It would form a negative ion... it lacks 1 e in its valence shell., it is easier for F to accept an e than to shed all existing 7.
e= electron
Answer: 0.0014 atm
Explanation:
Given that,
Original pressure of air (P1) = 1.08 atm
Original volume of air (T1) = 145mL
[Convert 145mL to liters
If 1000mL = 1l
145mL = 145/1000 = 0.145L]
New volume of air (V2) = 111L
New pressure of air (P2) = ?
Since pressure and volume are given while temperature is held constant, apply the formula for Boyle's law
P1V1 = P2V2
1.08 atm x 0.145L = P2 x 111L
0.1566 atm•L = 111L•P2
Divide both sides by 111L
0.1566 atm•L/111L = 111L•P2/111L
0.0014 atm = P2
Thus, the new pressure of air when the volume is decreased to 111 L is 0.0014 atm
An atom is the smallest you can get to identify the element.
Answer: 317 joules
Explanation:
The quantity of heat energy (Q) gained by aluminium depends on its Mass (M), specific heat capacity (C) and change in temperature (Φ)
Thus, Q = MCΦ
In this case,
Q = ?
Mass of aluminium = 50.32g
C = 0.90J/g°C
Φ = (Final temperature - Initial temperature)
= 16°C - 9°C = 7°C
Then, Q = MCΦ
Q = 50.32g x 0.90J/g°C x 7°C
Q = 317 joules
Thus, 317 joules of heat is gained.
Answer:
1188.0 mL.
Explanation:
- We can use the general law of ideal gas: PV = nRT.
where, P is the pressure of the gas in atm.
V is the volume of the gas in L.
n is the no. of moles of the gas in mol.
R is the general gas constant,
T is the temperature of the gas in K.
- If n and P are constant, and have two different values of V and T:
<em>V₁T₂ = V₂T₁
</em>
V₁ = 900 mL, T₁ = 27.0°C + 273 = 300.0 K.
V₂ = ??? mL, T₂ = 123.0°C + 273 = 396.0 K.
<em>∴ V₂ = V₁T₂/T₁ </em>= (900 mL)(396 K)/(300.0 K) = <em>1188.0 mL.</em>
