A) James Cook.
B) He put his sailors on a strict diet to see if they would get scurvy.
C) Sauerkraut.
D) He told others of this diet and that none of his sailors died of scurvy.
E) Chemicals can be found almost anywhere and almost anyone can be a scientist.
(a) In this section, give your answers to three decimal places.
(i)
Calculate the mass of carbon present in 0.352 g of CO
2
.
Use this value to calculate the amount, in moles, of carbon atoms present in 0.240 g
of
A
.
(ii)
Calculate the mass of hydrogen present in 0.144 g of H
2
O.
Use this value to calculate the amount, in moles, of hydrogen atoms present in 0.240 g
of
A
.
(iii)
Use your answers to calculate the mass of oxygen present in 0.240 g of
A
Use this value to calculate the amount, in moles, of oxygen atoms present in 0.240 g
of
A
(b)
Use your answers to
(a)
to calculate the empirical formula of
A
thank you
hope it helpsss
Answer : The volume of the cube submerged in the liquid is, 29.8 mL
Explanation :
First we have to determine the mass of ice.
Formula used :

Given:
Density of ice = 
Volume of ice = 45.0 mL


The cube will float when 40.5 g of liquid is displaced.
Now we have to determine the volume of the cube is submerged in the liquid.



Thus, the volume of the cube submerged in the liquid is, 29.8 mL
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.
d. When aluminum-28 undergoes beta decay, silicon-28 is produced.
Explanation:
When the nuclei of aluminium-28 decays, it produces silicon- 28:
Aluminium ²⁸₁₃Al
Silicon 28 ²⁸₁₄Si
beta particle ⁰₋₁
²⁸₁₃Al → ²⁸₁₄Si + ⁰₋₁
This way, the mass and atomic number are conserved.
Conservation of mass number:
28 = 28 + 0, 28 = 28
13 = 14 -1 , 13 = 13
Learn more:
Balancing nuclear equations brainly.com/question/10094982
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