C programming IEEE 750 floating point standards
What is the IEEE 750 floating point standard?
Maybe you mean IEEE-754?
IEEE 754-1985 governs binary floating-point arithmetic. It specifies number formats, basic operations, conversions, and exceptional conditions. The related standard IEEE 854-1987 generalizes 754 to cover decimal arithmetic as well as binary.
Saturday, May 22, 2010
Calculate the standard entropy change (J/mol-K).?
For the reaction whose reactants and products are:
reactants: C (graphite) H2O (l)
products: CH3COOH(l)
calculate the standard entropy change (J/mol-K).
The answer depends on how you balance the equation. For this question balance the equation using the smallest ratio of WHOLE numbers.
Enter a numeric answer only
Calculate the standard entropy change (J/mol-K).?
Lancenigo di Villorba (TV), Italy
+2,03 CAL / (K * MOL) = +8.5 J / (MOL * K)
I started from the reaction
2 C(graphite) + 2 H2O(l) ---%26gt; CH3COOH(l)
thus I wrote
DeltaS = 38.19 - 2 * (1.36 + 16.72) = 2.03 cal /(K*mol)
As I think the experiment runs in a Mahler's bomb, I assume the adiabatic nature of it, so I write :
(DeltaS)system = (DeltaS)universe
I hope this helps you.
Reply:the balanced equation is
2C + 2H2O -%26gt; CH3COOH
and for calculating the enthalpy change you need to either the entropy change or the enthalpy of formation of products and enthalpy of dissociation of reactants.
if you are given entropy change that is S then you can use the formula S=H/T
T= TEMPERATURE IN KELVIN
H= ENTHALPY CHANGE
S=ENTROPY CHANGE
if you are given enthalpy of formation of products and enthalpy of dissociation of reactants then you will use the formula
H(total)= H(products) - H(reactants)
and if you are given the GIBBS FREE ENERGY CHANGE (G) then you will use
G = H - TS
G= energy change ( it is zero for elements)
Units of the mentioned parameters:-
H= Joules per mole (J/mol)
S=Joules per kelvin per mole (J/K mol)
T= kelvin (K)
G=Joules per mole (J/mol)
reactants: C (graphite) H2O (l)
products: CH3COOH(l)
calculate the standard entropy change (J/mol-K).
The answer depends on how you balance the equation. For this question balance the equation using the smallest ratio of WHOLE numbers.
Enter a numeric answer only
Calculate the standard entropy change (J/mol-K).?
Lancenigo di Villorba (TV), Italy
+2,03 CAL / (K * MOL) = +8.5 J / (MOL * K)
I started from the reaction
2 C(graphite) + 2 H2O(l) ---%26gt; CH3COOH(l)
thus I wrote
DeltaS = 38.19 - 2 * (1.36 + 16.72) = 2.03 cal /(K*mol)
As I think the experiment runs in a Mahler's bomb, I assume the adiabatic nature of it, so I write :
(DeltaS)system = (DeltaS)universe
I hope this helps you.
Reply:the balanced equation is
2C + 2H2O -%26gt; CH3COOH
and for calculating the enthalpy change you need to either the entropy change or the enthalpy of formation of products and enthalpy of dissociation of reactants.
if you are given entropy change that is S then you can use the formula S=H/T
T= TEMPERATURE IN KELVIN
H= ENTHALPY CHANGE
S=ENTROPY CHANGE
if you are given enthalpy of formation of products and enthalpy of dissociation of reactants then you will use the formula
H(total)= H(products) - H(reactants)
and if you are given the GIBBS FREE ENERGY CHANGE (G) then you will use
G = H - TS
G= energy change ( it is zero for elements)
Units of the mentioned parameters:-
H= Joules per mole (J/mol)
S=Joules per kelvin per mole (J/K mol)
T= kelvin (K)
G=Joules per mole (J/mol)
Standard form?
I don't know how to use standard form. The formula for it is
ax+by=c
I have to solve 2 problems using it.
A horizontal line intersects with a vertical line as (-3,7). Give the equation of each line in standard form.
What are all the steps?
Write an equation in standard form for the line parallel to
x-2y+7=0 and contains the point (-4,0).
Please help me with both of these. I need to show all steps and please explain.
Standard form?
Any horizontal line will be y = number. As you want it to intersect another at (-3, 7) it must be y = 7. This is in standard form. Remember that any of a, b, c can be zero (or one which you don't write in).
Lines which are parallel differ only in the c number so for your second one you want a line parallel to x - 2y = -7 so it will be
x - 2y = number. I'll leave you to find the number knowing that
x = -4, y = 0 must make the equation correct.
