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fgiga [73]
3 years ago
5

What is 8 correct things you do in the science LAB

Chemistry
2 answers:
liraira [26]3 years ago
7 0
1. Conduct yourself in a responsible manner at all times in the laboratory.

<span>2. Follow all written and verbal instructions carefully.  If you do not understand a direction or part of a procedure, ASK YOUR TEACHER BEFORE PROCEEDING WITH THE ACTIVITY.

</span><span>3. Never work alone in the laboratory.  No student may work in the science classroom without the presence of the teacher.

</span>4. When first entering a science room, <span>do not touch any equipment, chemicals, or other materials in the laboratory area until you are instructed to do so.

</span><span>5. Perform only those experiments authorized by your teacher.  Carefully follow all instructions, both written and oral.  Unauthorized experiments are not allowed.

</span>6<span>.<span> Do not eat food, drink beverages, or chew gum in the laboratory.</span>  Do not use laboratory glassware as containers for food or beverages.

</span>7<span>. Be prepared for your work in the laboratory.  Read all procedures thoroughly before entering the laboratory.  Never fool around in the laboratory.  Horseplay, practical jokes, and pranks are dangerous and prohibited. 
</span>

<span><span>8.<span> Always work in a well-ventilated area.  </span></span>
</span>

<span>9. Observe good housekeeping practices.  <span>Work areas should be kept clean and tidy at all times. 


</span></span>
Rina8888 [55]3 years ago
3 0
1. put goggles on
2.but gloves on
3.put a lab coat on

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A 20.0 g piece of aluminum at 5.00 C is dropped into 20.2 g of water at 90.00 C. The final temperature is 75.00 C. Use the First
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Answer:

The specific heat of aluminium is 0.906 J/g°C

Explanation:

Step 1: data given

Mass of aluminium = 20.0 grams

Temperature = 5.00 °C

Mass of water = 20.2 grams

Temperature of water = 90.00 °C

The final temperature = 75.00 °C

Specific heat of water = 4.184 J/g°C

Step 2: calculate the specific heat of aluminium

heat won = heat lost

Qaluminium = -Qwater

Q = m*c* ΔT

m(aluminium * c(aluminium) *ΔT(aluminium = -m(water) * c(water) *ΔT(water)

⇒with m(aluminium) = mass of aluminium = 20.0 grams

⇒with c(aluminium) = the specific heat of aluminium = TO BE DETERMINED

⇒with ΔT(aluminium) = the change of temperature = T2 - T1 = 75.00 °C - 5.00 °C = 70.00 °C

⇒with m(water) = the mass of water = 20.2 grams

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

⇒with ΔT(water) = T2 - T1 = 75.00°C - 90.00 °C = -15.00 °C

20.0 * c(aluminium) * 70.00 = -20.2 * 4.184 * -15.00

c(aluminium) = 0.906 J/g°C

The specific heat of aluminium is 0.906 J/g°C

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