Answer:
8.45 moles are produced
Explanation:
CaCl₂ + Na₂CO₃ → CaCO₃ + 2 NaCl
From the equation, we can see that for every 1 mole of CaCl₂ and 1 mole Na₂CO₃ will give 1 mole of CaCO₃ and 2 moles of NaCl
to calculate how many moles of CaCO₃ ,we simply multiply multiply each by the 8.45 moles of CaCl₂ which will reacts
these is because for every 1 mole of CaCl₂ and 1 mole Na₂CO₃ will give 1 mole of CaCO₃ and 2 moles of NaCl
therefore we have every 1x8.45(8.45) mole of CaCl₂ and 1x8.45(8.45) mole Na₂CO₃ will give 1x8.45(8.45) mole of CaCO₃ and 2x8.45(16.9) moles of NaCl
8.45 moles are produced in the reaction
Answer:
<h2>67%</h2>
Explanation:
<h2>Thus the % composition of glucose by mass is carbon 40.0 % oxygen 53.3 % hydrogen 6.7 % in this way, the % composition by mass of any compound can be calculated provided that is formed is known. </h2>
Basically this is used in calculating the nuclear binding energy by converting the mass defect (calculated first) to energy and if we recall, Einstein's equation E=mc2 is the perfection equation to use because E=mc2 in which E represents units of energy, m represents units of mass, and c 2 is the speed of light squared.
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.
