Nitrogen dioxide, a major air pollutant, can be produced by the combustion of nitrogen oxide as shown. 2NO + O2 Right arrow. 2NO2 In a plant, 1,500 kg of nitrogen oxide is consumed per day to produce 1,500 kg of nitrogen dioxide per day. What is the percent yield? Use Percent yield equals StartFraction actual yield over theoretical yield EndFraction times 100.. 21.7% 32.6% 43.5% 65.2% Mark this and return Save and Exit Next Submit

65.2%

Explanation:

It is given that :

Therefore, 60 g of nitrogen oxide will produce 92 g of nitrogen dioxide

or 60 kg of nitrogen oxide will produce 92 kg of nitrogen dioxide

or 1500 kg of nitrogen oxide will produce of nitrogen dioxide

Therefore the percentage yield = (Actual yield / Expected yield) x 100

percentage yield =

Percentage yield = 65.217 %

assuming that your periodic table has eighteen columns, you should highlight   the four elements in the third column from the left of the table.

neutral atoms of elements after calcium contain d electrons. the energy of d orbitals of a particular main shell lies between the energy of the s and p orbitals of the next main shell. for example, electrons in the have energy higher than those in the orbital while lower than those of the orbital.

by the aufbau principle, electrons fill orbitals one at a time, starting from those of the lowest energy. the orbital is therefore filled right after orbitals and before orbitals.  

potassium has electron configuration and calcium . the s orbital of a main shell holds a maximum of two electrons. the next element would thus have the electron configuration and satisfy the requirements.

a neutral atom of the element contains three electrons on top of the most recent noble gas element. it should thus be placed in the third column of the periodic table. both scandium and yttrium would satisfy the conditions. an aufbau diagram indicates that the 4f orbital is filled before the 5d. however, it appears that exceptions exists for both lanthanum and actinium. neutral atoms of the two d-block elements also demonstrate the and should be highlighted as well.

the density of gold = 19.3 g/cm3

the relation between density and mass and volume is

density = mass / volume

given :

mass of gold = 13 g

density = 13g / volume = 19.3

volume = 13 /19.3 = 0.67 cm^3

thus volume of gold nugget = 0.67 cm^3