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s344n2d4d5 [400]
3 years ago
5

You have at your disposal 3 1-pound bags of various pure salts that dissolve readily in water. You can add one of the bags to a

55 gallon barrel of water. Which bag should you use to lower the freezing point by the greatest amount? (1 lb. = 0.454 kg)
Chemistry
1 answer:
Sedaia [141]3 years ago
6 0

Answer:

This question appears incomplete

Explanation:

However, it should be noted that addition of soluble salts generally lowers the freezing point of water hence after the addition, water will no longer freeze at 0°C but lower.

Soluble salts tend to form more ions in water, it is these ions that are responsible for interfering with the hydrogen bonds hence lowering the freezing. Thus, (since each bag are of the same weight) <u>the bag that contains the salt that ionizes more in water will lower the freezing point by the greatest amount</u>.

NOTE: Different weight of the salts could lead to more ions been formed in the water by some salts as against the other.

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Potassium chlorate (kclo3) decomposes in a reaction described by this chemical equation:2kclo3(s) → 2kcl(s) + 3o2(g)if 36.0 gram
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<span>If 36 gm of potassium chlorate enter into the reaction, the total mass of the two products will still be 36 gm because if there is only one reactant, the mass of the compounds after the reaction will be same that reactant based on the law of conservation of matter.</span>
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What is the mass number of an isotope with an atomic number of 57 and that has 69 neutrons?
Karolina [17]

Answer:

mass number=126

Explanation:

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7 0
2 years ago
a chemistry teacher adds 50.0 ml of 1.50 m h2so4 solution to 200 ml of water. What is the concentration of the final solution?
nevsk [136]

Answer:0.300M

Explanation:1) Data:

a) Initial solution

M = 1.50M

V = 50.0 ml = 0.050 l

b) Solvent added = 200 ml = 0.200 l

2) Formula:

Molarity: M = moles of solute / volume of solution is liters

3) Solution:

a) initial solution:

Clearing moles from the molarity formula: moles = M × V

moles of H₂SO₄ = M × V = 1.5M × 0.050 l = 0.075 mol

b) final solution:

i) Volumen of solution = 0.050 l + 0.200l = 0.250l

ii) M = 0.075 mol / 0.250 l = 0.300M ← answeer

5 0
2 years ago
Hydrogenation reactions are used to add hydrogen across double bonds in hydrocarbons and other organic compounds. use average bo
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The bond which breaks energy is positive.
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7 0
3 years ago
What is the pH of a 0.640 M solution of C₅H₅NHBr (Kb of C₅H₅N is 1.7 × 10⁻⁹)?
Elden [556K]

The pH of a 0.64 M solution of pyridine (C₅H₅N) is 9.52.  

<h3>What is pH ?</h3>

A figure expressing the acidity or alkalinity of a solution on a logarithmic scale on which 7 is neutral, lower values are more acid and higher values more alkaline.

The equation for the protonation of the base pyridine is the following:

C₅H₅N + H₂O ⇄ C₅H₅NH⁺ + OH⁻   (1)

Kb = 1.7 × 10⁻⁹ (Given)

To calculate the pH of the solution we need to use the following equation:

pH + pOH = 14

<em>pH = 14 - pOH</em>

     =14 - [-log[OH⁻]]

    = 14 + log[OH⁻]

Now, we need to find the concentration of the OH⁻ ions. Since pyridine is a weak base, at the equilibrium we have (eq 1):

C₅H₅N  +  H₂O  ⇄  C₅H₅NH⁺  +  OH⁻

0.64 - x                          x              x

After entering the values of [C₅H₅N] = 0.64-x, [C₅H₅NH⁺] = x, and [OH⁻] = x, into equation (2) we can find the concentration of OH⁻:

1.7 × 10⁻⁹  =[C₅H₅NH⁺]  [OH⁻]  /  [C₅H₅N]

                = x . x / 0.64-x

1.7 × 10⁻⁹ (0.64-x) - x² = 0

Solving the above quadratic equation for x, we have :

  • x₁ = -3.32 x 10⁻⁵
  • x₂ = 3.32 x 10⁻⁵

Now, We can calculate the pH, after taking the positive value, x₂, (concentrations cannot be negative) and entering into above equation :

<em />

<em>pH = </em>14 + log[OH⁻]

     = 14 + log (3.32 x 10⁻⁵)

 

     = 9.52

Therefore, the pH of the solution of pyridine is 9.52.

Find more about pH here:

brainly.com/question/8834103?referrer=searchResults

#SPJ1

3 0
2 years ago
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