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STatiana [176]
3 years ago
14

9) After lab, all of Darrel’s friends looked at his data and laughed and laughed. They told him that he was 30.8% too low in the

boiling point he had just recorded. He had recorded a boiling point of 50o C on his data sheet. What is the correct boiling point of the liquid he was working with in lab?
Chemistry
2 answers:
aliya0001 [1]3 years ago
4 0

Answer:

\boxed{72 \, ^{\circ}\text{C}}

Explanation:

\text{Percent error} = \dfrac{\lvert \text{Measured - Actual}\lvert}{ \text{Actual}} \times100 \,\%

Data:

Percent error = 30.8 %

    Measured = 50 °C

Calculation:

Let x = the actual value

\begin{array}{rcl}-30.8 \,\% & = &\dfrac{50 - x}{x} \times 100 \, \%\\\\-30.8x & = & 100(50 - x)\\-30.8x & = & 5000 - 100x\\69.2x & = & 5000\\x & = & \dfrac{5000}{69.2}\\ & = & \mathbf{72}\end{array}

\text{The correct boiling point is }\boxed{\mathbf{72 \, ^{\circ}\text{C}}}\\\text{Note: The answer can have only two significant figures, because that is all} \\\text{you gave for the measured value.}

zaharov [31]3 years ago
3 0
If he was 30.8% too low, it means that he was at 69.2% of the boiling point needed. So 50o C is 69.2% of total.

In order to know what 100% is, you can divide the number by it's percentage and then multiply it by a hundred.

So: 50/30.8=1.623
1.623*100=162.3
So the correct boiling point of the liquid he was working with in the lab is 162.3 oC
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Answer:

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Suppose you have just added 100 ml of a solution containing 0.5 mol of acetic acid per liter to 400 ml of 0.5 m naoh. what is th
Tpy6a [65]

pH = 13.5

Explanation:

Sodium hydroxide completely ionizes in water to produce sodium ions and hydroxide ions. Hydroxide ions are in excess and neutralize all acetic acid added by the following ionic equation:

\text{HAc} + \text{OH}^{-} \to \text{Ac}^{-} + \text{H}_2\text{O}

The mixture would contain

  • 0.4 \times 0.5 - 0.1 \times 0.5 = 0.15 \; \text{mol} of \text{OH}^{-} and
  • 0.1 \times 0.5 = 0.05 \; \text{mol} of \text{Ac}^{-}

if \text{Ac}^{-} undergoes no hydrolysis; the solution is of volume 0.1 + 0.4 = 0.5 \; \text{L} after the mixing. The two species would thus be of concentration 0.30 \; \text{mol} \cdot \text{L}^{-1} and 0.10 \; \text{mol} \cdot \text{L}^{-1}, respectively.

Construct a RICE table for the hydrolysis of \text{Ac}^{-} under a basic aqueous environment (with a negligible hydronium concentration.)

\begin{array}{cccccccc} \text{R} & \text{Ac}^{-}(aq) &+ & \text{H}_2\text{O}(aq) & \leftrightharpoons & \text{HAc}(aq) & + & \text{OH}^{-} (aq)\\ \text{I} & 0.10 \; \text{M} & & & & & &0.30 \; \text{M}\\ \text{C} & -x \; \text{M}& & & & +x \; \text{M}& & +x \; \text{M} \\ \text{E} & (0.10 - x) \; \text{M} & & & & x \; \text{M} & & (0.30 +x) \; \text{M} \end{array}

The question supplied the <em>acid</em> dissociation constant pK_afor acetic acid \text{HAc}; however, calculating the hydrolysis equilibrium taking place in this basic mixture requires the <em>base</em> dissociation constant pK_b for its conjugate base, \text{Ac}^{-}. The following relationship relates the two quantities:

pK_{b} (\text{Ac}^{-}) = pK_{w} - pK_{a}( \text{HAc})

... where the water self-ionization constant pK_w \approx 14 under standard conditions. Thus pK_{b} (\text{Ac}^{-}) = 14 - 4.7 = 9.3. By the definition of pK_b:

[\text{HAc} (aq)] \cdot [\text{OH}^{-} (aq)] / [\text{Ac}^{-} (aq) ] = K_b =  10^{-pK_{b}}

x \cdot (0.3 + x) / (0.1 - x) = 10^{-9.3}

x = 1.67 \times 10^{-10} \; \text{M} \approx 0 \; \text{M}

[\text{OH}^{-}] = 0.30 +x \approx 0.30 \; \text{M}

pH = pK_{w} - pOH = 14 + \text{log}_{10}[\text{OH}^{-}] = 14 + \text{log}_{10}{0.30} = 13.5

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Answer:

Explanation:

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A synthesis reaction takes place when carbon monoxide (CO) and hydrogen gas (H2) react to form methanol (CH3OH). How many grams
jek_recluse [69]

The mass of methanol produced is 8.0 g.

We have the masses of two reactants, so this is a <em>limiting reactant</em> problem.

We know that we will need a <em>balanced equation</em> with masses, moles, and molar masses of the compounds involved.

<em>Step 1</em>. <em>Gather all the informatio</em>n in one place with molar masses above the formulas and everything else below them.

MM: ___28.01  2.016 ___32.04

_______CO + 2H_2 → CH_3OH

Mass/g: 7.0 __2.5

<em>Step 2</em>. Calculate the <em>moles of each reactant</em>

Moles of CO = 7.0 g CO × (1 mol CO/28.01g CO) = 0.250 mol CO

Moles of H_2 =2.5 g H_2 × (1 mol H_2/2.016 g H_2) = 1.24 mol H_2

<em>Step 3. </em>Identify the<em> limiting reactan</em>t

Calculate the <em>moles of CH_3OH</em> we can obtain from each reactant.

<em>From CO</em>: Moles of CH_3OH = 0.250 mol CO  × (1 mol CH_3OH /1 mol CO)

= 0.250 mol CH_3OH

<em>From H_2</em>: Moles of CH_3OH = 1.24 mol H_2 × (1 mol CH_3OH /2 mol H_2)

= 0.620 mol CH_3OH

<em>CO is the limiting reactant</em> because it gives the smaller amount of CH_3OH.

<em>Step 4</em>. Calculate the <em>mass of CH_3OH</em>

Mass of CH_3OH = 0.250 mol CH_3OH × (32.04 g CH_3OH /1 mol CH_3OH) = 8.0 g CH_3OH

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3 years ago
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