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3 years ago

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Products and reactants are used and created at the same rate when a reaction is at this stage A. Le Chateliers principle B. Cata

lyst C. Balanced Equation D. Chemical equilibrium

1 answer:

Alex787 [66]3 years ago

4 0

Answer:

A. Le Chatelier's principle

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The molality of a solution that was made by dissolving NaCl in 1.5 kg of water is 0.40 m. If the molar mass of NaCl is 58.44 g/m

kirill115 [55]

Molality is one way of expressing concentration of a solute in a solution. It is expressed as the mole of solute per kilogram of the solvent. To calculate for the mass of the solute in a given solution, we need to calculate first for the number of moles of the substance by the product of the molality and the mass of the solvent. Then, we convert the number of moles into mass by using the molar mass. <span>
</span>

moles NaCl = 0.40 m (1.5 kg water) = 0.6 mol NaCl

mass NaCl = 0.6 mol NaCl ( 58.44 g / mol ) = 35.06 g NaCl

Therefore, the correct answer would be the third option.

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A 1 l container contains O2 gas at stp. if the diameter of the O2 molecule is 3.75 * 10^-10 m the mean free path of the o2 molec

Luba_88 [7]

The mean free path of the O₂ molecule is $5.97*10^{-5}$ m.

The mean free path of the O₂ molecule is calculated by the formula -

λ = $\frac{1}{\sqrt{2}*\pi *d^{2}*n}$

In this formula, λ represents mean free path of gas, d represents diameter and n is number of molecules.

As per the known fact, 1 mole of gas occupies 22.4 liter at STP.

So. 1 liter of O₂ gas will have number of moles = $\frac{1*1}{22.4}$

Number of moles = 0.045

Also, we are aware that 1 mole contains $6.02*10^{23}$ molecules.

So, 0.045 moles will contain number of molecules = $6.02*10^{23}*0.045$

Number of molecules = $2.7*10^{22}$ molecules

Now, keeping the values in formula to find mean free path of gas -

λ = $\frac{1}{1.4*3.14*(3.75*10^{-10})^{2} *2.7*10^{22} }$

λ = $\frac{1}{16746.7}$

λ = $5.97*10^{-5}$ m

Hence, the mean free path of the O₂ molecule is $5.97*10^{-5}$ m.

Learn more about calculation of mean free path -

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Answer:

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Explanation:

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