Answer:the temperature might increase because of the humidity in the room so everything gets goofy and moist
Explanation:
(2) a zinc strip and 3.0 M HCl(aq)
(3) zinc powder and 1.0 M HCl(aq)
(4) zinc powder and 3.0 M HCl(aq)
Answer:
(4) zinc powder and 3.0 M HCl(aq)
Explanation:
First consider the equation of reaction:
1 mole of Zinc requires 2 moles of HCl for reaction.
mole of zinc = mass/molar mass = 5.0/65.38 = 0.0765 mole
0.0765 mole of zinc will require: 0.0765 x 2 = 0.1530 mole HCl
Mole of 1.0 M HCl = Molarity x volume = 1 x 0.05 = 0.05 mole
Mole of 3.0 M HCl = 3 x 0.05 = 0.15
Hence, the 3.0 M HCl is what will give the required number of mole.
Zinc in powder form will react faster compared to in strip because the former has higher surface area for reaction.
Therefore, zinc powder and 3.0 M HCl will give the fastest reaction rate.
The correct option is option 4.
What volume of oxygen gas is released at STP if 10.0 g of potassium chlorate is decomposed? (The molar mass of KClO3 is 122.55 g/mol.)
0.914 L
1.83 L
2.74 L
3.66 L
1. The balanced reaction would be:
2KClO3 --> 2KCl + 3O2
We are given the amount of potassium chlorate for the reaction. This will be the starting point of our calculation.
10.0 g KClO3 ( 1 mol KClO3 / 122.55 g KClO3) (3 mol O2 / 2 mol KClO3) = 0.1224 mol O2
V = nRT/P =0.1224 mol O2 x 273 K x 0.08206 atm L/mol K / 1 atm
V=2.74 L
Answer:
third option 2.74L
The law that can be applied here is Charle’s Law which states that for a fixed mass of gas and at constant pressure, the volume is directly proportional with the absolute temperature. It is represented by the equation of V1/T1=V2/T2. Since there is no negative volume, we will change the temperature into Kelvin, (-123+273=150K) and (70+273=300K). plugging in the values, V2 = (300K)(120mL/150K) we get a volume of 240mL.
A) Pentanoid acid
B) Pentanal
C) Pentan-2-one
D) Propanoic acid and ethanoic acid
Answer:
C) Pentan-2-one
Explanation:
The product of the reaction of pentan-2-ol with acidified potassium dichromate is pentan-2-one.
Pentan-2-ol is a secondary alcohol, and secondary alcohols are oxidized to ketones by acidified potassium dichromate. The reaction is as follows:
CH3CH(OH)CH3 + K2Cr2O7 + H2SO4 → CH3COCH3 + Cr2(SO4)3 + K2SO4 + H2O
Propanoic and ethanoic acids are not products of this reaction.
Therefore, the answer is (C).
bardAI