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maw [93]
2 years ago
8

A 35.66g sample of copper is heated using 600j of energy. if the original temperature of the copper is 85C what is its final tem

perature?
Specific heat of copper: 0.38j/gC

A -6.39
B 6.39
C -78.61
D 91.4
Chemistry
1 answer:
ira [324]2 years ago
8 0

This problem is providing the mass, energy, initial temperature and specific heat of a sample of copper that is required to calculate the final temperature.

Thus, we recall the general heat equation:

Q=mC(T_f-T_i)\\

Which has to be solved for the final temperature, T_f as follows:

T_f=T_i+\frac{Q}{mC}

Finally, we plug in the numbers to obtain:

T_f=85\°C+\frac{600J}{35.66g*0.38\frac{J}{g\°C} } \\\\T_f=129.3\°C

However, this result is not given in the choices.

Learn more:

  • brainly.com/question/14383794
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Does adding salt to water lower its freezing point science project
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Yes, it does. The freezing point of the salt solution is lower than the freezing point of pure water since there will be more particles, which are water molecules plus salt molecules, present in the salt solution. The salt molecules will slow down the amount of crystals formed resulting to a lower freezing point.                                              
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4 years ago
What volume of each solution contains 0.14 mol of KCl? Express your answer using two significant figures.
trasher [3.6K]

Answer:

Solution given:

1 mole of KCl\rightarrow22.4l

1 mole of KCl\rightarrow74.55g

we have

0.14 mole of KCl\rightarrow74.55*0.14=10.347g

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5 0
3 years ago
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The density of dry air measured at 25 Celsius is 1.19 times 10 to the negative 3 what is the volume of 50.0g of air
Oksana_A [137]

Answer:

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Explanation:

Data Given:

Density (d) of air =  1.19 x 10⁻³g/mL

Mass of the air (m) = 50 g

Volume of the air (V) = ?

Solution:

Formula will be used

                d = m/V

As we have to find volume so rearrange the above equation

               V = m/d . . . . . . . . . . . (1)

Put values in above equation 1

                V = 50 g / 1.19 x 10⁻³g/mL

                V = 4.2 x 10⁴ mL

So,

volume of dry air = 4.2 x 10⁴ mL

4 0
3 years ago
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