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kupik [55]
3 years ago
15

Aluminum chloride is added to cesium does a single replacement reaction occur, and if so what is the correct chemical equation f

or this reaction
A. AlCL3 + 3Cs > No REaction
B. AlCl2 + 2Cs > 2CsCl + Al
C.AlCl3+ 3Cs > 3CsCl + Al
D. 2AlCl3 + 6Cs > 2AlCs3 + 3Cl2
Chemistry
2 answers:
krek1111 [17]3 years ago
5 0
The answer is C

I hope that helped
ss7ja [257]3 years ago
5 0

Answer: Option (C) is the correct answer.

Explanation:

When aluminium chloride reacts with cesium, then the reaction will be as follows.

        AlCl_{3} + Cs \rightarrow CsCl + Al

Number of reactant atoms are as follows.

  • Al = 1
  • Cl = 3
  • Cs = 1

Number of product atoms are as follows.

  • Al = 1
  • Cl = 1
  • Cs = 1

Thus, in order to balance the equation, multiply Cs by 3 on reactant side and multiply CsCl by 3 on the product side.

Therefore, the balanced chemical equation will be as follows.

     AlCl_{3} + 3Cs \rightarrow 3CsCl + Al    

Hence, we can conclude that out of the given options, AlCl3+ 3Cs > 3CsCl + Al is the correct chemical equation.

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The range in size of most atomic radii is approximately _____.
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Calculate the solubility of hydrogen in water at an atmospheric pressure of 0.380 atm (a typical value at high altitude).
Pani-rosa [81]

The question is incomplete, here is the complete question:

Calculate the solubility of hydrogen in water at an atmospheric pressure of 0.380 atm (a typical value at high altitude).

Atmospheric Gas         Mole Fraction      kH mol/(L*atm)

           N_2                         7.81\times 10^{-1}         6.70\times 10^{-4}

           O_2                         2.10\times 10^{-1}        1.30\times 10^{-3}

           Ar                          9.34\times 10^{-3}        1.40\times 10^{-3}

          CO_2                        3.33\times 10^{-4}        3.50\times 10^{-2}

          CH_4                       2.00\times 10^{-6}         1.40\times 10^{-3}

          H_2                          5.00\times 10^{-7}         7.80\times 10^{-4}

<u>Answer:</u> The solubility of hydrogen gas in water at given atmospheric pressure is 1.48\times 10^{-10}M

<u>Explanation:</u>

To calculate the partial pressure of hydrogen gas, we use the equation given by Raoult's law, which is:

p_{\text{hydrogen gas}}=p_T\times \chi_{\text{hydrogen gas}}

where,

p_A = partial pressure of hydrogen gas = ?

p_T = total pressure = 0.380 atm

\chi_A = mole fraction of hydrogen gas = 5.00\times 10^{-7}

Putting values in above equation, we get:

p_{\text{hydrogen gas}}=0.380\times 5.00\times 10^{-7}\\\\p_{\text{hydrogen gas}}=1.9\times 10^{-7}atm

To calculate the molar solubility, we use the equation given by Henry's law, which is:

C_{H_2}=K_H\times p_{H_2}

where,

K_H = Henry's constant = 7.80\times 10^{-4}mol/L.atm

p_{H_2} = partial pressure of hydrogen gas = 1.9\times 10^{-7}atm

Putting values in above equation, we get:

C_{H_2}=7.80\times 10^{-4}mol/L.atm\times 1.9\times 10^{-7}atm\\\\C_{CO_2}=1.48\times 10^{-10}M

Hence, the solubility of hydrogen gas in water at given atmospheric pressure is 1.48\times 10^{-10}M

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So, average atomic mass of neon = \frac{(90.51 X 19.99) + (0.27 X 20.99) + (9.22 X 21.99)}{100} = 20.177 amu.

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