Answer:
As for your question, I know to forget to put the options, specifically that your question is incomplete.
Explanation:
Although it could help you by telling you that always a reaction that seeks to balance the pH, and achieve neutrality ... It is necessary to achieve a concentration of OH equal to that of H +, in this way the hydroxyl and the protons.
False. The number of neutrons can be more or less than the number of protons.
Answer:
Chemical Bond (False)
Explanation:
A mutual electrical attraction between the nuclei and valence electrons of different atoms that binds the atoms together
5.75g
Explanation:
Given parameters:
Number of atoms = 1.505 x 10²³atoms
let us find the number of moles and mass of this atom.
Solution:
The mole is convenient unit for reporting measurements in chemistry. It allows us to make mathematical calculations:
one mole of substance contains 6.02 x 10²³atoms
Number of moles of Na given = 
= 0.25moles
Mass of Na atom given = number of moles x molar mass
Molar mass of Na= 23g/mole
Mass of Na = 0.25 x 23 = 5.75g
learn more:
Number of moles brainly.com/question/1841136
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Both of these questions can be solved using the equation M1V1 = M2V2, where M is concentration anf V is volume.
For the first case, M2 = 0.2 mol/L, M1 = 3 mol/L, and V2 = 250mL. So now you want V1. Solving for V1, V1 = (M2 / M1)V2 =
(0.2 / 3)(250) = 16.7 mL. So what that means is that you need 16.7 mL of 3M HCl, and the rest of the 250 mL (which would be 250 - 16.7 = 233.3 mL) would be water, with which you're diluting the HCl.
Same principle for the second problem, except now we have percentages and not mol/L. You can treat the percentages as concentrations. Since you're starting with pure isopropyl alcohol, M1 = 100%. You want a final volume of 500 mL and a final concentration of 70%. To find the volume of isopropyl alcohol you need to start with, solve for V1. So V1 = (M2 / M1)V2 = (70 / 100)(500) = 350 mL. So you need 350 mL of isopropyl alcohol and the rest of the 500 mL (that is, 150 mL) you can fill with water.
