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Crank
3 years ago
15

CaF2 + (NH4)2O --> 2 NH4F + CaO How many grams of NH4F can be formed from 34.6 grams of CaF2? (AKS 4f)

Chemistry
1 answer:
bixtya [17]3 years ago
5 0

Answer:

32.83 grams of NH₄F can be formed from 34.6 grams of CaF₂

Explanation:

The balanced reaction is:

CaF₂ + (NH₄)₂O ⇒ 2 NH₄F + CaO

By stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction), the following quantities participate in the reaction:

  • CaF₂: 1 mole
  • (NH₄)₂O: 1 mole
  • NH₄F: 2 mole
  • CaO: 1 mole

Being the molar mass of the compounds:

  • CaF₂: 78 g/mole
  • (NH₄)₂O: 52 g/mole
  • NH₄F: 37 g/mole
  • CaO: 56 g/mole

Then by stoichiometry, the following amounts of mass participate in the reaction:

  • CaF₂: 1 mole* 78 g/mole= 78 g
  • (NH₄)₂O: 1 mole* 52 g/mole= 52 g
  • NH₄F: 2 mole* 37 g/mole= 74 g
  • CaO: 1 mole* 56 g/mole= 56 g

You can apply the following rule of three: if 78 grams of CaF₂ form 74 grams of NH₄F by stoichiometry, 34.6 grams of CaF₂ will form how much mass of NH₄F?

mass of NH_{4} F=\frac{34.6 grams of CaF_{2} *74 grams ofNH_{4} F}{78grams of CaF_{2} }

mass of NH₄F= 32.83 grams

<u><em>32.83 grams of NH₄F can be formed from 34.6 grams of CaF₂</em></u>

<u><em></em></u>

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A gas bottle contains 8. 61×1023 oxygen molecules at a temperature of 359. 0 k then the thermal energy of the gas is 6.35KJ.

<h3>What is thermal energy? </h3>

Thermal energy is the energy contained within a system which is responsible for its temperature. Heat is also termed as flow of thermal energy.

Thermal energy is directly proportional to the temperature. As temperature increase thermal energy increases.

Given,

temperature = 395K

Number of oxygen molecules= 8.61 × 10^(23)

Firstly we will calculate the number of moles = N/A

where A is the avagadro number = 6.022 × 10^(23)

number of moles = 8.61 × 10^(23)/6.022 ×10^(23)

number of moles = 1.42moles

Using the energy equation,

E = 3/2 × nRT

where R is the gas constant = 8.314J/molK

E = 3/2 × 1.42 × 8.314 × 359

= 6357.4J

= 6.35KJ

Thus, we found that the thermal energy of the gas is 6.35KJ.

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A buffer consists of 0.120 M HNO2 and 0.150 M NaNO2 at 25°C. pka of HNO2 is 3.40. a. What is the pH of the buffer? b. What is th
Mashcka [7]

Explanation:

It is known that K_{a} of HNO_{2} = 4.5 \times 10^{-4}.

(a)  Relation between K_{a} and pK_{a} is as follows.

                       pK_{a} = -log (K_{a})

Putting the values into the above formula as follows.

                      pK_{a} = -log (K_{a})

                                    = -log(4.5 \times 10^{-4})

                                     = 3.347

Also, relation between pH and  pK_{a} is as follows.

              pH = pK_{a} + log\frac{[conjugate base]}{[acid]}

                     = 3.347+ log \frac{0.15}{0.12}

                    = 3.44

Therefore, pH of the buffer is 3.44.

(b)   No. of moles of HCl added = Molarity \times volume

                                            = 11.6 M \times 0.001 L

                                             = 0.0116 mol

In the given reaction, NO^{-}_{2} will react with H^{+} to form HNO_{2}

Hence, before the reaction:

No. of moles of NO^{-}_{2} = 0.15 M \times 1.0 L

                                           = 0.15 mol

And, no. of moles of HNO_{2} = 0.12 M \times 1.0 L

                                               = 0.12 mol

On the other hand, after the reaction :  

No. of moles of NO^{-}_{2} = moles present initially - moles added

                                          = (0.15 - 0.0116) mol

                                          = 0.1384 mol

Moles of HNO_{2} = moles present initially + moles added

                               = (0.12 + 0.0116) mol

                                = 0.1316 mol

As, K_{a} = 4.5 \times 10^{-4}

           pK_{a} = -log (K_{a})

                         = -log(4.5 \times 10^{-4})

                         = 3.347

Since, volume is both in numerator and denominator, we can use mol instead of concentration.

As, pH = pK_{a} + log \frac{[conjugate base]}{[acid]}

            = 3.347+ log {0.1384/0.1316}

            = 3.369

            = 3.37 (approx)

Thus, we can conclude that pH after the addition of 1.00 mL of 11.6 M HCl to 1.00 L of the buffer solution is 3.37.

6 0
3 years ago
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