CH₇ is the empirical formula of the car fuel.
Explanation:
To find the empirical formula we use the following algorithm.
First divide each mass the the molar weight of each element:
for carbon 2.87 / 12 = 0.239
for hydrogen 3.41 / 2 = 1.705
And now divide each quantity by the lowest number which is 0.239:
for carbon 0.239 / 0.239 = 1
for hydrogen 1.705 / 0.239 = 7.13 ≈ 7
The empirical formula of the car fuel is CH₇.
I have to tell you that in reality this formula is wrong because is not possible to exist. However the algorithm for finding the empirical formula is right, the problem may reside in the amounts of carbon and hydrogen given.
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<span>The molecular formula is 8(CH2O)= C8H16O8.</span>
Answer: ORGANIC ACIDS
Explanation:
CAM PLANTS CARBOXYLATE ORGANICS ACIDS through the addition of CO2 to PEP Carboxylase( a phosphoenolpyruvate carboxylase enzyme present in the mesophyll cells of the cytoplasm in a green plant) to produce Oxaloacetate (organic compound).
CO2 + PEP ⇒ C4H4O5 (oxaloacetate)
Oxaloacetate is then converted to a similar molecule, Malate (C4H6O5, another form of organic compound) that can be transported in to the bundle-sheath cells. Malate enters the plasmodesmata and releases the CO2. The CO2 then fixed by rubisco and made into sugars via the Calvin cycle.
Answer:
Explanation:
<u>1. Molecular chemical equation:</u>
- 2 KClO₃(s) → 2 KCl(s) + 3 O₂(g)
<u>2. Mole ratios:</u>
- 2 mol KClO₃ : 2 mol KCl : 3 mol O₂
<u>3. Number of moles of KClO₃</u>
- Number of moles = mass in grams / molar mass
- Molar mass of KClO₃ = 122.55 g/mol
- Number of moles of KClO₃ = 54.3 g / 122.5 g/mol ≈ 0.44308 mol
<u>3. Number of moles of O₂</u>
As per the theoretical mole ratio 2 mol of KClO₃ produce 3 mol of O₂, then set up a proportion to determine how many moles of O₂ will be produced from 0.44038 mol of KClO₃.
- 3 mol O₂ / 2 mol KClO₃ = x / 0.44038 mol KClO₃
- x = (3 / 2) × 0.44308 mol O₂ = 0.6646 mol O₂
Round to 3 significant figures: 0.665 mol of O₂ ← answer
Lead is an E (element) and can be found on the periodic table.