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Andre45 [30]
2 years ago
14

What kind of weather does a cold front usually bring?

Chemistry
2 answers:
svlad2 [7]2 years ago
7 0

Answer:

Sunny and windy but mostly windy

aleksandr82 [10.1K]2 years ago
7 0
I believe the answer is C
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If 42.9 mL of rubbing alcohol is
lisov135 [29]

Answer:

19.8 %

Explanation:

V% = ( V of solute \ 100 ml  of solution ) ×100%  =

   the volume of the solute in 100ml solition = 20 ml

  42.9 ml of solute ----------215 ml solution

     ×                     -----------  100 ml solution

42.6 ×   100 ml  ÷ 215 ml

= 19.8 ml of solute

the V% =   (19.8 ÷ 100)  ×100% = 19.8%

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3 years ago
WILL GIVE BRAINLYIST HELPPPPP PLEASE Which is the second stage of mitosis?
stich3 [128]
Metaphase is the answer
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Sometimes the oxidation state of an analyte must be adjusted before it can be titrated.
alexdok [17]

Answer:

Explanation:

True

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Calculate the amount of heat released when 27.0 g H2O is cooled from a liquid at 314 K to a solid at 263 K. The melting point of
BlackZzzverrR [31]

Answer : The amount of heat released, 45.89 KJ

Solution :

Process involved in the calculation of heat released :

(1):H_2O(l)(314K)\rightarrow H_2O(l)(273K)\\\\(2):H_2O(l)(273K)\rightarrow H_2O(s)(273K)\\\\(3):H_2O(s)(273K)\rightarrow H_2O(s)(263K)

Now we have to calculate the amount of heat released.

Q=[m\times c_{p,l}\times (T_{final}-T_{initial})]+\Delta H_{fusion}+[m\times c_{p,s}\times (T_{final}-T_{initial})]

where,

Q = amount of heat released = ?

m = mass of water = 27 g

c_{p,l} = specific heat of liquid water = 4.184 J/gk

c_{p,s} = specific heat of solid water = 2.093 J/gk

\Delta H_{fusion} = enthalpy change for fusion = 40.7 KJ/mole = 40700 J/mole

conversion : 0^oC=273k

Now put all the given values in the above expression, we get

Q=[27g\times 4.184J/gK\times (314-273)k]+40700J+[27g\times 2.093J/gK\times (273-263)k]

Q=45896.798J=45.89KJ     (1 KJ = 1000 J)

Therefore, the amount of heat released, 45.89 KJ

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3 years ago
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What's a system that removes solid and liquid wastes
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The excretory system.
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