Answer:
the number of neutrons in the nucleus
Explanation:
The mass number 205 is the total number of protons and neutrons in the nucleus. Lead has 82 protons, so a mass number of 205 indicates a nucleus with 123 neutrons.
Similarly, the mass number 208 indicates a lead nucleus with 126 neutrons.
The difference in atomic structure is the difference in the number of neutrons in the nucleus.
Answer:
Option A. FeCl3
Explanation:
The following data were obtained from the question:
Mass of iron (Fe) = 6.25g
Mass of the compound formed = 18g
From the question, we were told that the compound formed contains chlorine. Therefore the mass of chlorine is obtained as follow
Mass of chlorine (Cl) = Mass of compound formed – Mass of iron.
Mass of chlorine (Cl) = 18 – 6.25
Mass of chlorine (Cl) = 11.75g
The compound therefore contains:
Iron (Fe) = 6.25g
Chlorine (Cl) = 11.75g
The empirical formula for the compound can be obtained by doing the following:
Step 1:
Divide by their molar mass
Fe = 6.25/56 = 0.112
Cl = 11.75/35.5 = 0.331
Step 2:
Divide by the smallest
Fe = 0.112/0.112 = 1
Cl = 0.331/0.112 = 3
The empirical formula for the compound is FeCl3
Answer:
The specific heat of the metal is 2.09899 J/g℃.
Explanation:
Given,
For Metal sample,
mass = 13 grams
T = 73°C
For Water sample,
mass = 60 grams
T = 22°C.
When the metal sample and water sample are mixed,
The addition of metal increases the temperature of the water, as the metal is at higher temperature, and the addition of water decreases the temperature of metal. Therefore, heat lost by metal is equal to the heat gained by water.
Since, heat lost by metal is equal to the heat gained by water,
Qlost = Qgain
However,
Q = (mass) (ΔT) (Cp)
(mass) (ΔT) (Cp) = (mass) (ΔT) (Cp)
After mixing both samples, their temperature changes to 27°C.
It implies that
, water sample temperature changed from 22°C to 27°C and metal sample temperature changed from 73°C to 27°C.
Since, Specific heat of water = 4.184 J/g°C
Let Cp be the specific heat of the metal.
Substituting values,
(13)(73°C - 27°C)(Cp) = (60)(27°C - 22℃)(4.184)
By solving, we get Cp =
Therefore, specific heat of the metal sample is 2.09899 J/g℃.
What is the question that needs an answer?
Answer:
The answer to your question is: 0.25 l
Explanation:
Data
P1 = 1 atm
V1 = 0.5 l
P2 =2 atm
V2 = ?
T = constant
Formula
V1P1 = V2P2
Clear V2 from the formula
V2 = V1P1/P2
Substitution
V2 = (0.5)(1)/2 substitution
= 0.25 l result