He could go at any of the first 3 times because he won't get back if he goes on the last. We need more information to help answer it.
Since it talks about the number of particles, we would need the Avogadro's number which is an empirical value equal to 6.022×10²³ particles/mole. The other information that we have to know is the molar mass of Na₂SO₄ which is 142.04 g/mol.
Mass of a single particle of Na₂SO₄:
(142.04 grams Na₂SO₄/mol) * (1 mol/6.022×10²³ particles) = 2.359×10⁻²² g
Thus, each Na₂SO₄ weighs 2.359×10⁻²² g.
Number of Na₂SO₄ particles in the mixture:
(11.53 g Na₂SO₄) * (1 mol/142.04 g) * (6.022×10²³ particles/mol) = 4.89×10²² particles
Thus, the mixture contains 4.89×10²² particles is Na₂SO₄.
Answer: Glucose is a simple sugar with six carbon atoms and one aldehyde group.
Explanation: hope this helps you
The heat of a chemical reaction (change of enthalý) may be calcualted as the heat of formation of the products less the heat of formation of the reactants.
Then, for the given equation, the total heat of the combustion of pentnae is:
<span>ΔH°f = 5* ΔH°f CO2(g) + 6ΔH°f H2O(g) - ΔH°f C5H12(g) - 8ΔH°f O2(g)
</span>The standard heat of formation of O2(g) is zero because that is its natural state.
Now you can replace in the equation the values given:
<span>ΔH° = 5*(-393.5 kJ/mol) + 6*(-241.8kj/mol) - (-35.1kJ/mol) - 0
ΔH° = -3,383.2 kJ/mol
</span>
Then, the heat of the combustion of pentane is -3,383.2 kj/mol