Juan and Tanisha wanted to build a track for their marbles. They wanted to see how fast their marbles would run down one

Snowcat [4.5K]

Answer:

Compress
Fixed
Melts
Melting Point
Freezing Point
High
Crystalline
Lattice
Unit cell
Amorphous solids

Explanation:

Solids tend to be dense and difficult to compress.

They do not flow or take the shape of their containers, like liquids do, because the particles in solids vibrate around fixed points.

When a solid is heated until its particles vibrate so rapidly that they are no longer held in fixed positions, the solid melts.

Melting point is the temperature at which a solid changes to a liquid. The melting and freezing point of a substance are at the same temperature.

In general, ionic solids tend to have relatively high melting points, while molecular solids tend to have relatively low melting points.

Most solids are crystalline

The particles are arranged in a pattern known as a crystal lattice

The smallest subunit of a crystal lattice is the unit cell

Some solids lack an ordered internal structure and are called amorphous solids.

7 0

3 years ago

Consider the equilibrium

vladimir1956 [14]

Answer:

$Kp^{1000K}=0.141\\Kp^{298.15K}=2.01x10^{-18}$

$\Delta _rG=1.01x10^5J/mol$

Explanation:

Hello,

In this case, the undergoing chemical reaction is:

$C_2H_6(g)\rightleftharpoons H_2(g)+C_2H_4(g)$

Thus, Kp for this reaction is computed based on the given molar fractions and the total pressure at equilibrium, as shown below:

$p_{C_2H_6}^{EQ}=2bar*0.592=1.184bar\\p_{C_2H_4}^{EQ}=2bar*0.204=0.408bar\\p_{H_2}^{EQ}=2bar*0.204=0.408bar$

$Kp=\frac{p_{C_2H_4}^{EQ}p_{H_2}^{EQ}}{p_{C_2H_6}^{EQ}}=\frac{(0.408)(0.408)}{1.184}=0.141$

Now, by using the Van't Hoff equation one computes the equilibrium constant at 298.15K assuming the enthalpy of reaction remains constant:

$Ln(Kp^{298.15K})=Ln(Kp^{1000K})-\frac{\Delta _rH}{R}*(\frac{1}{298.15K}-\frac{1}{1000K} )\\\\Ln(Kp^{298.15K})=Ln(0.141)-\frac{137000J/mol}{8.314J/mol*K} *(\frac{1}{298.15K}-\frac{1}{1000K} )\\\\Ln(Kp^{298.15K})=-40.749\\\\Kp^{298.15K}=exp(-40.749)=2.01x10^{-18}$

Finally, the Gibbs free energy for the reaction at 298.15K is:

$\Delta _rG=-RTln(Kp^{298.15K})=8.314J/mol*K*298.15K*ln(2.01x10^{-18})\\\Delta _rG=1.01x10^5J/mol$

Best regards.

3 0

4 years ago

Which determines the reactivity of an alkali metal?

adelina 88 [10]

Answer:

its ability to lose electron

3 0

3 years ago

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Glycolic acid, which is a monoprotic acid and a constituent in sugar cane, has a pKa of 3.9. A 25.0 mL solution of glycolic acid

Phoenix [80]

Answer:

pH = 8.0

Explanation:

First, we have to calculate the moles of NaOH.

$35.8 \times 10^{-3}L.\frac{0.020mol}{L} =7.2\times 10^{-4}mol$

Let's consider the balanced equation.

C₂H₄O₃ + NaOH ⇒ C₂H₃O₃Na + H₂O

The molar ratio C₂H₄O₃: NaOH: C₂H₃O₃Na is 1: 1: 1. So, when 7.2 × 10⁻⁴ moles of NaOH react completely with 7.2 × 10⁻⁴ moles of C₂H₄O₃ they form 7.2 × 10⁻⁴ moles of C₂H₃O₃Na.

The concentration of C₂H₃O₃Na is:

$\frac{7.2\times 10^{-4}mol}{60.8 \times 10^{-3}L} =0.012M$

C₂H₃O₃Na dissociates according to the following equation:

C₂H₃O₃Na(aq) ⇒ C₂H₃O₃⁻(aq) + Na⁺(aq)

C₂H₃O₃⁻ comes from a weak acid so it undergoes basic hydrolisis.

C₂H₃O₃⁻ + H₂O ⇄ C₂H₄O₃ + OH⁻

If we know that pKa for C₂H₄O₃ is 3.9, we can calculate pKb for C₂H₃O₃⁻ using the following expression:

pKa + pKb = 14

pKb = 14 -3.9 = 10.1

10.1 = -log Kb

Kb = 7.9 × 10⁻¹¹

We can calculate [OH⁻] using the following expression:

[OH⁻] = √(Kb.Cb) where Cb is the initial concentration of the base

[OH⁻] = √(7.9 × 10⁻¹¹ × 0.012M) = 9.7 × 10⁻⁷ M

Now, we can calculate pOH and pH.

pOH = -log [OH⁻] = -log (9.7 × 10⁻⁷) = 6.0

pH + pOH = 14

pH = 14 - pOH = 14 - 6.0 = 8.0

7 0

3 years ago

How many atoms of nitrogen are in the chemical formula Ni(NO3)2?

katen-ka-za [31]

Answer:

2 atoms of nitrogen are present.

4 0

3 years ago

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