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3 years ago

Energy flows from colder objects to warmer objects. is it true or false?

Chemistry

2 answers:

Roman55 [17]3 years ago

8 0

Answer:

False

Explanation:

Heat energy will always be transferred from the hotter to the cooler object. The objects will exchange thermal energy, until thermal equilibrium is reached, i.e. until their temperatures are equal. We say that heat flows from the hotter to the cooler object.

mixas84 [53]3 years ago

7 0

Answer:

true. j

Explanation:

I think its true

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Answer:

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3 years ago

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Lena [83]

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3 years ago

What is the difference between ethanol and methanol.

KiRa [710]

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2 years ago

The oxygen atom in a water molecule A is electrically neutral. B carries a negative electrical charge. C carries a positive elec

Klio2033 [76]

Answer: The oxygen atom in a water molecule carries a negative electrical charge.

Explanation:

A polar covalent bond is defined as the bond which is formed when there is a difference of electronegativities between the atoms.

Hydrogen bonding (H-bonding) is an intermolecular force having partial ionic-covalent character. H-bonding takes place between a hydrogen atom (attached with an electronegative atom e.g. O, N and F) and an electronegative atom (O,N and F).

In , $H_2O$ , O is a highly electronegative atom attached to a H atom through a covalent bond. Therefore O atom gets partial negative charge and H atoms get partial positive charge.

7 0

3 years ago

Acetic acid, CH3CO2H, reacts with ethanol, C2H5OH, to form water and ethyl acetate, CH3CO2C2H5. The equilibrium constant for thi

Irina-Kira [14]

Explanation:

The initial concentrations for a mixture :

Acetic acid at equilibrium = 0.15 M

Ethanol at equilibrium = 0.15 M

Ethyl acetate at equilibrium = 0.40 M

Water at equilibrium = 0.40 M

$CH3COOH + C_2H_5OH\rightleftharpoons CH_3CO_2C_2H_5+H_2O$

Initially:

0.15 M 0.15 M 0.40 M 0.40 M

At equilibrium

(0.15-x)M (0.15-x) M (0.40+x) M (0.40+x) M

The equilibrium constant is given by expression

$K_c=\frac{[CH_3CO_2C_2H_5][H_2O]}{[CH_3COOH][C_2H_5OH]}$

$4.0=\frac{(0.40-x)\times (0.40-x)}{(0.15+x)\times (0.15+x)}$

Solving for x:

x = 0.0333

The equilibrium concentrations for a mixture :

Acetic acid at equilibrium = (0.15-x)M = (0.15-0.033) M = 0.117 M

Ethanol at equilibrium = (0.15-x)M = (0.15-0.033) M = 0.117 M

Ethyl acetate at equilibrium = (0.40+x)M = (0.40+0.033) M = 0.433 M

Water at equilibrium = (0.40+x)M = (0.40+0.033) M = 0.433 M

4 0

4 years ago