1. The number density of conduction electrons in pure silicon at room temperature is about 11016m3. By doping with phosphorous, you desire to increase this number by a factor of 1 million(106). At room temperature, assume that the thermal energy is sufficient that all of the extraelectrons from the P doping enter the conduction band. What fraction of silicon atoms must youreplace with phosphorous atoms

This question is incomplete, the complete question is;

The number density of conduction electrons in pure silicon at room temperature is about 10¹⁶ m⁻³. By doping with phosphorous, you desire to increase this number by a factor of 1 million (10⁶). At room temperature, assume that the thermal energy is sufficient that all of the extra electrons from the P doping enter the conduction band. What fraction of silicon atoms must you replace with phosphorous atoms

the  fraction of silicon atoms we must replace with phosphorous atoms is 4( 10⁶  - 1 )

Explanation:

Given that;  

the number density no =  10¹⁶ m⁻³

Required no =  (10¹⁶ × 10⁶) = 10²² m⁻³

Each silicon atom has 4 valence electron,

Hence; Silicon density =  10¹⁶/4

now, if x = No of silicon atoms to be replaced with phosphorous, then;

(10¹⁶/4 - x)4 + 5x = 10²²

10¹⁶ - 4x + 5x = 10²²

10¹⁶ + x = 10²²

x = 10²² - 10¹⁶

Fraction of silicon atoms will be;

⇒ 10²² - 10¹⁶ / 10¹⁶/4

⇒ 4( 10²² - 10¹⁶ / 10¹⁶ )

⇒ 4( 10²²/10¹⁶  - 10¹⁶/10¹⁶ )

⇒ 4( 10⁶  - 1 )

Therefore, the  fraction of silicon atoms we must replace with phosphorous atoms is 4( 10⁶  - 1 )

when 25.0 ml of 0.700 mol/l naoh was mixed in a calorimeter with 25.0 ml of 0.700 mol/l hcl, both initially at 20.0 °c, the temperature increased to 22.1 °c. the heat capacity of the calorimeter is 279 j/°c.

the equation for the reaction is:

naoh + hcl → nacl + h₂o

moles of hcl = 0.0250 l hcl ×

0.700

mol hcl

1 l hcl = 0.0175 mol hcl

volume of solution = (25.0 + 25.0) ml = 50.0 ml

mass of solution = 50.0 ml soln ×

1.00

ml soln

= 50.0 g soln

δt=t2-t1

= (22.1 – 20.0) °c = 2.1 °c

the heats involved are

heat from neutralization + heat to warm solution + heat to warm calorimeter = 0

q1+q2+q3=0

nδh+mcδt+cδt=0

0.0175 mol ×

δh

+ 50.0 g × 4.184 j·g⁻¹°c⁻¹ × 2.1 °c + 279 j°c⁻¹ × 2.1 °c = 0

0.0175 mol ×

δh

+ 439.32 j + 585.9 j = 0

0.0175 mol ×

δh

= -1025.22 j

δh=1025.22j0.0175 mol= -58 600 j/mol = -58.6 kj/mol

it's c (engineer's do their work with mechanical stuff not on any living being)