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4 years ago

Identify the molecule with the highest boiling point. NH3 CO2 C2H2 C3H8

Chemistry

1 answer:

Black_prince [1.1K]4 years ago

4 0

Answer:

NH3

Explanation:

NH3 molecules make strong intermolecular H bondings and also because of the increase in the size of atomic radius.

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what happens to the yield of hydrogen if the reaction is carried out at a higher pressure but at the same temperature

ikadub [295]

Answer:

The yield of hydrogen decreases

Explanation:

The Haber process is exothermic. The equation of the Haber process is;

N2(g) + 3H2(g) --------> 2NH3(g) ΔH = -ve

Now, if we maintain a constant temperature, the number of volumes will affect the yield of products as follows;

The higher the pressure the lesser yield of hydrogen because the increase in pressure shifts the equilibrium position towards the right hand side.

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3 years ago

Solubility is a measure of how well a solute can dissolve in a solvent at a given temperature. true or false

Ray Of Light [21]

True
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3 years ago

Which element belongs to the p6-block of the periodic table?

-Dominant- [34]

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2 years ago

CaCO3(s) ∆→CaO(s) + CO2(g).

sveta [45]

Answer:

74.9%.

Explanation:

Relative atomic mass data from a modern periodic table:

Ca: 40.078;
C: 12.011;
O: 15.999.

What's the <em>theoretical</em> yield of this reaction?

In other words, what's the mass of the CO₂ that should come out of heating 40.1 grams of CaCO₃?

Molar mass of CaCO₃:

$M(\text{CaCO}_3) = 40.078 + 12.011 + 3 \times 15.999 = 100.086\;\text{g}\cdot\text{mol}^{-1}$ .

Number of moles of CaCO₃ available:

$\displaystyle n(\text{CaCO}_3) =\frac{m}{M} = \frac{40.1}{100.086} = 0.400655\;\text{mol}$ .

Look at the chemical equation. The coefficient in front of both CaCO₃ and CO₂ is one. Decomposing every mole of CaCO₃ should produce one mole of CO₂.

$n(\text{CO}2) = n(\text{CaCO}_3)= 0.400655\;\text{mol}$ .

Molar mass of CO₂:

$M(\text{CO}_2) = 12.011 + 2\times 15.999 = 44.009\;\text{g}\cdot\text{mol}^{-1}$ .

Mass of the 0.400655 moles of $\text{CO}_2$ expected for the 40.1 grams of CaCO₃:

$m(\text{CO}_2) = n\cdot M = 0.400655 \times 44.009 = 17.632\;\text{g}$ .

What's the <em>percentage</em> yield of this reaction?

$\displaystyle \textbf{Percentage}\text{ Yield} = \frac{\textbf{Actual}\text{ Yield}}{\textbf{Theoretical}\text{ Yield}}\times 100\%\\\phantom{\textbf{Percentage}\text{ Yield}} = \frac{13.2}{17.632}\times 100\%\\\phantom{\textbf{Percentage}\text{ Yield}} =74.9\%$ .

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3 years ago

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Nata [24]

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3 years ago