Answer:
Al
Explanation:
4 Al + 3 O₂ → 2 Al₂O₃
You need to figure out which one has the smaller mole ratio. Convert both substances from grams to moles.
(10.0 g Al)/(26.98 g/mol) = 0.3706 mol Al
(19.0 g O₂)/(32.00 g/mol) = 0.5938 mol O₂
Now, use the mole ratios of reactant to product to see which substance produces the least amount of product.
(0.3706 mol Al) × (2 mol Al₂O₃/4 mol Al) = 0.1853 mol Al₂O₃
(0.5938 mol O₂) × (2 mol Al₂O₃/3 mol O₂) = 0.3958 mol Al₂O₃
Since aluminum produces the least amount of product, this is the limiting reagent.
Answer:
A. Yes, there is more than enough sodium carbonate.
Explanation:
Hello,
In this case, based on the given reaction which is:

By stoichiometry, one computes the grams of sodium carbonate that will neutralize 1,665 g of sulfuric acid as shown below:

Thus, the available mass is 2.0 kg so 0.2 kg are in excess, therefore: A. Yes, there is more than enough sodium carbonate.
Best regards.
Vanillin is the common name for 4-hydroxy-3-methoxy-benzaldehyde.
See attached figure for the structure.
Vanillin have 3 functional groups:
1) aldehyde group: R-HC=O, in which the carbon is double bonded to oxygen
2) phenolic hydroxide group: R-OH, were the hydroxyl group is bounded to a carbon from the benzene ring
3) ether group: R-O-R, were hydrogen is bounded through sigma bonds to carbons
Now for the hybridization we have:
The carbon atoms involved in the benzene ring and the red carbon atom (from the aldehyde group) have a <u>sp²</u> hybridization because they are involved in double bonds.
The carbon atom from the methoxy group (R-O-CH₃) and the blue oxygen's have a <u>sp³</u> hybridization because they are involved only in single bonds.
<u>Answer:</u>
Specific heat of a substance is the value that describe how the added heat energy of substance has the impact on its temperature.
Unit is <em>
</em>
<em>C = Q/m. ∆T</em>
<em>C – Specific heat
</em>
<em>Q- heat energy (J)</em>
<em>M – Mass (Kg)</em>
<em>∆T- change in temperature (K) </em>
<u>Explanation:</u>
<em>Given data:</em>
<em>M= 140 g = 0.14 Kg</em>
<em>Q – 1080 Joules.</em>
<em>∆T – 98.4 – 62.2 = 36.2</em>
Substituting the given data in Equation
<em>Specific heat of Aluminium =
</em>
Divide the mass by the density