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ser-zykov [4K]
3 years ago
15

An empty beaker weighs 25.91 g. When completely filled with water, the beaker and its contents have a total mass of 333.85 g. Wh

at volume does the beaker hold?
Chemistry
1 answer:
Finger [1]3 years ago
3 0

Answer:

The beaker holds 307.94  mL

Explanation:

As we know that the volume that beaker hold is the volume of water that occupied by it.

For this first we have to find mass of the water in the beaker

This can be calculated by the subtraction of beaker's weight from the weight of beaker and water.

     weight of water (m) = total weight - weight of beaker

Empty weight of beaker = 25.91 g

Weight of beaker with water = 333.85 g

Weight of water = 333.85 - 25.91 = 307.94 g

Density of water = 1 g/mL

We have

      Mass = Volume x density

      307.94  = Volume x 1

      Volume = 307.94  mL

The beaker holds 307.94  mL

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Write the chemical equation that represents the process of lattice energy for the case of nacl.
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Answer: Na^+(g)+Cl^-(g)\rightarrow NaCl(s),\Delta H_{Lattice}=-786kJ/mol

Explanation:

Lattice energy : It is defined energy released when ions combine together in a gaseous phase to form a compound. it is energy possessed by the crystal lattice of a compound. Denoted by symbol \Delta H_{Lattice}.

\Delta H_{Lattice}=positive , energy is absorbed while forming of the lattice

\Delta H_{Lattice}=negative, energy is release while forming of the lattice

Na^+(g)+Cl^-(g)\rightarrow NaCl(s),\Delta H_{Lattice}=-786kJ/mol

One mole of sodium ion when combines with one mole chloride ion release  786 kJ of energy.

4 0
3 years ago
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The standard enthalpy of formation of BrCl(g) is 14.7 kJmol-1 . The standard enthalpies for the atomization of Br2(l) and Cl2(g)
natita [175]

Explanation:

Equation of the reaction:

Br2(l) + Cl2(g) --> 2BrCl(g)

The enthalpy change for this reaction will be equal to twice the standard enthalpy change of formation for bromine monochloride, BrCl.

The standard enthalpy change of formation for a compound,

ΔH°f, is the change in enthalpy when one mole of that compound is formed from its constituent elements in their standard state at a pressure of 1 atm.

This means that the standard enthalpy change of formation will correspond to the change in enthalpy associated with this reaction

1/2Br2(g) + 1/2Cl2(g) → BrCl(g)

Here, ΔH°rxn = ΔH°f

This means that the enthalpy change for this reaction will be twice the value of ΔH°f = 2 moles BrCl

Using Hess' law,

ΔH°f = total energy of reactant - total energy of product

= (1/2 * (+112) + 1/2 * (+121)) - 14.7

= 101.8 kJ/mol

ΔH°rxn = 101.8 kJ/mol.

8 0
4 years ago
How does the mass of the water compare to the mass of the ice?
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When 3243. grams of iron (III) chloride are reacted with 511.8 grams of hydrosulfuric acid, which is the limiting reactant?
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Answer:

Hydrosulfuric acid will act as limiting reactant.

Explanation:

Given data:

Mass of iron(III) chloride = 3243.0 g

Mass of hydrosulfuric acid = 511.8 g

Limiting reactant = ?

Solution:

Chemical equation:

2FeCl₃ + 3H₂S       →       Fe₂S₃ + 6HCl

Number of moles of iron(III) chloride:

Number of moles = mass/molar mass

Number of moles = 3243.0 g/ 162.2 g/mol

Number of moles = 20 mol

Number of moles of hydrosulfuric acid:

Number of moles = mass/molar mass

Number of moles = 511.8 g/ 34.1 g/mol

Number of moles = 15 mol

Now we will compare the moles of both reactant with products

                      FeCl₃          :          Fe₂S₃

                       2                :            1

                      20               :          1/2 ×20 = 10

                      FeCl₃          :            HCl

                       2                :              6

                      20               :          6/2 ×20 = 60

                      H₂S             :          Fe₂S₃

                       3                :            1

                      15               :          1/3 ×15 = 5

                      H₂S            :            HCl

                       3                :              6

                      15                :          6/3 ×15 = 30

Hydrosulfuric acid producing less number of moles of product thus, it will act as limiting reactant.

 

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