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3 years ago

The liquid dispensed from a burette is called ___________.

Chemistry

2 answers:

11Alexandr11 [23.1K]3 years ago

8 0

The liquid that is been dispensed during titration as regards this question is Titrant.

Titration can be regarded as common laboratory method that is been carried out during quantitative chemical analysis.

This analysis helps to know the concentration of an identified analyte.

Burette can be regarded as laboratory apparatus.

It is used in the in measurements of variable amounts of liquid ,this apparatus helps in dispensation of liquid, especially when performing titration.

The specifications is been done base on their volume, or resolution.

The liquid that comes out of this apparatus is regarded as Titrant, and this is gotten during titration process, which is usually carried out during volumetric analysis.

Therefore, burrete is used in volumetric analysis.

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zysi [14]3 years ago

8 0

Answer:

Titrant.

Explanation:

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What transition energy corresponds to an absorption line at 527 nm?

Ede4ka [16]

Answer:

E = 3.77×10⁻¹⁹ J

Explanation:

Given data:

Wavelength of absorption line = 527 nm (527×10⁻⁹m)

Energy of absorption line = ?

Solution:

Formula:

E = hc/λ

h = planck's constant = 6.63×10⁻³⁴ Js

c = speed of wave = 3×10⁸ m/s

by putting values,

E = 6.63×10⁻³⁴ Js × 3×10⁸ m/s / 527×10⁻⁹m

E = 19.89×10⁻²⁶ Jm /527×10⁻⁹m

E = 0.0377×10⁻¹⁷ J

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The atmosphere's variable gases that most influence the greenhouse effect are _____. nitrogen and oxygen argon and carbon dioxid

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The correct answer is carbon dioxide and water vapor

These negative gasses get modified and then remain in the atmosphere without the possibility of leaving, which is why the greenhouse effect occurs.

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Which statement is true regarding a titration? It measures the amount of heat given off in combustion reactions. It determines h

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C) It determines the concentration of an unknown substance in neutralization reactions.

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The combustion of palmitic acid is represented by the chemical equation: C16H32O2(s) + 23O2(g) → 16 CO2(g) + 16 H2O(l) The magni

solniwko [45]

Answer:

The correct option is C

Explanation:

From the question we are told that

The reaction is

$C_{16}H_{32}O_2(g) + 23O_2(g) \to 16 CO_2(g) + 16 H_2O(l)$

Generally $\Delta H = \Delta U + \Delta N* RT$

Here $\Delta H$ is the change in enthalpy

$\Delta U$ is the change in the internal energy

$\Delta N$ is the difference between that number of moles of product and the number of moles of reactant

Looking at the reaction we can discover that the elements that was consumed and the element that was formed is $O_2$ and $CO_2$ and this are both gases so the change would occur in the number of moles

$\Delta H = \Delta U + [16 -23]* RT$

$\Delta H = \Delta U -7RT$

The negative sign in the equation tell us that the enthalpy $\Delta_r H$ would be less than the Internal energy $\Delta_r U$

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3 years ago

A chemistry graduate student is given 125.mL of a 1.00M benzoic acid HC6H5CO2 solution. Benzoic acid is a weak acid with =Ka×6.3

lubasha [3.4K]

Answer:

53.9 g

Explanation:

When talking about buffers is very common the problem involves the use of the Henderson Hasselbach formula:

pH = pKa + log [A⁻]/[HA]

where [A⁻] is the concentration of the conjugate base of the weak acid HA, and [HA] is the concentration of the weak acid.

We can calculate pKₐ from the given kₐ ( pKₐ = - log Kₐ ), and from there obtain the ratio [A⁻]/HA].

Since we know the concentration of HC6H5CO2 and the volume of solution, the moles and mass of KC6H5CO2 can be determined.

So,

4.63 = - log ( 6.3 x 10⁻⁵ ) + log [A⁻]/[HA] = - (-4.20 ) + log [A⁻]/[HA]

⇒ log [A⁻]/[HA] = 4.63 - 4.20 = log [A⁻]/[HA]

0.43 = log [A⁻]/[HA]

taking antilogs to both sides of this equation:

10^0.43 = [A⁻]/[HA] = 2.69

[A⁻]/ 1.00 M = 2.69 ⇒ [A⁻] = 2.69 M

Molarity is moles per liter of solution, so we can calculate how many moles of C6H5CO2⁻ the student needs to dissolve in 125. mL ( 0.125 L ) of a 2.69 M solution:

( 2.69 mol C6H5CO2⁻ / 1L ) x 0.125 L = 0.34 mol C6H5CO2⁻

The mass will be obtained by multiplying 0.34 mol times molecular weight for KC6H5CO2 ( 160.21 g/mol ):

0.34 mol x 160.21 g/mol = 53.9 g

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3 years ago