#1
<span>The rate of diffusion of a gas is inversely proportional to the square root of its molecular weight.
#2
shape, molecular weight. I don't know for sure though
#3
</span>O2, N2, Ar, H2O vapor
Answer : The thermal energy produced during the complete combustion of one mole of cymene is -7193 kJ/mole
Explanation :
First we have to calculate the heat released by the combustion.
where,
q = heat released = ?
= specific heat of calorimeter = 
= change in temperature = 
Now put all the given values in the above formula, we get:
Thus, the heat released by the combustion = 70.43 kJ
Now we have to calculate the molar enthalpy combustion.
where,
= molar enthalpy combustion = ?
q = heat released = 70.43 kJ
n = number of moles cymene = 
Therefore, the thermal energy produced during the complete combustion of one mole of cymene is -7193 kJ/mole
0.007 - ?micrograms
0.007= 700micrograms
multiply the mass value by 100000
The identity of an alpha particle is a <span>Helium nucelus.
Good luck~ Sans
</span>
Answer:
4.31 atm.
Explanation:
- We can use the general law of ideal gas: <em>PV = nRT.</em>
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 V are constant, and have different values of P and T:
<em>(P₁T₂) = (P₂T₁)</em>
<em></em>
P₁ = 3.81 atm, T₁ = 25°C + 273 = 298 K,
P₂ = ??? atm, T₂ = 50°C + 273 = 323 K,
- Applying in the above equation
(P₁T₂) = (P₂T₁)
<em>∴ P₂ = (P₁T₂)/(T₁)</em> = (3.81 atm)(323 K)/(298 K) = <em>4.31 atm.</em>
