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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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A rectangular open box, 25 ft by 10 ft in plan and 12 ft deep weighs 40 tons. Sufficient amount of stones is placed in the box a
dimulka [17.4K]

Answer:

44.95 tonnes

Explanation:

According to principle of buoyancy the object will just sink when it's weight is more than the weight of the liquid it displaces

It is given that empty weight of box = 40 tons

Let the mass of the stones to be placed be = M tonnes

Thus the combined mass of box and stones = (40+M) tonnes..........(i)

Since the box will displace water equal to it's volume V we have volume of box = 25ft*10ft*12ft= 3000ft^{3}

Volume= 84.95m^{3}

Since 1ft^{3} =0.028m^{3}

Now the weight of water displaced = Weight =\rho \times Volumewhererho is density of water = 1000kg/m^{3}

Thus weight of liquid displaced = \frac{84.95X1000}{1000}tonnes=84.95 tonnes..................(ii)

Equating i and ii we get

40 + M = 84.95

thus Mass of stones = 44.95 tonnes

3 0
3 years ago
HELP PLS
Angelina_Jolie [31]

Answer:

The correct option is;

B) Metamorphic Rocks

Explanation:

Zoisite, which is also referred to saualpite, is a metamorphic rock which is a hydroxy sorosilicate mineral formed from other types of rocks such as sedimentary, metamorphic and ingenious rocks in the process of their metamorphism under the presence high temperatures and pressures and mineral fluids which are hot

Zoiste is named after Sigmund Zois by Abraham Gottlob Werner in 1805 when Sigmund Zois sent Abraham Gottlob Werner the mineral specimen from Saualpe in 1805

6 0
3 years ago
The design specifications of a 1.2-m long solid circular transmission shaft require that the angle of twist of the shaft not exc
Verizon [17]

Answer:

c = 18.0569 mm

Explanation:

Strategy  

We will find required diameter based on angle of twist and based on shearing stress. The larger value will govern.  

Given Data  

Applied Torque

T = 750 N.m

Length of shaft

L = 1.2 m

Modulus of Rigidity

G = 77.2 GPa

Allowable Stress

г = 90 MPa

Maximum Angle of twist  

∅=4°

∅=4*\pi/180

∅=69.813 *10^-3 rad

Required Diameter based on angle of twist  

∅=TL/GJ

∅=TL/G*\pi/2*c^4

∅=2TL/G*\pi*c^4

c=\sqrt[4]{2TL/\pi G }∅

c=18.0869 *10^-3 rad

Required Diameter based on shearing stress

г = T/J*c

г = [T/(J*\pi/2*c^4)]*c

г =[2T/(J*\pi*c^4)]*c

c=17.441*10^-3 rad

Minimum Radius Required  

We will use larger of the two values  

c= 18.0569 x 10^-3 m  

c = 18.0569 mm  

3 0
3 years ago
DRIVERS ED
forsale [732]

Answer:

b

Explanation:

only if there signal is turned on

8 0
3 years ago
Read 2 more answers
Write a program that prompts the user to enter time in 12-hour notation. The program then outputs the time in 24-hour notation.
Juliette [100K]

Answer:

THE CODE FOR THE PROGRAM IS GIVEN BELOW:

#include <iostream>

#include "ConvertTimeHeader.h"

using namespace std;

int main()

{

convertTime convert;

int hr, mn, sc = 0;

 

cout << "Please input hours in 12 hr notation: ";

cin >> hr;

cout << "Please input minutes: ";

cin >> mn;

cout << "Please input seconds: ";

cin >> sc;

 

convert.invalidHr(hr);

convert.invalidMin(mn);

convert.invalidSec(sc);

convert.printMilTime();

 

system("Pause");

 

return 0;  

 

}

#include <iostream>

#include "ConvertTimeHeader.h"

using namespace std;

int convertTime::invalidHr (int hour)

{

try{

 if (hour < 13 && hour > 0)

  {hour = hour + 12;

  return hour;}

 else{

 

  cin.clear();

  cin.ignore();

  cout << "Invalid input! Please input hour again in correct 12 hour format: ";

  cin >> hour;

  invalidHr(hour);

  throw 10;

 }

   

}

catch (int c) { cout << "Invalid hour input!";}

}

int convertTime::invalidMin (int min)

{

try{

 if (min < 60 && min > 0)

  {return min;}

 else{

 

  cin.clear();

  cin.ignore();

  cout << "Invalid input! Please input minutes again in correct 12 hour format: ";

  cin >> min;

  invalidMin(min);

  throw 20;

  return 0;

 }

   

}

catch (int e) { cout << "Invalid minute input!" << endl;}

}

int convertTime::invalidSec(int sec)

{

try{

 if (sec < 60 && sec > 0)

  {return sec;}

 else{

 

  cin.clear();

  cin.ignore();

  cout << "Invalid input! Please input seconds again in correct 12 hour format: ";

  cin >> sec;

  invalidSec(sec);

  throw 30;

  return 0;

 }

   

}

catch (int t) { cout << "Invalid second input!" << endl;}

}

void convertTime::printMilTime()

{

cout << "Your time converted: " << hour << ":" << min << ":" << sec;

}

Explanation:

4 0
3 years ago
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