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3 years ago

Based on your graph in the Periodic Trends Lab, what is the relationship between atomic radius and ionization energy ?

1 answer:

6 0

As the atomic radius decreases, it becomes harder to remove an electron that is closer to a more positively charged nucleus. ... They experience a weaker attraction to the positive charge of the nucleus. Ionization energy increases from left to right in a period and decreases from top to bottom in a group.

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The osmotic pressure of a solution containing 2.04 g of an unknown compound dissolved in 175.0 mLof solution at 25 ∘C is 2.13 at

kherson [118]

Answer: The molecular formula of the compound is $C_4H_{10}O_4$

Explanation:

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

$\pi=iMRT$

Or,

$\pi=i\times \frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}\times RT$

where,

$\pi$ = osmotic pressure of the solution = 2.13 atm

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

Given mass of compound = 2.04 g

Volume of solution = 175.0 mL

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

T = temperature of the solution = $25^oC=[273+25]=298K$

Putting values in above equation, we get:

$2.13atm=1\times \frac{2.04\times 1000}{\text{Molar mass of compound}\times 175.0}\times 0.0821\text{ L.atm }mol^{-1}K^{-1}\times 298K\\\\\text{Molar mass of compound}=\frac{1\times 2.04\times 1000\times 0.0821\times 298}{2.13\times 175.0}=133.9g/mol$

Calculating the molecular formula:

The chemical equation for the combustion of compound having carbon, hydrogen and oxygen follows:

$C_xH_yO_z+O_2\rightarrow CO_2+H_2O$

where, 'x', 'y' and 'z' are the subscripts of carbon, hydrogen and oxygen respectively.

We are given:

Mass of $CO_2=36.26g$

Mass of $H_2O=14.85g$

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44 g of carbon dioxide, 12 g of carbon is contained.

So, in 36.26 g of carbon dioxide, $\frac{12}{44}\times 36.26=9.89g$ of carbon will be contained.

For calculating the mass of hydrogen:

In 18 g of water, 2 g of hydrogen is contained.

So, in 14.85 g of water, $\frac{2}{18}\times 14.85=1.65g$ of hydrogen will be contained.

Mass of oxygen in the compound = (22.08) - (9.89 + 1.65) = 10.54 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{9.89g}{12g/mole}=0.824moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{1.65g}{1g/mole}=1.65moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{10.54g}{16g/mole}=0.659moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.659 moles.

For Carbon = $\frac{0.824}{0.659}=1.25\approx 1$

For Hydrogen = $\frac{1.65}{0.659}=2.5$

For Oxygen = $\frac{0.659}{0.659}=1$

Converting the mole fraction into whole number by multiplying the mole fraction by '2'

Mole fraction of carbon = (1 × 2) = 2

Mole fraction of oxygen = (2.5 × 2) = 5

Mole fraction of hydrogen = (1 × 2) = 2

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : O = 2 : 5 : 2

The empirical formula for the given compound is $C_2H_5O_2$

For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.

The equation used to calculate the valency is:

$n=\frac{\text{Molecular mass}}{\text{Empirical mass}}$

We are given:

Mass of molecular formula = 133.9 g/mol

Mass of empirical formula = 61 g/mol

Putting values in above equation, we get:

$n=\frac{133.9g/mol}{61g/mol}=2$

Multiplying this valency by the subscript of every element of empirical formula, we get:

$C_{(2\times 2)}H_{(5\times 2)}O_{(2\times 2)}=C_4H_{10}O_4$

Hence, the molecular formula of the compound is $C_4H_{10}O_4$

4 0

3 years ago

3. What is the energy of a photon whose frequency is 5.2 x 1015 Hz? Use the equation: E = hxv

anzhelika [568]

Answer:

3. 3.45×10¯¹⁸ J.

4. 1.25×10¹⁵ Hz.

Explanation:

3. Determination of the energy of the photon.

Frequency (v) = 5.2×10¹⁵ Hz

Planck's constant (h) = 6.626×10¯³⁴ Js

Energy (E) =?

The energy of the photon can be obtained by using the following formula:

E = hv

E = 6.626×10¯³⁴ × 5.2×10¹⁵

E = 3.45×10¯¹⁸ J

Thus, the energy of the photon is 3.45×10¯¹⁸ J

4. Determination of the frequency of the radiation.

Wavelength (λ) = 2.4×10¯⁵ cm

Velocity (c) = 3×10⁸ m/s

Frequency (v) =?

Next, we shall convert 2.4×10¯⁵ cm to metre (m). This can be obtained as follow:

100 cm = 1 m

Therefore,

2.4×10¯⁵ cm = 2.4×10¯⁵ cm × 1 m /100 cm

2.4×10¯⁵ cm = 2.4×10¯⁷ m

Thus, 2.4×10¯⁵ cm is equivalent to 2.4×10¯⁷ m

Finally, we shall determine the frequency of the radiation by using the following formula as illustrated below:

Wavelength (λ) = 2.4×10¯⁷ m

Velocity (c) = 3×10⁸ m/s

Frequency (v) =?

v = c / λ

v = 3×10⁸ / 2.4×10¯⁷

v = 1.25×10¹⁵ Hz

Thus, the frequency of the radiation is 1.25×10¹⁵ Hz.

8 0

3 years ago

There are two isotopes of chlorine. below is a list of each and their mass and abundance. calculate the atomic mass of chlorine.

Vesna [10]

Chlorine is very lequid structure

6 0

3 years ago

For a particular redox reaction, NO-2 is oxidized to NO-3 and Ag+ is reduced to Ag . Complete and balance the equation for this

Salsk061 [2.6K]

The following are the steps to complete and balnce the equation for the given reaction

Explanation:

We are given, NO2– is oxidized to NO3– and Ag is reduced to Ag

NO2– + Ag+ -----> NO3– + Ag(s)

Step 1) Assign the oxidation state to each element reaction

NO2– + Ag+ -----> NO3– + Ag(s)

N= +3 N = +5

O = -2 O = -2

Ag = +1 Ag = 0

NO2– -----> NO3– ………oxidation half reaction

Ag+ -----> Ag(s) ……….reduction half reaction

Step 2) Balance the element other than O and H

NO2– -----> NO3–

Ag+ -----> Ag(s)

Step 3) Balance the O by adding 1 H2O for 1 O

NO2– + H2O -----> NO3–

Ag+ -----> Ag(s)

Step 4) Balance the H by adding H+

NO2– + H2O -----> NO3– + 2H+

Ag+ -----> Ag(s)

Step 5) Balance the charge by adding electron

NO2– + H2O -----> NO3– + 2H+ + 2e-

Ag+ + 1e------> Ag(s)

Step 6) Balance the electron in both half reaction

NO2– + H2O -----> NO3– + 2H+ + 2e-

2 Ag+ + 2e------> 2 Ag(s)

3 0

3 years ago

List three common states of matter and give definition

Wewaii [24]

Solid- particles are packed tightly together so they don’t move much
Liquid- particles are still close together but move freely
Gas- particles are neither close together nor fixed in place

6 0

3 years ago