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4 years ago

when wax is heated, it turns into a liquid. how are the wax molecules affected by this change of state?

Chemistry

2 answers:

ZanzabumX [31]4 years ago

5 0

Explanation:

Wax is a solid, so it's molecules will be held closer to each other by strong intermolecular forces of attraction.

As a result, when cooled down then wax obtains a define shape and volume.

On the other hand, when wax is heated then its molecules gain kinetic energy due to which there occurs more number of collisions between its particles. As a result, its molecules will move away from each other.

This will lead to change in state of wax from solid to liquid.

Vladimir [108]4 years ago

4 0

Wax molecules have most of their weak van der waals interactions and other hydrophobic interactions decrease and thus most of them are broken between these lipid substances, allowing for the increased molecular movement of these substances. As there are fewer interactions between similar compounds.

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Identify the limiting reactant in the reaction of nitrogen and hydrogen to form NH3 if 5.23 g of N2 and 5.52 g of H2 are combine

Marrrta [24]

Answer:

1) The limiting reactant is N₂ because it is present with the lower no. of moles than H₂.

2) The amount (in grams) of excess reactant H₂ = 4.39 g.

Explanation:

Firstly, we should write the balanced equation of the reaction:

N₂ + 3H₂ → 2NH₃.

1) To determine the limiting reactant of the reaction:

From the stichiometry of the balanced equation, 1.0 mole of N₂ reacts with 3.0 moles of H₂ to produce 2.0 moles of NH₃.

This means that N₂ reacts with H₂ with a ratio of (1:3).

We need to calculate the no. of moles (n) of N₂ (5.23 g) and H₂ (5.52 g) using the relation: n = mass / molar mass.

The no. of moles of N₂ in (5.23 g) = mass / molar mass = (5.23 g) / (28.00 g/mol) = 0.1868 mol.

The no. of moles of H₂ (5.52 g) = mass / molar mass = (5.52 g) / (2.015 g/mol) = 2.74 mol.

From the stichiometry, N₂ reacts with H₂ with a ratio of (1:3).

The ratio of the reactants of N₂ (5.23 g, 0.1868 mol) to H₂ (5.52 g, 2.74 mol) is (1:14.67).

∴ The limiting reactant is N₂ because it is present with the lower no. of moles than H₂.

0.1868 mol of N₂ react completely with 0.5604 mol of H₂ and the remaining of H₂ is in excess.

2) To determine the amount (in grams) of excess reactant of the reaction:

As showed in the part 1, The limiting reactant is N₂ because it is present with the lower no. of moles than H₂.

Also, 0.1868 mol of N₂ react completely with 0.5604 mol of H₂ and the remaining of H₂ is in excess.

The no. of moles are in excess of H₂ = 2.74 mol - 0.5604 mol (reacted with N₂) = 2.1796 mol.

∴ The amount (in grams) of excess reactant H₂ = n (excess moles) x molar mass = (2.1796 mol)((2.015 g/mol) = 4.39 g.

4 0

3 years ago

What is invection or the definition

LUCKY_DIMON [66]

Answer:

here are some synonyms of invective are abuse, billingsgate, obloquy, and vituperation.

5 0

3 years ago

At 1000 K, a sample of pure NO2 gas decomposes. 2 NO2(g) equilibrium reaction arrow 2 NO(g) + O2(g) The equilibrium constant KP

kirill115 [55]

Answer: The pressure of NO and $NO_2$ in the mixture is 0.58 atm and 0.024 atm respectively.

Explanation:

We are given:

Equilibrium partial pressure of $O_2$ = 0.29 atm

For the given chemical equation:

$2NO_2(g)\rightleftharpoons 2NO(g)+O_2(g)$

Initial: a

At eqllm: a-2x 2x x

Calculating for the value of 'x'

$\Rightarrow x=0.29$

Equilibrium partial pressure of NO = 2x = 2(0.29) = 0.58 atm

Equilibrium partial pressure of $NO_2$ = a - 2x = a - 2(0.29) = a - 0.58

The expression of $K_p$ for above equation follows:

$K_p=\frac{p_{O_2}\times (p_{NO})^2}{(p_{NO_2})^2}$

We are given:

$K_p=158$

Putting values in above expression, we get:

$158=\frac{0.29\times (0.58)^2}{(a-0.58)^2}\\\\a=0.555,0.604$

Neglecting the value of a = 0.555 because it cannot be less than the equilibrium concentration.

So, $a=0.604$

Equilibrium partial pressure of $NO_2$ = (a - 0.58) = (0.604 - 0.58) = 0.024 atm

Hence, the pressure of NO and $NO_2$ in the mixture is 0.58 atm and 0.024 atm respectively.

4 0

3 years ago

How do geologists monitor volcanoes?

Radda [10]

Answer:

<h2>Scientists use a wide variety of techniques to monitor volcanoes, including seismographic detection of the earthquakes and tremor that almost always precede eruptions, precise measurements of ground deformation that often accompanies the rise of magma, changes in volcanic gas emissions, and changes in gravity.</h2>

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3 years ago

Using the following equation how many grams of pure oxygen do you need to comsume 635.2g of glucose: C6H12O6 + 6O2 → 6CO2 + 6H2O

aalyn [17]

Answer:

677.76 g of oxygen needed.

Explanation:

Given data:

Mass of glucose = 635.2 g

Mass of oxygen needed = ?

Solution:

Chemical equation:

C₆H₁₂O₆ + 6O₂ → 6CO₂ + 6H₂O + energy

Number of moles of glucose:

Number of moles = mass/ molar mass

Number of moles = 635.2 g / 180.156 g/mol

Number of moles = 3.53 mol

Now we will compare the moles of glucose with oxygen.

C₆H₁₂O₆ : O₂

1 : 6

3.53 : 6×3.53 = 21.18 mol

Mass of oxygen:

Mass = number of moles × molar mass

Mass = 21.18 g × 32 g/mol

Mass = 677.76 g

4 0

3 years ago