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borishaifa [10]
2 years ago
6

What is the change in enthalpy when 11. 0 g of liquid mercury is heated by 15°c?

Chemistry
1 answer:
Vlad1618 [11]2 years ago
6 0

Enthalpy change is the difference between energy used and energy gained. The change in enthalpy of the liquid mercury is 0.0231 kJ.

<h3>What is the enthalpy change?</h3>

Enthalpy change is the difference between the energy used to break chemical bonds and the energy gained by the products formed in a chemical reaction.

The enthalpy change is given by,

\rm \Delta H_{rxn} = \rm q_{rxn}

and,

\rm q = mc\Delta T

Given,

Mass of the liquid mercury (m) = 11.0 gm

The specific heat of mercury (c) = 0.14 J per g per degree Celsius

Temperature change = 15 degrees Celsius

Enthalpy change is calculated as:

\begin{aligned} \rm q &= \rm mc\Delta T\\\\&= 11 \times 0.14 \times 15\\\\&= 23.1 \;\rm J\end{aligned}

Therefore, 0.0231 kJ is the change in enthalpy.

Learn more about enthalpy change here:

brainly.com/question/10932978

#SPJ4

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Answer:

( °F − 32) × 5/9 =  °C

Explanation:

Also there is a mental calculation to convert from Fahrenheit to Celsius. The ratio 5/9 is approximately equal 0.55555….

Subtract 32º to adapt the equivalent in the Fahrenheit scale.

Divide the degrees Celsius by 2 (multiply by 0.5).

Take 1/10 of this number (0.5 * 1/10 = 0.05) and add it to the number obtained previously.

Example: Convert 98.6º F to Centigrade.

98.6 - 32 = 66.6

66.6 * 1/2 = 33.3

33.3 * 1/10 = 3.3

33.3 + 3.3 = 36.6 which is an approximation in degrees Centigrade

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Wastewater discharged into a stream by a sugar refinery contains 3.40 g of sucrose (C12H22O11) per liter. A government-industry
Ipatiy [6.2K]

<u>Answer:</u> The pressure that must be applied to the apparatus is 0.239 atm

<u>Explanation:</u>

To calculate the osmotic pressure, we use the equation for osmotic pressure, which is:

\pi=iMRT

or,

\pi=i\times \frac{m_{solute}}{M_{solute}\times V_{solution}\text{ (in L)}}}\times RT

where,

\pi = osmotic pressure of the solution

i = Van't hoff factor = 1 (for non-electrolytes)

m_{solute} = mass of sucrose = 3.40 g

M_{solute} = molar mass of sucrose = 342.3 g/mol

V_{solution} = Volume of solution = 1 L

R = Gas constant = 0.0821\text{ L atm }mol^{-1}K^{-1}

T = temperature of the solution = 20^oC=[20+273]K=293K

Putting values in above equation, we get:

\pi =1\times \frac{3.40g}{342.3g/mol\times 1}\times 0.0821\text{ L. atm }mol^{-1}K^{-1}\times 293K\\\\\pi =0.239atm

Hence, the pressure that must be applied to the apparatus is 0.239 atm

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3 years ago
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