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Answer:
The mass of water = 219.1 grams
Explanation:
Step 1: Data given
Mass of aluminium = 32.5 grams
specific heat capacity aluminium = 0.921 J/g°C
Temperature = 82.4 °C
Temperature of water = 22.3 °C
The final temperature = 24.2 °C
Step 2: Calculate the mass of water
Heat lost = heat gained
Qlost = -Qgained
Qaluminium = -Qwater
Q = m*c*ΔT
m(aluminium)*c(aluminium)*ΔT(aluminium) = -m(water)*c(water)*ΔT(water)
⇒with m(aluminium) = the mass of aluminium = 32.5 grams
⇒with c(aluminium) = the specific heat of aluminium = 0.921 J/g°C
⇒with ΔT(aluminium) = the change of temperature of aluminium = 24.2 °C - 82.4 °C = -58.2 °C
⇒with m(water) = the mass of water = TO BE DETERMINED
⇒with c(water) = 4.184 J/g°C
⇒with ΔT(water) = the change of temperature of water = 24.2 °C - 22.3 °C = 1.9 °C
32.5 * 0.921 * -58.2 = -m * 4.184 * 1.9
-1742.1 = -7.95m
m = 219.1 grams
The mass of water = 219.1 grams
Cadmium chloride is a highly soluble compound. The equation for its dissolution is:
CdCl₂(s) → Cd⁺²(aq) + 2Cl⁻(aq)
This dissociation in water allows for the cadmium and chlorine ions to take part in reactions. This is the reason that solutions of chemicals are prepared when a reaction needs to take place.
Explanation:
Average atomic mass of the vanadium = 50.9415 amu
Isotope (I) of vanadium' s abundance = 99.75 %= 0.9975
Atomic mass of Isotope (I) of vanadium ,m= 50.9440 amu
Isotope (II) of vanadium' s abundance =(100%- 99.75 %) = 0.25 % = 0.0025
Atomic mass of Isotope (II) of vanadium ,m' = ?
Average atomic mass of vanadium =
m × abundance of isotope(I) + m' × abundance of isotope (II)
50.9415 amu =50.9440 amu× 0.9975 + m' × 0.0025
m'= 49.944 amu
The atomic mass of isotope (II) of vanadium is 49.944 amu.
Answer: The substance is in the gas phase only in region 5
The substance is in both the liquid and the solid phase in region 2
The substance is in only the liquid phase in region 3
The melting point is the temperature at region 2
The boiling point is the temperature at region 4
Explanation: