Answer:
CaCO3 is the limiting reactant
55 g of CO2 is made
Explanation:
First we must put down the reaction equation;
CaCO3(s) + 2HCl(aq) ---------> CaCl2(s) + H2O(l) + CO2(g)
Number of mole of CaCO3 = 125g/100gmol-1 = 1.25 moles
From the reaction equation;
1 mole of CaCO3 yields 1 mole of CO2
Hence 1.25 moles of CaCO3 yields 1.25 moles of CO2
For HCl;
number of moles of HCl = 125g/36.5 g mol-1 = 3.42 moles
From the reaction equation;
2 moles of HCl yields 1 mole of CO2
3.42 moles of HCl yields 3.42 * 1/2 = 1.71 moles of CO2
Hence CaCO3 is the limiting reactant.
Mass of CO2 produced = 1.25g * 44 gmol-1 = 55 g of CO2
16-18= -2 so it has a negative charge. Just subtract the electrons from the protons if you get a positive number it will have a positive charge and vice versa.
Answer:
Forming a problem requires the scientist to use creativity to imagine new solutions.
Explanation:
Albert Einstein remains a critically prominent figure who conducted remarkable, ground-breaking research that not only formed the foundations of modern physics but also strongly affected the scientific world. It is difficult to teach imagination but it can be harnessed and accepted. Nothing incites our imaginative impulses we love more than the prospect of immediate creative inspiration. And creativity hits its full potential when paired with the experience, insights, and skills people gained by questioning the real-life problems.
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.