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aleksandrvk [35]
2 years ago
6

THURSDAY Mr. Pate's science class performed an experiment involving the chemical reaction of sodium bicarbonate (baking soda) an

d vinegar. Students poured 50 mL vinegar into a small bottle, then added sodium bicarbonate to a balloon. They attached the balloon to the bottle and dumped the sodium bicarbonate into the bottle. The resulting reaction produced carbon dioxide gas, which filled the balloon. Lanie hypothesized that the more sodium bicarbonate that was added to the vinegar, the bigger the balloon would get. Victor hypothesized that more than 8 g of sodium bicarbonate would cause the balloon to pop. . The Effects of Sodium Bicarbonate Levels on Gas Production in 1. Was Lanie's hypothesis correct? a Chemical Reaction Explain your answer. Trial 1 Trial 2 Trial 3 Trial 4 2. Was Victor's hypothesis correct? Explain your answer, Mass of Sodium Bicarbonate 59 7. 5 9 2003. What could the students expect Added the balloon circumference to be if Balloon 12. 5 g of sodium bicarbonate were 30 cm 38 cm 44cm 51 cm Circumference added? 2. 5 g 10 9​
Chemistry
1 answer:
Kitty [74]2 years ago
7 0

The balloon will burst if the volume of the balloon is less than 1.68 L since 8 g of sodium bicarbonate produces 1.68 L of carbon dioxide gas.

<h3>What is the volume of carbin dioxude produced from the chemical reaction between sodium bicarbonate (baking soda) and vinegar?</h3>

The volume of carbon dioxide produced is obtained from the equation below:

CH_3COOH + NaCO_3 \rightarrow CH_3COONa + H_2O + CO_2 \\

1 mole of sodium bicarbonate reacts with one mole of acetic acid in vinegar to produce 1 mole of carbon dioxide.

The volume of 1 mole of carbon dioxide is 22.4 L

Moles of sodium bicarbonate in 8 g = mass/molar mass

molar mass of sodium bicarbonate = 106 g

moles of sodium bicarbonate = 8/106 = 0.075 moles

Moles of carbon dioxide produced = 0.075 moles

Volume of carbon dioxide = 0.075 × 22.4 = 1.68 L

Therefore, the balloon will burst if the volume of the balloon is less than 1.68 L.

Learn more about gas volume at: brainly.com/question/24263739

#SPJ1

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The copper(I) ion forms a chloride salt (CuCl) that has Ksp = 1.2 x 10-6. Copper(I) also forms a complex ion with Cl-:Cu+ (aq) +
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Answer: (a) The solubility of CuCl in pure water is 1.1 \times 10^{-3} M.

(b) The solubility of CuCl in 0.1 M NaCl is 9.5 \times 10^{-3} M.

Explanation:

(a)  Chemical equation for the given reaction in pure water is as follows.

           CuCl(s) \rightarrow Cu^{+}(aq) + Cl^{-}(aq)

Initial:                         0            0

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Equilibm:                   x             x

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Hence, the solubility of CuCl in pure water is 1.1 \times 10^{-3} M.

(b)  When NaCl is 0.1 M,

       CuCl(s) \rightarrow Cu^{+}(aq) + Cl^{-}(aq),  K_{sp} = 1.2 \times 10^{-6}

   Cu^{+}(aq) + 2Cl^{-}(aq) \rightleftharpoons CuCl_{2}(aq),  K = 8.7 \times 10^{4}

Net equation: CuCl(s) + Cl^{-}(aq) \rightarrow CuCl_{2}(aq)

               K' = K_{sp} \times K

                          = 0.1044

So for, CuCl(s) + Cl^{-}(aq) \rightarrow CuCl_{2}(aq)

Initial:                     0.1                 0

Change:                -x                   +x

Equilibm:            0.1 - x                x

Now, the equilibrium expression is as follows.

              K' = \frac{CuCl_{2}}{Cl^{-}}

         0.1044 = \frac{x}{0.1 - x}

              x = 9.5 \times 10^{-3} M

Therefore, the solubility of CuCl in 0.1 M NaCl is 9.5 \times 10^{-3} M.

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