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Cerrena [4.2K]
3 years ago
13

When exchanging information with anyone involved in the collision, you should

Chemistry
1 answer:
postnew [5]3 years ago
7 0

Answer:

  • Try to be as relax as possible.
  • Provide names of all parties involved.
  • Provide vehicle information and identification details.
  • Provide full names, address, registration numbers and insurance company details.

Explanation:

After a collision one may be confused, afraid and have no attention about the details that what happened because all the collision event happens in a short interval of time. So the first thing one should do during information exchange is to sit back and relax and be calm so that one can remind the things at some extent. After that provide all the details about injured people and the involved vehicles.

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solong [7]
Hey there!

Rocks and sand are nonliving. All organisms are living.
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Hope this helps! Have a Brainly day! :D
5 0
3 years ago
A chemical reaction occurs between Calcium, Oxygen, and Chlorine. The reaction is demonstrated:
ad-work [718]

Answer:

D.

Double replacement, CaO + Cl2O

Explanation:

8 0
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Read 2 more answers
A 0.150-kg sample of a metal alloy is heated at 540 Celsius an then plunged into a 0.400-kg of water at 10.0 Celsius, which is c
Zarrin [17]

Answer:

C_{alloy}=0.497\frac{J}{g\°C}

Explanation:

Hello there!

In this case, according to this calorimetry problem on equilibrium temperature, it is possible for us to infer that the heat released by the metal allow is absorbed by the water for us to write:

Q_{allow}=-(Q_{water}+Q_{Al})

Thus, by writing the aforementioned in terms of mass, specific heat and temperature, we have:

m_{alloy}C_{alloy}(T_{eq}-T_{alloy})=-(m_{water}C_{water}(T_{eq}-T_{water})+m_{Al}C_{Al}(T_{eq}-T_{Al})

Then, we solve for specific heat of the metallic alloy to obtain:

C_{alloy}=\frac{-(m_{water}C_{water}(T_{eq}-T_{water})+m_{Al}C_{Al}(T_{eq}-T_{Al})}{m_{alloy}(T_{eq}-T_{alloy})}

Thereby, we plug in the given data to obtain:

C_{alloy}=\frac{-(400g*4.184\frac{J}{g\°C} (30.5\°C-10.0\°C)+200g*0.900\frac{J}{g\°C}(30.5\°C-10.0\°C)}{150g(30.5\°C-540\°C)} \\\\C_{alloy}=0.497\frac{J}{g\°C}

Regards!

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3 years ago
Please help me with letter a question
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28: Which of the following alloys does not contain copper? 1.Duralumin 2. Bronze 3.Steel 4.Brass 5.Solder​
Alja [10]

Answer:

<h2>5.Solder........</h2><h3>Hope it helps you!!</h3>
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