Answer:
This process is known as doping. It can be done by adding either of two types of impurity to the crystal.
(A) By adding electron rich impurities i.e., group 15 elements to the silicon and germanium of group 14 elements.
hope it's helpful
Answer:
Explanation:
Given that:
Pressure = 791 mmHg
Temperature = 20.0°C
The conversion of T( °C) to T(K) is shown below:
T(K) = T( °C) + 273.15
So,
T = (20 + 273.15) K = 293.15 K
T = 293.15 K
Volume = 100 L
Using ideal gas equation as:
PV=nRT
where,
P is the pressure
V is the volume
n is the number of moles
T is the temperature
R is Gas constant having value = 62.3637 L.mmHg/K.mol
Applying the equation as:
791 mmHg × 1.14 L = n × 62.3637 L.mmHg/K.mol × 293.15 K
⇒n of 
produced = 0.0493 moles
According to the reaction:-
1 mole of carbon dioxide is produced 1 mole of calcium carbonate reacts
0.0493 mole of carbon dioxide is produced 0.0493 mole of calcium carbonate reacts
Moles of calcium carbonate reacted = 0.0493 moles
Molar mass of 
= 100.0869 g/mol
The formula for the calculation of moles is shown below:
Thus,
Impure sample mass = 5.28 g
Percent mass is percentage by the mass of the compound present in the sample.
Answer:
Total worth of gold in the ocean = $5,840,000,000,000,000
Explanation:
As stated in the question above, 4.0 x 10^-10 g of gold was present in 2.1mL of ocean water.
Therefore, In 1 L of ocean water there will be,
(4.0 x 10^-10)/0.0021
= 1.9045 x 10^-7 g of gold per Liter of ocean water.
So in 1.5 x 10^-21 L of ocean water, there will be
(1.9045 x 10^-7) * (1.5 x 10^-21)
= 2.857 x 10^14 g of gold in the ocean.
1 gram of gold costs $20.44, that is 20.44 dollars/gram. The total cost of the gold present in the ocean is
20.44 * (2.857 x 10^14)
= $5,840,000,000,000,000
Answer:
119 kCal per serving.
Explanation:
The heat energy necessary to elevates water's temperature from 23.4°C to 37.9°C can be calculated by the equation below:
Q = mcΔT
Q: heat energy
m: mass in g
c: specific heat capacity in cal/g°C
ΔT = temperature variation in °C
m is the mass of water, considering the density of water to be 1g/mL, 100 mL of water weights 100g. Therefore:
Q = 100 g x 1.00 cal/g°C x (37.9 - 23.4)°C
Q = 1450 cal
1450 cal ____ 0.341 g peanuts
x ____ 28 g peanuts
x = 119061.58 cal
This means that the cal from fat per serving of peanuts is at least 119 kCal.
