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Leya [2.2K]
2 years ago
7

The superheated water vapor is at 15 MPa and 350°C. The gas constant, the critical pressure, and the critical temperature of wat

er are R = 0.4615 kPa·m3/kg·K, Tcr = 647.1 K, and Pcr = 22.06 MPa. Determine the specific volume of superheated water based on the ideal-gas equation. (You must provide an answer before moving to the next part)
Chemistry
1 answer:
Vesnalui [34]2 years ago
4 0

Answer:

0.01917 m^3/kg.

Explanation:

Given:

P = 15 MPa

= 1.5 × 10^4 kPa

T = 350 °C

= 350 + 273

= 623 K

Molar mass of water, m = (2 × 1) + 16

= 18 g/mol

= 0.018 kg/mol

R = 0.4615 kPa·m3/kg·K

Using ideal gas equation,

P × V = n × R × T

But n = mass/molar mass

V = (R × T)/P

V/M = (R × T)/P × m

= (0.4615 × 623)/1.5 × 10^4

= 0.01917 m^3/kg.

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For the chemical reaction, 2HCl + Ca(OH)2 ➡️ CaCl2 + 2H2O what mass of calcium hydroxide in grams is needed to produce 3.63 mol
wlad13 [49]

Answer:

2HCl (aq) + Ca

(

O

H

)

2

(aq) ---------> Ca

C

l

2

(ppt) + 2

H

2

O (aq)

let us calculate the number of moles , as per the chemical reactions;

2 moles of HCl solution reacts with one mole Calcium Hydroxide

Ca

(

O

H

)

2

One mole of HCl has mass : 36.5 g/mol, two moles of HCl will have mass, 73 g.

One mole of Ca

(

O

H

)

2

has mass 74.1 g

as per equation; 73 g of HCl reacts with 74.1 g of Ca

(

O

H

)

2

1g of HCl reacts with 74.1g / 73 of Ca

(

O

H

)

2

1g of HCl reacts with 1.015 g of Ca

(

O

H

)

2

NOW AS PER THE QUESTION MOLARITY AND VOLUME OF HYDROCHLORIC ACID IS GIVEN, IT CAN BE USED TO CALCULATE THE MASS OF HYDROCHLORIC ACID IN THE SOLUTION.

NUMBER OF MOLES of HCl ; Molarity of solution x Volume of Solution

# of moles of HCl = (0.40 mol/L ) x 350 mL

= (0.40 mol/L ) x 0.350 L = 0.14 mol

mass of HCl that makes 0.14 mol of HCl = # of moles x molar mass of HCl

mass of HCl = 0.14 mol x 36.5 g/ mol

mass of HCl = 5.11g

As per Stoichiometry , 1g of HCl reacts with 1.015 g of Ca

(

O

H

)

2

, 5.11g of HCl can react with 5.11 x 1.015 = 5.1865 g or 5.2 g of

Ca

(

O

H

)

2

.

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