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Deffense [45]
3 years ago
6

When determining the melting point of a substance, should one use a large or a small sample in the capillary? Should the sample

be firmly or loosely packed? Explain?
Chemistry
1 answer:
REY [17]3 years ago
4 0

The glass capillary having one end closed and one open end is generally used for the determination of melting point of the sample. From the open end, the sample is put into the capillary, the sample must be firmly packed as the melting point is an intrinsic property that means it is independent of sample size. So, in order to determine the melting point of the sample small sample in the capillary is sufficient to measure the melting point of the sample. To obtain the more consistent value of melting point one must pack the sample firmly in the capillary.

Hence, when determining the melting point of a substance, one should use a small sample in the capillary and the sample should be firmly packed.

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In a chemical change energy can be ___?
qaws [65]

Answer:

chemical change energy can neither be created nor destroyed.

Explanation:

the rule of law of conversion states that energy can only be converted from said energy to another. it cannot be created and destroyed.

7 0
3 years ago
Read 2 more answers
Will a precipitate (ppt) form when 20.0 mL of 1.1 × 10 –3 M Ba(NO 3) 2 are added to 80.0 mL of 8.4 × 10 –4 M Na 2CO 3?
erastova [34]

Answer:

A precipitate will form, BaCO₃

Explanation:

When Ba²⁺ and CO₃²⁻ ions are in an aqueous media, BaCO₃(s), a precipitate, is produced following its Ksp expression:

Ksp = 5.1x10⁻⁹ = [Ba²⁺] [CO₃²⁻]

<em>Where the concentrations of the ions are the concentrations in equilibrium</em>

<em />

For actual concentrations of a solution, you can define Q, <em>reaction quotient, </em>as:

Q = [Ba²⁺] [CO₃²⁻]

<em>If Q > Ksp, the ions will react producing BaCO₃, if not, no precipitate will form</em>.

Actual concentrations of Ba²⁺ and CO₃²⁻ are:

[Ba²⁺] = [Ba(NO₃)₂] = 1.1x10⁻³ × (20.0mL / 100.0mL) = 2.2x10⁻⁴M

[CO₃²⁻] = [Na₂CO₃] = 8.4x10⁻⁴ × (80.0mL / 100.0mL) = 6.72x10⁻⁴M

<em>100.0mL is the volume of the mixture of the solutions</em>

<em />

Replacing in Q expression:

Q = [Ba²⁺] [CO₃²⁻]

Q = [2.2x10⁻⁴M] [6.72x10⁻⁴M]

Q = 1.5x10⁻⁷

As Q > Ksp

<h3>A precipitate will form, BaCO₃</h3>

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8 0
3 years ago
1. Consider the chemical equation. If there are 40 mol of NBr3 and 48 mol
marusya05 [52]

Answer:

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Explanation:

We'll begin by writing the balanced equation for the reaction. This is given below:

2NBr3 + 3NaOH —> N2 + 3NaBr + 3HOBr

The limiting reactant can be obtained as follow:

From the balanced equation above,

2 moles of NBr3 reacted with 3 moles of NaOH.

Therefore, 40 moles of NBr3 will react with = (40 x 3)/2 = 60 moles of NaOH.

From the above calculations,

A higher amount of NaOH i.e 60 moles than what was given from the question i.e 48 moles is required to react completely with 40 moles of NBr3. Therefore, NaOH is the limiting reactant and NBr3 is the excess reactant.

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