Reply:A horizontal line intersects with a vertical line as (-3,7). Give the equation of each line in standard form.
What are all the steps?
Both lines go through the point (-3,7)
where x = -3, the equation is the vertical line is x=-3
where y= 7, the equation is the horizontal line is y=7
--------------------------------------...
Write an equation in standard form for the line parallel to
x-2y+7=0 and contains the point (-4,0).
-2y=-x-7
y=1/2 x +7/2
so the slope you need is 1/2
y=1/2x+b
0=1/2(-4)+b
0=-2+b
2=b
y=1/2 x+2
2y=x+4
-x+2y=4
x-2x=-4
Reply:A linear equation in the standard form is Ax + By + C = 0, where A, B and C are integers, while A%26gt;0
A horzontal line has the form y = a, where a is a real number. i.e, everywhere on this line y has the same value of a, no matter what value of x you pick.
If a horizontal line intersects another line at (-3, 7), then the horiz. must have the equation y = 7.
or, In the standard form, 0x + 1y -7 = 0.
The other line is vertical, that means it has the same x value everywhere, no matter what y value you pick. So the equation of the vertical line must be x = -3, as it passes through x = -3.
In the standard form, x + 0y +3 = 0.
The Secomd Problem:
Two parallel lines have the same slope. So the key here is to find the slope of the given line. The given line is in the standard form, we have to re-arrange that in the slope-intercept form. i.e., y = (1/2 x) + (7/2)
This implies that both lines will have the slope of 1/2, as both are parallel.
So the equation for the new line could be written as
y = (1/2 x) + b. To find b, plug in the coordinates (-4, 0) in the equation. This gives 0 = -2 + b; or, b = 2
So the equation for the new line is y = 1/2x + 2. We need to write that in the standard form. Since only integer coefficients are allowed, let us multiply both sides by 2:
2y = x + 4
Re-arrange to the standard form:
x -2y + 4 = 0.
That is all.
avender
ax+by=c
I have to solve 2 problems using it.
A horizontal line intersects with a vertical line as (-3,7). Give the equation of each line in standard form.
What are all the steps?
Write an equation in standard form for the line parallel to
x-2y+7=0 and contains the point (-4,0).
Please help me with both of these. I need to show all steps and please explain.
Standard form?
Any horizontal line will be y = number. As you want it to intersect another at (-3, 7) it must be y = 7. This is in standard form. Remember that any of a, b, c can be zero (or one which you don't write in).
Lines which are parallel differ only in the c number so for your second one you want a line parallel to x - 2y = -7 so it will be
x - 2y = number. I'll leave you to find the number knowing that
x = -4, y = 0 must make the equation correct.
Reply:A horizontal line intersects with a vertical line as (-3,7). Give the equation of each line in standard form.
What are all the steps?
Both lines go through the point (-3,7)
where x = -3, the equation is the vertical line is x=-3
where y= 7, the equation is the horizontal line is y=7
--------------------------------------...
Write an equation in standard form for the line parallel to
x-2y+7=0 and contains the point (-4,0).
-2y=-x-7
y=1/2 x +7/2
so the slope you need is 1/2
y=1/2x+b
0=1/2(-4)+b
0=-2+b
2=b
y=1/2 x+2
2y=x+4
-x+2y=4
x-2x=-4
Reply:A linear equation in the standard form is Ax + By + C = 0, where A, B and C are integers, while A%26gt;0
A horzontal line has the form y = a, where a is a real number. i.e, everywhere on this line y has the same value of a, no matter what value of x you pick.
If a horizontal line intersects another line at (-3, 7), then the horiz. must have the equation y = 7.
or, In the standard form, 0x + 1y -7 = 0.
The other line is vertical, that means it has the same x value everywhere, no matter what y value you pick. So the equation of the vertical line must be x = -3, as it passes through x = -3.
In the standard form, x + 0y +3 = 0.
The Secomd Problem:
Two parallel lines have the same slope. So the key here is to find the slope of the given line. The given line is in the standard form, we have to re-arrange that in the slope-intercept form. i.e., y = (1/2 x) + (7/2)
This implies that both lines will have the slope of 1/2, as both are parallel.
So the equation for the new line could be written as
y = (1/2 x) + b. To find b, plug in the coordinates (-4, 0) in the equation. This gives 0 = -2 + b; or, b = 2
So the equation for the new line is y = 1/2x + 2. We need to write that in the standard form. Since only integer coefficients are allowed, let us multiply both sides by 2:
2y = x + 4
Re-arrange to the standard form:
x -2y + 4 = 0.
