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3 years ago

Write an equation showing how this buffer neutralizes added base (csoh).

1 answer:

5 0

Here, the buffer is NH₄Cl + NH₃
The strong base is added to the buffer solution, then the salt reacts with base to form weak base and water, which does not alter the pH

NH₄Cl + CsOH → NH₃ + CsCl + H₂O
NH₄⁺ + OH⁻ → NH₃ + H₂O

When we add a strong acid to the buffer solution, the base NH₃ reacts with acid and forms salt, which does not alter the pH
NH₃ + H⁺ → NH₄⁺

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tamaranim1 [39]

$a) \: n = \frac{N}{NA} \\ 1.5 = \frac{x}{6.02 \times \: {10}^{23} } \\ x = 1.5 \times 6.02 \times {10}^{23} \\ x = 9.03 \times {10}^{23} \: atoms \: of \: Na$

$b) \: n = \frac{N}{NA} \\ 0.25 = \frac{x}{6.02 \times \: {10}^{23} } \\ x = 0.25 \times 6.02 \times {10}^{23} \\ x = 1.505 \times {10}^{23} \: atoms \: of \: Si \:$
$c) \: n = \frac{N}{NA} \\ 6.755 = \frac{x}{6.02 \times \: {10}^{23} } \\ x = 6.755 \times 6.02 \times {10}^{23} \\ x = 40,6651 \times {10}^{23} \\ x = 4.07 \times {10}^{24} \: atoms \: of \: Pb \:$

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3 years ago

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Burka [1]

Answer:

Solubility of O₂(g) in 4L water = 3.42 x 10⁻² grams O₂(g)

Explanation:

Graham's Law => Solubility(S) ∝ Applied Pressure(P) => S =k·P

Given P = 0.209Atm & k = 1.28 x 10⁻³mol/L·Atm

=> S = k·P = (1.28 x 10⁻³ mole/L·Atm)0.209Atm = 2.68 x 10⁻³ mol O₂/L water.

∴Solubility of O₂(g) in 4L water at 0.209Atm = (2.68 x 10⁻³mole O₂(g)/L)(4L)(32 g O₂(g)/mol O₂(g)) = <u>3.45 x 10⁻² grams O₂(g) in 4L water. </u>

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Answer:

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The attached file it displayed the response along with the mechanism, please find the.

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Answer:

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