The effect of an insoluble impurity, such as sand, on the observed melting point of a compound would be none. It will not depress or elevate the melting point of the compound. Instead, it would affect the reading if you are trying to determine the melting point of the compound. This is because you might be missing the actual melting point of the compound since you will be waiting for the whole sample to liquify. You would not be able to determine exactly that temperature because of the insoluble impurity would have a different melting point than that of the compound.
Answer:
See explanation
Explanation:
We must first write the equation of the reaction as follows;
C3H8 + 5O2 ----> 3CO2 + 4H2O
Now;
We obtain the number of moles of C3H8 = 132.33g/44g/mol = 3 moles
So;
1 mole of C3H8 yields 3 moles of CO2
3 moles of C3H8 yields 3 × 3/1 = 9 moles of CO2
We obtain the number of moles of oxygen = 384.00 g/32 g/mol = 12 moles
So;
5 moles of oxygen yields 3 moles of CO2
12 moles of oxygen yields 12 × 3/5 = 7.2 moles of CO2
We can now decide on the limiting reactant to be C3H8
Therefore;
Mass of CO2 produced = 9 moles of CO2 × 44 g/mol = 396 g of CO2
Again;
1 moles of C3H8 yields 4 moles of water
3 moles of C3H8 yields 3 × 4/1 = 12 moles of water
Hence;
Mass of water = 12 moles of water × 18 g/mol = 216 g of water
In order to obtain the percentage yield from the reaction, we have;
b) Actual yield = 269.34 g
Theoretical yield = 396 g
Therefore;
% yield = actual yield/theoretical yield × 100/1
Substituting values
% yield = 269.34 g /396 g × 100
% yield = 68%
They can form acids or ‘acid rain’ like nitric acid or sulfuric acid which can damage buildings and structures.
70.0 g. The decomposition of 125 g CaCO3 produces 700 g CaO.
MM = 100.09 56.08
CaCO3 → CaO + CO2
Mass 125 g
a) Moles of CaCO3 = 125 g CaCO3 x (1 mol CaCO3/100.09 g CaCO3)
= 1.249 mol CaCO3
b) Moles of CaO = 1.249 mol CaCO3 x (1 mol CaO/1 mol CaCO3)
= 1.249 mol CaO
c) Mass of CaO = 1.249 mol CaO x (56.08 g CaO/1 mol CaO) = 70.0 g
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