To determine the empirical formula for the compound that contains <span>0.979 g Na, 1.365 g S, and 1.021 g O, we convert these to mole units. The molar masses to be used are:
Molar mass of Na = 23 g/mol
</span>Molar mass of S = 32 g/mol
Molar mass of O = 16 g/ mol
The number of moles is obtained using the molar mass for each element.
moles Na = 0.979 g Na/ 23 g/mol Na = 0.04256
moles S = 1.365 g Na/ 32 g/mol Na = 0.04265
moles O = 1.021 g O/ 16 g/mol Na = 0.06326
We then divide each with the smallest number of moles obtained.
Na: 0.04256/ 0.04256 = 1
S: 0.04265/ 0.04256 = 1.002 ≈ 1
O: 0.06326/ 0.04256 = 1.49 ≈ 1.5
We then have an empirical formula of NaSO₁.₅. However, chemical formulas must have only integers as subscripts, thus, we multiply each to 2. The empirical formula is then Na₂S₂O₃ also known as sodium thiosulfate.
The answer would be:
A. Cancel out CO because it appears as a reactant in one intermediate reaction and a product in the other intermediate reaction.
In this question, there are two half-reaction equations. To merge them up, you need to add the reactant with the reactant, then the product with the product. If there is a molecule on both side, you can cancel them. The full reaction would be:
C+ 1/2 O2 + CO + 1/2O2 ==>CO+ CO2 -----> remove CO from both side
C+ O2 ==>CO2
Answer:
C. 80g Nano3 at 10 degrees Celsius
Explanation:
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