That is all.
avender
Slope Intercept Form to Standard form?
I know how to get Slope Intercept Form (Y=mx+b) but I can't change to standard form I know the formula Ay+By=C
Slope Intercept Form to Standard form?
If you have it in the form (I assume you have an error up there) Ay + Bx = C,
then,
Ay = -Bx + C
y = -B/Ax + C/A
So m = -B/A
and b = C/A
Reply:all you have to do is get the x and y on the same side
ex.
y= 3x+4 slope intercept
subract the 3x to get the y and x on the same side
y-3x=4 standard form
choose as best answer if this helps please
Reply:The formula y-ysub1=m(x-xsub1) is the converter. y and x in this case are the same as the y and the x in the slope intercept form.
Slope Intercept Form to Standard form?
If you have it in the form (I assume you have an error up there) Ay + Bx = C,
then,
Ay = -Bx + C
y = -B/Ax + C/A
So m = -B/A
and b = C/A
Reply:all you have to do is get the x and y on the same side
ex.
y= 3x+4 slope intercept
subract the 3x to get the y and x on the same side
y-3x=4 standard form
choose as best answer if this helps please
Reply:The formula y-ysub1=m(x-xsub1) is the converter. y and x in this case are the same as the y and the x in the slope intercept form.
I am in a need of C++ project on Airline Reservation System.Please send me some and kindly help me?
i have been given a project on C++ for my vacations .I am in 12th standard. This project goes for board evaluation .So any computer savvy who can easily post me a project can help me. i will be very gateful to him/her. i am from Bhopal
I am in a need of C++ project on Airline Reservation System.Please send me some and kindly help me?
you should really research about this kind of stuff. This sites should help you on your way www.cplusplus.com/doc/tutorial/
www.cprogramming.com/tutorial.html
www.glenmccl.com/tutor.htm
oh yeah, if its a major project you should post on some forums for help on client server side c++ category
Reply:u can buy it from software shop.I am from Gwalior
I am in a need of C++ project on Airline Reservation System.Please send me some and kindly help me?
you should really research about this kind of stuff. This sites should help you on your way www.cplusplus.com/doc/tutorial/
www.cprogramming.com/tutorial.html
www.glenmccl.com/tutor.htm
oh yeah, if its a major project you should post on some forums for help on client server side c++ category
Reply:u can buy it from software shop.I am from Gwalior
Determine the following in the standard galvanic cell?
The following standard redox couples are connected to make a galvanic cell.
Sn/Sn2+ E1o = -0.14
Hg/Hg2+ E2o = 0.85
Determine the following for the standard galvanic cell:
a) Eo =
b) delta G =
c) log K =
Determine the following in the standard galvanic cell?
Since you have written;
Sn/Sn2+ E1o = -0.14
Hg/Hg2+ E2o = 0.85
this implies that the values are the oxidation potentials, however I've checked and found that these are the reduction potentials.
Therefore, the reduction potential of Hg2+ is greater than that of Sn2+, Hg2+ will be reduced and Sn will be oxidized.
(a) Anode half-reaction: Sn ----%26gt; Sn2+ + 2e- ..... E° = 0.14 V
Cathode half-reaction: Hg2+ + 2e- ----%26gt; Hg ....... E° = 0.85 V
Cell reaction: Sn + Hg2+ ----%26gt; Sn2+ + Hg .. E°cell = 0.99 V
(b) ΔG° = - n F E°
n = 2 mol e-
F = 96,485 C/mol e- (charge of 1 mol e-)(C = coulomb)
E° = 0.99 V = 0.99 J/C
ΔG° = - n F E°
ΔG° = - (2 mol e-)(96,485 C/mol e-)(0.99 J/C)
ΔG° = - 191,040 J
Positive sign of E° and negative sign of ΔG° indicate that the the process is spontaneous.
(c) log K = n E° / 0.0591 = (2)(0.99) / 0.0591 = 33.5
K = 10^33.5 = 3.16x10^33
Sn/Sn2+ E1o = -0.14
Hg/Hg2+ E2o = 0.85
Determine the following for the standard galvanic cell:
a) Eo =
b) delta G =
c) log K =
Determine the following in the standard galvanic cell?
Since you have written;
Sn/Sn2+ E1o = -0.14
Hg/Hg2+ E2o = 0.85
this implies that the values are the oxidation potentials, however I've checked and found that these are the reduction potentials.
Therefore, the reduction potential of Hg2+ is greater than that of Sn2+, Hg2+ will be reduced and Sn will be oxidized.
