Y=15 x=60
•1/3 •1/3
y=5 x=45
(60)/(60+5.05)=.922367 C
1-0.922367=0.07763259 H
(0.922367)(78.12)=72.05534204 C
(0.07763259)(78.12)=6.06 H
72.05534204/(12.01)=6 C
6.06/1.01=6 H
Empirical= CH
Molecular=C6H6
Answer: 1). The temperature of this gas is 1032.88 K.
2). There are 21.48 moles of gas exist at a pressure of 14.5 atm, a volume of 45 liters, and a temperature of
.
Explanation:
1). Given: No. of moles = 6 moles
Pressure = 10.6 atm
Volume = 48 L
Formula used to calculate temperature is as follows.
PV = nRT
where,
P = pressure
V = volume
n = no. of moles
R = gas constant = 0.0821 atm
T = temperature
Substitute the values into above formula as follows.
![PV = nRT\\10.6 atm \times 48 L = 6 mol \times 0.0821 L atm/mol K \times T\\T = \frac{10.6 atm \times 48 L}{6 mol \times 0.0821 L atm/mol K}\\= \frac{508.8}{0.4926} K\\= 1032.88 K](https://tex.z-dn.net/?f=PV%20%3D%20nRT%5C%5C10.6%20atm%20%5Ctimes%2048%20L%20%3D%206%20mol%20%5Ctimes%200.0821%20L%20atm%2Fmol%20K%20%5Ctimes%20T%5C%5CT%20%3D%20%5Cfrac%7B10.6%20atm%20%5Ctimes%2048%20L%7D%7B6%20mol%20%5Ctimes%200.0821%20L%20atm%2Fmol%20K%7D%5C%5C%3D%20%5Cfrac%7B508.8%7D%7B0.4926%7D%20K%5C%5C%3D%201032.88%20K)
Hence, temperature of this gas is 1032.88 K.
2). Given: Pressure = 14.5 atm
Volume = 45 L
Temperature = ![97^{o}C = (97 + 273) K = 370 K](https://tex.z-dn.net/?f=97%5E%7Bo%7DC%20%3D%20%2897%20%2B%20273%29%20K%20%3D%20370%20K)
Formula used to calculate number of moles is as follows.
![PV = nRT\\14.5 atm \times 45 L = n \times 0.0821 L atm/mol K \times 370 K\\n = \frac{14.5 atm \times 45 L}{0.0821 L atm/mol K \times 370 K}\\= \frac{652.5}{30.377} mol\\= 21.48 mol](https://tex.z-dn.net/?f=PV%20%3D%20nRT%5C%5C14.5%20atm%20%5Ctimes%2045%20L%20%3D%20n%20%5Ctimes%200.0821%20L%20atm%2Fmol%20K%20%5Ctimes%20370%20K%5C%5Cn%20%3D%20%5Cfrac%7B14.5%20atm%20%5Ctimes%2045%20L%7D%7B0.0821%20L%20atm%2Fmol%20K%20%5Ctimes%20370%20K%7D%5C%5C%3D%20%5Cfrac%7B652.5%7D%7B30.377%7D%20mol%5C%5C%3D%2021.48%20mol)
Hence, there are 21.48 moles of gas exist at a pressure of 14.5 atm, a volume of 45 liters, and a temperature of
.
Answer:
An electric current is a stream of charged particles, such as electrons or ions, moving through an electrical conductor or space. It is measured as the net rate of flow of electric charge through a surface or into a control volume. The moving particles are called charge carriers, which may be one of several types of particles, depending on the conductor. An electrical insulator is a material in which the electron does not flow freely or the atom of the insulator have tightly bound electrons whose internal electric charges do not flow freely; very little electric current will flow through it under the influence of an electric field.
Answer:
The Kinetic Energy is approximately 3 times decreased
Explanation:
A baseball weighs 5.13 oz.
a)What is the kinetic energy, in joules, of this baseball when it is thrown by a major league pitcher at 95.o mi/h?
b) By what factor will the kinetic energy change if the speed of the baseball is decreased to 54.8 mi/h? Express your answer as an integer.
Kinetic Energy (KE)=0.5×mass×velocity ^ 2
Kinetic Energy (KE)=0.5×mass × velocity ^ 2
Joules = kg×m^2/s^2
1 mile = 1609.344 meters
1 hour = 3600 sec
1 Oz = 28.34952 g = 0.02834952 kg
a) KE=0.5×m×v^2
=0.5×(5.13 oz × 0.02834952 kg/1 ounce)×(95 miles/h × 1609.344 m/1 mile × 1 hr/3600 s)^2
=130.761 kg×m^2/s^2 = 130.761 Joules
b) KE=0.5×m×v^2
=0.5×(5.13 oz × 0.02834952 kg/1 ounce)×(54.8 miles/h × 1609.344 m/1 mile × 1 hr/3600 s)^2
=43.51028 kg×m^2/s^2 = 43.51028 Joules
= 130.761 / 43.51028 = 3.00528,
As such the Kinetic Energy is approximately 3 times decreased