How do u name this aromatic organic?

What happens to the particles in matter when hear energy is added to them

Ipatiy [6.2K]

There are two possible situations.
1) If a phase change is not occurring, then the heat added contributes to increased translational energy of the particles. What that means is the particles move/vibrate faster.
2) If a phase change is occurring, then the heat added contributes to the breaking of bonds or intermolecular forces (depending on the chemical nature of the matter you're dealing with).

4 0

4 years ago

250.0 mL of 0.250 M calcium chloride is mixed with 440.0 mL of 0.155 M sodium hydroxide and a precipitation reaction occurs. Wha

Vladimir [108]

Answer:

Solid: 2.52 g

Concentrations: [CaCl₂] = 0.041 M, [NaCl] = 0.100 M

Explanation:

When calcium chloride (CaCl₂) reacts with sodium hydroxide (NaOH), a double replacement reaction occurs, forming NaCl and Ca(OH)₂. NaCl is a soluble salt, and Ca(OH)₂ is a little soluble base, thus, Ca(OH)₂ will be the precipiate.

The balanced reaction equation is:

CaCl₂(aq) + 2NaOH(aq) → 2NaCl(aq) + Ca(OH)₂(s)

The number of moles of the reactants mixed are their volume multiplied by their concentration:

nCaCl₂ = 0.250 L * 0.25 mol/L = 0.0625 mol

nNaOH = 0.440 L*0.155 mol/L = 0.0682 mol

One of the reactants is limiting, and the other is in excess. Let's suppose that CaCl₂ is limiting, then, by the stoichiometry:

1 mol of CaCl₂ ------------- 2 moles of NaOH

0.0625 mol ------------ x

By a simple direct three rule:

x = 0.125 mol of NaOH

Because there's less NaOH than the value found, NaOH must be the limiting reactant and CaCl₂ is in excess. Thus, by the stoichiometry:

1 mol of CaCl₂ ------------- 2 moles of NaOH

x ------------- 0.0682 mol

By a simple direct three rule:

2x = 0.0682

x = 0.0341 mol of CaCl₂ reacts

The number of moles of CaCl₂ that remains is: 0.0625 - 0.0341 = 0.0284 mol. The final volume is 250.0 mL + 440.0 mL = 690. mL = 0.69 L

[CaCl₂] = 0.0284/0.69 = 0.041 M

For the solube product:

2 moles of NaOH ------------ 2 moles of NaCl

0.0682 mol ------------ x

x = 0.0682 mol of NaCl formed

[NaCl] = 0.0682/0.69 = 0.100 M

For the precipitate:

2 moles of NaOH ----------- 1 mol of Ca(OH)₂

0.0682 mol ---------- x

x = 0.0341 mol of Ca(OH)₂ formed

The molar of Ca(OH)₂ is 74.0 g/mol, and the mass is the number of moles multiplied by the molar mass:

mCa(OH)₂ = 0.0341*74 = 2.52 g

8 0

3 years ago

A student placed 19.0 g of glucose (c6h12o6) in a volumetric flask, added enough water to dissolve the glucose by swirling, then

EleoNora [17]

<span>moles glucose = 19 g / 180 g/mol= 0.105
M = 0.105 / 0.100 L = 1.05

moles in 20.0 mL = 1.05 M x 0.0200 L = 0.0216

New concentration = 0.0216 /0.500 L = 0.0432 M
moles in 100 mL = 0.100 L x 0.0432 = 0.00432

mass = 0.00432 x 180 g/mol= 0.778 g</span>

7 0

3 years ago

Kc for the reaction of hydrogen and iodine to produce hydrogen iodide, H2(g) I2(g) ⇌ 2HI(g) is 54.3 at 430°C. Determine the init

Jlenok [28]

Answer : The initial concentration of HI and concentration of $HI$ at equilibrium is, 0.27 M and 0.386 M respectively.