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valina [46]
3 years ago
7

A. Convert the mass of the Salt weighed out to fg.

Chemistry
1 answer:
omeli [17]3 years ago
3 0

Answer:

6.564×10¹⁶ fg.

Explanation:

The following data were obtained from the question:

Mass of beaker = 76.9 g

Mass of beaker + salt = 142.54 g

Mass of salt in fg =?

Next, we shall determine the mass of the salt in grams (g). This can be obtained as follow:

Mass of beaker = 76.9 g

Mass of beaker + salt = 142.54 g

Mass of salt =?

Mass of salt = (Mass of beaker + salt) – (Mass of beaker)

Mass of salt = 142.54 – 76.9

Mass of salt = 65.64 g

Finally, we shall convert 65.64 g to femtograms (fg) as illustrated below:

Recall:

1 g = 1×10¹⁵ fg

Therefore,

65.64 g = 65.64 g × 1×10¹⁵ fg / 1g

65.64 g = 6.564×10¹⁶ fg

Therefore, the mass of the salt is 6.564×10¹⁶ fg.

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Taking into account the definition of calorimetry, the specific heat of metal is 0.165 \frac{cal}{gC}.

<h3>Definition of calorimetry</h3>

Calorimetry is the measurement and calculation of the amounts of heat exchanged by a body or a system.

Sensible heat is defined as the amount of heat that a body absorbs or releases without any changes in its physical state (phase change).

So, the equation that allows to calculate heat exchanges is:

Q = c× m× ΔT

where:

  • Q is the heat exchanged by a body of mass m.
  • C is the specific heat substance.
  • ΔT is the temperature variation.

<h3>Specific heat capacity of the metal</h3>

In this case, you know:

For metal:

  • Mass of metal = 50 g
  • Initial temperature of metal= 45 °C
  • Final temperature of metal= 11.08 ºC
  • Specific heat of metal= ?

For water:

  • Mass of water = 250 g
  • Initial temperature of water= 10 ºC
  • Final temperature of water= 11.08 ºC
  • Specific heat of water = 1.035 \frac{cal}{gC}

Replacing in the expression to calculate heat exchanges:

For metal: Qmetal= Specific heat of metal× 50 g× (11.08 C - 45 C)

For water: Qwater=  1.035 \frac{cal}{gC} × 250 g× (11.08 C - 10 C)

If two isolated bodies or systems exchange energy in the form of heat, the quantity received by one of them is equal to the quantity transferred by the other body. That is, the total energy exchanged remains constant, it is conserved.

Then, the heat that the gold gives up will be equal to the heat that the water receives. Therefore:

- Qmetal = + Qwater

- Specific heat of metal× 50 g× (11.08 C - 45 C)= 1.035 \frac{cal}{gC} × 250 g× (11.08 C - 10 C)

Solving:

- Specific heat of metal× 50 g× (-33.92 C)= 1.035 \frac{cal}{gC} × 250 g× 1.08 C

Specific heat of metal× 1696 g×C= 279.45 cal

Specific heat of metal= \frac{279.45 cal}{1696 gC}

<u><em>Specific heat of metal= 0.165 </em></u>\frac{cal}{gC}

Finally, the specific heat of metal is 0.165 \frac{cal}{gC}.

Learn more about calorimetry:

brainly.com/question/11586486

brainly.com/question/24724338

brainly.com/question/14057615

brainly.com/question/24988785

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