(a) Anode half-reaction: Sn ----%26gt; Sn2+ + 2e- ..... E° = 0.14 V
Cathode half-reaction: Hg2+ + 2e- ----%26gt; Hg ....... E° = 0.85 V
Cell reaction: Sn + Hg2+ ----%26gt; Sn2+ + Hg .. E°cell = 0.99 V
(b) ΔG° = - n F E°
n = 2 mol e-
F = 96,485 C/mol e- (charge of 1 mol e-)(C = coulomb)
E° = 0.99 V = 0.99 J/C
ΔG° = - n F E°
ΔG° = - (2 mol e-)(96,485 C/mol e-)(0.99 J/C)
ΔG° = - 191,040 J
Positive sign of E° and negative sign of ΔG° indicate that the the process is spontaneous.
(c) log K = n E° / 0.0591 = (2)(0.99) / 0.0591 = 33.5
K = 10^33.5 = 3.16x10^33
Sunday, August 2, 2009
Math Question.. standard form?
I have to find the equation of a line and the slope is -3/2 and y intercept is 3
so y = -3/2x + 3
how would I put that into standard form? (i think it's ax+by=c.)
x = 6/5
y= 6/5
thanks
Math Question.. standard form?
The standard form equation for a line is
y = mx + b. Where m is the slope and b the y intercept.
y = (-3/2)x + 3 is already the standard form of the equation. Dont worry about the fact that -3/2 is a negative fraction. it simply says that m = -3/2 which is true.
you can multiply throughout by 2 to get rid of the fraction, and rearrange to put the equation in the form ax+by=c but that IS NOT the standard form of the equation for a line.
Reply:It is the standard form... Report It
Reply:3x+2y=6
proof
3x+2y=6
-3x -3x
2y=-3x divid both sides by 2
y=6-3x/2
y=-3x/2+3
Reply:Muultiply y = -3/2x + 3 by 2. That is 2y = -3x + 6.Move the 3x to the other side of the equal. That is 3x+2y=6.
Reply:what's that x and y thing?
Anyways, to put it in standard form you should just multiply everything by 2 to get rid of that denominator and move it around so it would be 3x + 2y = 3
Reply:3x+2y=6
Reply:Wouldn't it = 6? 2y + 3x=6
Last poster i think forgot to mult 3 by 2 as well.
Here... another answer off I'll step it off:
-3/2X +3 = Y
(2)(-3/2X+3)=(2)(Y)
-3X+6 = 2Y
3X+2Y=6
Y=mX + B is 'slope-intercept' form
I posted a good source for Slope-intercept form, Point-Slope Form, and Standard form... last edit i promise :)
Response to additional info:
(1) ax+by+c= 0 is different than (2) ax+by=c because you'd have to subtract c from both sides of equation (2) to get equation (1). In your example problem just subtract 6 from both sides to get 3x+2y-6=0 [a=3,b=2,c=6]
violet
so y = -3/2x + 3
how would I put that into standard form? (i think it's ax+by=c.)
x = 6/5
y= 6/5
thanks
Math Question.. standard form?
The standard form equation for a line is
y = mx + b. Where m is the slope and b the y intercept.
y = (-3/2)x + 3 is already the standard form of the equation. Dont worry about the fact that -3/2 is a negative fraction. it simply says that m = -3/2 which is true.
you can multiply throughout by 2 to get rid of the fraction, and rearrange to put the equation in the form ax+by=c but that IS NOT the standard form of the equation for a line.
Reply:It is the standard form... Report It
Reply:3x+2y=6
proof
3x+2y=6
-3x -3x
2y=-3x divid both sides by 2
y=6-3x/2
y=-3x/2+3
Reply:Muultiply y = -3/2x + 3 by 2. That is 2y = -3x + 6.Move the 3x to the other side of the equal. That is 3x+2y=6.
Reply:what's that x and y thing?
Anyways, to put it in standard form you should just multiply everything by 2 to get rid of that denominator and move it around so it would be 3x + 2y = 3
Reply:3x+2y=6
Reply:Wouldn't it = 6? 2y + 3x=6
Last poster i think forgot to mult 3 by 2 as well.
Here... another answer off I'll step it off:
-3/2X +3 = Y
(2)(-3/2X+3)=(2)(Y)
-3X+6 = 2Y
3X+2Y=6
Y=mX + B is 'slope-intercept' form
I posted a good source for Slope-intercept form, Point-Slope Form, and Standard form... last edit i promise :)
Response to additional info:
(1) ax+by+c= 0 is different than (2) ax+by=c because you'd have to subtract c from both sides of equation (2) to get equation (1). In your example problem just subtract 6 from both sides to get 3x+2y-6=0 [a=3,b=2,c=6]
violet
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