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2 years ago

To see if it works like they want it to, engineers need to Question 3 options:

Chemistry

2 answers:

ziro4ka [17]2 years ago

4 0

Answer:

they need to do their research

MissTica2 years ago

3 0

Answer:

test their solution, so they'd be sure it would solve what they want it to solve and not cause more problems.

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You have 363 mL of a 1.25M potassium chloride solution, but you need to make a 0.50M potassium chloride solution. How many milli

maxonik [38]

Answer:- 544.5 mL of water need to be added.

Solution:- It is a dilution problem. The equation used for solving this type of problems is:

$M_1V_1=M_2V_2$

where, $M_1$ is initial molarity and $M_2$ is the molarity after dilution. Similarly, $V_1$ is the volume before dilution and $V_2$ is the volume after dilution.

Let's plug in the values in the equation:

$1.25M(363mL)=0.50M(V_2)$

$V_2=\frac{1.25M(363mL)}{0.50M}$

$V_2=907.5mL$

Volume of water added = 907.5mL - 363mL = 544.5 mL

So, 544.5 mL of water are need to be added to the original solution for dilution.

3 0

2 years ago

What is the main material for b horizon

STatiana [176]

i is dont no

koooooo

8 0

2 years ago

Glycolysis is the process by which energy is harvested from glucose by living things. Several of the reactions of glycolysis are

pantera1 [17]

Answer:

1. Options A and B

2. Options B and C

3.. B. Net ∆G = -16.7 KJ/mol; C. Net ∆G = -14.2 KJ/mol

Explanation:

1. The spontaneity of a chemical reaction depends on its standard free energy change, ∆G. If ∆G is negative, the reaction is favourable, but when it is positive, the reaction is unfavorable.

Therefore, since reaction A and B have ∆G to be positive, they are unfavorable

2. Coupling an unfavorable reaction to a favourable reaction can help the reaction to proceed in the forward direction as long as the net free energy change is negative.

Coupling reaction A and C, as well as reaction B and C will make the reactions to become favourable as net ∆G is negative in both instances.

3. A and C: net ∆G = 13.8 - 30.5 = -16.7 KJ/mol

B and C: net ∆G = 16.3- 30.5 = -14.2 KJ/mol

4 0

2 years ago

A sample from one of Earth's oceans has a salinity of 34. What is the concentration of dissolved salts in this sample of seawate

11Alexandr11 [23.1K]

Answer:

Concentration of dissolved salts = 34,038.76 ppm

Explanation:

Given:

Salinity of ocean water = 34

Find:

Concentration of dissolved salts

Computation:

Salinity of ocean water = 34 g/l

1g/l = 1001.14 ppm

Concentration of dissolved salts = 1001.14 ppm x 34

Concentration of dissolved salts = 34,038.76 ppm

3 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

2 years ago