<span>Answer:
Moles Ca(NO3)2 = 100 x 0.250 / 1000 = 0.025
Ca(NO3)2 >> Ca2+ + 2NO3-
Moles NO3- = 2 x 0.025 = 0.05
Moles HNO3 = 400 x 0.100 / 1000 = 0.04
Total moles = 0.05 + 0.04 = 0.09
Total volume = 500 ml = 0.500 L
M = 0.09 / 0.500 = 0.18</span>
Doping Se (group VI elements) with P(group V)elements would produce a P-TYPE semiconductor with HIGHER conductivity compared to pure Se
the reason is P dopant will introduce holes in the Se as P has lesser valence electron
A. Fe2O3 + 3CO= 2Fe+3CO2
Here element oxidised is CO or Carbon Monoxide, since oxygen is added.
B. 2HCl+2KMnO4+3H2C2O4=6CO2+2MnO2+2KCl+4H2O
Here Element reduced is 3H2C2O4, since Hydrogen is being added. Also KMnO4 is reduced, since Oxygen is removed.
Since both atoms are the same and are both nonmetals, they would form a Nonpolar covalent bond. This bond occurs when usually atoms of the same element or atoms of propriety electronegativity differences are sharing electrons to form bonds. There is an equal sharing of valence electrons in this chemical bond.
Answer:
Mass of chemical = 1.5 mg
Explanation:
Step 1: First calculate the concentration of the stock solution required to make the final solution.
Using C1V1 = C2V2
C1 = concentration of the stock solution; V1 = volume of stock solution; C2 = concentration of final solution; V2 = volume of final solution
C1 = C2V2/V1
C1 = (6 * 25)/ 0.1
C1 = 1500 ng/μL = 1.5 μg/μL
Step 2: Mass of chemical added:
Mass of sample = concentration * volume
Concentration of stock = 1.5 μg/μL; volume of stock = 10 mL = 10^6 μL
Mass of stock = 1.5 μg/μL * 10^6 μL = 1.5 * 10^6 μg = 1.5 mg
Therefore, mass of sample = 1.5 mg
