Answer:
2.83 g
Explanation:
At constant temperature and pressure, Using Avogadro's law
Given ,
V₁ = 2.12 L
V₂ = 3.12 L
n₁ = 0.120 moles
n₂ = ?
Using above equation as:



n₂ = 0.17660 moles
Molar mass of methane gas = 16.05 g/mol
So, Mass = Moles*Molar mass = 0.17660 * 16.05 g = 2.83 g
<u>2.83 g are in the piston.</u>
Answer:
a. withdraws electrons inductively
b. donates electrons by hyperconjugation
c. donates electrons by resonance
d. withdraws electrons inductively
Explanation:
a. The bromide ion is a highly electronegative ion (in the halide series). Electronegative substituents on acids increase the acidity by inductive electron withdrawal method. The higher the electronegativity of a substance, the greater the acidity. The halogens have this order of electronegativity:
F > Cl > Br>I
b. The carboxyl groups have a stabilization of the sigma and pi bonds. This is achieved through a special delocalization of electrons. Because of the delocalization, hyperconjugation is the result effect.
c. The NHCH₃ group has a highly electonegative nitrogen atom that pulls the electron cloud towards itself. In this case, it withdraws electrons inductively. As a result, it donates electrons by resonance.
d. The OCH₃ group has a highly electonegative oxygen atom. This oxygen atom withdraws electron cloud towards itself. As a result, it withdraws electrons inductively.
Answer:
The reason for the suspicion was because the manner in which iodine reacted chemically as well as its other chemical properties, indicated that it belonged in the same group as chlorine and bromine, while the much heavier tellurium should be placed in the previous group
The suspicion was proved to be correct when the atomic number of tellurium was found to be 52 and that of iodine was found to be 53 by later scientists
Explanation:
I think the answer is C for this question
Answer: a) 
b) 
Explanation:
If percentage are given then we are taking total mass is 100 grams.
So, the mass of each element is equal to the percentage given.
a) Mass of Ba= 66.06 g
Mass of Cl = 34.0 g
Step 1 : convert given masses into moles.
Moles of Ba =
Moles of Cl = \frac{\text{ given mass of Cl}}{\text{ molar mass of Cl}}= \frac{34g}{35.5g/mole}=0.96moles[/tex]
Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.
For Ba =
For O =
The ratio of Ba: Cl= 1:2
Hence the empirical formula is 
b) Mass of Bi= 80.38 g
Mass of O= 18.46 g
Mass of H = 1.16 g
Step 1 : convert given masses into moles.
Moles of Bi =
Moles of O=
Moles of H=
Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.
For Bi=
For O =
For H=
The ratio of Bi: O: H= 1:3: 3
Hence the empirical formula is 