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3 years ago

7

Benzoic acid 1.35 g, is reacted with oxygen in a constant volumecalorimeter to form H2 O(l) and CO2 (g)at 298 K. the mass of the

water volume in the inner bath is1.376*103 g. The temperature of the calorimeter and itsconstant rises 3.11 K as a result of this reaction. Calculate thecalorimeter constant.

1 answer:

brilliants [131]3 years ago

6 0

Answer:

Explanation:C7H6O2 + (15/2) O2 = > 3H2O + 7CO2 delH = Σ stoichiometric coefficient* enthalpy of formation delH = (7*-393.509 kJ/mol) + (3*-285.83 kJ/mol) - (15/2 * 0) -(1*-385.2 kJ/mol) delH = -3226.853 kJ/mol benzoic acid ...

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If it took 2 hours to clean 4 houses how many houses could be cleaned in 3 hours

Lena [83]

Answer:

6 houses

Explanation:

because

2hrs=4 houses which means you are cleaning 2houses in one hour

so in 3 hours you will houses because you will clean 2 houses in one hour

I hope this helped you sorry if I am wrong

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3 years ago

Write a statement to print the data members of InventoryTag. End with newline. Ex: if itemID is 314 and quantityRemaining is 500

Advocard [28]

Answer:

#include <stdio.h>

typedef struct InventoryTag_struct {

int itemID;

int quantityRemaining;

} InventoryTag;

int main(void) {

InventoryTag redSweater;

redSweater.itemID = 314;

redSweater.quantityRemaining = 500;

/* Your solution goes here */

printf("Inventory ID: %d, Qty: %d\n",redSweater.itemID,redSweater.quantityRemaining);

getchar();

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Explanation:

7 0

3 years ago

A rotating beam is subject to an alternating stress of 48 kpsi and a mean stress of 24 kpsi. The ultimate strength of the materi

Alisiya [41]

Answer:

goodman = 0.694

life of beam = 211597

Explanation:

alternating stress = 48 kpsi

mean stress = 24 kpsi

ultimate strength = 100 kpsi

endurance limit = 40 kpsi

goodman:

= $\frac{mean stress}{ultimate stress} +\frac{alternating stress}{endurance limit} =\frac{1}{N}$

= $\frac{24}{100} +\frac{48}{40} =\frac{1}{N}$

= 0.24 + 1.2 = $\frac{1}{N}$

N = 1/1.44

N = 0.694

2. check attachment for diagram

Log(N)-3/3 = log90 - log48/log90 - log40

Log(N)-3/3 = 0.77517

Log N = 5.325509

N = 10^(5.325509)

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3 years ago

A 50 Hz, four pole turbo-generator rated 100 MVA, 11 kV has an inertia constant of 8.0 MJ/MVA. (a) Find the stored energy in the

raketka [301]

Given Information:

Frequency = f = 60 Hz

Complex rated power = G = 100 MVA

Intertia constant = H = 8 MJ/MVA

Mechanical power = Pmech = 80 MW

Electrical power = Pelec = 50 MW

Number of poles = P = 4

No. of cycles = 10

Required Information:

(a) stored energy = ?

(b) rotor acceleration = ?

(c) change in torque angle = ?

(c) rotor speed = ?

Answer:

(a) stored energy = 800 Mj

(b) rotor acceleration = 337.46 elec deg/s²

(c) change in torque angle (in elec deg) = 6.75 elec deg

(c) change in torque angle (in rmp/s) = 28.12 rpm/s

(c) rotor speed = 1505.62 rpm

Explanation:

(a) Find the stored energy in the rotor at synchronous speed.

The stored energy is given by

$E = G \times H$

Where G represents complex rated power and H is the inertia constant of turbo-generator.

$E = 100 \times 8 \\\\E = 800 \: MJ$

(b) If the mechanical input is suddenly raised to 80 MW for an electrical load of 50 MW, find rotor acceleration, neglecting mechanical and electrical losses.

The rotor acceleration is given by

$P_a = P_{mech} - P_{elec} = M \frac{d^2 \delta}{dt^2}$

Where M is given by

$M = \frac{E}{180 \times f}$

$M = \frac{800}{180 \times 50}$

$M = 0.0889 \: MJ \cdot s/ elec \: \: deg$

So, the rotor acceleration is

$P_a = 80 - 50 = 0.0889 \frac{d^2 \delta}{dt^2}$

$30 = 0.0889 \frac{d^2 \delta}{dt^2}$

$\frac{d^2 \delta}{dt^2} = \frac{30}{0.0889}$

$\frac{d^2 \delta}{dt^2} = 337.46 \:\: elec \: deg/s^2$

(c) If the acceleration calculated in part(b) is maintained for 10 cycles, find the change in torque angle and rotor speed in revolutions per minute at the end of this period.

The change in torque angle is given by

$\Delta \delta = \frac{1}{2} \cdot \frac{d^2 \delta}{dt^2}\cdot (t)^2$

Where t is given by

$1 \: cycle = 1/f = 1/50 \\\\10 \: cycles = 10/50 = 0.2 \\\\t = 0.2 \: sec$

So,

$\Delta \delta = \frac{1}{2} \cdot 337.46 \cdot (0.2)^2$

$$ \Delta \delta = 6.75 \: elec \: deg$

The change in torque in rpm/s is given by

$\Delta \delta = \frac{337.46 \cdot 60}{2 \cdot 360\circ }$

$\Delta \delta =28.12 \: \: rpm/s$

The rotor speed in revolutions per minute at the end of this period (10 cycles) is given by

$Rotor \: speed = \frac{120 \cdot f}{P} + (\Delta \delta)\cdot t$

Where P is the number of poles of the turbo-generator.

$Rotor \: speed = \frac{120 \cdot 50}{4} + (28.12)\cdot 0.2$

$Rotor \: speed = 1500 + 5.62$

$$ Rotor \: speed = 1505.62 \:\: rpm$

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3 years ago

Test question for my account

Ierofanga [76]

Answer:

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3 years ago