After Frida stops exercising, she continues to breathe heavily. What is most likely occurring in her body? Heavy breathing durin
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Answer:
295.7 mL of 24% trichloroacetic acid (tca) is needed .
Explanation:
Let the volume of 24% trichloroacetic acid solution be x
Volume of required 10% trichloroacetic acid solution =8 bottles of 3 ounces
= 24 ounces = 709.68 mL
(1 ounces = 29.57 mL)
Amount of trichloroacetic acid in 24% solution of x volume of solution will be equal to amount of trichloroacetic acid in 10% solution of volume 709.68 mL.
x = 295.7 mL
295.7 mL of 24% trichloroacetic acid (tca) is needed .
We can first find the number of moles of gas in the container using the ideal gas law equation
PV = nRT
where P - pressure - 1.10 atm x 101 325 Pa/atm = 111 458 Pa
V - volume - 10.0 x 10⁻⁶ m³
n - number of moles
R - universal gas constant - 8.314Jmol⁻¹K⁻¹
T - temperature - 289.0 K
substituting these values in the equation
111 458 Pa x 10.0 x 10⁻⁶ m³ = n x 8.314Jmol⁻¹K⁻¹ x 289.0 K
n = 0.463 x 10⁻³ mol
molar mass of the compound = 0.04403 g/ (0.463 x 10⁻³ mol)
molar mass = 95.1 g/mol
first we need to find the empirical formula of the compound, this is the simplest ratio of whole numbers of components in the compound.
assuming 100 g of the compound is present
element C Cl
mass present 25.305 g 74.695 g
number of moles
= 2.11 = 2.10
divide by the least number of moles
2.11/2.10 = 1.00 = 2.10/2.10 = 1.00
therefore C - 1
Cl - 1
empirical formula - CCl
molecular formula is actual composition of components in the compound
mass of empirical unit - 12 + 35.5 = 47.5
mass of molecular formula = 95.1 g/mol
number of empirical units = 95.1 / 47.5 = 2.00
therefore molecular formula = 2(CCl)
molecular formula = C₂Cl₂
PV = nRT
where P - pressure - 1.10 atm x 101 325 Pa/atm = 111 458 Pa
V - volume - 10.0 x 10⁻⁶ m³
n - number of moles
R - universal gas constant - 8.314Jmol⁻¹K⁻¹
T - temperature - 289.0 K
substituting these values in the equation
111 458 Pa x 10.0 x 10⁻⁶ m³ = n x 8.314Jmol⁻¹K⁻¹ x 289.0 K
n = 0.463 x 10⁻³ mol
molar mass of the compound = 0.04403 g/ (0.463 x 10⁻³ mol)
molar mass = 95.1 g/mol
first we need to find the empirical formula of the compound, this is the simplest ratio of whole numbers of components in the compound.
assuming 100 g of the compound is present
element C Cl
mass present 25.305 g 74.695 g
number of moles
= 2.11 = 2.10
divide by the least number of moles
2.11/2.10 = 1.00 = 2.10/2.10 = 1.00
therefore C - 1
Cl - 1
empirical formula - CCl
molecular formula is actual composition of components in the compound
mass of empirical unit - 12 + 35.5 = 47.5
mass of molecular formula = 95.1 g/mol
number of empirical units = 95.1 / 47.5 = 2.00
therefore molecular formula = 2(CCl)
molecular formula = C₂Cl₂