Answer:
Final Volume = 5.18 Liters
Explanation:
Initial Condition:
P1 = 789 mm Hg x (1/760) atm /mm Hg = 1.038 atm
T1 = 22° C = 273 + 22 = 295 K
V1 = 4.7 L
Final Condition:
P2 = 755 mm Hg x (1/760) atm /mm Hg = 0.99 atm
T2 = 37° C = 273 + 37 = 310 K
V2 = ?
Since, (P1 x V1) / T1 = (P2 x V2) / T2,
Therefore,
⇒ (1.038)(4.7) / 295 = (0.99)(V2) / 310
⇒ V2 = 5.18 L (Final Volume)
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Answer:
Explanation:
- For the balanced reaction:
<em>4Fe(s) + 3O₂(g) → 2Fe₂O₃(s).</em>
It is clear that 4 mol of Fe react with 3 mol of O₂ to produce 2 mol of Fe₂O₃.
- Firstly, we need to calculate the no. of moles of 35.8 grams of Fe metal:
no. of moles of Fe = mass/molar mass = (35.8 g)/(55.845 g/mol) = 0.64 mol.
- Now, we can find the no. of moles of O₂ is needed to react with the proposed amount of Fe:
<em><u>Using cross multiplication:</u></em>
4 mol of Fe is needed to react with → 3 mol of O₂, from stichiometry.
0.64 mol of Fe is needed to react with → ??? mol of O₂.
∴ The no. of moles of O₂ needed = (3 mol)(0.64 mol)/(4 mol) = 0.48 mol.
- Finally, we can get the volume of oxygen using the information:
<em>It is known that 1 mole of any gas occupies 22.4 L at standard P and T (STP).</em>
<em></em>
<em><u>Using cross multiplication:</u></em>
1 mol of O₂ occupies → 22.4 L, at STP conditions.
0.48 mol of O₂ occupies → ??? L.
∴ The no. of liters of O₂ = (0.48 mol)(22.4 L)/(1 mol) = 10.752 L.
Answer:
Three possible blood type alleles are Iᴬ, Iᴮ and i
Explanation:
Iᴬ, Iᴮ and i are three possible blood type alleles.
Iᴬ and Iᴮ are known as co-dominant, and The i allele is recessive.
Thus, Three possible blood type alleles are Iᴬ, Iᴮ and i
<u>-TheUnknownScientist</u>
the answer is d. when more solute is added
