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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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The two boxcars A and B have a weight of 20000lb and 30000lb respectively. If they coast freely down the incline when the brakes
kolbaska11 [484]

Answer:

T=5.98 kips

Explanation:

First, introduce forces, acting on both cars:

on car A there are 4 forces acting: gravity force mA*g, normal reaction force, friction force and force T- it represents the interaction between cars A and B. On car B, there are three forces acting: gravity force, normal reaction force and force T. Note, that force T is acting on both cars, but it has opposite direction. Force T, acting on car A has direction, opposite to the friction force, whether the T, acting on B, is directed backwards- in the same direction with the friction force. Note, that both cars have the same acceleration, which is directed backwards.

Once the forces were established, we can write components of the Second Newtons Law on vertical and horizontal axes, considering that horizontal axis is directed backwards- in the same direction with the acceleration:

For car A on the vertical axis the equation is: -mAg+NA=0

For car A on the horizontal axis, the equation is: Ffr-T=mAa

For car B, on the vertical axis the equation is: -mBg+NB=0

For car B, on the horizontal axis, the equation is: T=mBa

We need to solve these equations to find force T, knowing that Ffr=μmAg, where

After the transformations, the equations for acceleration and force in the coupling will be:

a=(μmAg)/(mA+mB)=6.43 ft/s2- note, that the given answer is not correct for the given numerical values;

and force T: T=μmAmBg/(mA+mB)=6.0 kips- note, that the force answer is in line with the given numerical value

5 0
3 years ago
A certain piece of property is assessed at $150,000. If the tax rate is $2.50 per $100, what is the tax on this property?
stiks02 [169]

Answer:

The tax on this property is 3750 dollars

Explanation:

Given

Tax on per $100 is $2.50

Tax on every $1 is \frac{2.5}{100} = 0.025 dollars

Tax on property of value $150,000 is

150,000 * 0.025 = 3750 dollars

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2 years ago
Which of the following can effect LRO?
aleksley [76]

Answer:

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2 years ago
Calculate the osmotic pressure of seawater containing 3.5 wt % NaCl at 25 °C . If reverse osmosis is applied to treat seawater,
AlladinOne [14]

Answer:

Highest osmotic pressure that membrane may experience is

' =58.638 atm

Explanation:

Suppose sea-water taken is M= 1 kg

Density of water = 1000 kg/m3

Therefore Volume of water= Mass,M/Density of water

V= 1 kg/(1000 kg/m3)

V= 10-3 m3= 1 Litre

Since mass of Nacl is 3.5 wt%,Therefore in 1 kg of water

Mass present of NaCl= m= 0.035*1000 g

m= 35 g

Since molecular weight of NaCl= 58.44 g/mol =M.W.

Thus its Number of moles of Nacl= m/M.W

nNaCl= 35g/58.44 gmol-1

= 0.5989 mol

ans since volume of solution is 1 L thus concentration of NaCl is ,C= number of moles/Volume of solution in Litres

C= 0.5989mol/ 1L

=0.5989 M

Since 1 mol NaCL disssociates to form 2 moles of ions of Na+ andCl- Thus van't hoff factor i=2

And osmotic pressure  = iCRT ------------------------------(1)( Where R= 0.0821 L.atm/mol.K and T= 25oC= 298.15 K)

Putting in equation 1 ,we get  = 2*(0.5989 mol/L)*(0.0821 L.atm/mol.K)*298.15 K

=29.319 atm

Now as the water gets filtered out of the membrane,the water's volume decreases and concentration C of NacL increases, thus osmotic pressure also increases.Thus, at 50% water been already filtered out, the osmotic pressure at the membrane will be maximum

Thus Volume of water left after 50% is filtered out as fresh water= 0.5 L (assuming no salt passes through semi permeable membrane)

Thus New concentration of NaCl C'= 2*C

C'=2*0.5989 M

=1.1978 M

and Since Osmotic pressure is directly proportional to concentration, Thus As concentration C doubles to C', Osmotic Pressure  ' also doubles from  ,

Thus,Highest osmotic pressure that membrane may experience is,  '=2*  

=2*29.319 atm

' =58.638 atm

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3 years ago
What are the constraints of the problem for thermoforming
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Answer:

here are four options so you can choose the ones of your choice.

Explanation:

1. warpage

2 Dimensional Inconsistencies

3 Part Thickness Inconsistencies

4 Lack of Detail in Part Geometry and Aesthetics

please rate brainliest if helps and follow

6 0
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