Answer:
1.403x10²⁴ molecules
Explanation:
In order to calculate how many molecules of CO₂ are there in 102.5 g of the compound, we first<u> convert grams to moles</u> using its <em>molar mass</em>:
- 102.5 g ÷ 44 g/mol = 2.330 mol CO₂
Now we <u>convert moles into molecules </u>using <em>Avogadro's number</em>:
- 2.330 mol * 6.023x10²³ molecules/mol = 1.403x10²⁴ molecules
Answer:
18.9 moles
Explanation:
We have the following data:
V = 50 L
P = 12.4 atm
T= 127°C + 273 = 400 K
R = 0.082 L.atm/K.mol (it is the gas constant)
We use the ideal gas equation to calculate the number of moles n of the gas:
PV = nRT
⇒ n = PV/RT = (12.4 atm x 50 L)/(0.082 L.atm/K.mol x 400 K) = 18.9 mol
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Mass of BaO in initial mixture = 3.50g
Explanation:
Let mass of BaO in mixture be x g
mass of MgO in mixture be (6.35 - x) g
Initially CO_2
Volume = 3.50 L
Temp = 303 K
Pressure = 750 torr = 750 / 760 atm
Applying ideal gas equation
PV = nRT
n = PV / RT
(n)_CO_2 = ((750/760)* 3.50) / 0.0821 * 303
(n)_CO_2 = 0.139 mole
Finally; mole of CO_2
n= PV /RT
((245/760) *3.5) / 303* 0.0821
(n)_CO_2 = 0.045 mole
Mole of CO_2 reacted = 0.139 - 0.045
=0.044 mole
BaO + CO_2 BaCO_3
Mgo + CO_2 MgCO_3
moles of CO_2 reacted = ( moles of BaO + moles of MgO)
moles of BaO in mixture = x / 153 mole
moles of MgO in mixture = 6.35 - x mole / 40
Equating,
x/ 153 +6.35/40 = 0.094
= x/153 + 6.35 / 40 - x/40 =0.094
= x (1/40 - 1153) = (6.35/40 - 0.094)
= x * 10.018464
= 0.06475
mass of BaO in mixture = 3.50g
