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BaLLatris [955]
3 years ago
15

How isa molecule of mercury different from a molocue of water

Chemistry
1 answer:
Genrish500 [490]3 years ago
8 0
<h2>Answer:</h2>

The density of mercury molecule is higher than water.

<h3>Explanation:</h3>

Density is defined as mass per unit volume.In other words, density is the amount of matter within a given amount of space. water has the density of 1.0 gram per milliliter whereas the mercury has a density of 13.6 grams per centimeter squared.

One reason for the differences in density between mercury and water is that the atomic mass of mercury is 200.59 grams per mole. The atomic mass of water is 18.0 grams per mole. This is because mercury has a larger nucleus than hydrogen or water.

Additionally, there are strong inter-molecular forces (hydrogen bonds) between water molecules. hydrogen molecules do not stack upon one another as nicely as mercury atoms. Thus, there is additional empty spaces between the water molecules leading to its lower mass per volume(density)

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100 ml of a 0.300 m solution of agno3 reacts with 100 ml of a 0.300 m solution of hcl in a coffee-cup calorimeter and the temper
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100 ml of a 0.300 m solution of agno3 reacts with 100 ml of a 0.300 m solution of hcl in a coffee-cup calorimeter and the temperature rises from 21.80 °c to 23.20 °c. Assuming the density and specific heat of the resulting solution is 1.00 g/ml and 4.18 j/g ∙ °c respectfully, what is the ΔH°rxn?

39.013 kJ/mol.

Explanation:

AgNO3(aq) + HCl(aq) --------------> AgCl(s) + HNO3(aq)

We can calculate the amount of heat (Q) released from the solution using the relation:

Q = m.c.ΔT,

Where, Q is the amount of heat released from the solution (Q = ??? J).

m is the mass of the solution (m of the solution = density of the solution x volume of the solution = (1.0 g/mL)(200 mL) = 200 g.

c is the specific heat capacity of the solution (c = 4.18 J/g∙°C).

ΔT is the difference in the T (ΔT = final temperature - initial temperature = 23.20 °C - 21.80 °C = 1.4 °C).

∴ Q = m.c.ΔT = (200 g)(4.18 J/g∙°C)(1.4 °C) = 1170.4 J.

∵ ΔH°rxn = Qrxn/(no. of moles of AgNO₃).

Molarity (M) is defined as the no. of moles of solute dissolved in a 1.0 L of the solution.

M = (no. of moles of AgNO₃)/(Volume of the solution (L)).

∴ no. of moles of AgNO₃

               = (M)(Volume of the solution (L))

               = (0.3 M)(0.1 L) = 0.03 mol.

∴ ΔH°rxn

           = Qrxn/(no. of moles of AgNO₃)

            = (1170.4 J)/(0.03 mol)

            = 39013.33 J/mol

           = 39.013 kJ/mol.

7 0
3 years ago
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