What’s the oxidation number of k2so4

Serggg [28]

Option B, Bronze is made of the metals copper and tin.

Alloys are essentially materials that consist of two or more metal elements.

-T.B.

3 0

2 years ago

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Which of the following became a major drawback to using ice blocks for cooling food? availability expense unsanitary ice storage

son4ous [18]

<span> Ice was soon being harvested from contaminated water sources. Microbes in the ice invaded the food and people became ill
</span>so <span>unsanitary ice is your answer i hope this helps</span>

7 0

3 years ago

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What is the percentage composition of each element in dinitrogen monoxide, N2O?

Kitty [74]

The correct answer is C :)

4 0

2 years ago

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How many molecules of co2 form when 2.00 g of c2h5oh are produced?

Ronch [10]

The reaction involved in this problem is called the combustion reaction where a hydrocarbon reacts with oxygen to product carbon dioxide and water. The reaction of C2H5OH would be as follows:

C2H5OH + 3O2 = 2CO2 + 3H2O

To determine the number of molecules of CO2 that is formed, we need to determine the number of moles produced from the initial amount of C2H5OH and the relation from the reaction. Then we multiply avogadros number which is equal to 6.022x10^23 molecules per mole.

2.00 g C2H5OH ( 1 mol C2H5OH / 46.08 g C2H5OH ) ( 2 mol CO2 / 1 mol C2H5OH ) = 0.0868 mol CO2
0.0868 mol CO2 ( 6.022x10^23 molecules / mol ) = 5.23x10^22 molecules CO2

4 0

3 years ago

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