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algol13
3 years ago
9

Part of Dr. Ritchey’s work in her previous job involved boiling and condensing hydrocarbon mixtures. In one of her experiments,

she used a binary mixture of n-pentane (density 39.1 lbm/ft^3) and p-xylene (density 53.8 lbm/ft^3). One outlet stream of her condenser was measured to have a flow rate of 56 ft^3/min and a concentration of 30/70 n-pentane/p-xylene by volume. The other outlet stream was measured to have a flow rate of 16 ft^3/min and a concentration of 50/50 n-pentane/p-xylene by volume. If the system operates at steady state and the only inlet to her condenser has a flow rate of 72 ft^3/min, what is the inlet concentration of n-pentane in % by volume? Your answer should be between 0 and 100. Round your answer to 1 decimal place for entry into eCampus. Do not enter units. Example: 12.3

Chemistry
1 answer:
Hitman42 [59]3 years ago
8 0

Answer:

what is the inlet concentration of n-pentane in % by volume? = 34.4%

Explanation:

check below for explanation in the attachment

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A sample of gas contains 3.0L of nitrogen at 320kPa. What volume would be necessary to decrease the pressure at 110kPa
Phoenix [80]

Answer:

V_{2} = 8.92 L

Explanation:

We have the equation for ideal gas expressed as:

PV=nRT

Being:

P = Pressure

V = Volume

n = molar number

R = Universal gas constant

T = Temperature

From the statement of the problem I infer that we are looking to change the volume and the pressure, maintaining the temperature, so I can calculate the right side of the equation with the data of the initial condition of the gas:

P_{1} V_{1} =nRT

320Kpa*0.003m^{3} =nRT

1000L = 1m^{3}

So

nRT= 0.96

Now, as for the final condition:

P_{2}V_{2}=nRT

P_{2} V_{2} =0.96

clearingV_{2}

V_{2} =\frac{0.96}{P_{2} }

V_{2} =0.00872m_{3}

V_{2} = 8.92 L

6 0
3 years ago
Calculate the moles and grams of solute in each solution. D. 2.0 L of 0.30M Na2SO4. I already have A, B, and C. Thanks!
e-lub [12.9K]
Find the number of moles
C = n / V
C(Concentration) = 0.30 moles / L
V ( Volume) = 2 L
n = ??
n = C * V
n = 0.30 mol / L * 2 L
n = 0.60 mol


Find the molar mass
2Na = 23 * 2 = 46 grams
1S   = 32 * 1 = 32 grams
O4   = 16 * 4 = 64 grams
Total =            142 grams / mol

Find the mass
n = given mass / molar mass
n = 0.06 mol
molar Mass = 142 grams / mol
given mass = ???

given mass = molar mass * mols
given mass = 142 * 0.6
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85.2 are in a 2 L solution that has a concentration of 0.6 mol/L


4 0
3 years ago
Read 2 more answers
What type of reaction is Zn + CuCl2 → Cu + ZnCl2?​
Shalnov [3]

Answer:

Its single - displacement.

Explanation:

4 0
3 years ago
6. What is the oxidation number for the atom indicated in the following compounds.
Hatshy [7]

Answer:

a. +6;

b. +5;

c. +3.

Explanation:

Start with elements with well-known oxidation states.

The oxidation state on oxygen O in compounds is mostly -2. Common exceptions include:

  • -1 in peroxides and
  • positive when oxygen bonds to fluorine.

The oxidation state on group 1 metals (Li, Na, K, etc.) in compounds is mostly +1.

The oxidation state on group 2 metals (Be, Mg, Ca, etc.) in compounds is mostly +2.

Barium Ba is a group 2 metal. The oxidation state on Ba in the compound BaSO₄ is expected to be +2.

The oxidation state on hydrogen H in compounds is mostly +1. The oxidation state on H might be negative when it is bonded to metals.  

The oxidation state on halogens (F, Cl, Br, etc.) is mostly -1. The oxidation state may vary when the halogen is bonded to oxygen or another halogen element.

Compounds are neutral. The oxidation state on all atoms in a compound shall add up to 0. Both BaSO₄ and HClO₂ are neutral.

<h3>BaSO₄</h3>

Oxidation states:

  • Ba: +2;
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  • O: -2.

Let the oxidation state on S be x.

2 + x + 4 × (-2) = 0;

x = 6.

Hence, the oxidation state on S in BaSO₄ is +6.

<h3>HClO₂</h3>

Oxidation states:

  • H: +1;
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  • O: -2.

Let the oxidation state on Cl be x.

<em>Refer to the equation in BaSO₄ as an example. Try setting up the equation on your own. </em>

x = 3.

Hence, the oxidation state on Cl is +3.

<h3>PO₄³⁻</h3>

Ions carry charge. Oxidation states on atoms in an ion shall add up to the charge of the ion. The superscript of an ion shows its charge. The superscript 3- in the phosphate ion shows that the ion carries a charge of -3.

Oxidation states:

  • The oxidation state on P is to be found;
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Let the oxidation state on P be x.

x + 4 × (-2) = -3;

x = 5.

Hence, the oxidation state on P is +5.

4 0
3 years ago
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