Answer:
0.296 J/g°C
Explanation:
Step 1:
Data obtained from the question.
Mass (M) =35g
Heat Absorbed (Q) = 1606 J
Initial temperature (T1) = 10°C
Final temperature (T2) = 165°C
Change in temperature (ΔT) = T2 – T1 = 165°C – 10°C = 155°C
Specific heat capacity (C) =..?
Step 2:
Determination of the specific heat capacity of iron.
Q = MCΔT
C = Q/MΔT
C = 1606 / (35 x 155)
C = 0.296 J/g°C
Therefore, the specific heat capacity of iron is 0.296 J/g°C
Answer:
The answer is a. I learned that they can multiply by using your cells
1 m = 0,001 km
1m³ = 0,000000001 km³
278 m³ = 0,000000278 km³ = 2,78×10^(-7) km³
:•)
They define acids as proton donors, and bases as proton acceptors
If you were to have:
HNO3 + H2O -> H3O+. + NO3-
You can see that the nitric acid (HNO3) gave a hydrogen ion which has 1 proton, 0 neutrons and 0 electrons to the water so we just say that it gave a proton.
Now let's see a base
NH3 + H2O -> NH4+ + OH-
Now, you can see that the ammonia (NH3) gained a hydrogen ion (proton) from the water to become ammonium(NH4). which means it accepted a proton
That's basically it. Feel free to ask if you have any further questions
Answer:
What procedure did you use to complete the lab?
→Procedures that needs to be considered to complete the lab are- a thorough knowledge of lab assignments, knowledge about safety equipment, reviewing the MSDS of chemicals for lab experiment etc.
<h3>Explanation:</h3>
To be lab prepared one must follow these procedures-
1. One should have the knowledge of lab assignments to make the lab experiment easier.
2. To be aware about safety equipment and their uses in lab, like- the location of fire extinguisher in lab.
3. To know the steps of experiments to be performed
4. To fill notebook of lab with information regarding the experiment
5. One should review the data sheets of chemicals material safety.
6. To put on all the necessary dressings to perform experiment.
7. To have complete understanding about the experiment disposals.
