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Tatiana [17]
3 years ago
6

Write the total ionic equation for the reaction of hydrofluoric acid with potassium hydroxide. 1. koh(aq) hf(aq) → kf(s) h2o(ℓ)

2. k (aq) oh−(aq) hf(aq) → k (aq) f−(aq) h2o(ℓ) 3. koh(s) h (aq) f−(aq) → k (aq) f−(aq) h2o(ℓ) 4. k (aq) oh−(aq) h (aq) f−(aq) → k (aq) f−(aq) h2o(ℓ) 5. koh(aq) hf(aq) → kf(aq) h2o(ℓ) 6. k (aq) o2−(aq) h (aq) h (aq) f−(aq) → k (aq) f−(aq) o2−(aq) 2 h (aq) 7. k (aq) oh−(aq) h (aq) f−(aq) → kf(s) h2o(ℓ)
Chemistry
1 answer:
melisa1 [442]3 years ago
3 0

2. HF(aq) + K^(+)(aq) +OH^(-)(aq) → K^(+)(aq) + F^(-)(aq) + H_2O(ℓ)

<em>Step 1</em>. Write the <em>molecular equation </em>

HF(aq) + KOH(aq) → KF(aq) + H_2O(ℓ)

<em>Step 2</em>. Write the <em>total ionic equation </em>

HF and H_2O are <em>weak electrolytes</em>, so we write them as <em>molecules</em>.

HF(aq) + K^(+)(aq) +OH^(-)(aq) → K^(+)(aq) + F^(-)(aq) + H_2O(ℓ)

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A sample of sulfur hexafluoride gas occupies a volume of 5.10 L at 198 ºC. Assuming that the pressure remains constant, what tem
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Answer:

When the volume will be reduced to 2.50 L, the temperature will be reduced to a temperature of 230.9K

Explanation:

Step 1: Data given

A sample of sulfur hexafluoride gas occupies a volume of 5.10 L

Temperature = 198 °C = 471 K

The volume will be reduced to 2.50 L

Step 2 Calculate the new temperature via Charles' law

V1/T2 = V2/T2

⇒with V1 = the initial volume of sulfur hexafluoride gas = 5.10 L

⇒with T1 = the initial temperature of sulfur hexafluoride gas = 471 K

⇒with V2 = the reduced volume of the gas = 2.50 L

⇒with T2 = the new temperature = TO BE DETERMINED

5.10 L / 471 K = 2.50 L / T2

T2 = 2.50 L / (5.10 L / 471 K)

T2 = 230.9 K = -42.1

When the volume will be reduced to 2.50 L, the temperature will be reduced to a temperature of 230.9K

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Explanation:

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Answer :

Charles's Law : It is defined as the volume is directly proportional to the temperature of the gas at constant pressure and number of moles.

Mathematically,

V\propto T

                                Boiling water bath        Cool bath 1       Cool bath 2

Temperature (⁰C)                  99                              17                       2

Temperature (K)(T)    273+99=372             273+17=290      273+2=275

Volume of water                  0.0                             27.0                34.0

in cool flask (mL)

Volume of water=              135.8                           135.8               135.8

Air in flask (mL)

Volume of air                    135.8                           108.8               101.8

in cool flask (V)

\frac{V}{T}                                \frac{135.8}{372}=0.365             \frac{108.8}{290}=0.375         \frac{101.8}{275}=0.370

The graph volume versus temperature for a gas is shown below.

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