The enthalpy of combustion of 1 mole of benzene is 3169 kJ/mol .
The first step in answering this question is to obtain the balanced thermochemical equation of the reaction. The thermochemical equation shows the amount of heat lost or gained.
The thermochemical equation for the combustion of benzene is;
2 C6H6(l) + 15 O2(g) → 12 CO2(g) + 6 H2O(g) ΔrH° = -3169 kJ/mol
We can see that 1 mole of benzene releases about 3169 kJ/mol of heat.
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An osmolarity of saline solution is 308 mosmol/L.
m(NaCl) = 9 g; the mass of sodium chloride
V(solution) = 1 L; the volume of the saline solution
n(NaCl) = 9 g ÷ 58.44 g/mol
n(NaCl) = 0.155 mol; the amount of sodium chloride
number of ions = 2
Osmotic concentration (osmolarity) is a measure of how many osmoles of particles of solute it contains per liter.
The osmolarity = n(NaCl) ÷ V(solution) × 2
The osmolarity = 0.154 mol ÷ 1 L × 2
The osmolarity = 0.154 mol/L × 1000 mmol/m × 2
The osmolarity of the saline solution = 308 mosm/L.
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Answer:
Candy
Explanation:
its sweet and has a lot of sugar and acid
Answer:
[C₆H₁₂O₆] = 0.139 M
Explanation:
Molarity si defined as a sort of concentration. It indicates the moles of solute that are contained in 1 L of solution.
We can also say, that molarity are the mmoles of solute contained in 1 mL of solution.
For this case, the solute is sugar (glucose). Let's determine M (mmol/mL)
(3.95 g . 1mol / 180g) . (1000 mmol / 1mol) / 158 mL
We determine moles, we convert them to mmoles, we divide by mL
M = 0.139 M
Moles = 3.95 g . 1mol / 180g → 0.0219 mol
We convert mL to L → 158 mL . 1L/1000mL = 0.158L
M = 0.0219 mol / 0.158L = 0.139 M
