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4 years ago

PLEASE HELP !!!

Chemistry

1 answer:

garri49 [273]4 years ago

7 0

According to the Law of Definite Proportions from Dalton's Atomic Theory, each compound is composed of a fixed ratio of each of its individual elements. So, the number of individual elements per 1 particle of that compound is represented by the subscripts. The answers are as follows:

Table sugar: 12 atoms of carbon, 22 atoms of hydrogen; 11 atoms of oxygen; 45 total atoms
Marble: 1 atom of calcium, 1 atom of carbon; 3 atoms of oxygen; 5 total atoms
Natural gas: 1 atom of carbon, 4 atoms of hydrogen; 5 total atoms
Rubbing alcohol: 3 atoms of carbon, 8 atoms of hydrogen; 1 atom of oxygen; 12 total atoms
Table sugar: 1 atom of silicon; 2 atoms of oxygen; 3 total atoms

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If a temperature increase from 21.0 ∘c to 35.0 ∘c triples the rate constant for a reaction, what is the value of the activation

Airida [17]

Answer:

59.077 kJ/mol.

Explanation:

From Arrhenius law: <em>K = Ae(-Ea/RT)</em>

where, K is the rate constant of the reaction.

A is the Arrhenius factor.

Ea is the activation energy.

R is the general gas constant.

T is the temperature.

At different temperatures:

<em>ln(k₂/k₁) = Ea/R [(T₂-T₁)/(T₁T₂)]</em>

k₂ = 3k₁ , Ea = ??? J/mol, R = 8.314 J/mol.K, T₁ = 294.0 K, T₂ = 308.0 K.

ln(3k₁/k₁) = (Ea / 8.314 J/mol.K) [(308.0 K - 294.0 K) / (294.0 K x 308.0 K)]

∴ ln(3) = 1.859 x 10⁻⁵ Ea

∴ Ea = ln(3) / (1.859 x 10⁻⁵) = 59.077 kJ/mol.

4 0

3 years ago

At 298 K, the osmotic pressure of a glucose solution (C6H12O6 (aq)) is 12.1 atm. Calculate the freezing point of the solution. T

Anarel [89]

<u>Answer:</u> The freezing point of solution is -0.974°C

<u>Explanation:</u>

To calculate the concentration of solute, we use the equation for osmotic pressure, which is:

$\pi=iMRT$

where,

$\pi$ = osmotic pressure of the solution = 12.1 atm

i = Van't hoff factor = 1 (for non-electrolytes)

M = molarity of solute = ?

R = Gas constant = $0.0821\text{ L atm }mol^{-1}K^{-1}$

T = temperature of the solution = 298 K

Putting values in above equation, we get:

$12.1atm=1\times M\times 0.0821\text{ L.atm }mol^{-1}K^{-1}\times 298K\\\\M=\frac{12.1}{1\times 0.0821\times 298}=0.495M$

This means that 0.495 moles of glucose is present in 1 L or 1000 mL of solution

To calculate the mass of solution, we use the equation:

$\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}$

Density of solution = 1.034 g/mL

Volume of solution = 1000 mL

Putting values in above equation, we get:

$1.034g/mL=\frac{\text{Mass of solution}}{1000mL}\\\\\text{Mass of solution}=(1.034g/mL\times 1000mL)=1034g$

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Moles of glucose = 0.495 moles

Molar mass of glucose = 180.16 g/mol

Putting values in above equation, we get:

$0.495mol=\frac{\text{Mass of glucose}}{180.16g/mol}\\\\\text{Mass of glucose}=(0.495mol\times 180.16g/mol)=89.18g$

Depression in freezing point is defined as the difference in the freezing point of pure solution and freezing point of solution.

The equation used to calculate depression in freezing point follows:

$\Delta T_f=\text{Freezing point of pure solution}-\text{Freezing point of solution}$

To calculate the depression in freezing point, we use the equation:

$\Delta T_f=iK_fm$

Or,

$\text{Freezing point of pure solution}-\text{Freezing point of solution}=i\times K_f\times \frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams)}}$

where,

Freezing point of pure solution = 0°C

i = Vant hoff factor = 1 (For non-electrolytes)

$K_f$ = molal freezing point elevation constant = 1.86°C/m

$m_{solute}$ = Given mass of solute (glucose) = 89.18 g

$M_{solute}$ = Molar mass of solute (glucose) = 180.16 g/mol

$W_{solvent}$ = Mass of solvent (water) = [1034 - 89.18] g = 944.82 g

Putting values in above equation, we get:

$0-\text{Freezing point of solution}=1\times 1.86^oC/m\times \frac{89.18\times 1000}{180.16g/mol\times 944.82}\\\\\text{Freezing point of solution}=-0.974^oC$

Hence, the freezing point of solution is -0.974°C

8 0

3 years ago

Can someone explain this to me, please?

zimovet [89]

Answer:

Isotopes have same atomic numbers, no. of protons and no. of electrons. Only their no. of neutrons and atomic mass are changed.

Atomic Mass = 24

Atomic No. = 11

Hence,

No. of protons in Na-24 = 11

No. of neutrons = Atomic Mass - Atomic Number

No. of neutrons = 24 - 11

No. of neutrons = 13

Atomic Number = 11

$\rule[225]{225}{2}$

Hope this helped!

<h3>~AH1807</h3><h3>Peace!</h3>

5 0

3 years ago

An experiment has been set up to determine if different types of insulation wrap will affect the temperature of water in a conta

kozerog [31]

Insulation wraps because independent is the variable you are changing to affect the dependent variable (what you are measuring)

3 0

4 years ago

What elements can join wit other elements o form a covalent bond

Bas_tet [7]

Answer:

Non metals join to form covalent bond.

Explanation:

Covalent bond:

It is formed by the sharing of electron pair between bonded atoms.

The atom with larger electronegativity attract the electron pair more towards it self and becomes partial negative while the other atom becomes partial positive.

For example:

In water the electronegativity of oxygen is 3.44 and hydrogen is 2.2. That's why electron pair attracted more towards oxygen, thus oxygen becomes partial negative and hydrogen becomes partial positive.

Both atoms bonded through covalent bond.

In Cl₂ both chlorine atoms are bonded through the covalent bond.

8 0

3 years ago