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aleksandrvk [35]
2 years ago
11

What is the molality of a solution that contains 418.8 g HCl in 4.62 kg water?

Chemistry
2 answers:
LenKa [72]2 years ago
6 0
The answer is c. 15.3
dsp732 years ago
5 0
The answer to this is 15.3
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The shape and the color
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Ammonia can be produced in the laboratory by heating ammonium chloride
AnnyKZ [126]

Answer:

Mass = 2.89 g

Explanation:

Given data:

Mass of NH₄Cl = 8.939 g

Mass of Ca(OH)₂ = 7.48 g

Mass of ammonia produced = ?

Solution:

2NH₄Cl   +  Ca(OH)₂     →    CaCl₂ + 2NH₃ + 2H₂O

Number of moles of NH₄Cl:

Number of moles = mass/molar mass

Number of moles = 8.939 g / 53.5 g/mol

Number of moles = 0.17 mol

Number of moles of Ca(OH)₂ :

Number of moles = mass/molar mass

Number of moles = 7.48 g / 74.1 g/mol

Number of moles = 0.10 mol

Now we will compare the moles of ammonia with both reactant.

                      NH₄Cl          :          NH₃

                          2              :           2

                         0.17          :          0.17

                   Ca(OH)₂         :          NH₃

                        1                :           2

                    0.10              :          2/1×0.10 = 0.2 mol

Less number of moles of ammonia are produced by ammonium chloride it will act as limiting reactant.

Mass of ammonia:

Mass = number of moles × molar mass

Mass = 0.17 mol × 17 g/mol

Mass = 2.89 g

6 0
2 years ago
A few pieces of dry ice, CO2(s), at -78°C are placed in a flask that contains air at 21°C. The flask is sealed by placing an uni
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3 years ago
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What is the de Broglie wavelength (in meters) of a 45-g golf ball traveling at 72 m/s?
Cerrena [4.2K]

Answer: The de broglie wavelength is 2.037 \times 10^{-34} m.

Explanation:

Calculate  \lambda = \frac{h}{p}as follows.

          \lambda = \frac{h}{p}

where,

          h = plank's constant = 6.6 \times 10^{-34} m^{2} kg/s

         p = momentum = mass \times velocity

Putting the values in the formula as follows.

        \lambda = \frac{h}{mass \times velocity}

                               =  \frac {6.6 \times 10^{-34} m^{2} kg/s}{0.045 kg \times 72 m/s}                        

                               =  2.037 \times 10^{-34} m

Thus, the de broglie wavelength is 2.037 \times 10^{-34} m.

                               

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3 years ago
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