Using deep-sea sediment cores found that Milankovitch cycles correspond with periods of major climate change over the past 450,000 years, with Ice Ages occurring when Earth was undergoing different stages of orbital variation.
Solution of 0.25 M is prepared in two steps,
1) Calculate Amount of Solute:
Molar Mass of Solute: 342.3 g/mol
As we know,
Molarity = Moles / 1 dm³
or,
Moles = Molarity × 1 dm³
Putting Values,
Moles = 0.25 mol.dm⁻³ × 1 dm³
Moles = 0.25 moles
Now, find out mass of sucrose,
As,
Moles = Mass / M.mass
or,
Mass = Moles × M.mass
Putting Values,
Mass = 0.25 mol × 342.3 g.mol⁻¹
Mass = 85.57 g
2) Prepare Solution:
Take Volumetric flask and add 85.57 g of sucrose in it. Then add distilled water up to the mark of 1 dm³. Shake well! The solution prepared is 0.25 M in 1 Liter.
6.11% w/v of Cu2+ implies that 6.11 g of Cu2+ is present in 100 ml of the solution
therefore, 250 ml of the solution would have: 250 ml * 6.11 g/100 ml = 15.275 g
# moles of Cu2+ = 15.275 g/63.546 g mole-1 = 0.2404 moles
1 mole of CuCl2 contain 1 mole of Cu2+ ion
Hence, 0.2404 moles of Cu2+ would correspond to 0.2404 moles of CuCl2
Molar mass of CuCl2 = 134.452 g/mole
The mass of CuCl2 required = 0.2404 moles * 134.452 g/mole = 32.32 grams
Answer:
The metal has a heat capacity of 0.385 J/g°C
This metal is copper.
Explanation:
<u>Step 1</u>: Data given
Mass of the metal = 21 grams
Volume of water = 100 mL
⇒ mass of water = density * volume = 1g/mL * 100 mL = 100 grams
Initial temperature of metal = 122.5 °C
Initial temperature of water = 17°C
Final temperature of water and the metal = 19 °C
Heat capacity of water = 4.184 J/g°C
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<u>Step 2: </u>Calculate the specific heat capacity
Heat lost by the metal = heat won by water
Qmetal = -Qwater
Q = m*c*ΔT
m(metal) * c(metal) * ΔT(metal) = - m(water) * c(water) * ΔT(water)
21 grams * c(metal) *(19-122.5) = -100 * 4.184 * (19-17)
-2173.5 *c(metal) = -836.8
c(metal) = 0.385 J/g°C
The metal has a heat capacity of 0.385 J/g°C
This metal is copper.
