Explanation:
The required concentration of M1 =0.222 M.
The required volume of is V1 =225 mL.
The standard solution of is M2 =16 M.
The volume of standard solution required can be calculated as shown below:
Since the number of moles of solute does not change on dilution.
The number of moles
Hence, 3.12 mL of 16 m nitric acid is required to prepare 0.222 M and 225 mL of nitric acid.
Answer:
True
Explanation:
In 1909 Ernest Rutherford proposed to Geiger and Marsden to conduct an experiment in which they would have to launch alpha particles from a radioactive source against a gold foil a few atoms thick.
The aim of the experiment was to corroborate Rutherford's idea that the particles would pass through the metal foil with little deviation.
Rutherford's interpretation of the experimental results gave rise to a new atomic model, and concluded that the mass of the atom was concentrated in a small region of positive charges that impeded the passage of alpha particles. He suggested a new model in which the atom had a nucleus or center in which the mass and the positive charge are concentrated, and that in the extra nuclear zone are the negatively charged electrons.
Answer: yes
Explanation: When you mix an acid an a base the color it showing a chemical change.
Answer:
Meiosis is a process where a single cell divides twice to produce four cells containing half the original amount of genetic information. These cells are our sex cells – sperm in males, eggs in females. ... These four daughter cells only have half the number of chromosomes? of the parent cell – they are haploid.
Explanation:
Basically haploid cells
If you have 8.51x10-3 g of O₂, and excess ZnS, moles of ZnO form is 0.174×10⁻³ moles.
The balanced equation is given as :
2ZnS + 3O₂ ------> 2ZnO + 2SO₂
number of moles of O₂ = mass / molar mass
= 8.51 × 10⁻³ g / 31.998 g/mol
= 0.265× 10⁻³ moles
now, 3 moles of O₂ form 2 moles of ZnO
0.265× 10⁻³ moles of O₂ form( 2/3) × 0.265× 10⁻³ = 0.174×10⁻³ moles
The number of moles of ZnO is 0.174×10⁻³ moles
Thus, If you have 8.51x10-3 g of O₂, and excess ZnS, moles of ZnO form is 0.174×10⁻³ moles.
To learn more about no. of moles here
brainly.com/question/26416088
#SPJ1