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inna [77]
3 years ago
9

Liquid n-pentane and liquid n-octane are mixed to form a stream flowing at a rate of 1,496.5 lbm/hr. An in-line density measurem

ent determines the mixture density is 0.654 g/mL. Calculate the volumetric flow rate of the liquid n-pentane feed in cm3/sec. The specific gravities of n-pentane and n-octane are 0.631 and 0.708, respectively. Assume ideal mixing.

Chemistry
1 answer:
Ivenika [448]3 years ago
4 0

Answer:

202.20 cm^3/s

Explanation:

Hello,

I'm sending a photo showing the solution for this exercise.

The solution of the 2x2 system could be achieved by using a calculator or any  available algebraic method.

Best regards.

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The mass percentage of hydrochloric acid within a solution is 28.00%28.00% . Given that the density of this solution is 1.1411.1
sashaice [31]

Answer:

8.76M

Explanation:

Given that

Mass from the density = 1141g

According to the given situation the computation of molarity of the solution is shown below:-

we will took HCL solution which is 1000mL

HCl = 28% by mass

So,

Mass of HCl in 1-litre solution is

= \frac{28}{100} \times 1141

Which gives the result of molar mass HCI is

= 319.48g /mol

Now,

Molarity is

= \frac{319.48}{36.45}

Which gives results of molarity is

= 8.76M

8 0
3 years ago
How many liters of P, are in 3.5x1022 molecules at STP?
suter [353]

Answer: they giving you some hard question

Explanation:

i dont know  know what the answer is do you have any answer options

4 0
2 years ago
!!!!!PLEASE HELP!!!!!
ollegr [7]

Answer:

Difference in the potential energy of the reactants and products

Explanation:

 

The products have a lower potential energy than the reactants, and the sign of ΔH is negative. In an endothermic reaction, energy is absorbed. The products have a higher potential energy than the reactants, and the sign of ΔH is positive.

3 0
2 years ago
Calculate the solubility of zn(oh)2(s) in 2.0 m naoh solution. (hint: you must take into account the formation of zn(oh)2−4, whi
Brilliant_brown [7]
When we have:

Zn(OH)2 → Zn2+ 2OH-  with Ksp = 3 x 10 ^-16

and:

Zn2+ + 4OH- → Zn(OH)4 2-  with Kf = 2 x 10^15
 
by mixing those equations together:

Zn(OH)2 + 2OH- → Zn(OH)4 2- with K = Kf *Ksp = 3 x 10^-16 * 2x10^15 =0.6

by using ICE table:

         Zn(OH)2 + 2OH- → Zn(OH)4 2-

initial                     2m              0

change                  -2X                +X     

Equ                       2-2X                 X

when we assume that the solubility is X

and when K = [Zn(OH)4 2-] / [OH-]^2

        0.6 = X / (2-2X)^2    by solving this equation for X

∴ X = 0.53 m

∴ the solubility of Zn(OH)2 = 0.53 M
4 0
3 years ago
CO2(g)+CCl4(g)⇌2COCl2(g) Calculate ΔG for this reaction at 25 ∘C under these conditions: PCO2PCCl4PCOCl2===0.140 atm0.185 atm0.7
padilas [110]

<u>Answer:</u> The \Delta G for the reaction is 54.425 kJ/mol

<u>Explanation:</u>

For the given balanced chemical equation:

CO_2(g)+CCl_4(g)\rightleftharpoons 2COCl_2(g)

We are given:

\Delta G^o_f_{CO_2}=-394.4kJ/mol\\\Delta G^o_f_{CCl_4}=-62.3kJ/mol\\\Delta G^o_f_{COCl_2}=-204.9kJ/mol

To calculate \Delta G^o_{rxn} for the reaction, we use the equation:

\Delta G^o_{rxn}=\sum [n\times \Delta G_f(product)]-\sum [n\times \Delta G_f(reactant)]

For the given equation:

\Delta G^o_{rxn}=[(2\times \Delta G^o_f_{(COCl_2)})]-[(1\times \Delta G^o_f_{(CO_2)})+(1\times \Delta G^o_f_{(CCl_4)})]

Putting values in above equation, we get:

\Delta G^o_{rxn}=[(2\times (-204.9))-((1\times (-394.4))+(1\times (-62.3)))]\\\Delta G^o_{rxn}=46.9kJ=46900J

Conversion factor used = 1 kJ = 1000 J

The expression of K_p for the given reaction:

K_p=\frac{(p_{COCl_2})^2}{p_{CO_2}\times p_{CCl_4}}

We are given:

p_{COCl_2}=0.735atm\\p_{CO_2}=0.140atm\\p_{CCl_4}=0.185atm

Putting values in above equation, we get:

K_p=\frac{(0.735)^2}{0.410\times 0.185}\\\\K_p=20.85

To calculate the gibbs free energy of the reaction, we use the equation:

\Delta G=\Delta G^o+RT\ln K_p

where,

\Delta G = Gibbs' free energy of the reaction = ?

\Delta G^o = Standard gibbs' free energy change of the reaction = 46900 J

R = Gas constant = 8.314J/K mol

T = Temperature = 25^oC=[25+273]K=298K

K_p = equilibrium constant in terms of partial pressure = 20.85

Putting values in above equation, we get:

\Delta G=46900J+(8.314J/K.mol\times 298K\times \ln(20.85))\\\\\Delta G=54425.26J/mol=54.425kJ/mol

Hence, the \Delta G for the reaction is 54.425 kJ/mol

7 0
3 years ago
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