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garri49 [273]
2 years ago
11

Washing soda, a compound used to prepare hard water for washing laundry, is a hydrate, which means that a certain number of wate

r molecules are included in the solid structure. Its formula can be written as Na2CO3⋅xH2O, where x is the number of moles of H2O per mole of Na2CO3. When a 3.070 −g sample of washing soda is heated at 25 ∘C, all the water of hydration is lost, leaving 1.14 g of Na2CO3.
What is the value of x?

Chemistry
1 answer:
emmainna [20.7K]2 years ago
3 0

Answer: 10

Explanation:

The detailed solution is contained in the image attached. The molar mass of hydrated and anhydrous salts are obtained and the number of moles of hydrated and hydrated salts are equated. The masses of hydrated and anhydrous salts are gives in the question and are simply substituted accordingly. This can now be used to obtain the number of molecules of water of crystallization as required in the question.

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Oxalic Acid, a compound found in plants and vegetables such as rhubarb, has a mass percent composition of 26.7% C, 2.24% H, and
blondinia [14]

Answer:

HCO₂

Explanation:

From the information given:

The mass of the elements are:

Carbon C = 26.7 g;     Hydrogen H = 2.24 g     Oxygen O = 71.1 g

To determine the empirical formula;

First thing is to find the numbers of moles of each atom.

For Carbon:

=26.7 \ g\times \dfrac{1 \ mol }{12.01 \ g} \\ \\ =2.22 \ mol \ of \ Carbon

For Hydrogen:

=2.24 \ g\times \dfrac{1 \ mol }{1.008 \ g} \\ \\ =2.22 \ mol \ of \ Hydrogen

For Oxygen:

=71.1 \ g\times \dfrac{1 \ mol }{1.008 \ g} \\ \\ =4.44 \ mol \ of \ oxygen

Now; we use the smallest no of moles to divide the respective moles from above.

For carbon:

\dfrac{2.22 \ mol \ of \ carbon}{2.22} =1 \ mol \ of \ carbon

For Hydrogen:

\dfrac{2.22 \ mol \ of \ carbon}{2.22} =1 \ mol \ of \ hydrogen

For Oxygen:

\dfrac{4.44 \ mol \ of \ Oxygen}{2.22} =2 \ mol \ of \ oxygen

Thus, the empirical formula is HCO₂

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A 27 kg iron block initially at 375 C is quenched in an insulated tank that contains 130kg of water at 26 C. Assume the water th
Bess [88]

Solution :

a). Applying the energy balance,

$\Delta E_{sys}=E_{in}-E_{out}$

$0=\Delta U$

$0=(\Delta U)_{iron} + (\Delta U)_{water}$

$0=[mc(T_f-T_i)_{iron}] + [mc(T_f-T_i)_{water}]$

$0 = 27 \times 0.45 \times (T_f - 375) + 130 \times 4.18 \times (T_f-26)$

$t_f=33.63^\circ C$

b). The entropy change of iron.

$\Delta s_{iron} = mc \ln\left(\frac{T_f}{T_i} \right)$

           $ = 27 \times 0.45\ \ln\left(\frac{33.63 + 273}{375 + 273} \right)$

           = -9.09 kJ-K

Entropy change of water :

$\Delta s_{water} = mc \ \ln\left(\frac{T_f}{T_i} \right)$

           $ = 130 \times 4.18\ \ln\left(\frac{33.63 + 273}{26 + 273} \right)$

           = 10.76 kJ-K

So, the total entropy change during the process is :

$\Delta s_{tot} = \Delta s_{iron} + \Delta s_{water} $

        = -9.09 + 10.76

         = 1.67 kJ-K

c). Exergy of the combined system at initial state,

$X=(U-U_{0}) - T_0(S-S_0)+P_0(V-V_0)$

$X=mc (T-T_0) - T_0 \ mc \ \ln \left(\frac{T}{T_0} \right)+0$

$X=mc\left((T-T_0)-T_0 \ ln \left(\frac{T}{T_0} \right)\right)$

$X_{iron, i} = 27 \times 0.45\left(((375+273)-(12+273))-(12+273) \ln \frac{375+273}{12+273}\right)$

$X_{iron, i} =63.94 \ kJ$

$X_{water, i} = 130 \times 4.18\left(((26+273)-(12+273))-(12+273) \ln \frac{26+273}{12+273}\right)$

$X_{water, i} =-13.22 \ kJ$

Therefore, energy of the combined system at the initial state is

$X_{initial}=X_{iron,i} +X_{water, i}$

            = 63.94 -13.22

            = 50.72 kJ

Similarly, Exergy of the combined system at initial state,

$X=(U_f-U_{0}) - T_0(S_f-S_0)+P_0(V_f-V_0)$

$X=mc\left((T_f-T_0)-T_0 \ ln \left(\frac{T_f}{T_0} \right)\right)$

$X_{iron, f} = 27 \times 0.45\left(((33.63+273)-(12+273))-(12+273) \ln \frac{33.63+273}{12+273}\right)$

$X_{iron, f} = 216.39 \ kJ$

$X_{water, f} = 130 \times 4.18\left(((33.63+273)-(12+273))-(12+273) \ln \frac{33.63+273}{12+273}\right)$

$X_{water, f} =-9677.95\ kJ$

Thus, energy or the combined system at the final state is :

$X_{final}=X_{iron,f} +X_{water, f$

            = 216.39 - 9677.95

            = -9461.56 kJ

d). The wasted work

$X_{in} - X_{out}-X_{destroyed} = \Delta X_{sys}$

$0-X_{destroyed} = $

$X_{destroyed} = X_{initial} - X_{final}$

                = 50.72 + 9461.56

                = 9512.22 kJ

6 0
3 years ago
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