Calculate the specific heat of the substance if 373 J is required to raise the temperature of a 312 gram sample by 25°C?

Answers

Answer 1

Explanation:

Q = 373J

∆∅ = 25°C

m = 312g

c = x

Q = mc∆∅

373 = 312(x)(25)

373 = 7800x

x = 373/7800

x = 0.0478J/(g°C)

Answer 2

The specific heat of the substance if 373 J is required to raise teh temperature of a 312 gram sample by 25°C is 0.0478 J/g

What is specific heat?

The amount of energy needed to raise the temperature of one gram of a substance by one degree Celsius.

By the formula of specific heat

Q = mc∆∅

Q = 373J

m = 312g

∆∅ = 25° C

Putting the values in the equation

[tex]373 = 312\times c \times (25)\\\\373 = 7800c\\c= \dfrac{373}{7800} = 0.0478J/(g \circ C)[/tex]

Thus, the specific heat of the substance if 373 J is required to raise teh temperature of a 312 gram sample by 25°C is 0.0478 J/g

Learn more about specific heat

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Explanation:

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Answer:

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Answers

Answer:

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Explanation:

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Answer:

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Answer:

D

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Answers

Answer:

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Explanation:

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Answers

Answer:

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Explanation:

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Answers

Answer:

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Explanation:

HELLP PLEASE
_ Na+Cl 2 -> _NaCl
reactant side number of atoms?
Na = ?
Cl = ?
Product side number of atoms?
Na = ?
Cl = ?
what is the final equation?
_ Na+Cl 2 -> _NaCl

Answers

Answer:

2Na+Cl2 -> 2NaCl

Explanation:

1Na+1Cl 2 -> 1NaCl (i'm assuming they want you to balance it so ill start with 1)

Na = 1

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Na = 1

Cl = 1

We need to have 2 Cl on the product side

1Na+1Cl 2 -> 2NaCl

Now we need 2 Na on the reactants side

The final equation is

2Na+Cl2 -> 2NaCl

Now it is balanced

Na = 2

Cl = 2

--

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Cl = 2

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Answer:

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Answers

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Explanation:

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Answers

Answer:

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Answers

Answer:

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Answers

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Answers

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