Each nitrogen molecule consists of two atoms of nitrogen that are bonded by a triple covalent bond. This is a direct consequence of the fact that each nitrogen atom has 5 valence electrons. Each atom can thus complete its octet by sharing three electrons.

Explanation:

Sorry i had to look it up i didn't know this answer

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6 0

3 years ago

How many grams of nitric acid will react completely with a block of iron metal that is 4.5 cm by 3.0 cm by 3.5 cm, if the densit

denpristay [2]

Balance the equation first:

2 Fe+6 HNO3→2 Fe(NO3)3+3H2

Then calculate mass of Iron :

4.5×3.0×3.5 cm3(1 mL1 cm3)(7.87 g Fe1 ml)=371.86 g Fe

Now use Stoichiometry:

371.86 g Fe×(1 mol Fe55.85 g Fe)×(6 mol HNO32 mol Fe)=19.97 mol HNO3

Convert moles of nitric acid to grams

19.97 mol HNO3×(63.01 g HNO31 mol HNO3)=1258.3 g HNO3

7 0

3 years ago

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What volume (in mL) of a 0.200 MHNO3 solution is required to completely react with 27.6 mL of a 0.100 MNa2CO3 solution according

ladessa [460]

Answer:

There is 27.6 mL of a 0.200 M HNO3 solution required

Explanation:

Step 1: The balanced equation is:

Na2CO3(aq)+2HNO3(aq)→2NaNO3(aq)+CO2(g)+H2O(l)

This means for 1 mole Na2CO3 consumed, there is consumed 2 mole of HNO3 and there is produced 2 moles of NaNO3, 1 mole of CO2 and 1 mole of H2O

Step 2: Calculating moles of Na2CO3

moles of Na2CO3 =volume of Na2CO3 * Molarity of Na2CO3

moles of Na2CO3 = 27.6 *10^-3 * 0.1 M = 0.00276 moles

Step 3: Calculating moles of HNO3

In the balanced equation, we can see that for 1 mole of Na2CO3 consumed, there are consumed 2 moles of HNO3.

So for 0.00276 moles consumed of Na2CO3, there are consumed 0.00552 moles of HNO3.

This means 0.00276 moles of the base Na2CO3 would react with 0.00552 moles of the acid HNO3

Step 4: Calculating the volume of HNO3

volume of HNO3 = moles of HNO3 / Molarity of HNO3

volume of HNO3 = 0.00552 moles / 0.200 M = 0.0276 L

0.0276 L = 27.6 ml

There is 27.6 mL of a 0.200 M HNO3 solution required

4 0

3 years ago