Asample of 0.020 mole of oxygen gas is confined at 227°c and 0.50 atmospheres. what would be the pressure of this sample at 27°c and the same volume?

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0.3 atm

Explanation:-

Number of moles n = 0.020 mol

Pressure P 1 = 0.5 atm

Temperature T 1 = 227 C + 273 = 500 K

We know the value of universal Gas constant R = 0.082 L atm K−1 mol−1

We use the relation PV = nRT

V = nRT /P

Plugging in the values

V = 0.020 mol x (0.082 L atm K−1 mol−1 ) x 500 K / (0.5 atm)

= 1.64 L

Now the new temperature T 2 = 27 C + 273 = 300 K.

Volume is same = 1.64 L

Number of moles n is same = 0.020 mol

Once again we use the relation

PV = nRT

New Pressure P = nRT / V

= 0.020 x 0.082 L atm K−1 mol−1 x 300 K / 1.64 L

= 0.3 atm

i believe it is 40–40–20%

explanation:

2fe + 3h2so4 -> fe2(so4)3 + 3h2

it is necessary to balance equations in order to get the proper amount of reactants as well as products in order not to make a miscalculation.

you cannot change the subscripts as that would change the substance itself. for instance, diatomic elements such as h2 only have the subscript two when a gas. another example would be so4, if you were to change the subscript, the element would no longer be sulfate.

we change the coefficients instead as that only changes the number of molecules rather than the substance itself.

hope this !