The formula to find yield is
(Actual Yield)/(Theorectical Yield) x100
Just do the math.
85.22% x 113 = 96.2986
Convert it to 3 significant figures
Ans: 96.3g
Answer:
The specific heat capacity of the metal is 0.843J/g°C
Explanation:
Hello,
To determine the specific heat capacity of the metal, we have to work on the principle of heat loss by the metal is equals to heat gained by the water.
Heat gained by the metal = heat loss by water + calorimeter
Data,
Mass of metal (M1) = 512g
Mass of water (M2) = 325g
Initial temperature of the metal (T1) = 15°C
Initial temperature of water (T2) = 98°C
Final temperature of the mixture (T3) = 78°C
Specific heat capacity of metal (C1) = ?
Specific heat capacity of water (C2) = 4.184J/g°C
Heat loss = heat gain
M2C2(T2 - T3) = M1C1(T3 - T1)
325 × 4.184 × (98 - 78) = 512 × C1 × (78 - 15)
1359.8 × 20 = 512C1 × 63
27196 = 32256C1
C1 = 27196 / 32256
C1 = 0.843J/g°C
The specific heat capacity of the metal is 0.843J/g°C
So you have a balloon rising through the atmosphere. use the formula p1/v1=p2/v2 and add the variables into the equation, leaving 295/52.5=252/x. multiply 252 by 52.5 and divide that number by 295.
52.5*252=13230. divide by 295 =44.9 L
Answer: acetone molecule ( CH₃-CO-CH₃)
Explanation:
1) Acetone is CH₃-CO-CH₃
2) That is a molecule (build up of covalent bonds).
3) When dissolved, covalent bonded compounds remain as separate molecules, then it is said that the major species present in the solution is the molecule. The molecules of acetone are surrounded (sovated) by the molecules of water.
This as opposed to the case of ionic compounds that ionize. When a compound as NaCl dissolves in water, it ionizes completely, so the major speceies are not NaCl formulas, but the ions Na⁺ and Cl⁻, not molecules.
That leads to the answer: the major species present when acetone is dissolved in water is the molecules of acetone (you do not need to state the fact that the molecules of water are part of the solution, because that is not the target of the question).
Let the mass of CaO = x grams
So mass of BaO = 5.14 -x grams
moles of CaO = mass / molar mass = x / 56
Moles of BaO = mass / molar mass = 5.14-x / 153
initial moles of CO2 = PV / RT = 750 X 1.50 / 760 X 0.0821 X 303 = 0.06
final mole sof CO2 = PV / RT = 230 X 1.50 / 760 X 0.0821 X 303 = 0.018
So moles of BaCO3 and CaCO3 formed = 0.06 - 0.018 = 0.042 moles
x / 56 + (5.14-x) /153 = 0.042
on solving
x = 0.68
So mass of CaO = 0.68 g
So percentage of CaO = 0.68 X 100 / 5.14 = 13.4 %
Percentage of BaO = 86.6%
