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attashe74 [19]
4 years ago
8

a beaker with water and the surrounding air are all at degrees Celsius. After ice cubes are placed in the water, heat is transfe

rred from?
Chemistry
2 answers:
Yuki888 [10]4 years ago
8 0

(3) the water to the ice cubes
  This is so because the ice cubes are colder compared to their surrounding objects.
They absorb the heat from the water, cooling it, therefore vanishing(melting).

Another way to put it would be

The answer is (3) the water to ice cubes.

As the ice cubes should be at a temperature of about 0 degree ( freezing point) , at the same time the temperature of water is 24 degree. Thus, heat is transferred from water to ice cubes.

AfilCa [17]4 years ago
4 0

Answer:

(3) the water to the ice cubes.

Explanation:

There is some info missing. I believe this is the original question:

<em>A beaker with water and the surrounding air are all at 24°C. After ice cubes are placed in the water, heat is transferred from:(1) the ice cubes to the air(2) the beaker to the air(3) the water to the ice cubes(4) the water to the beaker.</em>

<em />

Heat is transferred from bodies at higher temperatures to bodies at lower temperatures.

<em>Heat is transferred from:</em>

<em>(1) the ice cubes to the air.</em> NO. Heat will be transferred from air (24°C) to the ice cubes (0°C).

<em>(2) the beaker to the air.</em> NO. The beaker and the air are at the same temperature.

<em>(3) the water to the ice cubes.</em> YES. Water (24°C) is at a higher temperature than ice (0°C).

<em>(4) the water to the beaker.</em> NO. The water and the beaker are at the same temperature.

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Answer:

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Explanation:

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3 years ago
In which pair would both compounds have the same empirical formula? A. H2O and H2O2 B. BaSO4 and BaSO3 C. FeO and Fe2O3 D. C6H12
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4 years ago
Calculate the ph of a 0.60 m h2so3, solution that has the stepwise dissociation constants ka1 = 1.5 × 10-2 and 1.82 1.06 1.02 2.
Vsevolod [243]
Missing in your question Ka2 =6.3x10^-8
From this reaction:
 H2SO3 + H2O ↔ H3O+  + HSO3-
by using the ICE table :
                H2SO3     ↔    H3O     +    HSO3- 
intial         0.6                     0                  0
change     -X                      +X                +X
Equ         (0.6-X)                  X                   X

when Ka1 = [H3O+][HSO3-]/[H2SO3]
So by substitution:
1.5X10^-2 = (X*X) / (0.6-X) by solving this equation for X
∴ X = 0.088
∴[H2SO3] = 0.6 - 0.088 = 0.512
[HSO3-] = [H3O+] = 0.088

by using the ICE table 2:
                 HSO3-     ↔   H3O     +     SO3-
initial        0.088              0.088              0
change    -X                      +X                   +X
Equ         (0.088-X)          (0.088+X)          X

Ka2= [H3O+] [SO3-] / [HSO3-]
we can assume [HSO3-] =  0.088 as the value of Ka2 is very small
 6.3x10^-8 = (0.088+X)*X / 0.088
X^2 +0.088 X - 5.5x10^-9= 0 by solving this equation for X
∴X= 6.3x10^-8
∴[H3O+] = 0.088 + 6.3x10^-8
               = 0.088 m ( because X is so small)
∴PH= -㏒[H3O+]
       = -㏒ 0.088 = 1.06 
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