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4 years ago

What did the early atomic theory

Chemistry

2 answers:

Virty [35]4 years ago

7 0

Answer:B

Explanation:

The early theory says that atom Is the smallest indivisible particle. Which was later proven to contain electron neutron and proton

solmaris [256]4 years ago

7 0

Answer:

the right answer is " it started scientific through about the structure of the atom"

Explanation:

just took the quiz on apex

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Can formaldehyde molecules dissolve well in water?

Mice21 [21]

Answer:

Formaldehyde is highly soluble in water.

Explanation:

Formaldehyde is the simplest of the aldehydes (R−CHO) it is an organic compound which occurs naturally with the formula H−CHO. . The IUPAC name of formaldehyde is methanal.

Formaldehyde at room temperature is a colorless gas and the liquid is invisible or like white-water. Formaldehyde has an irritating and very pungent like odour. Formaldehyde is very soluble in water.

Now according to Henry's constant law (3 × 10-5 kPa·m3/mol) Formaldehyde will be very unlikely to volatilize from water.and hence, Formaldehyde is considered to be highly soluble in water.

3 0

4 years ago

Why can't we use chromatography paper to separate different colored pencils?

Sveta_85 [38]

Chromatography is a way of separating a mixture of chemicals which are in gas or liquid form. when the paper absorbs the liquid, the different colours from the colouring pencils will not be separated because they are insoluble solid pieces,also because they are solid the colours will not be able to creep up the paper which is why chromatography can not be used to separate different coloured pencils.

3 0

3 years ago

For most solids at room temperature, the specific heat is determined by oscillations of the atom cores in the lattice (each osci

Cloud [144]

Answer:

The specific heat of copper is $C= 392 J/kg\cdot ^o K$

Explanation:

From the question we are told that

The amount of energy contributed by each oscillating lattice site is $E =3 kT$

The atomic mass of copper is $M = 63.6 g/mol$

The atomic mass of aluminum is $m_a = 27.0g/mol$

The specific heat of aluminum is $c_a = 900 J/kg-K$

The objective of this solution is to obtain the specific heat of copper

Now specific heat can be defined as the heat required to raise the temperature of 1 kg of a substance by $1 ^o K$

The general equation for specific heat is

$C = \frac{dU}{dT}$

Where $dT$ is the change in temperature

$dU$ is the change in internal energy

The internal energy is mathematically evaluated as

$U = 3nk_BT$

Where $k_B$ is the Boltzmann constant with a value of $1.38*10^{-23} kg \cdot m^2 /s^2 \cdot ^o K$

T is the room temperature

n is the number of atoms in a substance

Generally number of atoms in mass of an element can be obtained using the mathematical operation

$n = \frac{m}{M} * N_A$

Where $N_A$ is the Avogadro's number with a constant value of $6.022*10^{23} / mol$

M is the atomic mass of the element

m actual mass of the element

So the number of atoms in 1 kg of copper is evaluated as

$m = 1 kg = 1 kg * \frac{10000 g}{1kg } = 1000g$

The number of atom is

$n = \frac{1000}{63.6} * (6.0*0^{23})$

$= 9.46*10^{24} \ atoms$

Now substituting the equation for internal energy into the equation for specific heat

$C = \frac{d}{dT} (3 n k_B T)$

$=3nk_B$

Substituting values

$C = 3 (9.46*10^{24} )(1.38 *10^{-23})$

$C= 392 J/kg\cdot ^o K$

5 0

3 years ago

Substance X is a compound containing 632mg of manganese and 368mg of oxygen. Substance X is shown

defon

The empirical formula : MnO₂.

<h3>Further explanation</h3>

Given

632mg of manganese(Mn) = 0.632 g

368mg of oxygen(O) = 0.368 g

M Mn = 55

M O = 16

Required

The empirical formula

Solution

You didn't include the pictures, but the steps for finding the empirical formula are generally the same

Find mol(mass : atomic mass)

Mn : 0.632 : 55 = 0.0115

O : 0.368 : 16 =0.023

Divide by the smallest mol(Mn=0.0115)

Mn : O =

$\tt \dfrac{0.0115}{0.0115}\div \dfrac{0.023}{0.0115}=1\div 2$

The empirical formula : MnO₂

8 0

3 years ago

A 35.0 mL sample of 0.150 M acetic acid (CH3COOH) is titrated with 0.150 M NaOH solution. Calculate the pH after the following v

marissa [1.9K]

