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Sedaia [141]
3 years ago
11

Ben and Jerry were testing powders in science lab. First they put the powder in a zip lock bag. Next they added some water. The

powder fizzed and bubbled; soon it disappeared. When the bubbling stopped the bag contained a clear liquid and a lot of gas.
What is the MOST OBVIOUS sign of a chemical change in this reaction?
Chemistry
2 answers:
ch4aika [34]3 years ago
6 0
The powder fizzled and bubbled; soon it disappeared.
fgiga [73]3 years ago
3 0
The most obvious one is that when the powder came into contact with the water it fizzed.
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Help please :&lt;<br> I just need the answer for this as quickly as possible.
11Alexandr11 [23.1K]

Answer:

I think its the 3th one

Explanation:

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A student is trying to calculate the density of a ball. She already knows the mass, but she needs to determine the volume as wel
murzikaleks [220]

Answer:

V equals four-thirds times pi times r cubed

Explanation:

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7 0
3 years ago
Calculate the energy that is required to change 50.0 g ice at -30.0°C to a liquid at 73.0°C. The heat of fusion = 333 J/g, the h
OverLord2011 [107]

Answer:

There is 3.5*10^4 J of energy needed.

Explanation:

<u>Step 1:</u> Data given

Mass of ice at -30.0 °C = 50.0 grams

Final temperature = 73.0 °C

The heat of fusion = 333 J/g

the heat of vaporization = 2256 J/g

the specific heat capacity of ice = 2.06 J/gK

the specific heat capacity of liquid water = 4.184 J/gK

<u>Step 2:</u> Calculate the heat absorbed by ice

q = m*c*(T2-T1)

⇒ m = the mass of ice = 50.0 grams

⇒ c = the heat capacity of ice = 2.06 J/gK = 2.06 J/g°C

⇒ T2 = the fina ltemperature of ice = 0°C

⇒ T1 = the initial temperature of ice = -30.0°C

q = 50.0 * 2.06 J/g°C * 30 °C

q = 3090 J

<u>Step 3:</u> Calculate heat required to melt the ice at 0°C:

q = m*(heat of fusion)

q = 50.0* 333J/g

q =  16650 J

<u> </u>

<u>Step 4</u>: Calculate the heat required to raise the temperature of water from 0°C to 73.0°C

q = m*c*(T2-T1)

 ⇒ mass = 50.0 grams

⇒ c = the specific heat of water = 4.184 J/g°C

⇒ ΔT = T2-T1 = 73.0 - 0  = 73 °C

q = 50.0 * 4.184 * 73.0 = 15271.6 J

<u>Step 5:</u> Calculate the total energy

qtotal = 3090 + 16650 + 15271.6 = 35011.6 J = 3.5 * 10^4 J

There is 3.5*10^4 J of energy needed.

8 0
3 years ago
Taking a test please help. Thank you.
Komok [63]
It’s the 3rd one obviously bro
3 0
3 years ago
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