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3 years ago

7

In a science lab, a student heats up a chemical from 10 °C to 25 °C which requires thermal energy of 30000 J. If mass of the obj

ect is 40 g, the specific heat capacity of the chemical would be
Group of answer choices

25 J /g* °C

75 J /g* °C

100 J /g* °C

50 J /g* °C

1 answer:

olganol [36]3 years ago

3 0

Answer:

The specific heat capacity of the object is 50 J/g°C ( option 4 is correct)

Explanation:

Step 1: Data given

Initial temperature = 10.0 °C

Final temperature = 25.0 °C

Energy required = 30000 J

Mass of the object = 40.0 grams

Step 2: Calculate the specific heat capacity of the object

Q = m* c * ΔT

⇒With Q = the heat required = 30000 J

⇒with m = the mass of the object = 40.0 grams

⇒with c = the specific heat capacity of the object = TO BE DETERMINED

⇒with ΔT = The change in temperature = T2 - T2 = 25.0 °C - 10.0°C = 15.0 °C

30000 J = 40.0 g * c * 15.0 °C

c = 30000 J / (40.0 g * 15.0 °C)

c = 50 J/g°C

The specific heat capacity of the object is 50 J/g°C ( option 4 is correct)

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Irina-Kira [14]

Answer : The enthalpy of formation of $H_2SO_4$ is, -812.4 kJ/mole

Explanation :

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According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

The formation of $H_2SO_4$ will be,

$S+H_2+2O_2\rightarrow H_2SO_4$ $\Delta H_{formation}=?$

The intermediate balanced chemical reaction will be,

(1) $S+O_2\rightarrow SO_2$ $\Delta H_1=-298.2$

(2) $SO_2+\frac{1}{2}O_2\rightarrow SO_3$ $\Delta H_2=-98.2$

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Now adding all the equations, we get the expression for enthalpy of formation of $H_2SO_4$ will be,

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Explanation :

Formula used :

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where,

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$M_2\text{ and }V_2$ are the final molarity and volume of diluted spinach solution.

We are given:

$M_1=3.0M\\V_1=?\\M_2=0.30M\\V_2=500.0mL$

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

$3.0M\times V_1=0.30M\times 500.0mL\\\\V_1=50mL$

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White phosphorus melts and then vaporizes at high temperatures. The gas effuses at a rate that is 0.404 times that of neon in th

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