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>
Answer:
0.019 moles of M2CO3
Explanation:
M2CO3(aq) + BaCl2 (aq) --> 2MCl (aq) + BaCO3(s)
From the equation above;
1 mol of M2CO3 reacts to produce 1 mol of BaCO3
Mass of BaCO3 formed = 3.7g
Molar mass of BaCO3 = 197.34g/mol
Number of moles = Mass / Molar mass = 3.7 / 197.34 = 0.0187 ≈ 0.019mol
Since 1 mol of M2CO3 reacts with 1 mol of BaCO3,
1 = 1
x = 0.019
x = 0.019 moles of M2CO3
<h3>
Answer:</h3>
19.3 g/cm³
<h3>
Explanation:</h3>
Density of a substance refers to the mass of the substance per unit volume.
Therefore, Density = Mass ÷ Volume
In this case, we are given;
Mass of the gold bar = 193.0 g
Dimensions of the Gold bar = 5.00 mm by 10.0 cm by 2.0 cm
We are required to get the density of the gold bar
Step 1: Volume of the gold bar
Volume is given by, Length × width × height
Volume = 0.50 cm × 10.0 cm × 2.0 cm
= 10 cm³
Step 2: Density of the gold bar
Density = Mass ÷ volume
Density of the gold bar = 193.0 g ÷ 10 cm³
= 19.3 g/cm³
Thus, the density of the gold bar is 19.3 g/cm³
