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2 years ago

Heat the oxygen to about 150 K. What state of matter is this?

Chemistry

1 answer:

adelina 88 [10]2 years ago

3 0

Answer:

Gas

Explanation:

It is easier if you convert the kelvin temperature into Celsius degrees:

ºC = T - 273.15 = 150 - 273.15 = -123.15ºC

Now, you know that that is a very cold temperature. Thus, may be the oxygen is not gas any more but it changed to liquid . . . or solid?

You must search for the boiling point and melting (freezing) point of oxygen in tables or the internet. At standard pressure (about 1 atm) they are:

Melting point: −218.79 °C,

Boiling point: −182.962 °C

That means that:

below -218.79ºC oxygen is solid (not our case).
between -218.79ºC and -182.962ºC oxygen is liquid (not our case)
over -182.962ºC oxygen is a gas. This is our case, because -123.15ºC is a higher temperature than -182.962ºC.

Hence, the state of matter of oxygen at 150K, and standard pressure, is gas.

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Convert 5.28 x 1019 molecules of C6H1206 to grams.

nlexa [21]

Answer:

$m=0.0158g$

Explanation:

Hello there!

In this case, it is possible to comprehend these mass-particles problems by means of the concept of mole, molar mass and the Avogadro's number because one mole of any substance has 6.022x10²³ particles and have a mass equal to the molar mass.

In such a way, for C₆H₁₂O₆, whose molar mass is about 180.16 g/mol, the referred mass would be:

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3 years ago

A scientist, Dr. Sigma, states that a chemical taken internally once a day will prevent the common cold. However, no other scien

pshichka [43]

Answer:

not valid

Explanation:

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After submitting an instrument for comparison to the consultation and expert opinion, it must meet two quality criteria: validity and reliability. The validity of content is often established based on two situations, one that concerns the design of a test and the other, the validation of an instrument subject to translation and standardization procedures to adapt it to different cultural meanings. It is here that the task of the expert becomes a fundamental task to eliminate irrelevant aspects, incorporate those that are essential and / or modify those that require it.

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3 years ago

Calculate the standard enthalpy of formation of NOCl(g) at 25 ºC, knowing that the standard enthalpy of formation of NO(g) at th

stepan [7]

Answer:

The standard enthalpy of formation of NOCl(g) at 25 ºC is 105 kJ/mol

Explanation:

The ∆H (heat of reaction) of the combustion reaction is the heat that accompanies the entire reaction. For its calculation you must make the total sum of all the heats of the products and of the reagents affected by their stoichiometric coefficient (number of molecules of each compound that participates in the reaction) and finally subtract them:

Enthalpy of the reaction= ΔH = ∑Hproducts - ∑Hreactants

In this case, you have: 2 NOCl(g) → 2 NO(g) + Cl₂(g)

So, ΔH= $2*H_{NO} +H_{Cl_{2} }-2*H_{NOCl}$

Knowing:

ΔH= 75.5 kJ/mol
$H_{NO}$ = 90.25 kJ/mol
$H_{Cl_{2} }$ = 0 (For the formation of one mole of a pure element the heat of formation is 0, in this caseyou have as a pure compound the chlorine Cl₂)
$H_{NOCl}$ =?