The combustion of ammonia in the presence of excess oxygenyields no2(g) and h2o(g): 4nh3(g) + 7o2(g) → 4no2(g) + 6h2o(g)the combustion of 43.9 mg of ammonia produces how many grams ofno2(g)? a. 178b. 0.119c. 43.9d. 0.954e. 2.58

There will be produced 0.119 grams of NO2 (option B)

Explanation:

Step 1: Data given

Mass of ammonia = 43.9 mg

Oxygen is in excess

Molar mass of NH3 = 17.03 g/mol

Step 2: The balanced equation

4NH3(g) + 7O2(g) → 4NO2(g) + 6H2O(g)

Step 3: Calculate moles NH3

Moles NH3 = Mass NH3 / molar mass NH3

Moles NH3 = 0.0439 g/ 17.03 g/mol

Moles NH3 = 0.00258 moles

Step 4: Calculate moles of NO2

For 4 moles NH3 we need 7 moles of O2 to produce 4 moles NO2 and 6 moles of H2O

For 0.00258 moles of NH3 we'll have 0.00258 moles of NO2

Step 5: Calculate mass of NO2

Mass of NO2 = moles NO2 * molar mas NO2

Mass NO2= 0.00258 moles * 46 g/mol

Mass NO2 = 0.119 grams

There will be produced 0.119 grams of NO2 (option B)

2b2t

explanation:

2b2t

metals are good conductor of electricity. it has free electrons on their surface. according to electron pool theory, each atom present in a metal crystal losses all of its valence electron and as a result a pool of free is form. these pool of electrons are free to move and   conduct electricity easily   while making metal as good conductor.

the correct answer is a. the ability of electrons to flow throughout the metal.

if it's not useful, feel free to report for mistake in answer.