The second one, I could be mistaken though
Reactants would be the missing word in the sentence because they make up the products in a chemical reaction
Answer:
A. 266g/mol
Explanation:
A colligative property of matter is freezing point depression. The formula is:
ΔT = i×Kf×m <em>(1)</em>
Where:
ΔT is change in temperature (0°C - -0,14°C = 0,14°C)i is Van't Hoff factor (1 for a nonelectrolyte dissolved in water), kf is freezing point molar constant of solvent (1,86°Cm⁻¹) and m is molality of the solution (moles of solute per kg of solution). The mass of the solution is 816,0g
Replacing in (1):
0,14°C = 1×1,86°Cm⁻¹× mol Solute / 0,816kg
<em>0,0614 = mol of solute</em>.
As molar mass is defined as grams per mole of substance and the compound weights 16,0g:
16,0g / 0,0614 mol = 261 g/mol ≈ <em>A. 266g/mol</em>
I hope it helps!
Given:
Ma = 31.1 g, the mass of gold
Ta = 69.3 °C, the initial temperature of gold
Mw = 64.2 g, the mass of water
Tw = 27.8 °C, the initial temperature of water
Because the container is insulated, no heat is lost to the surroundings.
Let T °C be the final temperature.
From tables, obtain
Ca = 0.129 J/(g-°C), the specific heat of gold
Cw = 4.18 J/(g-°C), the specific heat of water
At equilibrium, heat lost by the gold - heat gained by the water.
Heat lost by the gold is
Qa = Ma*Ca*(T - Ta)
= (31.1 g)*(0.129 J/(g-°C)(*(69.3 - T °C)-
= 4.0119(69.3 - T) j
Heat gained by the water is
Qw = Mw*Cw*(T-Tw)
= (64.2 g)*(4.18 J/(g-°C))*(T - 27.8 °C)
= 268.356(T - 27.8)
Equate Qa and Qw.
268.356(T - 27.8) = 4.0119(69.3 - T)
272.3679T = 7738.32
T = 28.41 °C
Answer: 28.4 °C
