If 276 g of dinitrogen tetroxide and 64.0 g of hydrazine are mixed, how many grams of nitrogen gas can you produce?

84 g of N₂        

Explanation:

We begin from the reaction:

N₂O₄  +  2N₂H₄ →  3N₂  + 4H₂O

  1st step: Find out the limiting reactant.

We convert the mass to moles.

       276 g . 1mol/ 92g = 3 moles of N₂O₄

64 g . 1mol / 32g = 2 moles of hidrazine  

Limiting reactant is the hydrazine. Ratio is 1:2, then for 3 moles of tetroixde I may need 6 moles of N₂H₄. (But I only have 2).

To determine the grams of produced nitrogen we see stoichiometry (2:3)    

2 moles of hydrazine can produce 3 moles of N₂

Definetaly our 2 moles make 3 moles of gas.

We convert the moles to mass.

3 mol . 28g /1mol = 84 g of N₂                                                                                                

atomic number of carbon is 6

electronic configuration of carbon is 1s²2s²2p²

orbital diagram of c is as shown here in the image.

we can conclude that the melting point of table salt is above room temperature (it's actually 801 deg c). the melting point of mercury must be below room temperature (- 39 deg c).