The answer will be 6CO2 + 6H2O ——-> 1 C6H12O6 + 6O2
Answer is: <span>he boiling point of a 1.5 m aqueous solution of fructose is </span>100.7725°C.
The boiling point
elevation is directly proportional to the molality of the solution
according to the equation: ΔTb = Kb · b.<span>
ΔTb - the boiling point
elevation.
Kb - the ebullioscopic
constant. of water.
b - molality of the solution.
Kb = 0.515</span>°C/m.
b = 1.5 m.
ΔTb = 0.515°C/m · 1.5 m.
ΔTb = 0.7725°C.
Tb(solution) = Tb(water) + ΔTb.
Tb(solution) = 100°C + 0.7725°C = 100.7725°C.
So you have to multiply 786,3 times 0.98 to get the mass the mass is 770.57 if u round its 770.6 or 771
<em>m Na₂CO₃: 23g×2 + 12g + 16g×3 = 106 g/mol</em>
------------------------------
1 mol ------- 106g
X ------------ 10,6g
X = 10,6/106
<u>X = 0,1 mol Na₂CO₃</u>
Answer:
Molarity of solution is 1.10x10⁻³ M
Explanation:
Solute NaOCl
7.4% by mass means, that in 100 grams of solution, we have 7.4 g of solute.
Molar mass of NaOCl = 74.45 g/m
Mol = Mass / Molar mass
7.4 g / 74.45 g/m = 0.099 moles
Density of solution = 1.12 g/mL
Density = Mass / volume
1.12g/mL = 100 g / volume
Volume = 100 g / 1.12g/mL = 89.3 mL
Molarity = mol /L
89.3 mL = 0.0893 L
0.099 moles / 0.0893 L = 1.10x10⁻³ M
