How much of NaCl is in 1.90 L of 0.300 M
NaCl?
Answer in units of mol.

The answer to your question is 0.57 moles

Explanation:

Data

volume = 1.9 L

[NaCl] = 0.3

moles of NaCl = ?

Process

1.- To solve this problem use the formula of Molarity.

Molarity = moles/ volume (l)

-Solve for moles

moles = Molarity x volume

-Substitution

 moles = 0.3 x 1.9

-Result

 moles = 0.57

-Conclusion

There are 0.57 moles of NaCl in 1.90 L of a 0.3 M solution.

for points guys

you need to take a look at the periodic system in order to sort these elements into pairs that would most likely exhibit similar chemical properties. first, you need to find elements that are found in the same group (groups are vertical in the periodic system), and then to see whether they have the same valence number. if they meet these criteria, they will have similar properties.

the first pair: ne (neon) and kr (krypton), are noble gases found in group 8

the second pair: li (lithium) and k (potassium), found in group 1

the third pair: s (sulfur) and te (tellurium), found in group 6