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Gwar [14]
4 years ago
7

A 0.21 m3 drum contains 100L of a mixture ofseveral degreasing solvents in water.The concentraton oftrichloroethylen in the head

space( gas phase) above the water wasmeasured to be 0.00301 atm.The Henry's constant fortrichloroethylene is 0.00985atm.m3.mol-1 at25o.What is the concentration of tricholoroethylene inthe water in units of molarity? ( assume that the temperature ofthe solution is 25o)
Chemistry
1 answer:
LiRa [457]4 years ago
5 0

Answer:

The concentration is 6.42×10^-4 M

Explanation:

Number of moles trichloroethylene = pressure × volume/Henry's constant = 0.00301×0.21/0.00985 = 0.0642 mol

Volume of mixture = 100 L

Concentration of trichloroethylene = number of moles of trichloroethylene/volume of solution = 0.0642/100 = 6.42×10^-4 M

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4 0
4 years ago
Consider the following reactants:
sdas [7]

Answer:

A. 3NaOH(aq)+Fe(NO3)3(aq) ---> Fe(OH)3 (solid)+ 3NaNO3(aq)

B. Double replacement reaction

AB +CD ---> AD + CB

4 0
3 years ago
Place the following in order of increasing X-Se-X bond angle, where X represents the outer atoms 7) in each molecule. SeO2 SeCl6
igomit [66]

Answer:

SeCl₆ < SeF₂ <  SeO₂

Explanation:

To determine the order of the bond angle of the compound, we need to determine the hybridization of each compound given.

For:

SeO₂, The hybridization is sp² because the Se is doubly bonded to O at both sides and Se contains a lone pair of electrons.

SeCl₆, Se is attached to six different Cl atoms via a single bond in a hexagonal shape. Thus, SeCl₆ hybridization is sp³d²

SeF₂, Here, Se is singly bonded to F atoms on both sides and Se contains a lone pair of electrons. The hybridization is sp³

Therefore, the bond angle in sp² = 120°, sp³ = 109.8°, sp³d² = 90°

Hence, the order is:  SeCl₆ < SeF₂ <  SeO₂

6 0
3 years ago
What is the stoichiometric ratio between BaCl2 and NaCl
bixtya [17]
<span>BaCl2+Na2SO4---->BaSO4+2NaCl There is 1.0g of BaCl2 and 1.0g of Na2SO4, which is the limiting reagent? "First convert grams into moles" 1.0g BaCl2 * (1 mol BaCl2 / 208.2g BaCl2) = 4.8 x 10^-3 mol BaCl2 1.0g Na2SO4 * (1 mol Na2SO4 / 142.04g Na2SO4) = 7.0 x 10^-3 mol Na2SO4 (7.0 x 10^-3 mol Na2SO4 / 4.8 x 10^-3 mol BaCl2 ) = 1.5 mol Na2SO4 / mol BaCl2 "From this ratio compare it to the equation, BaCl2+Na2SO4---->BaSO4+2NaCl" The equation shows that for every mol of BaCl2 requires 1 mol of Na2SO4. But we found that there is 1.5 mol of Na2SO4 per mol of BaCl2. Therefore, BaCl2 is the limiting reagent.</span>
7 0
4 years ago
1. A helium-filled balloon had a volume of 8.50 L on the ground at 20.0°C and a
maw [93]

Answer:

133.74 L

Explanation:

First we <u>convert the given pressures and temperatures into atm and K</u>, respectively:

  • 750.0 Torr ⇒ 750/760 = 0.9868 atm
  • 20°C ⇒ 20+273.16 = 293.16 K
  • 40°C ⇒ 40+273.16 = 313.16 K

Then we<u> use the PV=nRT formula to calculate the number of moles of helium in the balloon</u>, using<em> the data of when it was on the ground</em>:

  • 0.9868 atm * 8.50 L = n * 0.082 atm·L·mol⁻¹·K⁻¹ * 293.16 K
  • n = 2.866 mol

Then, knowing the value of n, we <u>use PV=nRT once again, this time to calculate V</u> using <em>the data of when the balloon was high up:</em>

  • 0.550 atm * V = 2.866 mol * 0.082 atm·L·mol⁻¹·K⁻¹ * 313.16 K
  • V = 133.74 L
5 0
3 years ago
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