
3 years ago
The burning of a sample of propane generated 1 04.6 kJ of heat. All of this heat was used to heat 500.0 g of water that had an i
Paul [167]

Answer: 70.0°C

Explanation:

Quantity of heat = Mass * Specific heat * Change in temperature

Quantity of heat = 104.6 KJ

Mass = 500.0 g

Specific heat of water is 4.18 J/g°C

Change in temperature assuming final temperature is x = x - 20

Units should be in grams and joules:

104,600 = 500 * 4.18 * (x - 20)

104,600 = 2,090 * (x - 20)

x - 20 = 104,600/2,090

x = 104,600/2,090 + 20

x = 69.8

= 70.0°C

3 0
3 years ago
How many moles are 21.67 L of NH4CI?<br> Type your answer...
NNADVOKAT [17]

Answer:

0.967mole

Explanation:

Given parameters:

Volume of NH₄Cl  = 21.67L

Unknown:

Number of moles  = ?

Solution:

We assume that the volume was taken at standard temperature and pressure,

 Then;

  Number of moles  = \frac{volume }{22.4L}  

 Number of moles  = \frac{21.67}{22.4}   = 0.967mole

6 0
3 years ago
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