Plzzz help will do anything ;) How many molecules are there in 2.25 moles of H2CO3

You can obtain a rough estimate of the size of a molecule with the following simple experiment: Let a droplet of oil spread out

mina [271]

Answer:

The diameter of the oil molecule is $4.4674\times 10^{-8} cm$ .

Explanation:

Mass of the oil drop = $m=9.00\times 10^{-7} kg$

Density of the oil drop = $d=918 kg/m^3$

Volume of the oil drop: v

$d=\frac{m}{v}$

$v=\frac{m}{d}=\frac{9.00\times 10^{-7} kg}{918 kg/m^3}$

Thickness of the oil drop is 1 molecule thick.So, let the thickness of the drop or diameter of the molecule be x.

Radius of the oil drop on the water surface,r = 41.8 cm = 0.418 m

1 cm = 0.01 m

Surface of the sphere is given as: a = $4\pi r^2$

$a=4\times 3.14\times (0.418 m)^2=2.1945 m^2$

Volume of the oil drop = v = Area × thickness

$\frac{9.00\times 10^{-7} kg}{918 kg/m^3}=2.1945 m^2\times x$

$x= 4.4674\times 10^{-10} m= 4.4674\times 10^{-8} cm$

The thickness of the oil drop is $4.4674\times 10^{-8} cm$ and so is the diameter of the molecule.

6 0

3 years ago

A buffer solution contains 0.345 M acetic acid and 0.377 M sodium acetate . If 0.0613 moles of potassium hydroxide are added to

melamori03 [73]

Answer:

pH = 5.54

Explanation:

The pH of a buffer solution is given by the Henderson-Hasselbach (H-H) equation:

pH = pKa + log $\frac{[CH_3COO^-]}{[CH_3COOH]}$

For acetic acid, pKa = 4.75.

We calculate the original number of moles for acetic acid and acetate, using the given concentrations and volume:

CH₃COO⁻ ⇒ 0.377 M * 0.250 L = 0.0942 mol CH₃COO⁻

CH₃COOH ⇒ 0.345 M * 0.250 L = 0.0862 mol CH₃COOH

The number of CH₃COO⁻ moles will increase with the added moles of KOH while the number of CH₃COOH moles will decrease by the same amount.

Now we use the H-H equation to calculate the new pH, by using the new concentrations:

pH = 4.75 + log $\frac{(0.0942+0.0613)mol/0.250L}{(0.0862-0.0613)mol/0.250L}$ = 5.54

6 0

3 years ago

What orbitals are used to form the 10 sigma bonds in propane (ch3ch2ch3)? Label each atom with the appropriate hybridization. Dr

SVETLANKA909090 [29]

Mixing of pure orbitals having nearly equal energy to form equal number of completely new orbitals is said to be hybridization.

For the compound, $CH_3CH_2CH_3$ the electronic configuration of the atoms, carbon and hydrogen are:

Carbon (atomic number=6): In ground state= $1s^{2}2s^{2}2p^{2}$

In excited state: $1s^{2}2s^{1}2p^{3}$

Hydrogen (atomic number=1): $1s^{1}$

All the bonds in the compound is single bond( $\sigma$ -bond) that is they are formed by head on collision of the orbitals.

The structure of the compound is shown in the image.

The Carbon-Hydrogen bond is formed by overlapping of s-orbital of hydrogen to p-orbital of carbon.

In order to complete the octet the required number of electrons for carbon is 4 and for hydrogen is 1. So, the electron in $1s^{1}$ of hydrogen will overlap to the 2p^{3}-orbital of carbon.

Thus, the hybridization of Hydrogen is $s$ -hybridization and the hybridization of Carbon is $sp^{3}$ -hybridization.

The hybridization of each atom is shown in the image.

3 0

3 years ago

What is the molality of a solution made by dissolving 15.20 g of i2 in 1.33 mol of diethyl ether, (ch3ch2)2o?

Paraphin [41]

The molarity of solution made by dissolving 15.20g of i2 in 1.33 mol of diethyl ether (CH3CH2)2O is =0.6M

calculation

molarity =moles of solute/ Kg of the solvent

mole of the solute (i2) = mass /molar mass
the molar mass of i2 = 126.9 x2 = 253.8 g/mol

moles is therefore= 15.2 g/253.8 g/mol = 0.06 moles

calculate the Kg of solvent (CH3CH2)2O
mass = moles x molar mass
molar mass of (CH3CH2)2O= 74 g/mol

mass is therefore = 1.33 moles x 74 g/mol = 98.42 grams
in Kg = 98.42 /1000 =0.09842 Kg

molarity is therefore = 0.06/0.09842 = 0.6 M

3 0

3 years ago

Which equation represents a double replacement reaction?

AveGali [126]

Answer:

C) LiOH + HCl → LiCl + H₂O

General Formulas and Concepts:

Chemistry - Reactions

Synthesis Reactions: A + B → AB
Decomposition Reactions: AB → A + B
Single-Replacement Reactions: A + BC → AB + C
Double-Replacement Reactions: AB + CD → AD + BC

Explanation:

Step 1: Define

RxN A: 2Na + 2H₂O → 2NaOH + H₂

RxN B: CaCO₃ → CaO + CO₂

RxN C: LiOH + HCl → LiCl + H₂O

RxN D: CH₄ + 2O₂ → CO₂ + 2H₂O

Step 2: Identify

RxN A: Single Replacement Reaction

RxN B: Decomposition Reaction

RxN C: Double Replacement Reaction

RxN D: Combustion Reaction

7 0

3 years ago