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3 years ago

17

Chemistry

1 answer:

leonid [27]3 years ago

8 0

Answer:

Which of the following

properties distinguishes a solution

oversaturated with a dilute?

The supersaturated solution is one in which the solvent has dissolved more solute than it can dissolve in the saturation equilibrium. The solute can be a solid, or a gas. The molecules of the solvent surround those of the solute and seek to open space between themselves to be able to harbor more amount of solute.

A dilute solution is a solution that has not reached the maximum concentration of solute dissolved in a solvent. The additional solute will dissolve when added in a dilute solution and will not appear in the aqueous phase. It is considered a state of dynamic equilibrium where the speeds in which the solvent dissolves the solute are greater than the recrystallization rate.

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Nataly [62]

Answer:

C. An organized way to test scientific ideas.

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3 years ago

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2 years ago

Please help me. How do I do ideal gas law?

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3 years ago

What is the product of the unbalanced equation below?

Stolb23 [73]

Answer:

Explanation:

The answer is D. I'm not sure that it is a solid. I don't think it is a ppte, which is the only way it can be a true solid. It is ionic if the reaction is taking place in water and there is someway to start the reaction. Be that as it may, the internal balace numbers of the chemical produced is the only possible answer. The balanced eq;uatioon is

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3 0

3 years ago

139%

GaryK [48]

<u>Answer:</u>

The percent composition of this compound is 94%

<u>Explanation:</u>

The reaction can be formed as

$2 \mathrm{Fe}+3 \mathrm{Cl}_{2} \rightarrow 2 \mathrm{FeCl}_{3}$

$\frac{\text { Weight of } \mathrm{Cl}_{2}}{\text { 3* Molar Mass of } \mathrm{Cl}_{2}}=\frac{\text { Weight of } \mathrm{Fe}}{2 * \text { Molar Mass of Fe }}$

$\frac{\text { Weight of } \mathrm{Cl}_{2}}{3 *(2 * 35.5)}=\frac{3.56}{2 * 55.8}$

$\text { Weight of } C l_{2}=\frac{3.56 * 3 * 71}{2 * 55.8}=6.79 \mathrm{g}$

$\mathrm{n}\left(\mathrm{Cl}_{2}\right)=\mathrm{m}\left(\mathrm{Cl}_{2}\right) / \mathrm{M}\left(\mathrm{Cl}_{2}\right)=6.79 / 71=0.1 \mathrm{m}$

$\mathrm{n}(\mathrm{Fe})=\mathrm{m}(\mathrm{Fe}) / \mathrm{M}(\mathrm{Fe})=3.56 / 55.8=0.06 \mathrm{m}$

Based on no. of iron reacted,

$\mathrm{n}(\text { moles of } \mathrm{Fe})=\mathrm{n}\left(\text { moles of } \mathrm{FeCl}_{3}\right)$

n = m/M

$\mathrm{m}\left(\mathrm{FeCl}_{3}\right)=\mathrm{n}^{*} \mathrm{M}=0.06^{*} 162.5=9.75 \mathrm{g}$

% composition of $FeCl_3$

= $(9.75 / 10.39)^{*} 100$

= 94%

6 0

3 years ago