How many molecules of sodium nitrate are produced when 17 g of sodium azide, nan3, react according to the following equation: nan3(aq)+agno3(aq) > agn3(s)+nano3(aq) question options: 3.7x1023 1.6 x 1023 1.85x1023 6.02x1023

 The number  of  molecules of  sodium  nitrate  that reacted is

1.6 x10^23  molecules

 calculation

NaN₃(aq) +AgNO₃(aq)  →     AgN₃(s)  +  NaNo₃ (aq)

step 1: find moles of NaN₃

 moles =  mass÷molar mass

from periodic table the  molar mass  of NaN₃ = 23 +( 3 x 14)= 65 g/mol

moles =17 g÷65 g/mol = 0.262  moles

Step 2: use the  mole ratio to calculate the  moles of NaNO₃

NaN₃:NaNO₃  is 1:1

Therefore the moles of NaNO₃  is also = 0.262  moles

Step 3: use the Avogadro's  law constant to calculate the number of molecules of NaNO₃

That  is according  to Avogadro's  law

1 mole = 6.02 x10^23 molecules

0.262 moles= ?   molecules

by cross multiplication

=[(0.262 moles x 6.02 x10^23  molecules) / 1 mole] = 1.6 x10^23   molecules

(b) which contains higher levels of vitamin b, oranges or spinach?

explanation:

you can eliminate a, c, and d because they are questions based on opinions.