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Yakvenalex [24]
3 years ago
11

What is the reason why carbonated beverages be kept cold?

Chemistry
1 answer:
Makovka662 [10]3 years ago
3 0
Option A) Decreased temperature keeps gases like carbon dioxide dissolved.

Carbonation is made with CO2 which is also know as liquid carbonic. The low temperature favors higher solubility of CO2 in water
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When a 1 kg ball is dropped from the roof of a building its speed increases from 0 m/s to 10 m/s just before it hits the ground.
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30 kinetic energy is the answer.
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At the boiling point, the vapor pressure of a liquid is equal to the _______.
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At the boiling point, the vapor pressure of a liquid is equal to the atmospheric pressure.
6 0
3 years ago
Calculate either [ H 3 O + ] or [ OH − ] for each of the solutions.
STALIN [3.7K]

Answer: Solution A : [H_3O^+]=0.300\times 10^{-7}M

Solution B : [OH^-]=0.107\times 10^{-5}M

Solution C : [OH^-]=0.177\times 10^{-10}M

Explanation:

pH or pOH is the measure of acidity or alkalinity of a solution.

pH is calculated by taking negative logarithm of hydrogen ion concentration and pOH is calculated by taking negative logarithm of hydroxide ion concentration.

pH=-\log[H_3O^+]

pOH=-log[OH^-}

pH+pOH=14

[H_3O^+][OH^-]=10^{-14}

a. Solution A: [OH^-]=3.33\times 10^{-7}M

[H_3O^+]=\frac{10^{-14}}{3.33\times 10^{-7}}=0.300\times 10^{-7}M

b. Solution B : [H_3O^+]=9.33\times 10^{-9}M

[OH^-]=\frac{10^{-14}}{9.33\times 10^{-9}}=0.107\times 10^{-5}M

c. Solution C : [H_3O^+]=5.65\times 10^{-4}M

[OH^-]=\frac{10^{-14}}{5.65\times 10^{-4}}=0.177\times 10^{-10}M

7 0
3 years ago
Oh, no! You just spilled 85.00 mL of 1.500 M sulfuric acid on your lab bench and need to clean it up immediately! Right next to
vredina [299]

Explanation:

We will balance equation which describes the reaction between sulfuric acid and sodium bicarbonate: as follows.

   H_2SO_4(aq) + 2NaHCO_3(s) \rightarrow Na2SO_4(aq) + 2H_2O(l) + 2CO_2(g)

Next we will calculate how many moles of H_2SO_4 are present in 85.00 mL of 1.500 M sulfuric acid.

As,       Molarity = \frac{\text{moles of solute}}{\text{liters of solution&#10;}}

            1.500 M = \frac{n}{0.08500 L&#10;}

                    n = 0.1275 mol H_2SO_4

Now set up and solve a stoichiometric conversion from moles of H_2SO_4  to grams of NaHCO_3. As, the molar mass of NaHCO_3 is 84.01 g/mol.

 0.1275 mol H_2SO_4 \times (\frac{2 mol NaHCO_3}{1 mol H_2SO_4}) \times (\frac{84.01 g NaHCO_3}{1 mol NaHCO_3})

                 = 21.42 g NaHCO_3

So unfortunately, 15.00 grams of sodium bicarbonate will "not" be sufficient to completely neutralize the acid. You would need an additional 6.42 grams to complete the task.

4 0
3 years ago
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