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DochEvi [55]
4 years ago
9

Are the particles in a solid in constant motion, or are they "frozen in place?"

Chemistry
2 answers:
babymother [125]4 years ago
4 0
Particles that are in a solid are tightly locked in place so the answer to your question would be  "Frozen in place"
Tems11 [23]4 years ago
3 0
The molecules are frozen in place but still vibrate
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Using the equation below, if you have 4.3 mol of nitrogen tribromide and
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Answer:

sodium hydroxide is the limiting reactant

Explanation:

The first step is usually to put down the balanced reaction equation. This is the first thing to do when solving any problem related to stoichiometry. The balanced reaction equation serves as a guide during the solution.

2NBr3 + 3NaOH = N2 + 3NaBr + 3HOBr

Let us pick nitrogen gas as our product of interest. Any of the reactants that gives a lower number of moles of nitrogen gas is the limiting reactant.

For nitrogen tribromide

From the balanced reaction equation;

2 moles of nitrogen tribromide yields 1 mole of nitrogen gas

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8 0
3 years ago
Determine the freezing point of an aqueous solution containing 10.50 g of magnesium bromide in 200.0 g of water.
Rudiy27
For an aqueous solution of MgBr2, a freezing point depression occurs due to the rules of colligative properties. Since MgBr2 is an ionic compound, it acts a strong electrolyte; thus, dissociating completely in an aqueous solution. For the equation:

                                ΔTf<span> = (K</span>f)(<span>m)(i)
</span>where: 
ΔTf = change in freezing point = (Ti - Tf)
Ti = freezing point of pure water = 0 celsius
Tf = freezing point of water with solute = ?
Kf = freezing point depression constant = 1.86 celsius-kg/mole (for water)
m = molality of solution (mol solute/kg solvent) = ?
i = ions in solution = 3

Computing for molality:
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m = 10.5g MgBr2 / 184.113/ 0.2 kg water = 0.285 mol/kg


For the problem, 
ΔTf = (Kf)(m)(i) = 1.86(0.285)(3) = 1.59 = Ti - Tf = 0 - Tf

Tf = -1.59 celsius
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