The empirical formula of the alcohol is C₄H₄O
To solve the question given above, we'll begin by calculating the percentage of oxygen in the compound. This can be obtained as follow:
Carbon (C) = 70.57 %
Hydrogen (H) = 5.935 %
<h3>Oxygen (O) =? </h3>
O = 100 – (C + H)
O = 100 – (70.57 + 5.935)
O = 100 – 76.505
<h3>O = 23.495%</h3>
Finally, we shall determine the empirical formula of the compound. This can be obtained as follow:
Carbon (C) = 70.57 %
Hydrogen (H) = 5.935 %
Oxygen (O) = 23.495%
<h3>Empirical formula =?</h3>
Divide by their molar mass
C = 70.57 / 12 = 5.881
H = 5.935 /1 = 5.935
O = 23.495 / 16 = 1.468
Divide by the smallest
C = 5.881 / 1.468 = 4
H = 5.935 / 1.468 = 4
O = 1.468 / 1.468 = 1
Therefore, the empirical formula of the alcohol is C₄H₄O
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Phosphoric acid is a weak acid, while rubidium hydroxide is a strong acid.
H₃PO₄ + RbOH --> Rb₃PO₄ + H₂O
We get Rb₃PO₄ because PO₄ has a charge of 3-, that is PO₄³⁻. Rb has a charge of 1+. You give the subscript of one the charge of the other as this is an ionic compound. So you end up with Rb₃PO₄, a neutral compound.
Now let's balance the equation:
H₃PO₄ + 3RbOH --> Rb₃PO₄ + 3H₂O
Answer 1:
For compound A:
<span>2.8 g of nitrogen for each 1.6 g of oxygen
</span>Atomic weight of N = 14
Atomic weight of O = 16
Thus, number of moles of N = 2.8/14 = 0.2
and number of moles of O = 1.6/16 = 0.1
Thus, molar ratio of N and O in compound is 2:1.
Therefore, <span>lowest whole-number mass ratio of nitrogen that combines with a given mass of oxygen is 2:1. And the compound formed is N2O.
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</span>Answer 2:
For compound B:
5.6 g of nitrogen for each 9.6 g of oxygen
Atomic weight of N = 14
Atomic weight of O = 16
Thus, number of moles of N = 5.6/14 = 0.4
and number of moles of O = 9.6/16 = 0.6
Thus, molar ratio of N and O in compound is 4:6 = 2:3.
Therefore, lowest whole-number mass ratio of nitrogen that combines with a given mass of oxygen is 2:3. And the compound formed is N2O3.
Cellular respiration is the process by which the chemicalenergy of "food" molecules is released and partially captured in the form of ATP. Carbohydrates, fats, and proteins can all be used as fuels in cellular respiration, but glucose is most commonly used as an example to examine the reactions and pathways involved.