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WINSTONCH [101]
2 years ago
14

How would you measure the melting point of a solid that melts above 100.

Chemistry
1 answer:
meriva2 years ago
7 0

Answer:

The determination of melting point can be done using the capillary tube method. ... A small capillary tube is filled with a small amount of the chemical component. The capillary is then heated in a chemical bath along with the thermometer to observe the point of melting of the solid component.Electrically heated apparatus is used to measure the melting point of a solid that melts above 100 degrees Celsius. This apparatus can also be used for even higher melting points.

Explanation:

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All of the following conditions of STP are true except A. 101.3 kPa B. 273.15 K C. 22.4 L D. 3.81 kPa
kaheart [24]
  3.81   kpa  is  the  condition   which  is  not  true   at  STP

According  to  IUPAC  the  standard  temperature  and  pressure  that  is  STP  the  temperature  is   273.15  k  or  0   degrees  celsius .  and  the  absolute  temperature   of   101.325 Kpa   or  1  atm.  In  addition at STP   the  volume of  ideal  gas  is  22.4 
8 0
3 years ago
Write and balance molecular equations for the following reactions between aqueous solutions. You will need to decide on the form
musickatia [10]

Answer:

This is the balanced equation:

Pb(NO₃)₂ (aq) + 2NaI (aq) → 2NaNO₃ (aq)  +  PbI₂ (s) ↓    

Explanation:

This are the reactants:

PbNO₃

NaI

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4 0
3 years ago
Give six examples of complex compounds.
ra1l [238]

Answer:

Examples of complex compound include potassium ferrocyanide K4[Fe(CN)6] and potassium ferricyanide K3[Fe(CN)6]. Other examples include pentaamine chloro cobalt(III) chloride [Co(NH)5Cl]Cl2 and dichlorobis platinum(IV) nitrate [Pt(en)2Cl2](NO3)2.

3 0
3 years ago
How does hydrogen bonding affect water as it becomes colder and eventually freezes? (1 point)
Sergio039 [100]
The 2nd one I think but I need some points
5 0
2 years ago
Help and show work please
olya-2409 [2.1K]
ANWERS ~
We know that :
1 cal (th) = 4.184 J
1 J = 0.2390057361 cal (th) , so :

•55.2 j to cal > 13.193116635 cal
•110 call > 460.24 joule
•65 kj > divide the energy value by 4.184
= 15.535 kilocalories calorie (IT)
——————
Converting form C to F > (F-32)*5/9Understand it better if we have Fahrenheit just add to the equation mentioned to find Celsius.
+to find F to C> (9/5*C)+32

•425 Fahrenheit = (425- 32) × 5/9 =218.33333333 Celsius

•1935 C = 3515 F
———————————-
Converting Celsius to kelvin,We know that :
K = C + 273.15
C = K - 273.15
And from F to K=9/5(F+459.67)
And K to F =(9/5 *k)-459.67
•39.4 Celsius = 312.55 kelvin
•337 Fahrenheit = (337+ 459.67) × 5/9 =442.594 kelvin





4 0
3 years ago
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