Which electron notation is incorrect?

How Many Atoms are present in this formula NH4NO3

Bess [88]

2 nitrogen atoms, 4 hydrogen atoms and 3 oxygen atoms

8 0

2 years ago

A solution was prepared by dissolving 0.800 g of sulfur S8, in 100.0 g of acetic acid, HC2H3O2. Calculate the freezing point and

sammy [17]

Answer: The freezing point of solution is 16.5°C and the boiling point of solution is 118.2°C

Explanation:

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

$Molality=\frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ in grams}}$

Where,

$m_{solute}$ = Given mass of solute $(S_8)$ = 0.800 g

$M_{solute}$ = Molar mass of solute $(S-8)$ = 256.52 g/mol

$W_{solvent}$ = Mass of solvent (acetic acid) = 100.0 g

Putting values in above equation, we get:

$\text{Molality of solution}=\frac{0.800\times 1000}{256.52\times 100.0}\\\\\text{Molality of solution}=0.0312m$

Calculation for freezing point of solution:

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

$\Delta T_f=\text{freezing point of acetic acid}-\text{Freezing point of solution}$

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

$\Delta T_f=iK_fm$

or,

$\text{Freezing point of acetic acid}-\text{Freezing point of solution}=iK_fm$

where,

Freezing point of acetic acid = 16.6°C

i = Vant hoff factor = 1 (for non-electrolyte)

$K_f$ = molal freezing point depression constant = 3.59°C/m

m = molality of solution = 0.0312 m

Putting values in above equation, we get:

$16.6^oC-\text{freezing point of solution}=1\times 3.59^oC/m\times 0.0312m\\\\\text{Freezing point of solution}=16.5^oC$

Hence, the freezing point of solution is 16.5°C

Calculation for boiling point of solution:

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

The equation used to calculate elevation in boiling point follows:

$\Delta T_b=\text{Boiling point of solution}-\text{Boiling point of acetic acid}$

To calculate the elevation in boiling point, we use the equation:

$\Delta T_b=iK_bm$

or,

$\text{Boiling point of solution}-\text{Boiling point of acetic acid}=iK_fm$

where,

Boiling point of acetic acid = 118.1°C

i = Vant hoff factor = 1 (for non-electrolyte)

$K_f$ = molal boiling point elevation constant = 3.08°C/m

m = molality of solution = 0.0312 m

Putting values in above equation, we get:

$\text{Boiling point of solution}-118.1^oC=1\times 3.08^oC/m\times 0.0312m\\\\\text{Boiling point of solution}=118.2^oC$

Hence, the boiling point of solution is 118.2°C

5 0

3 years ago

You have 8 moles of a gas at 250 K in a 6 L container. What is the pressure of the gas? Show your work

Bogdan [553]

Hello:

In this case, we will use the Clapeyron equation:

P = ?

n = 8 moles

T = 250 K

R = 0.082 atm.L/mol.K

V = 6 L

Therefore:

P * V = n * R * T

P * 6 = 8 * 0.082* 250

P* 6 = 164

P = 164 / 6

P = 27.33 atm

Hope that helps!

7 0

3 years ago

Which is true about metals used for jewelry, such as platinum and gold? A. They have low flammability. B. They have low reactivi

Morgarella [4.7K]

i think the answer is B. They have low reactivity.Hope this helped (:

8 0

2 years ago

Read 2 more answers

At a certain temperature the rate of this reaction is second order in NH4OH with a rate constant of 34.1 M^-1s^-1: Suppose a ves

Tom [10]

Answer:

Time = 0.929s = 0.93s (2 s.f)

Explanation:

Rate constant, k = 34.1 M^-1s^-1

Initial Concentration, [A]o = 0.100M

Time = ?

Final Concentration [A] = 0.0240M

The parameters are represented in the following equation as;

1/[A] = kt + 1/[A]o

kt = 1/[A] - 1/[A]o

kt = 1/0.0240 - 1/0.1

kt = 31.67

t = 31.67 / 34.1

t = 0.929s = 0.93s (2 s.f)

5 0

2 years ago