The question is incomplete. Complete question is:
Which list represents the classification of the
elements nitrogen, neon, magnesium, and
silicon, respectively?
(1) metal, metalloid, nonmetal, noble gas
(2) nonmetal, noble gas, metal, metalloid
(3) nonmetal, metalloid, noble gas, metal
(4) noble gas, metal, metalloid, nonmetal
.................................................................................................................
Answer: The correct answer is option 2) nonmetal, noble gas, metal, metalloid
Reason:
1) Non-metal are bad conductors of electricity. Also non-metal are electronegative in nature. Hence, among the provided elements, nitrogen satisfied the criteria. Hence, it is a non-metal.
2) Noble gas are the elements which have completely filled outermost shell. Hence, among the provided elements, Neon satisfied the criteria.
3) Metal are good conductors of electricity. Also metal are electropositive in nature. Hence, among the provided elements, magnesium satisfied the criteria. Hence, it is a metal.
4) Metalloid have conductivity better than non-metals but poor than conductors. Hence, among the provided elements, Silicon satisfied the criteria. Hence, it is a metalloid.
Answer:
copper I oxide
Explanation:
Copper is known to form two oxides; copper I oxide and copper II oxide.
In the question we have the ions; Cu1+ and O2-. Ionic compounds are formed by a combination of ions as typified below;
Cu^+ + O^2- -------> Cu2O
This compound is copper I oxide
Answer:
89.3 %
Explanation:
M(CH4) = 12+ 4*1 = 16 g/mol
M(O2) = 2*16 = 32 g/mol
M(CO2) = 12 + 2*16 = 44 g/mol
8.50 g * 1 mol/16 g = 0.5313 mol CH4
15.9 g * 1 mol/32 g = 0.4969 mol O2
9.77 g * 1 mol/44 g = 0.2220 mol CO2
1) CH4 + 2O2 -----> CO2 + 2H2O
from reaction 1 mol 2 mol
given 0.5313 mol (0.4969 mol)
1 mol CH4 --- 2 mol O2
0.5313 mol CH4 --- x mol O2
x= 2*0.5313 = 1.0626 mol O2
We can see that for given amount of CH4 we do not have enough O2, so O2 is a limiting reactant.
2) CH4 + 2O2 -----> CO2 + 2H2O
from reaction 2 mol 1 mol
given 0.4969 mol x mol
x = 0.4969*1/2 = 0.2485 mol CO2 theoretical yield
3)
Practical yield CO2 = 0.2220 mol
Theoretical yield CO2 = 0.2485 mol
% yield = (0.2220/0.2485)*100% = 89.3 %