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stealth61 [152]
3 years ago
9

Nitrate salts (NO3-), when heated, can produce nitrites (NO2-) plus oxygen (O2). A sample of potassium nitrate is heated, and th

e O2 gas produced is collected in a 720 mL flask. The pressure of the gas in the flask is 2.9 atm, and the temperature is recorded to be 319 K. The value of R= 0.0821 atm L/(mol K)
How many moles of O2 gas were produced?

After a few hours, the 720 mL flask cools to a temperature of 293K.

Pnew = atm

What is the new pressure due to the O2 gas?
Engineering
1 answer:
Mice21 [21]3 years ago
5 0

Answer:

0.07973 moles of O2 gas were produced

2.66 atm

Explanation:

Using the ideal gas equation

PV =nRT

P =pressure = 2.9 atm

V= volume = 720 mL =  720mL/1000= 0.72 L

n = number of moles = ?

R = the gas constant = 0.0821 atm L/(mol K)

T = temperature in Kelvin = 319 K

make n subject of the formula

n =PV/RT

n = 2.9 atm x 0.72 L / 0.0821 atm L/(mol K) x 319K = 0.07973 moles of O2 gas were produced

Pnew = atm?

P = nRT/V

 = 0.07973 x 0.0821 atm L/(mol K) x 293K/0.72 L = 2.66 atm

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Write the following decorators and apply them to a single function (applying multiple decorators to a single function): 1. The f
natita [175]

Answer:

Complete question is:

write the following decorators and apply them to a single function (applying multiple decorators to a single function):

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3. The second decorator is called emphasis and has an inner function called wrapper. The purpose of this decorator is to add the html tags of <em> and </em> to the argument of the decorator similar to step 1. The return value of the wrapper should look like: return “<em>” + func() + “</em>.

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Code :

def strong_decorator(func):

def func_wrapper(name):

return "<strong>{0}</strong>".format(func(name))

return func_wrapper

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def func_wrapper(name):

return "<em>{0}</em>".format(func(name))

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4 years ago
8. Two 40 ft long wires made of differing materials are supported from the ceiling of a testing laboratory. Wire (1) is made of
san4es73 [151]

Answer:

Material K has a modulus of elasticity E=3.389× 10¹¹ Pa

Material H has a modulus of elasticity E=1.009 × 10⁹ Pa

Material K has higher value of modulus of elasticity than material H

Material K is stiffer.

Explanation:

Wire 1 material H

Length=L = 40 ft =12.192 m

Diameter= 3/8 in = 0.009525 m

Area= A= πr²,where r=0.009525/2 =0.004763

A=3.142*0.004763² =0.00007126 m²

Force, F= 225 lb=  225*4.45 =1001.25 N

Change in length =Δ L= 0.10 in = 0.00254

To find modulus of elasticity apply'

E=F*L/A*ΔL

E=1001.25*12.192/(0.004763*0.00254)

E= 1009027923.58 Pa

E=1.009 × 10⁹ Pa

For Wire 2 material K

Length=L= 40 ft =12.192 m

Diameter = 3/16 in = 0.1875 in = 0.004763 m

Area= πr² = 3.142 * (0.004763/2)² = 0.00000567154 m²

Force, F= 225 lb=  225*4.45 =1001.25 N

Change in length =Δ L= 0.25 in =0.00635 m

To find modulus of elasticity apply'

E=F*L/A*ΔL

E= (1001.25*12.192)/(0.00000567154 * 0.00635 )

E=338955422575 Pa

E=3.389× 10¹¹ Pa

Material  K has a greater modulus of elasticity

The material with higher value of E is stiffer than that with low value of E.The stiffer material is K.

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mario62 [17]

Answer:

Explanation:

Ohms Law I=E/R (resistive requires no power factor correction)

150/25= 6 amps

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