Answer:
The heat capacity of the calorimeter is 5.11 J/g°C
Explanation:
Step 1: Given data
50.0 mL of water with temperature of 80.0 °C
Specific heat capacity of water = 4.184 J/g°C
Consider the density of water = 1g/mL
50.0 mL of water in a calorimeter at 20.0 °C
Final temperature = 47.0 °C
Step 2: Calculate specific heat capacity of the water in calorimeter
Q = Q(cal) + Q(water)
Q(cal) = mass * C(cal) * ΔT
Qwater = mass * Cwater * ΔT
Qcal = -Qwater
mass(cal) * C(cal) * ΔT(cal) = mass(water) * C(water) * ΔT(water)
50 grams * C(cal) * (47.0 - 20.0) =- 50grams * 4.184 J/g°C * (47-80)
1350 * C(cal) = 6903.6
C(cal) = 5.11 J/g°C
The heat capacity of the calorimeter is 5.11 J/g°C
The easiest way to approximate ph level is through the ph paper measurement system because using pH paper would be best for doing a quick measurement in the field. It does not essentially need to be calibrated or standardized and the papers make available an instant estimate.
I don't know what model you're referring to so I can't answer the question. However, upon researching, I found a similar problem. I posted it as an attached picture. Looking at the model, the amount of grams a herbivore eat each day corresponds to the arrow pointing inwards. Since the label says 4.0 g,
<em>the answer is 4 g per day</em>.
Some of the scientific questions that may be answered through the experiment are:
(1) What are the physical changes that may occur in the solution or the indicator when added with acidic/basic solution?
(2) How much of the indicator is needed in order to bring about a significant physical change in the solution to identify its H+ concentration?
if wrong i sry :(
Nitrogen has 5 valence electrons.