Answer:

a) 2.79

b) 4.75

c) 6.59

d) 8.81

e) 11.03

f) 12.42

Explanation:

Step 1: Data given

Volume of acetic acid = 35.0 mL

Molarity of acetic acid = 0.150 M

Molarity of NaOH = 0.150 M

Step 2: pH after adding 0 mL of NaOH

HA + H2O ⇆ A- + H3O+

Initial concentration

[HA] = 0.150 M

[A-] = 0M

[H3O+] = 0M

Concentration at the equilibrium

[HA] = 0.150 - x M

[A-] = xM

[H3O+] = xM

Ka = [H3O+][A-] / [HA] = x²/(0.150 -x) = 1.76*10^-5

0.150 >>> x

We can write it like this

x²/(0.150) = 1.76*10^-5

x² =0.00000264

x = 0.00162

[H3O+] = 0.00162 M

pH = -log [H3O+] = -log (0.00162) = 2.79

Step 3: pH after adding 17.5 mL of NaOH

This is the point of half-neutralization.

pH = pKa = -log (1.76*10^-5- 4.75

Step 4: pH after adding 34.5 mL of NaOH

HA + OH- ⇆ A- + H2O

Moles acetic acid = molarity * volume

Moles acetic acid = 0.150 M * 0.035 L

Moles acetic acid = 0.00525 moles

Moles NaOH = 0.150 M *0.0345 L

Moles NaOH = 0.005175 moles

Initial moles

[HA] = 0.00525 moles

[OH-] = 0.005175 moles

[A-] = 0 moles

[H2O] = 0 moles

Moles at the equilibrium

[HA] = 0.00525 - 0.005175 = 0.000075 moles

[OH-] = 0.005175 - 0.005175 = 0 moles

[A-] = 0.005175 moles

pH = pKa + log([A-]/[HA]

pH = 4.75 +log(0.005175/0.000075)

pH = 4.75 + 1.84 = 6.59

Step 4: pH after adding 35.0 mL of NaOH

HA + OH- ⇆ A- + H2O

Moles acetic acid = molarity * volume

Moles acetic acid = 0.150 M * 0.035 L

Moles acetic acid = 0.00525 moles

Moles NaOH = 0.150 M *0.035 L

Moles NaOH = 0.00525 moles

Initial moles

[HA] = 0.00525 moles

[OH-] = 0.00525 moles

[A-] = 0 moles

[H2O] = 0 moles

Moles at the equilibrium

[HA] = 0.00525 - 0.00525 = 0 moles

[OH-] = 0.00525 - 0.00525 = 0 moles

[A-] = 0.00525 moles

We have a solution of 0.00525 mol of A - in 70.0 mL

[A-] = 0.00525 moles / 0.070 L = 0.075 M

A- + H20 ⇆ HA + OH-

Initial moles A- = 0.0750 moles

initial moles HA = 0 moles

Initial moles OH- = 0 moles

Moles A- at the equilibrium = 0.0750 - x

Moles HA at the equilibrium = x

Moles OH- at the equilibrium = x

Kb = Kw/Ka = 10^-14 / 1.76*10^-5

Kb = 5.68 *10^-10

Kb = [HA][OH-] / [A-] = x² / 0.0750 - x = 5.68*10^-10

Since 0.0750 >>> x we can write

5.68*10^-10 = x²/0.0750

x = 6.53 *10^-6

[OH-] = x = 6.53 * 10^-6 M

pOH = - log (6.53*10^-6) = 5.19

pH = 14 - pOH = 14 - 5.19 = 8.81

Step 5: pH after adding 35.5 mL of NaOH

We add an excess moles of NaOH

We have an excess of 0.5 mL NaOH

Moles excess = 0.0005 L *0.150 M = 0.000075 moles

[OH-] = 0.000075 moles / (0.035+0.0355L)

[OH-] = 1.06 * 10^-3 M

pOH = -log( 1.06 *10^-3)

pOH = 2.97

pH = 14 -2.97 = 11.03

Step 6: pH after adding 50.0 mL of NaOH

We have an excess of 15.0 mL NaOH

Moles excess = 0.015 L *0.150 M = 0.00225 moles

[OH-] = 0.00225 moles / (0.035+0.050L)

[OH-] = 0.0265 M

pOH = -log( 0.0265M)

pOH = 1.58

pH = 14 -1.58 = 12.42

4 0

3 years ago