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Rus_ich [418]
3 years ago
7

What is the specific latent heat of fusion for a substance that takes 550 KJ to melt 14 KG at 262 K?

Chemistry
1 answer:
Arte-miy333 [17]3 years ago
5 0

Answer:- The right choice is A. 3.9*10^4\frac{J}{kg} .

Solution:- Latent heat of fusion means the heat required to melt the solid at constant temperature means there is no change in temperature only the solid changes to liquid. So, it is a solid to liquid phase change.

q=m*\Delta H_f_u_s

where q is the heat required to convert solid to liquid, m is the mass and \Delta H_f_u_s is the latent heat of fusion.

From given info, 550 kJ that is 550000 J of heat is required to melt 14 kg of solid at 262K temperature. Let's rearrange the equation for latent heat of fusion and plug in the values in it.

\Delta H_f_u_s=\frac{q}{m}

\Delta H_f_u_s=\frac{550000J}{14kg}

\Delta H_f_u_s = 39286\frac{J}{kg}

If we round this value to two sig figs and write in scientific notations then it becomes 3.9*10^4\frac{J}{kg} .

So, the right choice is A. 3.9*10^4\frac{J}{kg} .

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10.0 g of gaseous ammonia and 6.50 g of oxygen gas are introduced into a previously evacuated 5.50 L vessel. If the ammonia and
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Answer:

The density is 3g/L

Explanation:

The reaction that occurs in the vessel is:

4 NH₃ + 5 O₂ → 4 NO + 6 H₂O

10,0g of NH₃ are:

10,0g * \frac{1mol}{17,031g} = 0,587 moles

6,50 g of O₂ are:

6,50g * \frac{1mol}{32g} = 0,203 moles

For a complete reaction of O₂ there are necessaries:

0,203 mol * \frac{4molNH_{3}}{5molO_{2}}= 0,163 moles of NH_{3}

O₂ is limiting reactant. The excess moles of NH₃ are:

0,587 - 0,163 = <em>0,424 moles of NH₃</em>

These moles are:

0,424mol * \frac{17,031g}{1mol} = <em>7,22g of NH₃</em>

Knowing O₂ is limiting reactant, mass of NO and H₂O are:

0,203molO_{2}*\frac{4molNO}{5molO_{2}}*\frac{30,01g}{1molNO} = <em>4,87g of NO</em>

0,203molO_{2}*\frac{6molH_{2}O}{5molO_{2}}*\frac{18,02g}{1molH_{2}O} = <em>4,39g of H₂O</em>

The total mass is: 7,22g + 4,87g + 4,39g = 16,48g ≡ <em>16,5g </em>

<em>-</em><em>The same mass add in the first. By matter conservation law-</em>

As vessel volume is 5,50L, density is:

16,5g/5,50L = <em>3g/L</em>

I hope it helps!

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