Melting of wax is a physical change and chemical change is the burning of candle.
Chemical changes are those changes by which composition of the original matter changes and physical changes are those changes by which only appearance of the matter changes.
In the given activity:
Hence, Roger is correct ie. The melting wax is a physical change and the burning candle wick is a chemical change.
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Answer:
The answer is
Explanation:
The density of a substance can be found by using the formula
From the question
mass = 10 g
volume = 5 mL
The density of the substance is
We have the final answer as
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A. What is the limiting reagent?
B. How many grams of the excess reactant remains?
C. How many grams of each product is formed?
D. If 12 grams of NaNO3 actually formed in the reaction, what is the percent yield of this reaction?
Answer:
Explanation:
Pb(NO3)2 (aq) + 2 NaI (aq) --> PbI2 (s) + 2 NaNO3 (aq)
MM for each compound -
Pb(NO3): 207 + 14x2 + 16x3x2 = 331
PI2: 207 + 127x2 = 461
NaI: 23 + 127 = 150
NaNO3: 23 + 14 + 16x3 = 85
Moles of Pb(NO3)2 = 50/331 = 0.15
Moles of NaI = 30/150 = 0.2
Ratio of moles is 1:2
So NaI is limiting
Limited to 0.2/2 = 0.1 mole of Pb(NO3)2
Excess = 0.15 - 0.1 = 0.05 mole
Mass remains = 0.05x331 = 16.55 grams
Moles of NaNO3 formed = Moles of NaI reacted = 0.2
Mass = 0.2x85 = 17 grams
Moles of PbI2 formed = Moles of Pb(NO3)2 reacted = 0.1
Mass = 0.1x461 = 46.1 grams
If 12 grams of NaNO3 actually formed in the reaction,
percent yield = 12/17x100% = 70.6%
Answer:
Explanation:
Moles of Pb(NO3)2 = mass/molecular mass
= 50.0 grams/(207.20*1 + 14.01*2 + 16*6)
= 50.0 grams/331.22
= 0.15 moles
Moles of NaI
= 30/(22.99+126.9)
= 30/149.89
= 0.2 Moles
A. NaI is less 2x Pb(NO3)2 so NaI is the limiting reagent.
B. The ratio is 1 to 2 so there is 0.15 - 0.2/2 = 0.05 mole
or 16.78 grams of Pb(NO3)2 left.
C. As NaI is limiting, only 0.2 Moles of NaNO3 is formed.
Mass = Moles * Molecular Mass
Molecular Mass of NaNO3 can be calculated as:
Na - 22.99
N - 14.01
O - 3(16) = 48
23+14+48 = 85gram / mole
Thus, Mass = 0.2*85 = 17 gram of NaNO3
Mass is conserved in a chemical reaction.
Mass of PbI2 can be calculated as:
50+30-16.78-17
= 46.3 gram of PbI2
Mass =
12.75
Thus, 12.75g of Sodium Nitrate can be formed
Answer:
the change in internal energy per mole of sucrose is 5989 KJ/mol
Explanation:
Since the calorimeter is a bomb calorimeter , where the volume remains constant
ΔE= Qrx
And assuming no heat losses to the surroundings
Qv + Qrx = 0
Cc*ΔT - Qrx = 0 ,
Qrx = Cc * ΔT = 7.50 KJ/°C * 22 °C = 165 KJ
the change in internal energy is calculated dividing by the number of sucrose moles involved n. the molecular weight of sucrose is
Mw= 12* 12g/mol + 22* 1g/mol + 11* 16 g/mol = 342 g/mol
n = m / Mw = 10 g / 342 g/mol = 0.029 mol
the change in internal energy per mole of sucrose is
Δe= ΔE/n = 165 KJ/0.029 mol = 5989 KJ/mol