The ocean hold 97 percent of earths water.
<span>70.4 mg CO2 x 1.0 g /1000 mg x 1 mole CO2/ 44 gCO2 x 1 mole C/1 mole CO2 = 0.0016 moles C
14.4 mg H2O x 1.0 g/1000 mg x 1 mole H2O/18 g H2O x 2 moles H/ 1 mole H2O = 0.0016 moles O
molar mass of C=12 g/mole
molar mass of H=1 g/mole
0.0016 moles C x 12 g C/ 1 mole C = 0.0192 g C or 19.2 mg C
0.00156 moles H x 1 g H/1 mole H = 0.00156 g H or 1.56 mg H
mg O= 30.4 mg vanillin - 19.2 mg C – 1.56 mg H = 9.64 mg O
molar mass of O=16 g/mole
9.64 mg O x 1 g/1000 mg x 1 mole O/16.0 g = 0.000602
C.0016 H.0016 O.000602; divide all the moles by the smallest value of0.000602
C2.66H2.66O1 is the empirical formula;
to obtain whole numbers multiply by 3
3[C2.66H2.66O1] = C8H8O3
above formula weight: 8(C) + 8(H) + 3(O) = 8(12) + 8(1) + 3(16) = 152 amu
The empirical formula weight and the molecular formula weight are the same .
Molecular formula is C8H8O3.</span>
The answer is: the mass of carbon is 420.6 grams.
m(C₈H₁₈) = 500 g; mass of octane.
M(C₈H₁₈) = 114.22 g/mol; molar mass of octane.
n(C₈H₁₈) = m(C₈H₁₈) ÷ M(C₈H₁₈).
n(C₈H₁₈) = 500 g ÷ 114.22 g/mol.
n(C₈H₁₈) = 4.38 mol; amount of octane.
In one molecule of octane, there are eight carbon atoms:
n(C) = 8 · n(C₈H₁₈).
n(C) = 8 · 4.38 mol.
n(C) = 35.02 mol; amount of carbon.
m(C) = 35.02 mol · 12.01 g/mol.
m(C) = 420.6 g; mass of carbone.
Answer:
Explanation:
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In this case, since there is a 2:2 mole ratio between sodium peroxide and water according to the given reaction, it is possible to apply the following stoichiometric setup for the calculation of the required mass of water:
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Answer:
The time taken for the cross mark to disappear decreases steadily down the column.
Explanation:
Now if we look at the data provided, we will discover that the volume of the HCl was held constant while the volume of the thiosulphate was increased steadily and the volume of water decreased steadily.
Recall that a system is more concentrated when it contains less volume of water and more volume of reactants. Hence as the volume of water in the system is being reduced, the concentration of reactants is increased.
It has been established that an increase in the concentration of reactants lead to an increase in the rate of reaction. The disappearance of the cross shows the completion of the reaction between HCl and thiosulphate. The faster or slower the cross disappears, the faster or slower the rate of reaction.
Since increase in concentration of reactants increases the rate of reaction, it is observed that as the volume of the thiosulphate increases (reactant concentration increases) the cross disappears faster (rate of reactant increases). Hence as the volume of thiosulphate increases, it takes a shorter time for the cross to disappear. This implies that the time column in the table (refer to the question) will decrease steadily as the volume of thiosulphate increases.
