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Mamont248 [21]
2 years ago
11

18.74 g of P4 is reacted with 9.6 g of Cl2 balanced is P4 (s) + 6 Cl2 (g) → 4 PCl3 (l) how many PlC3 can I make?

Chemistry
1 answer:
Yanka [14]2 years ago
3 0

P4(s) + 6Cl2(g) → 4PCl3(l)

n P4 = m P4 / Mr P4

n P4 = 18.74 / 123.88

n P4 = 0.151 mol

n Cl2 = m Cl2 / Mr Cl2

n Cl2 = 9.6 / 70.9

n Cl2 = 0.135 mol

Cl2 act as <u>limiting reactant</u>

n PCl3 = (coef. PCl3)/(coef. Cl2) • n Cl2

n PCl3 = (4/6) • 0.135

n PCl3 = 0.09 mol

m PCl3 = n PCl3 • Mr PCl3

m PCl3 = 0.09 • 137.33

m PCl3 = 12.36 gr

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m=2.575g

Explanation:

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In this case, since the density is defined as the ratio of the mass and volume:

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7 0
2 years ago
How much (Q) heat is needed to melt 35 g of iodine? Hf = 61.7 J/g.
aev [14]

Taking into account the definition of calorimetry and latent heat, a heat of 2159.5 J is needed to melt 35 g of iodine.

<h3>Calorimetry</h3>

Calorimetry is the measurement and calculation of the amounts of heat exchanged by a body or a system.

<h3>Latent heat</h3>

Latent heat is defined as the energy required by a quantity of substance to change state.

When this change consists of changing from a solid to a liquid phase, it is called heat of fusion and when the change occurs from a liquid to a gaseous state, it is called heat of vaporization.

The heat Q that is necessary to provide for a mass m of a certain substance to change phase is equal to

Q = m×L

where L is called the latent heat of the substance and depends on the type of phase change.

<h3>Heat needed to melt iodine</h3>

In this case, you know:

  • m= 35 g
  • L=61.7 \frac{J}{g}

Replacing in the definition of latent heat:

Q= 35 g× 61.7 \frac{J}{g}

Solving:

<u><em>Q=2159.5 J</em></u>

Finally, a heat of 2159.5 J is needed to melt 35 g of iodine.

Learn more about calorimetry:

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D iced tea

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irinina [24]
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  • P=8969.14/300
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7 0
2 years ago
Why do you think the transition metals are 10 columns wide?
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The transition metals are 10 columns wide because they can hold no more than 10 electrons. This is why their width correlates with the number of electrons that are included in these metals. Transition metals include most commonly used metals, such as iron, copper, silver, and gold.
4 0
3 years ago
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