Answer:
A sample of pure NO2 is heated to 338 ∘C at which temperature it partially dissociates according to the equation 2NO2(g)⇌2NO(g)+O2(g) At equilibrium the density of the gas mixture is 0.515 g/L at 0.745 atm .
(4x^2)x
Kc= -----------
(A-2x)^2
PV=nRT
n/v = P/RT = .745/(0.0821)(334+273) = .01495
To Find the initial molarity of NO2
(mol/L)(g/mol) + (mol/L)(g/mol) + (mol/L)(g/mol)= g/L
Thus:
46(A-2x) + 2x(30) + 32x = .515 g/L
46A-92x+60x+32x = .515
46A=.515
A=.01120 M
Using the total molarity found
(A-2x)+2x+x = .01495 M
A+x=.01495
Plug in A found into the above equation:
.01120+x = .01495
x=.00375
Now Plug A and x into the original Equilibrium Constant Expression:
(4x^2)x
Kc= -----------
(A-2x)^2
Kc = 0.000014
Explanation:
Answer:
Has a positive charge
Explanation:
Protons inside the nucleus help bind nucleus together. They also attract negative charged electrons.
In science negative to negative/positive to positive repels and negative to positive or vice versa attract to each other.
B. 272
I think because it's 3 sig figs
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
Answer:
42 m/s
Explanation:
To we convert units for speed we can use dimensional analysis. First thing we do is seperate the measurement into a fraction. After this we can multiply by 1km over 0.62137 miles. We do this so that the miles cancel out.
×
= 
After this we can use a conversion factor and divide by 3.6.
÷ 3.6 = 42 m/s