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Margarita [4]
4 years ago
9

What’s the answer!?!!

Chemistry
1 answer:
EastWind [94]4 years ago
3 0

Answer: generic material and protein coat. Have a great day

Explanation:

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A 50.00 g sample of an unknown metal is heated to 45.00°C. It is then placed in a coffee-cup calorimeter filled with water. The
V125BC [204]

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:

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brainly.com/question/24988785

#SPJ1

7 0
2 years ago
What are subatomic particles?​
vova2212 [387]

Explanation:

a particle smaller than an atom (e.g., a neutron) or a cluster of such particles (e.g., an alpha particle).

6 0
2 years ago
A 2.0 L flask contains nitrogen and oxygen gases at 25°C. The total pressure is 0.91 atm and contains 0.050 mol of nitrogen. Cal
iVinArrow [24]

Answer:

.025 ml 02

Explanation:

I made a pdf and pasted it in and etc

7 0
3 years ago
What is the half-life of a first-order reaction if it takes 4.4 x 102 seconds for the concentration to decrease from 0.50 M to 0
elena55 [62]

Answer: The half-life of a first-order reaction is, 3.3\times 10^2s

Explanation:

All the radioactive reactions follows first order kinetics.

Rate law expression for first order kinetics is given by the equation:

k=\frac{2.303}{t}\log\frac{[A_o]}{[A]}

where,  

k = rate constant = ?

t = time taken = 440 s

[A_o] = initial amount of the reactant = 0.50 M

[A] = left amount =  0.20 M

Putting values in above equation, we get:

k=\frac{2.303}{440s}\log\frac{0.50}{0.20}

k=2.083\times 10^{-3}s^{-1}

The equation used to calculate half life for first order kinetics:

t_{1/2}=\frac{0.693}{k}

Putting values in this equation, we get:

t_{1/2}=\frac{0.693}{2.083\times 10^{-3}s^{-1}}=332.69s=3.3\times 10^2s

Therefore, the half-life of a first-order reaction is, 3.3\times 10^2s

4 0
3 years ago
How much heat is absorbed if 75.4 g of CO2 reacts?
iris [78.8K]

Answer:

6x

Explanation:

7 0
3 years ago
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