You would need 8.1 <span>moles of LiOH</span>
The complete question is
If the compound below is oxidized, the resulting product is ___.
the compound given is Butanal
Select one:
a. methane and propane
b. butanal
c. butanoic acid
d. butane
Answer:
C. Butanoic Acid
Explanation:
When Butanal is oxidized using reagents like KMnO4, Tollens reagent etc.
When Aldehyde is oxidized corresponding Carboxylic Acid is produced
Butanal → Butanoic Acid
Answer:
percentage by mass of each element in a compound.
Explanation:
Answer:
Lewis structure for nitrogen triiodide,
is given in the attachment.
Explanation:
Given:
The given compound is Nitrogen triiodide. In which 1 atom of Nitrogen combines with 3 atoms of Iodine. Both Nitrogen and Iodine are non-metals,So they form covalent bond by sharing of electrons.
The electron configuration of Nitrogen and Iodine is given below;
There are 5 electrons in valance shell of Nitrogen atom and 7 electrons in valance shell of Iodine atom.
So, 3 atom of Iodine shares 1 electron with 1 electrons of Nitrogen.
The Lewis dot Structure is in the attachment.
Answer:
(a) weight percent of Cu = 44.59%
(b) weight percent of Zn = 53.49%
(c) weight percent of Pb = 1.91%
Explanation:
Given: Alloy contains- Cu=93.1 g, Zn=111.7 g, Pb=4.0 g
Therefore, the total mass of the alloy (m) = mass of Cu (m₁) + mass of Zn (m₂)+ mass of Pb (m₃)
m= 93.1 g + 111.7 g + 4.0 g = 208.8 g
(a) weight percent of Cu = (m₁ ÷ m)× 100% = (93.1 g ÷ 208.8 g)× 100% =44.59%
(b) weight percent of Zn = (m₂ ÷ m)× 100% = (111.7 g ÷ 208.8 g)× 100% =53.49%
(c) weight percent of Pb = (m₃ ÷ m)× 100% = (4.0 g ÷ 208.8 g)× 100% =1.91%
