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3 years ago

I NEED HELP PLEASE!!

Chemistry

1 answer:

VikaD [51]3 years ago

5 0

Answer:

The amount of heat absorbed is <u>5.183889 kJ</u> .

Explanation:

In conversion of water to ice it rejects some heat while in conversion of ice to water it absorbs heat which is called latent heat which is given as 6.02 kJ/mol.

The amount of ice given is 15.5 g.

Converting it to moles as the latent heat is given in per moles:

$\frac{given\\weight \\ (in\\grams)}{molecular\\weight\\(in\\grams)}$

Molecular mass of Hydrogen (H) and Oxygen (O) is 1 u and 16 u respectively.

Molecular mass of water is 18 g ( $H_{2} O$ ⇒ 2*1+16=18 ).

mole = 15.5/18 ≈ 0.8611 moles

Therefore the amount of heat absorbed by 15.5 g of ice ( 0.8611 moles) = <em>Latent heat * moles </em>

Heat absorbed = 6.02*0.8611

= 6.02*(15.5/18)

≈ 5.183889 kJ

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Identify the covalent compounds based on the names of the compounds. barium nitrate dinitrogen tetroxide boron trifluoride ammon

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Covalent compounds are composed of atoms that are linked via covalent bonds i.e. bonds formed by mutual sharing of electrons. This is in complete contrast to ionic compounds which are held together by ionic bonds, i.e. bonds formed by complete transfer of electrons from one atom to the other.

In the given examples we have:

Barium nitrate: Ba(NO3)2 - Ionic

Dinitrogen tetroxide: N2O4- Covalent

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Carbon tetrachloride: CCl4- Covalent

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5 0

3 years ago

Read 2 more answers

A 237g sample of molybdnum metal is heated to 100.1 0C and then dropped into an insulated cup containing 244 g of water at 10.0

miv72 [106K]

Answer:

The specific heat of molybdenum is 0.254 joules per gram-Celsius.

Explanation:

We consider the system formed by the molybdenum metal and water as our system, a control mass inside an insulated cup, that is, a container that avoids any energy and mass interactions between system and surroundings.

From statement we notice that metal is cooled down whereas water is heated. According to the First Law of Thermodynamics, we know that:

$Q_{metal} - Q_{water} = 0$

$Q_{metal} = Q_{water}$

Where:

$Q_{water}$ - Heat received by water, measured in joules.

$Q_{metal}$ - Heat released by metal, measured in joules.

Now we expand this identity by definition of sensible heat:

$m_{metal}\cdot c_{metal}\cdot (T_{m,o}-T) = m_{water}\cdot c_{water}\cdot (T-T_{w,o})$

The specific heat of the metal is cleared within equation above:

$c_{metal} = \frac{m_{water}\cdot c_{water}\cdot (T-T_{w,o})}{m_{metal}\cdot (T_{m,o}-T)}$

If we know that $m_{water} = 0.237\,kg$ , $m_{metal} = 0.244\,kg$ , $c_{water} = 4186\,\frac{J}{kg\cdot ^{\circ}C}$ , $T_{w,o} = 10\,^{\circ}C$ , $T_{m,o} = 100.10\,^{\circ}C$ and $T = 15.30\,^{\circ}C$ , the specific heat of molybdenum is:

$c_{metal} = \frac{(0.237\,kg)\cdot \left(4186\,\frac{J}{kg\cdot ^{\circ}C} \right)\cdot (15.30\,^{\circ}C-10\,^{\circ}C)}{(0.244\,kg)\cdot (100.10\,^{\circ}C-15.30\,^{\circ}C)}$