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juin [17]
2 years ago
13

According to the following thermochemical equation, if 951.1 g of NO 2 is produced, how much heat is released at constant pressu

re? 2NO(g) + O2(g) 2NO 2(g); DH -114.4 kJ
Chemistry
1 answer:
nordsb [41]2 years ago
4 0

Answer:

The amount of released is 1,182 kJ.

Explanation:

When heat is released at constant pressure, this change in energy is known as enthalpy (ΔH°) of the reaction. Enthalpy is an extensive property, so it depends on the amount of reacting material. Let's take a look at the provided equation:

2 NO(g) + O₂ ⇄ 2 NO₂(g)   ΔH° = -114.4 kJ

Since this equation is balanced with 2 moles of NO₂(g), we can say that 114.4 kJ are released every 2 moles of NO₂(g) produced. By convention, when enthalpies are negative, it means that energy is <em>released</em> and the reaction is exothermic. Conversely, positive enthalpies mean energy is <em>absorbed</em> and the reaction is endothermic.

We can calculate the amount of energy released taking into account the previous relationship (-114.4 kJ/2 moles of NO₂(g)), the mass of NO₂(g) produced (951.1g) and its molar mass (46.00g/mol). The calculations would be:

951.1g.\frac{1molNO_{2} }{46.00g} .\frac{-114.4kJ}{2molesNO_{2} } =-1,182kJ

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How does cohesion affect the evaporation rate of water
wariber [46]

Answer:

Evaporation occurs because among the molecules near the surface of the liquid there are always some with enough heat energy to overcome the cohesion of their neighbors and escape. At higher temperatures the number of energetic molecules is greater, and evaporation is more rapid.

8 0
3 years ago
How many grams of sodium acetate are in solution in the third beaker?
Kipish [7]

Answer:

46g of sodium acetate.

Explanation:

The data is: <em>Precipitation from a supersaturated sodium acetate solution. The solution on the left was formed by dissolving 156g of the salt in 100 mL of water at 100°C and then slowly cooling it to 20°C. Because the solubility of sodium acetate in water at 20°C is 46g per 100mL of water, the solution is supersaturated. Addition of a sodium acetate crystal causes the excess solute to crystallize from solution.</em>

The third solution is the result of the equilibrium in the solution at 20°C. As the maximum quantity that water can dissolve of sodium acetate at this temperature is 46g per 100mL and the solution has 100mL <em>there are 46g of sodium acetate in solution. </em>The other sodium acetate precipitate because of decreasing of temperature.

I hope it helps!

6 0
3 years ago
Please help
sergey [27]

Answer:

The scientific method is an empirical method of acquiring knowledge that has characterized the development of science since at least the 17th century.

The basic steps of the scientific method are:

1) investigation

2) hypothesis

3) interpretation

4) conclusions

7 0
2 years ago
Rhodium crystallizes in a face-centered cubic unit cell. The radius of a rhodium atom is 135 pm. Determine the density of rhodiu
Deffense [45]

Answer:

Density of unit cell ( rhodium) = 12.279 g/cm³

Explanation:

Given that:

The radius (r) of a rhodium atom = 135 pm

The atomic mass of rhodium = 102.90 amu

For a face-centered cubic unit cell,

r = \dfrac{a}{2\sqrt{2}}

where;

a = edge length.

Making "a" the subject of the formula:

a = 2 \sqrt{2} \times r

a = 2 \times 1.414 \times 135 \ pm

a = 381.8 pm

to cm, we get:

a = 381.8 × 10⁻¹⁰ cm

However, recall that:

density \ of \ unit \ cell = \dfrac{mass \ of \ unit \ cell}{volume \ of \unit \ cell}

where;

mass of unit cell = mass of atom × numbers of atoms per unit cell

Also;

mass\  of\ atom =\dfrac{ atomic \ mass}{Avogadro  \  number}

mass\  of\ atom =\dfrac{ 102.9}{6.023 \times 10^{23}}

Recall also that number of atoms in a unit cell for a  face-centered cubic = 4

So;

mass \ of \ unit \ cell= \dfrac{102.90}{6.023 \times 10^{23}}\times 4

mass of unit cell = 6.83380375 × 10⁻²² g

Density  \ of  \ unit \  cell = \dfrac{6.83380375 \times 10^{-22}}{(381.8\times 10^{-10})^3}

Density of unit cell ( rhodium) = 12.279 g/cm³

3 0
2 years ago
Calculate the pH of a solution created by placing 2.0 grams of yttrium hydroxide, Y(OH)3, in 2.0 L of H2O. Ksp for Y(OH)3 is 6.0
oee [108]

Answer:

pH = 8.314

Explanation:

  • Y(OH)3(s) ↔ Y+  +  3OH-

equil:   S               S         3S

∴ Ksp = [ Y+ ] * [ OH- ]³ = 6.0 E-24

⇒ 6.0 E-24 = ( S )*( 3S )³

⇒ 6.0 E-24 = 27S∧4

⇒ 2.22 E-25 = S∧4

⇒ ( 2.22 E-25 )∧(1/4) = S

⇒ S = 6.866 E-7 M

⇒ [ OH- ] = 3*S =2.06 E-6 M

⇒ pOH = - Log [ OH- ]

⇒ pOH = - Log ( 2.06 E-6 )

⇒ pOH = 5.686

∴ pH = 14 - pOH

⇒ pH = 8.314

8 0
3 years ago
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