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3 years ago

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Consider a three sequentially stepped process named as Process A, Process B and Process C. Input comes into Process A. Output fo

rm A goes into Process B. Output from B goes into Process C. Output of C is the final output. Suppose that it takes 2 minutes per unit in Process A, 3 minutes per unit in Process B and 1 minute per unit in Process C. Suppose further that Process A receives input at the rate of 30 per hour. Where would you place a buffer?

1 answer:

weqwewe [10]3 years ago

3 0

Answer:B the buffer will be placed in process B

Explanation:

Buffer is the extra time added to a time estimate to keep a project on track. Since process B takes a longer time to process, the buffer will be added to prevent unforseen situations.

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Consider the following reaction: . SO2Cl2(g)⇌SO2(g)+Cl2(g) . A reaction mixture is made containing an initial [SO2Cl2] of 2.2×10

Alex Ar [27]

The equilibrium constant (Kc) is the product of the equilibrium concentrations of the products raised to their corresponding stoichiometric coefficients divided by the reactants as well. In this case the equilibrium concentration of Cl2 which also applies to SO2 is 1.3x10^-2. The final equilibrium concentration of SO2Cl2 is 9x10^-3. Kc is then equal to 0.0188.

3 0

3 years ago

Read 2 more answers

An ethylene glycol solution contains 21.4 g of ethylene glycol (C2H6O2) in 97.6 mL of water. (Assume a density of 1.00 g/mL for

8090 [49]

Answer: The freezing point and boiling point of the solution are $-6.6^0C$ and $101.8^0C$ respectively.

Explanation:

Depression in freezing point:

$T_f^0-T^f=i\times k_f\times \frac{w_2\times 1000}{M_2\times w_1}$

where,

$T_f$ = freezing point of solution = ?

$T^o_f$ = freezing point of water = $0^0C$

$k_f$ = freezing point constant of water = $1.86^0C/m$

i = vant hoff factor = 1 ( for non electrolytes)

m = molality

$w_2$ = mass of solute (ethylene glycol) = 21.4 g

$w_1$ = mass of solvent (water) = $density\times volume=1.00g/ml\times 97.6ml=97.6g$

$M_2$ = molar mass of solute (ethylene glycol) = 62g/mol

Now put all the given values in the above formula, we get:

$(0-T_f)^0C=1\times (1.86^0C/m)\times \frac{(21.4g)\times 1000}{97.6g\times (62g/mol)}$

$T_f=-6.6^0C$

Therefore,the freezing point of the solution is $-6.6^0C$

Elevation in boiling point :

$T_b-T^b^0=i\times k_b\times \frac{w_2\times 1000}{M_2\times w_1}$

where,

$T_b$ = boiling point of solution = ?

$T^o_b$ = boiling point of water = $100^0C$

$k_b$ = boiling point constant of water = $0.52^0C/m$

i = vant hoff factor = 1 ( for non electrolytes)

m = molality

$w_2$ = mass of solute (ethylene glycol) = 21.4 g

$w_1$ = mass of solvent (water) = $density\times volume=1.00g/ml\times 97.6ml=97.6g$

$M_2$ = molar mass of solute (ethylene glycol) = 62g/mol

Now put all the given values in the above formula, we get:

$(T_b-100)^0C=1\times (0.52^0C/m)\times \frac{(21.4g)\times 1000}{97.6g\times (62g/mol)}$

$T_b=101.8^0C$

Thus the boiling point of the solution is $101.8^0C$

4 0

3 years ago

How many moles of ammonium sulfate can be from the reaction of 30.0 mol of NH3 with H2SO4 according to the following equation: 2

marysya [2.9K]

Answer:

15 moles of ammonium sulfate would be formed from 30 moles of ammonia.

Explanation:

Given data:

Number of moles of ammonium sulfate formed = ?

Number of moles of ammonia = 30.0 mol

Solution:

Chemical equation:

2NH₃ + H₂SO₄ → (NH₄)₂SO₄

Now we will compare the moles of ammonium sulfate with ammonia.

NH₃ : (NH₄)₂SO₄

2 : 1

30.0 : 1/2×30.0 = 15.0 mol

So 15 moles of ammonium sulfate would be formed from 30 moles of ammonia.

7 0

3 years ago

Please help!!! (im really bad at chemistry)

Lubov Fominskaja [6]

Answer:

1000ml/1L

Explanation:

8 0

2 years ago

Balance the equation

andrezito [222]

Balanced equation: shown in photo

6 0

2 years ago