Answer: A-2 energy levels
Explanation:
Sorry if i’m wrong
Answer:
0.1593 L.
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>P₁V₁T₂ = P₂V₂T₁</em>
<em></em>
P₁ = 600 torr/760 = 0.789 atm, V₁ = 185.0 mL = 0.185 L, T₁ = 25.0°C + 273 = 298.0 K.
P₂ (at STP) = 1.0 atm, V₂ = ??? L, T₂ (at STP = 0.0°C) = 0.0°C + 273 = 273.0 K.
<em>∴ V₂ = P₁V₁T₂/P₂T₁</em> = (0.789 atm)(0.185 mL)(298.0 K)/(1.0 atm)(273.0 K) = <em>0.1593 L.</em>
Answer:
It will be reported too low.
Explanation:
To measure the specific heat of the metal (s), the calorimeter may be used. In it, the metal will exchange heat with the water, and they will reach thermal equilibrium. Because it can be considered an isolated system (there're aren't dissipations) the total amount of heat (lost by metal + gained by water) must be 0.
Qmetal + Qwater = 0
Qmetal = -Qwater
The heat is the mass multiplied by the specific heat multiplied by the temperature change. If c is the specific heat of the water:
m_metal*s*ΔT_metal = - m_water *c*ΔT_water
s = -m_water *c*ΔT_water / m_metal*ΔT_metal
So, if m_water is now less than it was supposed to be, s will be reported too low, because they are directly proportional.
Answer:
Longer hydrocarbon molecules have a stronger intermolecular force. More energy is needed to move them apart so they have higher boiling points . This makes them less volatile and therefore less flammable
