<span>70.4 mg CO2 x 1.0 g /1000 mg x 1 mole CO2/ 44 gCO2 x 1 mole C/1 mole CO2 = 0.0016 moles C
14.4 mg H2O x 1.0 g/1000 mg x 1 mole H2O/18 g H2O x 2 moles H/ 1 mole H2O = 0.0016 moles O
molar mass of C=12 g/mole
molar mass of H=1 g/mole
0.0016 moles C x 12 g C/ 1 mole C = 0.0192 g C or 19.2 mg C
0.00156 moles H x 1 g H/1 mole H = 0.00156 g H or 1.56 mg H
mg O= 30.4 mg vanillin - 19.2 mg C – 1.56 mg H = 9.64 mg O
molar mass of O=16 g/mole
9.64 mg O x 1 g/1000 mg x 1 mole O/16.0 g = 0.000602
C.0016 H.0016 O.000602; divide all the moles by the smallest value of0.000602
C2.66H2.66O1 is the empirical formula;
to obtain whole numbers multiply by 3
3[C2.66H2.66O1] = C8H8O3
above formula weight: 8(C) + 8(H) + 3(O) = 8(12) + 8(1) + 3(16) = 152 amu
The empirical formula weight and the molecular formula weight are the same .
Molecular formula is C8H8O3.</span>
Answer: 6.162g of Ag2SO4 could be formed
Explanation:
Given;
0.255 moles of AgNO3
0.155 moles of H2SO4
Balanced equation will be given as;
2AgNO3(aq) + H2SO4(aq) -> Ag2SO4(s) + 2HNO3(aq)
Seeing that 2 moles of AgNO3 is required to react with 1 moles of H2SO4 to produce 1 mole of Ag2SO4,
Therefore the number of moles of Ag2SO4 produced is given by,
n(Ag2SO4) = 0.255 mol of AgNO3 ×
[0.155mol H2SO4 ÷ 2 mol AgNO3] x
[ 1 mol Ag2SO4 ÷ 1 mol H2SO4]
= 0.0198 mol of Ag2SO4.
mass = no of moles x molar mass
From literature, molar mass of Ag2SO4 = 311.799g/mol.
Thus,
Mass = 0.0198 x 311.799
= 6.162g
Therefore, 6.162g of Ag2SO4 could be formed
<span> the line-emission spectrum of an atom is caused by the energies released when electrons. releases energy of only certain values. </span>
Answer:
Zirconium, Zr, 40
Explanation:
Protons can be found by looking at the atomic number. Neon's atomic number is 10, so just multiply 10 by 4 and you get 40.
