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
Answer:
Without dark matter galaxies would loose an extreme amount of gas required to create stars.
Without dark matter the universe wont have as many galaxies clumped together forming larger versions of those galaxies. This would cause a change in the structure of the "skeleton" of the web.
(Hope this can help, I didn't do exactly as it is said to because that is your job)
:)
Explanation:
Forbes gives somewhat of an explanation if you are curious.
(Ethan Siegal, "The Universe Would Be Very Different Without Dark Matter", Forbes)
Answer:
V₂ = 15.6 L
Explanation:
Given data:
Initial volume = 175 mL (0.175 L)
Initial pressure = 1 atm
Initial temperature = 273 K
Final temperature = -5°C (-5+273 = 268 K)
Final volume = ?
Final pressure = 1.16 kpa (1.16/101=0.011 atm)
Formula:
P₁V₁/T₁ = P₂V₂/T₂
P₁ = Initial pressure
V₁ = Initial volume
T₁ = Initial temperature
P₂ = Final pressure
V₂ = Final volume
T₂ = Final temperature
Solution:
V₂ = P₁V₁ T₂/ T₁ P₂
V₂ = 1 atm × 0.175 L × 268 K / 273 K × 0.011 atm
V₂ = 46.9 L / 3.003
V₂ = 15.6 L
In writing the formula for a salt the symbol of the cation is first then the anion is written second .
