Given that there is 48 liters of gasoline to be burned and that 45 kJ of energy is released per gram of gasoline burned, the amount of energy that the gasoline fuel produces can then be calculated, First, we convert 48 liters of gasoline to units of mass (grams) in order to use the given conversion of 45 kJ per gram of gasoline. To do this, we use the density of gasoline which is 0.77 g/mL. The following expression is then used:
48 L gasoline x 1000 mL/L x 0.77 g/mL x 45 kJ/g gasoline = 1663200 kJ
<span>The amount of energy produced by burning 48 L of gasoline was then determined to be 1663200 kJ. </span>
The number in isotope platinum-194 stands for the <em>total </em>amount of protons and neutrons. In other words, this is the mass. =)
Answer:
The volume of the sample is 17.4L
Explanation:
The reaction that occurs requires the same amount of CO and NO. As the moles added of both reactants are the same you don't have any limiting reactant. The only thing we need is the reaction where 4 moles of gases (2mol CO + 2mol NO) produce 3 moles of gases (2mol CO2 + 1mol N2). The moles produced are:
0.1800mol + 0.1800mol reactants =
0.3600mol reactant * (3mol products / 4mol reactants) = 0.2700 moles products.
Using Avogadro's law (States the moles of a gas are directly proportional to its pressure under constant temperature and pressure) we can find the volume of the products:
V1n2 = V2n1
<em>Where V is volume and n moles of 1, initial state and 2, final state of the gas</em>
Replacing:
V1 = 23.2L
n2 = 0.2700 moles
V2 = ??
n1 = 0.3600 moles
23.2L*0.2700mol = V2*0.3600moles
17.4L = V2
<h3>The volume of the sample is 17.4L</h3>
Answer;
1.6 kg.
Solution;
The density is 1.36 g/ml;
The volume is 1.25 qt
However; 1 qt = 946.35 ml
Mass is given by; density × volume;
= 1.25 qts × 946.25 ml/qt × 1.36 g/ml =1609 g
but; 1 kg = 1000 g
Hence the mass = 1609/1000 = 1.609 Kg or 1.61 (sig figs)
