2.10
The 0 after 1 is not significant and does not affect the value of the number; therefore, when the number will be rounded to two significant figures, it will become 2.1, so it will be the same.
Answer:
Hi, the question is incomplete. However, the question is about the calculation of volume of a product when the volume of one of the reactants is provided.
9.587 cm^3
Explanation:
The balanced equation for the chemical reaction is shown below:
⇒
In the chemical reaction above, 2 moles of water produced 4 moles of hydrogen fluoride. If 4.8 cm^3 of water were consumed, we can calculated the volume of hydrogen fluoride that would be produced as follow:
Using STP, 1 mole of gas has a volume of 22.4 L
Thus, 4.8 cm^3 = 0.0048 L is equivalent to 2.14*10^-4
since 2 moles of water produced 4 moles of hydrogen fluoride, therefore, 2.14*10^-4 would produced 2*2.14*10^-4 = 4.28*10^-4 moles
we can convert the moles to L by multiplying with 22.4
volume of hydrogen fluoride = 4.28*10^-4 * 22.4 = 0.009587 L = 9.587 cm^3
Answer:
The answer to your question is below
Explanation:
1)
Balanced chemical reaction
2CH₃OH + 3O₂ ⇒ 2 CO₂ + 4H₂O
Reactant Element Product
2 C 2
8 H 8
8 O 8
Molar mass of CH₃OH = 2[12 + 16 + 4]
= 2[32]
= 64 g
Molar mass of O₂ = 3[16 x 2] = 96 g
Theoretical proportion CH₃OH/O₂ = 64 g/96g = 0.67
Experimental proportion CH₃OH/O₂ = 60/48 = 1.25
Conclusion
The limiting reactant is O₂ because the Experimental proportion was higher than the theoretical proportion
2)
Balanced chemical reaction
S₈ + 12O₂ ⇒ 8SO₃
Reactant Elements Products
8 S 8
24 O 24
Molar mass of S₈ = 32 x 8 = 256 g
Molar mass of O₂ = 12 x 32 = 384 g
Theoretical proportion S₈ / O₂ = 256 / 384
= 0.67
Experimental proportion S₈ / O₂ = 40 / 35
= 1.14
Conclusion
The limiting reactant is O₂ because the experimental proportion was lower than the theoretical proportion.
Answer:
Molarity = 0.809 M
mole fraction = 0.047
Explanation:
The complete question is
Calculate the molarity and mole fraction of acetone in a 1.09-molal solution of acetone (CH3COCH3) in ethanol (C2H5OH). (Density of acetone = 0.788 g/cm3; density of ethanol = 0.789 g/cm3.) Assume that the volumes of acetone and ethanol add.
Solution -
Solution for molarity:
1.09-molal means 1.09 moles of acetone in 1.00 kilogram of ethanol.
1)
Mass of 1.09 mole of acetone
= 1.09 mol x 58.0794 g/mol = 63.306 g
Density of acetone = 0.788 g/cm3
Thus, volume of 1.09 moles of acetone = 63.306 g/0.788 g/cm3 = 80.34 cm3
For ethanol
1000 g divided by 0.789 g/cm3 = 1267.427 cm3
Total volume of the solution = Volume of acetone + Volume of ethanol = 80.34 cm3 + 1267.427 cm3 = 1347.765 cm3 = 1.347 L
a) Molarity:
1.09 mol / 1.347 L = 0.809 M
Mole Fraction
a) moles of ethanol:
1000 g / 46.0684 g/mol = 21.71 mol
b) moles of acetone:
1.09 / (1.09 + 21.71) = 0.047
Molar mass FeSO₄ =151.908 g/mol
Volume = 450 mL / 1000 = 0.45 L
1 mol ----------- 151.908 g
n mol ----------- 0.8 g
n = 0.8 * 1 / 151.908
n = 0.00526 moles
M = n / V
M = 0.00526 / 0.45
M = 0.01168 mol/L⁻¹
hope this helps!
