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pantera1 [17]
3 years ago
6

You pre-weigh a glass vial to hold your sample and find its mass to be 5.010 g. You add your sample to the vial and reweigh it o

n the same balance and find that the mass has increased to 6.130 g. What is the mass of the sample in grams
Chemistry
1 answer:
san4es73 [151]3 years ago
8 0
Answer : 1.12 grams

Yo find the mass of the sample, you take the increased mass and subtract the original mass.

6.130 - 5.010 = 1.12
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You are really and i mean very dumb and extremely 100 percent stupid.

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3 years ago
We mix 0.08 moles of chloroacetic acid (ClCH2COOH) and 0.04 moles of
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Answer:

A. pH using molar concentrations = 2.56

B. pH using activities                      = 2.46

C. pH of mixture                              = 2.56

Explanation:

A. pH using molar concentrations

ClCH₂COOH + H₂O ⇌ ClCH₂COO⁻ + H₃O⁺

        HA        + H₂O ⇌          A⁻         + H₃O⁺

We have a solution of 0.08 mol HA and 0.04 mol A⁻

We can use the Henderson-Hasselbalch equation to calculate the pH.

\begin{array}{rcl}\text{pH} & = & \text{pK}_{\text{a}} + \log \left(\dfrac{[\text{A}^{-}]}{\text{[HA]}}\right )\\\\& = & 2.865 +\log \left(\dfrac{0.04}{0.08}\right )\\\\& = & 2.865 + \log0.50 \\& = &2.865 - 0.30 \\& = & \mathbf{2.56}\\\end{array}

B. pH using activities

(i) Calculate [H⁺]

pH = -log[H⁺]

\text{[H$^{+}$]}  = 10^{-\text{pH}} \text{ mol/L} = 10^{-2.56}\text{ mol/L} = 2.73  \times 10^{-3}\text{ mol/L}

(ii) Calculate the ionic strength of the solution

We have a solution of 0.08 mol·L⁻¹ HA, 0.04 mol·L⁻¹ Na⁺, 0.04 mol·L⁻¹ A⁻, and 0.00273 mol·L⁻¹ H⁺.

The formula for ionic strength is  

I = \dfrac{1}{2} \sum_{i} {c_{i}z_{i}^{2}}\\\\I = \dfrac{1}{2}\left [0.04\times (+1)^{2} + 0.04\times(-1)^{2} +  0.00273\times(+1)^{2}\right]\\\\=  \dfrac{1}{2} (0.04 + 0.04 + 0.00273) = \dfrac{1}{2} \times 0.08273 = 0.041

(iii) Calculate the activity coefficients

\ln \gamma = -0.510z^{2}\sqrt{I} = -0.510(-1)^{2}\sqrt{0.041} = -0.510\times 0.20 = -0.10\\\gamma = 10^{-0.10} = 0.79

(iv) Calculate the initial activity of A⁻

a = γc = 0.79 × 0.04= 0.032

(v) Calculate the pH  

\begin{array}{rcl}\text{pH} & = & \text{pK}_{\text{a}} + \log \left(\dfrac{a_{\text{A}^{-}}}{a_{\text{[HA]}}}\right )\\\\& = & 2.865 +\log \left(\dfrac{0.032}{0.08}\right )\\\\& = & 2.865 + \log0.40 \\& = & 2.865 -0.40\\& = & \mathbf{2.46}\\\end{array}\\

C. Calculate the pH of the mixture

The mixture initially contains 0.08 mol HA, 0.04 mol Na⁺, 0.04 mol A⁻, 0.05 mol HNO₃, and 0.06 mol NaOH.

The HNO₃ will react with the NaOH to form 0.05 mol Na⁺ and 0.05 mol NO₃⁻.

The excess NaOH will react with 0.01 mol HA to form 0.01 mol Na⁺ and 0.01 mol A⁻.

The final solution will contain 0.07 mol HA, 0.10 mol Na⁺, 0.05 mol A⁻, and 0.05 mol NO₃⁻.

(i) Calculate the ionic strength

I = \dfrac{1}{2}\left [0.10\times (+1)^{2} + 0.05 \times(-1)^{2} +  0.05\times(-1)^{2}\right]\\\\=  \dfrac{1}{2} (0.10 + 0.05 + 0.05) = \dfrac{1}{2} \times 0.20 = 0.10

(ii) Calculate the activity coefficients

\ln \gamma = -0.510z^{2}\sqrt{I} = -0.510(-1)^{2}\sqrt{0.10} = -0.510\times 0.32 = -0.16\\\gamma = 10^{-0.16} = 0.69

(iii) Calculate the initial activity of A⁻:

a = γc = 0.69 × 0.05= 0.034

(iv) Calculate the pH

\text{pH} = 2.865 + \log \left(\dfrac{0.034}{0.07}\right ) = 2.865 + \log 0.49 = 2.865 - 0.31 = \mathbf{2.56}

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3 years ago
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Ionic formula for sodium oxide
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