Answer:
2.627775588 degrees Celsius (with 2 sig figs)
Explanation:
We need to use our equation q=m×s×(change in temp) and solve for our change in temperature. So divide both sides by m×s and you'll get:
Change in temp=q/(m×s) Where q= J of heat m=mass and s= specific heat
So now we just plug our numbers in:
56J/(23.6g×0.903J/g°C)= 2.627775588 and since 56 only has 2 sig figs, we round the final answer to 2.6°C
Answer:
709 g
Step-by-step explanation:
a) Balanced equation
Normally, we would need a balanced chemical equation.
However, we can get by with a partial equation, as log as carbon atoms are balanced.
We know we will need an equation with masses and molar masses, so let’s <em>gather all the information</em> in one place.
M_r: 30.07 236.74
C₂H₆ + … ⟶ C₂Cl₆ + …
m/g: 90.0
(i) Calculate the moles of C₂H₆
n = 90.0 g C₂H₆ × (1 mol C₂H₆ /30.07 g C₂H₆)
= 2.993 mol C₂H₆
(ii) Calculate the moles of C₂Cl₆
The molar ratio is (1 mol C₂Cl₆/1 mol C₂H₆)
n = 2.993 mol C₂H₆ × (1 mol C₂Cl₆/1 mol C₂H₆)
= 2.993 mol C₂Cl₆
(iii) Calculate the mass of C₂Cl₆
m = 2.993 mol C₂Cl₆ × (236.74 g C₂Cl₆/1 mol C₂Cl₆)
m = 709 g C₂Cl₆
The reaction produces 709 g C₂Cl₆.
Tin
Chemical Element
Tin is a chemical element with the symbol Sn and atomic number 50. It is a main group metal in group 14 of the periodic table. Wikipedia
Symbol: Sn
Electron configuration: [Kr] 4d105s25p2
Atomic number: 50
Melting point: 449.5°F (231.9°C)
Atomic mass: 118.71 u
Boiling point: 4,717°F (2,603°C)
Electrons per shell: 2, 8, 18, 18, 4
Answer:
68,2%
Explanation:
Supposing the initial salt concentration of lake Parsons is the same of non-isolated lakes, 6,67L, and the change of salt concentration in isolated lake is just for water evaporation it is possible to write:
6,67gL⁻¹×X = 21gL⁻¹×Y
<em>-Where X is the initial water and Y is the water that remains in the isolated lake-</em>
Thus:
6,67X = 21Y
0,318 = Y/X
0,318 is the ratio of water that remains between total water. To obtain the ratio of evaporated water:
1-0,318 = 0,682
In percentage: <em>68,2%</em>
<em />
I hope it helps!
<em />
