Chemistry Middle School

Which of the following is the name of the process scientists use to gain
knowledge about the physical world?

Answers

Answer 1

Answer:

The Scientific Method.

Related Questions

Need help balancing equations all 20 to 40 please attach work

Answers

Because start in the beginning of a long story in the year of 1990 it was a really beautiful day of summer in the big beach of santa monica

Answer: Solution attached.

Each equation is now balanced.

Explanation:

A 125G sample of water was heated to 100.0°C and then I borrow platinum 20.0°C is dropped into the beaker the temperature of the platinum in the beaker quickly rose 235.0°C the specific heat of platinum is 0.13 j/g°C. The specific heat of water is 4.184 J/g°C. What is the mass of platinum

Answers

Answer:

mass of platinum = 2526.12 g

Explanation:

Given data:

Mass of water = 125 g

Initial temperature of water= 100.0°C

Initial temperature of Pt = 20.0°C

Final temperature = 235°C

Specific heat of Pt = 0.13 j/g°C

Specific heat of water = 4.184 j/g°C

Mass of platinum = ?

Solution:

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT = T2 - T1

Q(w) = Q(Pt)

m.c.  (T2 - T1)   =   m.c.   (T2 - T1)

125 g × 4.184 j/g°C ×  (235°C - 100.0°C)  = m × 0.13 j/g°C ×  (235°C - 20°C)

125 g × 4.184 j/g°C × 135°C  = m × 0.13 j/g°C × 215°C

70605 j = m×27.95 j/g

m = 70605 j /27.95 j/g

m = 2526.12 g

Final answer:

To find the mass of platinum dropped into hot water, you would apply the conservation of energy principle and use the specific heat values for both substances. However, the provided temperatures suggest an error in the question, as platinum would not heat to a temperature higher than the water's temperature.

Explanation:

The question involves finding the mass of platinum, which was dropped into hot water, by using the concept of heat transfer between the metal and water. According to the law of conservation of energy, the heat lost by the water is equal to the heat gained by the platinum. To solve for the mass of the platinum, we use the formula q = mcΔT, where 'q' is the heat transfer, 'm' is the mass, 'c' is the specific heat, and 'ΔT' is the change in temperature.

However, the given information seems to contain a mistake, as platinum will not naturally heat up to 235.0°C when dropped into water that is at 100.0°C. There must be an error in the given temperatures. Assuming we had the correct temperature details and using the given specific heats for water (4.184 J/g°C) and platinum (0.13 J/g°C), we could set up the heat transfer equations and solve for the unknown mass of platinum.

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