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2 years ago

500 mL of He at 300 K is heated to 450 K. Find V2

Chemistry

1 answer:

ruslelena [56]2 years ago

8 0

Answer:

V2 = 0.75 L

Explanation:

assumin ideal gas:

PV = RTn

∴ T1 = 300K

∴ V1 = 0.500 L

∴ T2 = 450 K

⇒ V2 = ?

assuming P, n constant:

⇒ PV1 = RT1n

⇒ V1/T1 = Rn/P = K..........(1)

⇒ V2/T2 = RN/P = K..........(2)

(1) = (2):

⇒ V1/T1 = V2/T2

⇒ V2 = (V1/T1)T2

⇒ V2 = (0.500 L/300 K)(450 K)

⇒ V2 = 0.75 L

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Answer:

Concentration of sulfuric acid in the acid rain sample is 0.0034467 mol/L.

Explanation:

Volume of NaOH = 1.7 ml = 0.0017 L

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Moles of NaOH = n

$0.0811 M=\frac{n}{0.0017 L}$

n = 0.0001378 mol

$H_2SO_4+2NaOH\rightarrow Na_2SO_4+2H_2O$

According to reaction, 2 mol of NaOH neutralize 1 mol of sulfuric acid.

Then 0.0001378 mol of NaOH will neutralize:

$\frac{1}{2}\times 0.0001378 mol=6.8935\times 10^{-5} mol$ of sulfuric acid.

Concentration of sulfuric acid in the acid rain sample: x

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Concentration of sulfuric acid in the acid rain sample is 0.0034467 mol/L.

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2 years ago

Determine the poh of a 0.348 m ba(oh)2 solution at 25°c.

vladimir1956 [14]

Given:

Concentration of Ba(OH)2 = 0.348 M

To determine:

pOH of the above solution

Explanation:

Based on the stoichiometry-

1 mole of Ba(OH)2 is composed of 1 mole of Ba2+ ion and 2 moles of OH- ion

Therefore, concentration of OH- ion = 2*0.348 = 0.696 M

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Ans: pOH of 0.348M Ba(OH)2 is 0.157

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2 years ago

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A 20.0-mL sample of lake water was acidified with nitric acid and treated with excess KSCN to form a red complex (KSCN itself is

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Answer:

8.09x10⁻⁵M of Fe³⁺

Explanation:

Using Lambert-Beer law, the absorbance of a sample is proportional to its concentration.

In the problem, the Fe³⁺ is reacting with KSCN to produce Fe(SCN)₃ -The red complex-

The concentration of Fe³⁺ in the reference sample is:

4.80x10⁻⁴M Fe³⁺ × (5.0mL / 50.0mL) = 4.80x10⁻⁵M Fe³⁺

Because reference sample was diluted from 5.0mL to 50.0mL.

That means a solution of 4.80x10⁻⁵M Fe³⁺ gives an absorbance of 0.512

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0.345 × (4.80x10⁻⁵M Fe³⁺ / 0.512) = 3.23x10⁻⁵M.

As the solution was diluted from 20.0mL to 50.0mL, the concentration of Fe³⁺ in Jordan lake is:

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2 years ago

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n =?

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n = PV /RT

n = (1.5x12)/(0.082x280)

n = 0.784mol

Next, we convert this mole ( i.e 0.784mol) of F2 to mass

MM of F2 = 19x2 = 38g/mol

Mass conc of F2 = n x MM

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Equation for the reaction is given below

F2 + 2NaCl —> 2NaF + Cl2

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Next, we find the mass of NaCl that reacted with 29.792g of F2.

From the equation,

38g of F2 redacted with 117g of NaCl.

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Therefore, 91.73g of NaCl reacted with f2

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2 years ago

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