A is obviously out because it leads to a volume of 125.0 milliliters of the new solution and gives you a lower concentration than you were aiming for.
D is out because you are adding 75 milliliters of the stock solution, so your concentration would be too high. You only need 25.0 milometers of stock solution per 100 milliliters of the new solution.
C is also out because it leads to 50.0 milliliters stock solution per 100 milliliters of the new solution and hence the wrong concentration.
B is by default the correct answer. It also details the correct technique. First you add the stock solution (This you know from your calculations to be 25 milliliters.) then you add the water up to the volume you needed. (Because the calculations only tell you the total volume of water not what you need to add) You also add the water last so you can rinse the neck of the flask to make sure you also get all the stock solution residue into the stock solution.
I would add the final step of stirring, but B is the only answer that can be correct.
False that atom is the smallest identifiable unit of a compound.
The smallest identifiable unit of a compound is the Element. Element is the one which make up the compound and element is made up by atoms. Example of element is oxygen and hydrogen which make up water (H2O) which is a compound.
Answer:
NaClO + CH₃COOH ----> HClO + CH3CO- + Na
Explanation:
This reaction occurs between the combination of a salt and an acid, that is, an oxide-reduction reaction
<span>The symbol for the element whose atoms have 40 electrons each is Zr. This is the element zirconium. In the atoms of a pure element, the number of positively charged protons is normally equal to the number of negatively charged electrons. Hence, the number of electrons in the atom can be inferred from the atomic number, which corresponds to the number of protons in an atom. The atomic number of zirconium is 40.</span>
Answer:
N₂ = 6.022 × 10²³ molecules
H₂ = 18.066 × 10²³ molecules
NH₃ = 12.044 × 10²³ molecules
Explanation:
Chemical equation;
N₂ + 3H₂ → 2NH₃
It can be seen that there are one mole of nitrogen three mole of hydrogen and two moles of ammonia are present in this equation. The number of molecules of reactant and product would be calculated by using Avogadro number.
The given problem will solve by using Avogadro number.
It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance.
The number 6.022 × 10²³ is called Avogadro number.
For example,
Number of molecules of nitrogen gas:
1 mol = 6.022 × 10²³ molecules
Number of molecules of hydrogen:
3 mol × 6.022 × 10²³ molecules/ 1 mol
18.066 × 10²³ molecules
Number of molecules of ammonia:
2 mol × 6.022 × 10²³ molecules/ 1 mol
12.044 × 10²³ molecules
