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2 years ago

Write the electron configuration for the elements whose atomic numbers are the following:

Chemistry

1 answer:

andreev551 [17]2 years ago

6 0

Answer:

3= Lithium (Li) = [He] 2s1

6= Carbon (C) = [He] 2s2 2p2

8=Oxygen (O)= [He] 2s2 2p4

13=Aluminium (Al)= [Ne] 3s2 3p1

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A second-order reaction has a rate law: rate = k[a]2, where k = 0.150 m−1s−1. If the initial concentration of a is 0.250 m, what

Cerrena [4.2K]

Rate law for the given 2nd order reaction is:

Rate = k[a]2

Given data:

rate constant k = 0.150 m-1s-1

initial concentration, [a] = 0.250 M

reaction time, t = 5.00 min = 5.00 min * 60 s/s = 300 s

To determine:

Concentration at time t = 300 s i.e. $[a]_{t}$

Calculations:

The second order rate equation is:

$1/[a]_{t} = kt +1/[a]$

substituting for k,t and [a] we get:

1/[a]t = 0.150 M-1s-1 * 300 s + 1/[0.250]M

1/[a]t = 49 M-1

[a]t = 1/49 M-1 = 0.0204 M

Hence the concentration of 'a' after t = 5min is 0.020 M

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3 years ago

8 Write the equation

MAVERICK [17]

Answer:

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3 years ago

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3 years ago

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A sample of krypton gas with a volume of 10.00 L has a temp of 303 K and exerts

marshall27 [118]

Answer: The final volume will be 14.85 L.

Explanation:

To calculate the final volume when temperature increases, we use Charles' Law.

This law states that volume is directly proportional to the temperature of the gas if number of moles and pressure remains constant.

$V\propto T\\\\\frac{V_1}{T_1}=\frac{V_2}{T_2}$

where,

$V_1\text{ and }T_1$ = Initial volume and temperature

$V_2\text{ and }T_2$ = Final volume and temperature

We are given:

$V_1=10L\\
T_1=303K\\
V_2=?L\\
T_2=450K$

Putting values in above equation, we get:

$\frac{10L}{303K}=\frac{V_2}{450K}$

$V_2=14.85L$

Hence, the final volume of the gas is 14.85L

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3 years ago

A,B and C have the following solubilities in 100 mL of water at 25°C, respectively: 15g, 35 g and 65 g. If 75 g of each are adde

Pani-rosa [81]

Answer:

Only C can dissolve in water because its solubility is greater than 0.375 g/mL

C dissolves completely

22.5 g of A remain undissolved in 100 mL of water

2.5 g of B remain undissolved in 100 mL of water

Explanation:

Solubility : The amount of salt that can be dissolved in a given volume of solvent at a particular temperature .

$Solubility = \frac{mass}{volume}$

For each salt the solubility of 75 g in 200 mL water is

$Solubility = \frac{75}{200}$

0.375 g/mL

This mean 1 ml of water has 0.375 g of A , B and C

100 mL of water contain = 37.5 g of each salt A , B,C

In the question , the solubility of A , B and C in 100 are given as :

A = 15 g : It means maximum 15 g of A can dissolve in 100 mL

$Solubility = \frac{15}{100}$ = 0.15 g/mL

B = 35 g : It means maximum 35 g of B can dissolve in 100 mL

$Solubility = \frac{35}{100}$ = 0.35 g/mL

C = 65 g : It means maximum 65 g of A can dissolve in 100 mL

$Solubility = \frac{65}{100}$ = 0.65 g/mL

Since ,Solubility of A and B is less than 0.375 g/mL

So A and B can't dissolve in water at 25°C

A is excess by 37.5 - 15 = 22.5 g

B is in excess by 37.5 - 35 = 2.5 g

Hence 22.5 g of A remain undissolved

Hence 2.5 g of B remain undissolved

Only C can dissolve in water because its solubility is greater than 0.375 g/mL

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3 years ago