Answer:
b). silver (Ag)
Explanation:
If you look at the periodic table, you just need to look at the atomic number of the element, because the atomic number tells you how many protons there are in the nucleus of the element.
But do be careful because some periodic tables have the molar mass at the top left corner, but the one I use has the atomic number at the top left corner, so make sure you look for the atomic number and not the molar mass.
Answer:
C. Occurs where a plate is moving down.
Explanation:
Ocean trenches are steep depressions in the deepest parts of the ocean where old ocean crust from one tectonic plate is pushed beneath another plate, raising mountains, causing earthquakes, and forming volcanoes on the seafloor and on land.
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Answer:
The product is cyclohexanol
Explanation:
Firstly,
A ketone undergo a borohydride reduction reaction to form an alcohol as below,
R-CO-R' ⇒ R-CO(OH)-R'
- IR Spectrum confirms that alcohol group is existed with the peak at 3400 cm⁻¹
- From 1H-NMR, the product has 10 hydrogen atoms, the MS suggest that the formula is C₅H₁₀O (M = 86). With this formula, the alcohol is monosaturated. Since, the substance already underwent reduction reaction, the only way to suggest a monosaturated compound is a cyclic alcohol. So the compound is cyclopentanol.
- Check with other spectroscopic properties,
- 3 signals of 13C NMR confirms the structure is symmetrical, δ 24.2, (-<u>C</u>H₂-CH₂-CH(CH₂-)-OH), δ 35.5 (-CH₂-<u>C</u>H₂-CH(CH₂-)-OH), δ 73.3 (-CH₂-CH₂-<u>C</u>H(CH₂-)-OH).
1.56 δ (4H, triplet) - (-C<u>H</u>₂-CH₂-CH-OH) ; triplet as coupling with 2 H,
1.78 δ (4H, multiplet) - (-CH₂-C<u>H</u>₂-CH-OH); multiplet as coupling with 2H of CH₂, 1 H of CH
3.24 δ (1H, quintet); - (-CH₂-CH₂-C<u>H</u>(CH₂-)-OH), coupling with4 H of 2 group of CH₂
3.58 δ (1H, singlet); - (-CH₂-CH₂-CH(CH₂-)-O<u>H</u>), hydrogen of alcohol group, not tend to coupling with other hydrogen
Answer:
how can I solve this ?4Al+3O2 produce 2Al2O3 find a) oxygen atoms needed to react with 5.4 g of aluminium b) grams of oxygen needed to react with 0.6 mol of aluminium?
(A) n=m/M,
n(Al)=5.4/27=0.2 moles
n(O2)=n(Al)*3/4=0.2*3/4=0.15 moles
Number of oxygen atoms= n(O2)*Avogadro's number
=0.15*6.02*10^23=9.03*10^22 oxgyen atoms
(B)
n=m/M
n(Al)=0.6/27=0.02222 moles
n(O2)=n(Al)*3/4=0.016666 moles
m=n*M
m(O2)=0.0166666*32=0.53333 grams