The correct answer would be equation 4, because there are 4 Hydrogens and 2 Oxygens on each side of the equation making it balanced.
Answer is: <span>the molality of the glycerol solution is 10.03 m.
V(solution) = 100 mL.
m</span>(solution) = V(solution) · d(solution).
m(solution) = 100 mL ·1.120 g/mL.
m(solution) = 112 g.
m(glycerol) = ω(glycerol) · m(solution).
m(glycerol) = 0.48 · 112 g.
m(glycerol) = 53.76 g.
m(water) = 112 g - 53.76 g.
m(water) = 58.24 g ÷ 1000 g/kg = 0.05824 kg.
n(C₃H₈O₃) = m(C₃H₈O₃) ÷ M(C₃H₈O₃).
n(C₃H₈O₃) = 53.76 g ÷ 92 g/mol.
n(C₃H₈O₃) = 0.584 mol.
b(C₃H₈O₃) = n(C₃H₈O₃) ÷ m(H₂O).
b(C₃H₈O₃) = 0.584 mol ÷ 0.05824 kg.
m(C₃H₈O₃) = 10.027 mol/kg.
Everything is made of molecules so I don’t think there would be life without molecules..
Answer:
153.3 grams of ZrCl₄ are produced
Explanation:
The equation of the reaction is as follows:
ZrSiO₄ + Cl₂ ----> ZrCl₄ + SiO₂ + O₂
molar mass of ZrSiO₄ = (91 + 32 + 16 * 4) = 187.0 g/mol
molar mass of ZrCl₄ = (91 + 35.5 * 4) = 233.0 g/mol
molar mass of Cl₂ = (35.5 * 2) 71.0 g/mol
From the equation of reaction, 1 mole of ZrSiO₄ reacts with one mole of Cl₂ to produce one mole of ZrCl₄
number of moles of ZrSiO₄ present in 123 g = 123/187 = 0.65 moles
number of moles of Cl₂ present in 85.0 g = 85.0/71.0 = 1.19 moles
therefore, ZrSiO₄ is the limiting reactant
123.0 g of ZrSiO₄ will react with excess Cl₂ to produce 123 * 233/187 grams of ZrCl₄ = 153.3 grams of ZrCl₄
Therefore, 153.3 grams of ZrCl₄ are produced
First let us see what
kind of bonds are formed in the compound. By drawing the structure, we see that
the kind of bonds are:
N =- triple bond -= C –
O
<span>So there is only
single bond between C and O therefore the hybridization of C is sp.</span>
