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3 years ago

8

At this temperature, 0.300 mol H 2 0.300 mol H2 and 0.300 mol I 2 0.300 mol I2 were placed in a 1.00 L container to react. What

concentration of HI HI is present at equilibrium?

1 answer:

Alekssandra [29.7K]3 years ago

5 0

Answer:

The concentration of HI present at equilibrium is 0.471 M.

Explanation:

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The reaction between halogens and alkanes is very rapid. true false

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Halogens and Alkali react aggressively.

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3 years ago

The Kelvin temperature required for 0.0470 mol of helium gas to fill a balloon to 1.20 L under 0.878 atm is

4vir4ik [10]

From the ideal gas law, PV = nRT, we can rearrange the equation to solve for T given the other parameters.

T = PV/nR

where P = 0.878 atm, V = 1.20 L, n = 0.0470 moles, and R = 0.082057 L•atm/mol•K. Plugging in our values, we obtain the temperature in Kelvin:

T = (0.878 atm)(1.20 L)/(0.0470 mol)(0.082057 L•atm/mol•K)
T = 273 K

So, the second answer choice would be correct.

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3 years ago

What is the approximate size of solute particles in a solution?

jeyben [28]

Answer:

The answer is B

Explanation:

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3 years ago

Read 2 more answers

Cinnamon owes its flavor and odor to cinnamaldehyde (C9H8O). Determine the boiling point elevation of a solution of 92.7 mg of c

Flauer [41]

Answer:

The boiling point elevation is 3.53 °C

Explanation:

∆Tb = Kb × m

∆Tb is the boiling point elevation of the solution

Kb is the molal boiling point elevation constant of CCl4 = 5.03 °C/m

m is the molality of the solution is given by moles of solute (C9H8O) divided by mass of solvent (CCl4) in kilogram

Moles of solute = mass/MW =

mass = 92.7 mg = 92.7/1000 = 0.0927 g

MW = 132 g/mol

Moles of solute = 0.0927/132 = 7.02×10^-4 mol

Mass of solvent = 1 g = 1/1000 = 0.001 kg

m = 7.02×10^-4 mol ÷ 0.001 kg = 0.702 mol/kg

∆Tb = 5.03 × 0.702 = 3.53 °C (to 2 decimal places)

6 0

3 years ago

When 3.51 g of phosphorus was burned in chlorine, the product was a phosphorus chloride. Its vapor took 1.77 times as long to ef

kumpel [21]

Answer:

<em>molar mass of the phosphorus chloride = 138.06 g/mol</em>

<em></em>

Explanation:

<em>mass of phosphorus will be the same as mass of CO2, since it is stated that they are of equal amount.</em>

mass = 3.51 g

<em>lets assume that it took the CO2 1 sec to effuse, then the time taken by the phosphorus chloride will be 1.77 sec</em>

From this we can say that

rate of effusion of CO2 = 3.51/1 = 3.51 g/s

rate of effusion of the phosphorus chloride = 3.51/1.77 = 1.98 g/s

<em>From graham's equation of effusion</em>,

$\frac{Rc}{Rp}$ = $\sqrt{\frac{Mp\\}{Mc} }$

Rc = rate of effusion of CO2 = 3.51 g/s

Rp = rate of effusion of phosphorus chloride = 1.98 g/s

Mc = molar mass of CO2 = 44.01 g/mol

Mp = molar mass of the phosphorus chloride = ?

Imputing values into the equation, we have

$\frac{3.51}{1.98}$ = $\sqrt{\frac{Mp\\}{44.01} }$

1.77 = $\frac{\sqrt{Mp} }{6.64}$

11.75 = $\sqrt{Mp}$

Mp = $11.75^{2}$

Mp = <em>molar mass of the phosphorus chloride = 138.06 g/mol</em>

5 0

3 years ago