Not sure what you exactly mean, but I think what you are talking about are the spheres of earth (Lithosphere, Hydrosphere, Atmosphere, and Biosphere.)
Lithosphere (Terrestrial) is the land on earth. Hydrosphere (Hydrologic) is the water on earth.
I'm not sure if this is 100% correct but I'm guessing this is what you are referring to. It could be different across different science classes.
There should be 1025 grams
Answer:
An insulated beaker with negligible mass contains liquid water with a mass of 0.205kg and a temperature of 79.9 °C How much ice at a temperature of −17.5 °C must be dropped into the water so that the final temperature of the system will be 31.0 °C? Take the specific heat for liquid water to be 4190J/Kg.K, the specific heat for ice to be 2100J/Kg.K, and the heat of fusion for water to be 334000J/kg.
The answer to the above question is
Therefore 0.1133 kg ice at a temperature of -17.5 ∘C must be dropped into the water so that the final temperature of the system will be 31.0 °C
Explanation:
To solve this we proceed by finding the heat reaquired to raise the temperature of the water to 31.0 C from 79.9 C then we use tht to calculate for the mass of ice as follows
ΔH = m×c×ΔT
= 0.205×4190×(79.9 -31.0) = 42002.655 J
Therefore fore the ice, we have
Total heat = mi×L + mi×ci×ΔTi = mi×334000 + mi × 2100 × (0 -−17.5) = 42002.655 J
370750×mi = 42002.655 J
or mi = 0.1133 kg
Therefore 0.1133 kg ice at a temperature of -17.5 ∘C must be dropped into the water so that the final temperature of the system will be 31.0 °C
Answer:
Mass = 32.69 g
Explanation:
Given data:
Mass of copper = 32 g
Mass of zinc formed = ?
Solution:
Chemical equation:
Cu + Zn(NO₃)₂ → Cu(NO₃)₂ + Zn
Number of moles of copper:
Number of moles = mass/molar mass
Number of moles = 32 g/ 63.55 g/mol
Number of moles = 0.5 mol
now we will compare the moles of zinc with copper.
Cu : Zn
1 : 1
0.5 : 0.5
Mass of Zn:
Mass = number of moles × molar mass
Mass = 0.5 mol × 65.38 g/mol
Mass = 32.69 g
Answer:
Explanation: M(PCL5)= 31 + 5(35.5)
=208.5g/mol
M(H20)= 18g/mol
n(PCL5) = 75.5÷208.5
= 0.362mol
n(HCl)/n(PCL5)= 5/1
n(HCl)= 5×0.362
=1.81mol of HCl
