Which has more gravitational potential energy—a book sitting on a desk or the same
1 answer:
Answer:
The book sitting on the desk
Explanation:
The gravitational potential energy of an object is the energy possessed by the object due to its position relative to the ground.
It is calculated as:
where
m is the mass of the object
g is the acceleration due to gravity
h is the height of the object with respect to the ground
From the formula, we see that the GPE of an object is directly proportional to the heigth h: so, the higher the location of the object, the larger the GPE.
In this problem, we are comparing a book sitting on a desk and the same book sitting on the floor. In the two situations, the mass of the book is the same; however, in the first case, the value of the height is h, while in the second case, the value of h is lower (because the book is located at a lower height, being on the floor).
Therefore, we can conclude that the first book must have a larger GPE, since it has a larger value of h.
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Answer: 72 grams of are needed to completely burn 19.7 g
Explanation:
According to avogadro's law, 1 mole of every substance weighs equal to molecular mass and contains avogadro's number of particles.
To calculate the number of moles, we use the equation:
Putting in the values we get:
According to stoichiometry:
1 mole of requires 5 moles of oxygen
0.45 moles of require=
moles of oxygen
Mass of
72 grams of are needed to completely burn 19.7 g
Answer: 27.09 ppm and 0.003 %.
First, <u>for air pollutants, ppm refers to parts of steam or gas per million parts of contaminated air, which can be expressed as cm³ / m³. </u>Therefore, we must find the volume of CO that represents 35 mg of this gas at a temperature of -30 ° C and a pressure of 0.92 atm.
Note: we consider 35 mg since this is the acceptable hourly average concentration of CO per cubic meter m³ of contaminated air established in the "National Ambient Air Quality Objectives". The volume of these 35 mg of gas will change according to the atmospheric conditions in which they are.
So, according to the <em>law of ideal gases,</em>
PV = nRT
where P, V, n and T are the pressure, volume, moles and temperature of the gas in question while R is the constant gas (0.082057 atm L / mol K)
The moles of CO will be,
n = 35 mg x x
→ n = 0.00125 mol
We clear V from the equation and substitute P = 0.92 atm and
T = -30 ° C + 273.15 K = 243.15 K
V =
→ V = 0.0271 L
As 1000 cm³ = 1 L then,
V = 0.0271 L x = 27.09 cm³
<u>Then the acceptable concentration </u><u>c</u><u> of CO in ppm is,</u>
c = 27 cm³ / m³ = 27 ppm
<u>To express this concentration in percent by volume </u>we must consider that 1 000 000 cm³ = 1 m³ to convert 27.09 cm³ in m³ and multiply the result by 100%:
c = 27.09 x
x 100%
c = 0.003 %
So, <u>the acceptable concentration of CO if the temperature is -30 °C and pressure is 0.92 atm in ppm and as a percent by volume is </u>27.09 ppm and 0.003 %.