Answer:
a) 14.2 atm
b) 4.46 atm
c) 1.06 atm
Explanation:
For an ideal gas,
PV = nRT
P = pressure of the gas
V = volume occupied by the gas
n = number of moles of the gas
R = molar gas constant = 0.08206 L.atm/mol.K
T = temperature of the gas in Kelvin
a) For HF,
P =?, V = 2.5L, n = 1.35 moles, T = 320K
P = 1.35 × 0.08206 × 320/2.5
P = 14.2 atm
b) For NO₂
P =?, V = 4.75L, n = 0.86 moles, T = 300K
P = 0.86 × 0.08206 × 300/4.75
P = 4.46 atm
c) For CO₂
P =?, V = 5.5 × 10⁴ mL = 55L, n = 2.15 moles, T = 57°C = 330K
P = 2.15 × 0.08206 × 330/55
P = 1.06 atm
Well, they are equal and opposite, but the force is high enough to break up the snowball. That is where the energy is dissipated.
Answer: 8.1 x 10^24
Explanation:
I(t) = (0.6 A) e^(-t/6 hr)
I'll leave out units for neatness: I(t) = 0.6e^(-t/6)
If t is in seconds then since 1hr = 3600s: I(t) = 0.6e^(-t/(6 x 3600) ).
For neatness let k = 1/(6x3600) = 4.63x10^-5, then:
I(t) = 0.6e^(-kt)
Providing t is in seconds, total charge Q in coulombs is
Q= ∫ I(t).dt evaluated from t=0 to t=∞.
Q = ∫(0.6e^(-kt)
= (0.6/-k)e^(-kt) evaluated from t=0 to t=∞.
= -(0.6/k)[e^-∞ - e^-0]
= -0.6/k[0 - 1]
= 0.6/k
= 0.6/(4.63x10^-5)
= 12958 C
Since the magnitude of the charge on an electron = 1.6x10⁻¹⁹ C, the number of electrons is 12958/(1.6x10^-19) = 8.1x10^24 to two significant figures.
Answer:
2)primary mirror
Explanation:
A reflecting telescope contains mirrors and a refracting telescope contains lenses. There are two mirrors in basic reflecting telescope. The primary mirror gathers the light from the bodies at distance and reflects it towards the secondary mirror. The secondary mirror is placed at such an angle that it then reflects the light towards the eye piece and we can observe the faraway body.
