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AfilCa [17]
2 years ago
13

What is the temperature of a 100 liter container having 1 mole of an ideal gas at a pressure of 20 kilopascals? (Given: R = 8.31

4 l · kPa/K · mol)
Chemistry
1 answer:
Shalnov [3]2 years ago
6 0
For this case, we use the equation for an ideal gas which is expressed as PV=nRT where P is the pressure, V is the volume, n is the number of moles and T is the temperature. We calculate as follows:

PV = nRT
T = PV / nR
T = 20 kPa (100 L) / 1 mol (8.314) 
T = 240.56 K
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Find the pH of the equivalence point(s) and the volume (mL) of 0.0372 M NaOH needed to reach the point(s) in titrations of(a) 42
Neko [114]

The pH of the equivalence points is 8.54

Concept of pH

A solution's acidity or alkalinity can be determined based on the concentration of hydrogen ions in the solution, or pH. Acidic aqueous solutions at 25 °C have a pH under 7, while basic or alkaline aqueous solutions have a pH above 7. Since the concentration of H3O+ is equal to the concentration of OH in pure water, a pH level of 7.0 at 25°C is referred to as "neutral". Strong bases may have a pH above 14, while very strong acids may have a pH below 14.

0.0520 M CH3COOH in 42.2 mL of moles is as follows:

2.194x103 mol CH3COOH = 0.0422L (0.0520mol / L)

that react with NaOH, resulting in:

NaOH + CH3COOH = CH3COO + Na+ + H2O

Thus, 1 mole of acetic acid and 1 mole of NaOH react.

As a result, 2.194x103 mol of NaOH are required to reach the equivalence point in volume:

To attain the equivalency point, 2.194x103 mol (1L / 0.0372mol) = 0.05899L 58.99mL of 0.0372 M NaOH

You will only have CH3COO at the equivalency point because it is in equilibrium with water, so:

H2O(l) + CH3COO(aq) CH3COOH(aq) + OH (aq)

A definition of equilibrium is:

Kb = 5.6x1010 = [OH] / [CH3COO] / [CH3COOH]

2.194x103mol of CH3COO has a molarity of (0.05899L + 0.0422L) = 0.02168M.

Therefore, equilibrium concentrations are:

[CH3COO]=0.02168M-X [CH3COOH]=X [OH]=X

5.6x1010 = [X] [X] / [0.02168M - X] converts to Kb.

1.214x1011 - 5.6x1010X = X2 X2 + 5.6x1010X - 1.214x1011 = 0 Finding the value of X:

False response; there are no negative concentrations. X: -3.48x106

As [OH] = X, [OH] = 3.484x106M, X is 3.484x106.

As 14 = pOH + pH pH = 8.54, pOH = -log [OH], or 5.46.

To know more about pH visit :

brainly.com/question/12546875

#SP4

3 0
1 year ago
the term diffusion comes from a latin word meaning to spread apart. how is the term diffusion related to its latin word of origi
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it means basically the same thing but can be used in different context. like if someone says, "lets diffuse the situation." it means lets separate it and find out what's going on with it. Like if you don't understand a sentence when you were younger your teacher would tell you to break down the sentence, and try to use context clues. What you are doing is diffusing the sentence so that your not looking at it as one big confusing sentence, your "breaking it down" and that helps you a lot. I hope I helped you, if not let me know and I can diffuse the topic a little more to make it more approachable and easy to understand. :)

3 0
3 years ago
Read 2 more answers
Write the rates for the following reactions in terms of the disappearance of reactants and appearance of products: (a) 302 .....
Oduvanchick [21]

Answer:

Explanation:

\mathbf{From \  the  \ information \  given:} \\ \\ \mathbf{The \  rates \  of  \ the \ f ollowing \  reactions \  can \  be \  expressed  \ as \  follows:}

(a)

\mathbf{3O_2 \to 2O_3} \\ \\ \\ \mathbf{-\dfrac{1}{3}\dfrac{d[O_2]}{dt}=\dfrac{2}{3} \dfrac{d[O_3]}{dt}}

(b)

\mathbf{C_2H_6 \to C_2H_4 + H_2}  \\ \\ \\ \mathbf{  -\dfrac{d[C_2H_6]}{dt}= \dfrac{d[C_2H_4]}{dt}=\dfrac{d[H_2]}{dt}}

(c)

\mathbf{ClO^-+Br^- \to BrO^-+Cl^-} \\ \\ \\ \mathbf{ -\dfrac{d[ClO^-]}{dt}= -\dfrac{d[Br^-]}{dt} =  \dfrac{d[BrO^-]}{dt} = \dfrac{d[Cl^-]}{dt}   }

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\mathbf{(CH_3)_3 CCl+H_2O \to (CH_3)_3COH + H^+ + Cl^-} \\ \\ \\  \mathbf{- \dfrac{d[(CH_3)_3CCl}{dt}= - \dfrac{d[H_2O]}{dt}= \dfrac{d[CH_3)_3COH}{dt}= \dfrac{d[H^+]}{dt}= \dfrac{d[Cl^-]}{dt}}

(e)

\mathbf{2AsH_3 \to 2As + 3H_2} \\ \\ \\  \mathbf{-\dfrac{1}{2}\dfrac{d[AsH_3]}{dt}=\dfrac{1}{2}\dfrac{d[As]}{dt}=\dfrac{1}{3}\dfrac{d[H_2]}{dt}}

5 0
3 years ago
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Svetllana [295]

Answer: c , v, c, c, v

Explanation:

8 0
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How many atoms of lead are in 3.25 moles of lead
musickatia [10]

Answer: 1.96x10^24 atoms

Explanation:

3.25*6.02214076*10^23 atoms = 1.96x10^24

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