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3 years ago

Use the periodic table to identify each of the elements below as an alkali metal

Chemistry

1 answer:

Neporo4naja [7]3 years ago

3 0

$\huge{ \mathfrak{\underline{question : }}}$

Use the periodic table to identify each of the elements below as an alkali metal or an alkaline earth metal.

Rb = Rubidium

$\huge{ \mathfrak{\underline{Answer : }}}$

Rubidium (Rb) is an Alkali metal.

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2 years ago

What is the change in enthalpy when 180 g of water vapor condenses at 100°C? (AH, = 40.67 kJ/mol)

Snowcat [4.5K]

Answer:

$Q=-407kJ$

Explanation:

Hello there!

In this case, considering that the heat has two forms, sensible (variable temperature) and latent (constant temperature), we can notice that phase changes account for latent heat as the temperature remains the same. In such a way, given the enthalpy of vaporization of water, 40.67 kJ/mol, the enthalpy of condensation (reverse process) is the negative value, -40.67 kJ/mol; therefore, the associated latent heat would be:

$Q=180g*\frac{1mol}{18.02g} *-40.67\frac{kJ}{mol} \\\\Q=-407kJ$

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3 years ago

A chemistry graduate student is given of a pyridine solution. Pyridine is a weak base with . What mass of should the student dis

Molodets [167]

Answer:

34.1g of C₅H₅NHCl the student need to dissolve to the solution.

Explanation:

Full question is:

A chemistry graduate student is given 500.mL of a 0.20M pyridine C5H5N solution. Pyridine is a weak base with =Kb×1.7x10−9 . What mass of C5H5NHCl should the student dissolve in the C5H5N solution to turn it into a buffer with pH =4.76 ?

Using H-H expression for weak bases, it is possible to find pH of a buffer thus:

pOH = pKb + log [BH⁺] / [B]

<em>Where pKb is -log Kb = 8.77, </em>[BH⁺] concentration of C₅H₅NHCl and [B] concentration of C₅H₅N (It is possible to take the moles of both compounds and not its concentration.

As pH the student wants is 4.76, pOH is:

pOH = 14 - pH = 14 - 4.76 = 9.24

Replacing:

9.24 = 8.77 + log [C₅H₅NHCl] / [C₅H₅N]

Moles of C₅H₅N are:

0.500L × (0.20mol / L) = 0.10mol C₅H₅N

Replacing again:

9.24 = 8.77 + log [C₅H₅NHCl] / [0.10mol]

2.9512 = [C₅H₅NHCl] / [0.10mol]

0.29512 moles = [C₅H₅NHCl].

As molar mass of C₅H₅NHCl is 115.56g/mol, mass of 0.29512 moles are:

0.29512 moles C₅H₅NHCl × (115.56g / mol) =

<h3>34.1g of C₅H₅NHCl the student need to dissolve to the solution.</h3>

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