Explanation:
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The compound known as butylated hydroxytoluene, abbreviated as BHT, contains carbon, hydrogen, and oxygen. A 3.001 g sample of BHT was combusted in an oxygen rich environment to produce 8.990 g of CO2(g) and 2.944 g of H2O(g). Insert subscripts below to appropriately display the empirical formula of BHT
Answer:
Explanation:
A 3.001 g sample of BHT was combusted in an oxygen rich environment to produce 8.990 g of CO2(g) and 2.994 g of H2O(g).
If all the carbon in BHT is present in and also, all the hydrogen in BHT is present in
, Then we can determine for the corresponding numbers of moles of Carbon(C) and Hydrogen (H) respectively as:
moles of =
= 0.2043 moles
∴ moles of C = 0.2043 moles
moles of =
= 0.1635 moles
∴ moles of H = 2 × 0.1635 moles
= 0.327 moles
Since number of moles=
number of moles of H = 0.327 moles
molar mass of H = 1.008 g/mol
∴ mass of H in the sample = 0.327 moles × 1.008 g/mol
= 0.329616g
mass of C in the sample can be calculated as = 0.2043 moles ×
= 2.453643 g
mass of C+H in the sample = 2.453643g + 0.329616g
= 2.783259 g
mass of O can be calculated as = 3.001 g - 2.783259 g
= 0.217741 g
∴ moles of O = 0.217741g ×
= 0.0136 moles
Now, since; we've gotten our data, we can now proceed to calculate for the empirical formula.
C H O
0.2043 0.327 0.0136
Dividing by the least number (0.0136) , we have :
15.02 24.04 1
≅
15 24 1
Therefore, the empirical formula would be :
12 times breathe give 240 ml of pure . Each breathe gives 20 ml of
.
Let us consider, volume of air per breathe= x ml.
Pure from inhaled air=
ml and Pure
from exhaled air=
ml.
Pure from inhaled and exhaled air= 20 ml
So, +
= 20
Therefore, x = 55.5 ml
So, volume of air per breath= 55.5 ml.