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Semenov [28]
3 years ago
11

fluorine gas and water vapor react to form hydrogen fluorine gas and oxygen what volume of hydrogen fluorine would be produced b

y this reaction if 3.7L of fluorine was consumed?
Chemistry
1 answer:
DerKrebs [107]3 years ago
3 0

The volume of the gas mixture is 1.7 L and its total pressure is 0.810 atm. ... If the carbon and oxygen react completely to form CO(g), what will be the final .... acid to generate hydrogen gas, which is collected over a liquid whose vapor ..... It takes more energy to add an electron to fluorine than to oxygen because the radius

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A 100.0 mL solution containing 0.864 g of maleic acid (MW=116.072 g/mol) is titrated with 0.276 M KOH. Calculate the pH of the s
Lilit [14]

Answer:

pH = 1.32

Explanation:

                 H₂M + KOH ------------------------ HM⁻ + H₂O + K⁺

This problem involves a weak diprotic acid which we can solve by realizing they amount  to buffer solutions.  In the first  deprotonation if all the acid is not consumed we will have an equilibrium of a wak acid and its weak conjugate base. Lets see:

So first calculate the moles reacted and produced:

n H₂M = 0.864 g/mol x 1 mol/ 116.072 g  =  0.074 mol H₂M

54 mL x  1L / 1000 mL x 0. 0.276 moles/L = 0.015 mol KOH

it is clear that the maleic acid will not be completely consumed, hence treat it as an equilibrium problem of a buffer solution.

moles H₂M left = 0.074 - 0.015 = 0.059

moles HM⁻ produced = 0.015

Using the Henderson - Hasselbach equation to solve for pH:

ph = pKₐ + log ( HM⁻/ HA) = 1.92 + log ( 0.015 / 0.059) = 1.325

Notes: In the HH equation we used the moles of the species since the volume is the same and they will cancel out in the quotient.

For polyprotic acids the second or third deprotonation contribution to the pH when there is still unreacted acid ( Maleic in this case) unreacted.

           

3 0
3 years ago
Calculate the pH of 0.50M H2S
-BARSIC- [3]
I got that pH=3.65 using the fact that Ka=[H⁺][A⁻]/[HA] at equilibrium.  In the ice table, I stands for initial, C stands for change, and E stands for equilibrium.  

I hope this helps.  Let me know if anything is unclear.

5 0
3 years ago
In this experiment we will be using a 0.05 M solution of HCl to determine the concentration of hydroxide (OH-) in a saturated so
gulaghasi [49]

<u>Answer:</u> The moles of hydroxide ions present in the sample is 0.0008 moles

<u>Explanation:</u>

To calculate the concentration of acid, we use the equation given by neutralization reaction:

n_1M_1V_1=n_2M_2V_2

where,

n_1,M_1\text{ and }V_1 are the n-factor, molarity and volume of acid which is HCl.

n_2,M_2\text{ and }V_2 are the n-factor, molarity and volume of base which is Ca(OH)_2

We are given:

n_1=1\\M_1=0.05M\\V_1=16mL\\n_2=2\\M_2=?M\\V_2=36.0mL

Putting values in above equation, we get:

1\times 0.05\times 16=2\times M_2\times 36\\\\M_2=\frac{1\times 0.05\times 16}{2\times 36}=0.011M

To calculate the moles of hydroxide ions, we use the equation used to calculate the molarity of solution:

\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}

Molarity of solution = 0.011 M

Volume of solution = 36.0 mL

Putting values in above equation, we get:

0.011=\frac{\text{Moles of }Ca(OH)_2\times 1000}{36}\\\\\text{Moles of }Ca(OH)_2=\frac{0.011\times 36}{1000}=0.0004mol

1 mole of calcium hydroxide produces 1 mole of calcium ions and 2 moles of hydroxide ions.

Moles of hydroxide ions = (0.0004 × 2) = 0.0008 moles

Hence, the moles of hydroxide ions present in the sample is 0.0008 moles

8 0
3 years ago
How many kilograms of chlorine are in 21 kg of CF2Cl2
Alik [6]

21 kg x [(3 x 35.45)/(12.01 + 19.00 + (3 x 35.45))] =  

21 kg x (106.35/137.36) = 16.3 kg of chlorine  

You just multiply the weight of the material by the fraction of chlorine (by weight). The others are done the same way

5 0
3 years ago
Read 2 more answers
A 15.0 L tank of gas is contained at a high pressure of 8.20 x 10^4 torr. The tank is opened and the gas expands into an empty c
Inga [223]

Answer:

20.5torr

Explanation:

Given parameters:

V₁  = 15L

P₁  = 8.2 x 10⁴torr

V₂ = 6 x 10⁴L

Unknown:

P₂  = ?

Solution:

To solve this problem we have to apply the claims of Boyle's law.

Boyle's law is given mathematically as;

            P₁ V₁   = P₂V₂

where P₁ is the initial pressure

          V₁ is the initial volume

           P₂ is final pressure

           V₂ is final volume

   8.2 x 10⁴  x 15  = P₂  x 6 x 10⁴

         P₂  = 20.5torr

5 0
3 years ago
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