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Ainat [17]
3 years ago
13

A 215-g sample of copper metal at some temperature is added to 26.6 g of water. The initial water temperature is 22.22 oC, and t

he final temperature is 24.44 oC. If the specific heat of copper is 0.385 Jg-1oC-1, what was the initial temperature of the copper? Any additional constants needed can be found in your textbook.
Chemistry
1 answer:
andrezito [222]3 years ago
8 0

The initial temperature of the copper metal was 27.38 degrees.

Explanation:

Data given:

mass of the copper metal sample = 215 gram

mass of water = 26.6 grams

Initial temperature of water = 22.22 Degrees

Final temperature of water = 24.44 degrees

Specific heat capacity of water = 0.385 J/g°C

initial temperature of copper material , Ti=?

specific heat capacity of water = 4.186 joule/gram °C

from the principle of:

heat lost = heat gained

heat gained by water is given by:

q water = mcΔT

Putting the values in the equation:

qwater = 26.6 x 4.186 x (2.22)

qwater = 247.19 J

qcopper = 215 x 0.385 x (Ti-24.4)

              = 82.77Ti - 2019.71

Now heat lost by metal = heat gained by water

82.77Ti - 2019.71 = 247.19

Ti = 27.38 degrees

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The concentration of [H3O⁺]=2.86 x 10⁻⁶ M

<h3>Further explanation</h3>

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\large {\boxed {\bold {Ka \: = \: \frac {[H ^ +] [A ^ -]} {[HA]}}}}

Reaction

HC₂H₃O₂ (aq) + H₂O (l) ⇔  (aq) + H₃O⁺ (aq) Ka = 1.8 x 10⁻⁵

\tt Ka=\dfrac{[C_2H_3O^{2-}[H_3O^+]]}{[HC_2H_3O_2]}}\\\\1.8\times 10^{-5}=\dfrac{0.22\times [H_3O^+]}{0.035}

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5 0
3 years ago
A diver exhales a bubble with a volume of 250 mL at a pressure of 2.4 atm and a temperature of 15 °C. What is the volume of the
alexdok [17]

Answer : The volume of the bubble is, 625 mL

Explanation :

Combined gas law is the combination of Boyle's law, Charles's law and Gay-Lussac's law.

The combined gas equation is,

\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}

where,

P_1 = initial pressure of gas = 2.4 atm

P_2 = final pressure of gas = 1.0 atm

V_1 = initial volume of gas = 250 mL

V_2 = final volume of gas = ?

T_1 = initial temperature of gas = 15^oC=273+15=288K

T_2 = final temperature of gas = 27^oC=273+27=300K

Now put all the given values in the above equation, we get:

\frac{2.4atm\times 250mL}{288K}=\frac{1.0atm\times V_2}{300K}

V_2=625mL

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7 0
3 years ago
How many grams of C6H12O6 are needed to be dissolved in water to make 100. grams of a 250. ppm solution?
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Answer:

0.025 g C6H12O6

Explanation:

ppm = (g solute/ g solution)* 10^6

g solute= (ppm * g solution)/ 10^6

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g solute=0.025 g C6H12O6

4 0
3 years ago
g A radioactive isotope of mercury, 197Hg, decays to gold, 197Au, with a disintegration constant of 0.0108hrs.-1. What % of the
weqwewe [10]

Answer:

7.49% of Mercury

Explanation:

Let N₀ represent the original amount.

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From the question given above, the following data were obtained:

Rate of disintegration (K) = 0.0108 h¯¹

Time (t) = 10 days

Percentage of Mercury remaining =?

Next, we shall convert 10 days to hours. This can be obtained as follow:

1 day = 24 h

Therefore,

10 days = 10 day × 24 h / 1 day

10 days = 240 h

Thus, 10 days is equivalent to 240 h.

Finally, we shall determine the percentage of Mercury remaining as follow:

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Time (t) = 10 days

Percentage of Mercury remaining =?

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Log (N₀/N) = 2.592 / 2.303

Log (N₀/N) = 1.1255

Take the anti log of 1.1255

N₀/N = anti log 1.1255

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Invert the above expression

N/N₀ = 1/13.3506

N/N₀ = 0.0749

Multiply by 100 to express in percent.

N/N₀ = 0.0749 × 100

N/N₀ = 7.49%

Thus, 7.49% of Mercury will be remaining after 10 days

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3 years ago
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Pavel [41]
I think the answer is true if so i hope it helps

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