Sodium oxide (na2o) can react with hydrochloric acid (hcl) to produce sodium chloride (nacl) and water (h2o) according to the following equation: na2o + 2hcl -> 2nacl + h2o how many total atoms are present before and after the reaction

Na2O + 2HCl -> 2NaCl + H2O then it wil be 10.3 atoms left in the cycle

Step-by-step explanation: you would take

(Na2O) then (HCl) then (NaCl) and, (H2O)

And you would got to the table of atoms and put them together if you look at the name of a chemical it show a number well you would divide them in to fractions and the denominator you would need to focus on because that is going to be your answer.  

7 before and 7 after

Step-by-step explanation:

Before:

Na2O=3 atoms

2HCl=4 atoms

After:

2NaCl=4 atoms

H2O=3 atoms

v v cj van config k tbs eng r.c r.c fcb fb gb frfr

2/3x + 7 = 29

step-by-step explanation:

two thirds = 2/3

plus a number = x

at least 29 = 29

there for you should get 2/3x + 7 = 29