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2 years ago

How many kg moles of Sodium Sulphate will contain 10 kg of Sodium?

Engineering

2 answers:

IRISSAK [1]2 years ago

8 0

Answer:

70.40mol cuz

1g sodium sulfate = 0.00704mol

take 10kg × 1000 = 10,000g

10,000g × 0.00704

final answer 70.40mol

enyata [817]2 years ago

5 0

70.40mol cuz

1g sodium sulfate = 0.00704mol

take 10kg × 1000 = 10,000g

10,000g × 0.00704

final answer 70.40mol

(as per my thinking)

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almond37 [142]

Answer:

a)exit velocity of the steam, V2 = 2016.8 ft/s

b) the amount of entropy produced is 0.006 Btu/Ibm.R

Explanation:

Given:

P1 = 100 psi

V1 = 100 ft./sec

T1 = 500f

P2 = 40 psi

n = 95% = 0.95

a) for nozzle:

Let's apply steady gas equation.

$h_1 + \frac{(v_1) ^2}{2} = h_2 + \frac{(v_2)^2}{2}$

h1 and h2 = inlet and exit enthalpy respectively.

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s1 = 1.708 Btu/Ibm.R

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1193.5 Btu/Ibm

Let's find the actual h2 using the formula :

$n = \frac{h_1 - h_2*}{h_1 - h_2}$

$n = \frac{1278.8 - h_2*}{1278.8 - 1193.5}$

solving for h2, we have

$h_2 = 1197.77 Btu/Ibm$

Take Btu/Ibm = 25037 ft²/s²

Using the first equation, exit velocity of the steam =

$(1278.8 * 25037) + \frac{(100)^2}{2}= (1197.77*25037)+ \frac{(V_2)^2}{2}$

Solving for V2, we have

V2 = 2016.8 ft/s

b) The amount of entropy produced in BTU/ lbm R will be calculated using :

Δs = s2 - s1

Where s1 = 1.708 Btu/Ibm.R

At h2 = 1197.77 Btu/Ibm and P2 =40 psi,

S2 = 1.714 Btu/Ibm.R

Therefore, amount of entropy produced will be:

Δs = 1.714Btu/Ibm.R - 1.708Btu/Ibm.R

= 0.006 Btu/Ibm.R

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