Chemistry, 04.10.2019 18:30 treyvoniskewl

Suppose a mercury thermometer contains 3.250g of mercury and has a capillary that is 0.180mm in diameter.. how far does the mercury rise in the capillary when the temperature changes from 0.0 ∘c to 25.0 ∘c? the density of mercury at these temperatures is 13.596 g/cm3 and 13.534 g/cm3, respectively.

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1) You need to get volume of both temperatures by using first attached formula V= Mass/Density V_1 = (3.250 g)/(13.596 g/(cm^3)) = .2390 cm^3
At 25 degrees C (V_2e have:

V_2 = (3.250 g)/(13.534 g/(cm^3)) = .2401 cm^3

V_1 = (3.250 g)/(13.596 g/(cm^3)) = .2390 cm^3
At 25 degrees C (V_2e have:

V_2 = (3.250 g)/(13.534 g/(cm^3)) = .2401 cm^3

2) Using the second formula you get the height of 0 degree
h = V/(pi * r^2)

h_1 =(.2390 cm^3)/(pi*(.009 cm)^2) = 939.2 cm

(radius in cm is (0.180 mm) / 2 * (1 cm)/(10 mm) or .009 cm

3) Then with h1 you can easily get the height of 25 degrees
h_2 =(.2401 cm^3)/(pi*(.009 cm)^2) = 943.5 cm

Subtract 943.5 cm - 939.2 cm, and obtain a rise in mercury height of 4.3 cm

Answer from: Gimagg8195

The mercury will rise in the capillary to a height of 4.3 cm.FURTHER EXPLANATION

The steps in solving this problem are the following:

1. Sort the given from the problem.

2. Get the volume of mercury at 0.0 °C and 25.0 °C.

3. Get the difference in volume for the two temperatures.

4. Solve for the height of the mercury using the volume of a cylinder (shape of the thermometer).

STEP 1. Sort the given information in the problem.

Given:

mass of mercury = 3.250 g

diameter of capillary = 0.180 mm

density at 0.0°C = 13.596 g/cm³

density at 25.0° = 13.534 g/cm³

Find:

volume change from 0.0°C to 25.0° C

change in height from 0.0°C to 25.0°C

STEP 2: Using the equation for density, calculate the volume at 0.0°C and 25.0° C. The general equation to be used is:

$volume = \frac{mass}{density}$

For 0.0°C

$volume = \frac{3.250 \ g}{13.596 \frac{g}{cm^3}}\\\boxed {volume = 0.23904 \ cm^3}$

For 25.0°C

$volume = \frac{3.250 \ g}{13.534 \frac{g}{cm^3}}\\\boxed {volume = 0.24014 \ cm^3}$

STEP 3: Solve for the volume change.

$\Delta V = 0.24014 \ cm^3 - 0.23904 \ cm^3\\\boxed {\Delta V = 1.1 \times 10^{-3} \ cm^3}$

STEP 4: From the change in volume, the change in height can be calculated using the equation below:

$\Delta h = \frac{\Delta V}{\pi\ r^2}\\\Delta h = \frac{1.1 \times 10^{-3} \ cm^3}{\pi({9 \times 10^{-3} \ cm)}^2}\\\\\boxed {\Delta h = 4.32 \ cm}$

From the given in the problem, the least number of significant figures is 2, thus, the final answer must only have 2 significant figures as well.

Therefore,

$\boxed {\boxed {\Delta h = 4.3 \ cm}}$

LEARN MORELearn more about thermal expansion linkLearn more about capillary action link

Keywords: mercury thermometer

Answer from: Quest

the answer to your question is c

Answer from: Quest

Answer : option c) heat is released during the process.explanation :

the enthalpy of formation of water is given as -285.8kj/mol, which states that the enthalpy of formation is negative.

negative enthalpy of formation indicates that the heat is released during the reaction process. the heat is released from reaction to the surroundings the enthalpy is expressed with a negative sign with the energy given out in the reaction process.

the formation of water takes place by giving out heat which is -285.8 kj/mol.

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