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Kay [80]
2 years ago
6

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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Titration of a 20.0-mL sample of acid rain required 1.7 mL of 0.0811 M NaOH to reach the end point. If we assume that the acidit
Rasek [7]

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

Molarity of NaOH = 0.0811 M

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

x=\frac{6.8935\times 10^{-5}}{0.02 L}=0.0034467 mol/L

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

7 0
2 years ago
Determine the poh of a 0.348 m ba(oh)2 solution at 25°c.
vladimir1956 [14]

<u>Given:</u>

Concentration of Ba(OH)2 = 0.348 M

<u>To determine:</u>

pOH of the above solution

<u>Explanation:</u>

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

pOH = -log[OH-] = - log[0.696] = 0.157

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

6 0
2 years ago
Read 2 more answers
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
Novosadov [1.4K]

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⁺

<em>Because reference sample was diluted from 5.0mL to 50.0mL.</em>

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

Now, as the sample of the lake gives an absorbance of 0.345, its concentration is:

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

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

3.23x10⁻⁵M Fe³⁺ × (50.0mL / 20.0mL) = <em>8.09x10⁻⁵M of Fe³⁺</em>

4 0
2 years ago
A chemist reacted 12.0 liters of F2 gas with NaCl in the laboratory to form Cl2 gas and NaF. Use the ideal gas law equation to d
Nataly_w [17]

Answer: 91.73g of NaCl

Explanation:

First, we solve for the number of moles of F2 using the ideal gas equation

V = 12L

P = 1.5 atm

T = 280K

R = 0.082atm.L/mol/K

n =?

PV = nRT

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

= 0.784 x 38 = 29.792g

Equation for the reaction is given below

F2 + 2NaCl —> 2NaF + Cl2

Molar Mass of NaCl = 23 + 35.5 = 58.5g/mol

Mass conc. of NaCl from the equation = 2 x 58.5 = 117g

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.

Therefore, 29.792g of F2 will react with Xg of NaCl i.e

Xg of NaCl = (29.792 x 117)/38

= 91.73g

Therefore, 91.73g of NaCl reacted with f2

3 0
2 years ago
An acetaminophen suspension for infants contains 80 mg/0.80 mL. The recommended dose is 15 mg/kg body weight. How many mL of thi
omeli [17]
11 lb = 4.99 kg. 

So, 74.85 mg should be given to the infant. Thus 0.75 mL of the suspension should be given.
8 0
2 years ago
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