Answer:
Explanation:
a) for shifting reactions,
Kps = ph2 pco2/pcoph20
=[h2] [co2]/[co] [h2o]
h2 + co2 + h2O + co + c3H8 = 1
it implies that
H2 + 0.09 + H2O + 0.08 + 0.05 = 1
solving the system of equation yields
H2 = 0.5308,
H2O = 0.2942
B) according to Le chatelain's principle for a slightly exothermic reaction, an increase in temperature favors the reverse reaction producing less hydrogen. As a result, concentration of hydrogen in the reformation decreases with an increasing temperature.
c) to calculate the maximum hydrogen yield , both reaction must be complete
C3H8 + 3H2O ⇒ 3CO + 7H2( REFORMING)
CO + H2O ⇒ CO2 + H2 ( SHIFTING)
C3H8 + 6H2O ⇒ 3CO2 + 10 H2 ( OVER ALL)
SO,
Maximum hydrogen yield
= 10mol h2/3 molco2 + 10molh2
= 0.77
⇒ 77%
Answer:
City planner would be a good answer. Unless there are any options, this seems like the best bet. Hope this helps!
Answer:
<u>Yes</u>
Explanation:
In such a case, one way to check the <em>credit charge</em> is to <u>contact your bank, </u>doing so would allow the bank to check your account properly to determine where the transaction was originated from.
Another way you could check is to contact the online merchant where such a transaction was initiated.
Answer:
required feedback resistance ( R2 ) = 100 k Ω
Explanation:
Given data :
Voltage gain = 100
input resistance ( R1 ) = 1 k ohms
calculate feedback resistance required
voltage gain of differential amplifier
= Voltage gain = R2/R1
= 100 = R2/1
hence required feedback resistance ( R2 ) = 100 k Ω
Answer:
#include <iostream>
#include <iomanip>
using namespace std;
class pointType
{
public:
pointType()
{
x=0;
y=0;
}
pointType::pointType(double x,double y)
{
this->x = x;
this->y = y;
}
void pointType::setPoint(double x,double y)
{
this->x=x;
this->y=y;
}
void pointType::print()
{
cout<<"("<<x<<","<<y<<")\n";
}
double pointType::getX()
{return x;
}
double pointType::getY()
{return y;
}
private:
double x,y;
};
int main()
{
pointType p2;
double x,y;
cout<<"Enter an x Coordinate for point ";
cin>>x;
cout<<"Enter an y Coordinate for point ";
cin>>y;
p2.setPoint(x,y);
p2.print();
system("pause");
return 0;
}
