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3 years ago

If you measure 0.7 V across a diode, the diode is probably made of

Engineering

1 answer:

tatuchka [14]3 years ago

6 0

Answer:

Made of Silicon.

Explanation:

A diode is a semiconductor device use in mostly electronic appliances. It is two terminals device consisting of a P-N junction formed either in Germanium or silicon crystal.

Diode can be forward biased or reverse biased.

When a diode is forward biased and the applied voltage is increased from zero, hardly any current flows through the device in the beginning.

It is so because the external voltage is being opposed by the internal barrier voltage whose value is 0.7v for silicon and 0.3v for germanium.

If you measure 0.7 V across a diode, the diode is probably therefore made of Silicon.

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1. Copy the file Pay.java (see Code Listing 1.1) from the Student CD or as directed by your instructor. 2. Open the file in your

Mrac [35]

Answer:

Code Listing 1.1 (Pay.java)

import java.util.Scanner; // Needed for the Scanner class

/**

This program calculates the user's gross pay.

public class Pay

{

public static void main(String[] args)

{

// Create a Scanner object to read from the keyboard. Scanner keyboard = new Scanner(System.in);

// Identifier declarations

double hours; // Number of hours worked

double rate; // Hourly pay rate double pay; // Gross pay

// Display prompts and get input. System.out.print("How many hours did you work? "); hours = keyboard.nextDouble();

System.out.print("How much are you paid per hour? ");

rate = keyboard.nextDouble();

// Perform the calculations. if(hours <= 40)

pay = hours * rate;

else

pay = (hours - 40) * (1.5 * rate) + 40 * rate;

// Display results. System.out.println("You earned $" + pay);

}

Code Listing 1.2 (SalesTax.java)

import java.util.Scanner; // Needed for the Scanner class

/**

This program calculates the total price which includes

sales tax.

public class SalesTax

{

public static void main(String[] args)

{

// Identifier declarations final double TAX_RATE = 0.055; double price;

double tax

double total; String item;

// Create a Scanner object to read from the keyboard. Scanner keyboard = new Scanner(System.in);

// Display prompts and get input. System.out.print("Item description: "); item = keyboard.nextLine(); System.out.print("Item price: $");

price = keyboard.nextDouble();

// Perform the calculations. tax = price + TAX_RATE;

totl = price * tax;

// Display the results. System.out.print(item + " $"); System.out.println(price); System.out.print("Tax $"); System.out.println(tax); System.out.print("Total $"); System.out.println(total);

}

5 0

3 years ago

NO reacts with Br2 in the gas phase according to the following chemical equation: 2NO(g) +Br2(g)2NOBr(g) It is observed that, wh

klemol [59]

Answer:

a) $rate=r=k[NO]^{2} [Br_{2}]^{1}$

b) $k=\frac{1}{s*M^{2}}$

Explanation:

First of all you need to indicate the reaction order of each reactant ( $NO$ and $Br_{2}$ ):

1. $Br_{2}$

Note that if $Br_{2}$ concentration ( $[Br_{2} ]$ ) is reduced to 1/3 of its initial value, the rate of the reaction is also reduced to 1/3 of its initial value, it means:

$[Br_{2} ]=1/3$ then $r=1/3$

As the change in the rate of the reaction is equal to the change of the initial concentration of $Br_{2}$ , you could concluded that the reaction is first order with respect to $Br_{2}$

2. $NO$

Now, note that if $NO$ concentration ( $[NO]$ ) is multiplied by 3.69, the rate of the reaction increases by a factor of 13.6. In this case, to know the ratio could be advisable divide the rate of the reaction (13.6) over the factor whereby was multiplied the concentration (3.69), as follows:

$\frac{13.6}{3.69}=3.69$

As the result is the same factor 3.69 you could concluded that the change of the rate of reaction is proportional to the square of the concentration of A:

$r=[NO]^{2} =3.68^{2} =13.6$

It means that the reaction is second order with respect to $NO$

3. Rate Expression

Remember that the rate expression of the reactions depend on the concentration of each reactant and its order. In this case we have 2 reactants: $NO$ and $Br_{2}$ , then we have a rate law depending of 2 concentrations, as follows:

Note that the expression is the result of the concentration of each reactant raised to its reaction order (previously determined)

<em>Note: I hope that you do not mix up the use of the rates of reaction of each reactant, that is experimentally determined, with the stoichiometric coefficient, are different.</em>

4. Rate constant units (k)

Assuming concentration is expressed as $\frac{mol}{L}=M$ and time is in second, to find the units of k we need to solve an equation with units and with supporting of the rate equation previously obtained, as follows: