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Fynjy0 [20]
3 years ago
6

25.0 mL of a hydrofluoric acid solution of unknown concentration is titrated with 0.200 M NaOH. After 20.0 mL of the base soluti

on has been added, the pH in the titration flask is 3.00. What was the concentration of the original hydrofluoric acid solution. (Ka(HF) = 7.1 × 10–4)
Chemistry
1 answer:
lesantik [10]3 years ago
6 0

Answer:

[HF]₀ = 0.125M

Explanation:

NaOH + HF => NaF + H₂O

Adding 20ml of 0.200M NaOH into 25ml of HF solution neutralizes 0.004 mole of HF leaving 0.004 mole NaF in 0.045L with 0.001M H⁺ at pH = 3.   This is 0.089M NaF and 0.001M HF remaining.

=> 45ml of solution with pH = 3 and contains 0.089M NaF from titration becomes a common ion problem.

                HF  ⇄    H⁺    +      F⁻

C(eq)       [HF]     10⁻³M      0.089M (<= soln after adding 20ml 0.200M NaOH)

Ka = [H⁺][F⁻]/[HF]₀ => [HF]₀ = [H⁺][F⁻]/Ka

[HF]₀ = (0.001)(0.089)/(7.1 x 10⁻⁴) M = 0.125M

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Answer:

The empirical formula of ribose (a sugar) is CH2O. In a separate experiment, using a mass spectrometer, the molar mass of ribose was determined to be 150 g/mol.Explanation:

5 0
2 years ago
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I need to make up an experiment !! Please help
katen-ka-za [31]

Question: Baking a Cake Without Flour.

Hypothesis: I think that when I remove the flour from the standard cake recipe, I'll end up with a flat but tasty cake.

Procedure: I baked two cakes during my experiment. For my control, I baked a cake following a normal recipe. I used the Double Fudge Cake recipe on page 292 of the Betty Crocker Cookbook. For my experimental cake, I followed the same recipe but left out the flour. I first obtained a 2-quart mixing bowl.  

Results: My control cake, which I cooked for 25 minutes, measured 4 cm high.  Eight out of ten tasters that I picked at random from the class found it to be an acceptable dessert. After 25 minutes of baking, my experimental cake was 1.5 cm high and all ten tasters refused to eat it because it was burnt to a crisp.

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I hope this helped :))

7 0
2 years ago
Use the following equation to answer the questions below:
Gala2k [10]

Explanation:

The equation of the reaction is given as;

Be + 2HCl → BeCl2 + H2

What is the mass of beryllium required to produce 25.0g of beryllium chloride?

1 mol of Be produces 1 mol of BeCl2

Converting to mass;

Mass = Molar mass  *  Number of moles

9.01g of Be produces 79.92g of BeCl2

xg of Be produces 25g of BeCl2

Solving for x;

x = 25 * 9.01 / 79.92

x = 2.82 g

What is the mass of hydrochloric acid required to produce 25.0g of beryllium chloride? g

Converting 25.0g of beryllium chloride to moles;

Number of moles = Mass / Molar mass

Number of moles = 25 / 79.92 = 0.3128 mol

2 mol of HCl produces 1 mol of BeCl2

x mol of HCl would produce 0.3128 mol of BeCl2

solving for x;

x = 0.3128 * 2 = 0.6256 mol

Converting to mass;

Mass = 0.6256 * 36.5 = 22.83 g

What is the mass of hydrogen gas produced when 25.0g of beryllium chloride is also produced? g

25g of BeCl2 = 0.3128 mol of BeCl2

From the equation;

1 mol of H2 is produced alongside 1 mol of BeCl2

This means;

0.3128 mol of H2 would also be produced alongside 0.3128 mol of BeCl2

Mass = Number of moles * Molar mass

Mass = 0.3128mol * 2.0159 g/mol = 0.6306 g

3 0
3 years ago
The heat of fusion for water is 80. cal/g. How many calories of heat are needed to melt a 35 g ice cube that has a temperature o
HACTEHA [7]

Answer: 2800 calories

Explanation:

Latent heat of fusion is the amount of heat required to convert 1 mole of solid to liquid at atmospheric pressure.

Amount of heat required to fuse 1 gram of water = 80 cal

Mass of ice given = 35 gram

Heat required to fuse 1 g of ice at 0^0C = 80 cal

Thus Heat required to fuse 35 g of ice =\frac{80}{1}\times 35=2800cal

Thus 2800 calories of energy is required to melt 35 g ice cube

6 0
3 years ago
Which formula represents the compound commonly known as phosphine
Arlecino [84]
I looked up what is the molecular formula for Phosphine and got this: PH3
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8 0
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