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Citrus2011 [14]
3 years ago
9

A 45.0 g sample of a metal at 85.6 °C is placed in 150.0 g of water at 24.6 °C. The final temperature of the system is 28.3 °

C. Calculate the specific heat of the metal.
Chemistry
1 answer:
earnstyle [38]3 years ago
3 0

Answer:

904.014 j/kgk

Explanation:

Mass of metal = 45g

Temperature of metal = 85.6°

Mass of water = 150

Temperature of water = 24.6

Final temperature of system = 28.3

Heat lost by metal = Heat gained by water

m1 * c1 * dt = m2 * c2 * dt

Q = quantity of heat

Q = m*c*dt

dt = change in temperature

dt of water = 28.3 - 24.6 = 3.7

dt of metal = 85.6 - 28.3 = 57.3

Specific heat capacity of water, c = 4200

(45 * 10^-3) * c * 57.3 = (150 * 10^-3) * 4200 * 3.7

2.5785c1 = 2331

c1 = 2331 / 2.5785

= 904.01396

= 904.014 j/kgk

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Why resolution power in TEM microscope is higher than SEM microscope
sleet_krkn [62]

Answer:

Here's what I get.

Explanation:

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3 0
3 years ago
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What is the empirical formula of a substance that contains 2.64 g of c, 0.444 g of h, and 7.04 g of o?
vitfil [10]

To calculate empirical formula, number of moles of different atoms calculated first:

Number of moles of Carbon , n=  

=\frac{Given mass of carbon}{Molar mass of carbon}

=\frac{2.64 g}{12.011 g mol^{-1}}

= 0.220 mol  

Number of moles H =\frac{Given mass of hydrogen}{Molar mass of hydrogen}

=\frac{0.444 g}{1.008 g mol^{-1}}

= 0.440 mol

Number of moles of oxygen =\frac{Given mass of oxygen}{Molar mass of oxygen}

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= 0.440 mol

Ratio of carbon

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Ratio of hydrogen

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= 2  

Ratio of oxygen

=\frac{0.440}{0.220}

= 2  

So, the empirical formula is CH₂O₂

5 0
4 years ago
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