Describe one trait that you believe is unique to you that you think is

How many atoms of phosphorus are in 4.5 grams of tetraphosphorus decoxide?

Alexxx [7]

Answer:

In 4.5 grams of tetraphosphorus decoxide we have 3.85 * 10^22 phosphorus atoms

Explanation:

Step 1: Data given

tetraphosphorus decoxide = P4O10

Molar mass of P4O10 = 283.89 g/mol

Mass of P4O10 = 4.5 grams

Number of Avogadro = 6.022 * 10^23 / mol

Step 2: Calculate moles of P4O10

Moles P4O10 = mass P4O10 / molar mass P4O10

Moles P4O10 = 4.5 grams / 283.89 g/mol

Moles = 0.016 moles

Step 3: Calculate moles of P

For 1 mol P4O10 we have 4 moles of phosphorus

For 0.016 moles P4O10 we have 4*0.016 = 0.064 moles P

Step 4: Calculate number of P atoms

Number of P atoms = moles P * number of Avogadro

Number of P atoms = 0.064 moles * 6.022*10^23

Number of P atoms = 3.85 * 10^22 atoms

In 4.5 grams of tetraphosphorus decoxide we have 3.85 * 10^22 phosphorus atoms

3 0

2 years ago

Write the formula and determine the percent by mass of the salt in barium hydroxide octahydrate *

user100 [1]

Answer:

315.51g/mol

Explanation:

137(33 + (16.00 + 1.01) 2 + 8 [1.01 (2) + 16.00] = 315.51g/mol

7 0

2 years ago

Which formula can be used to calculate the molar mass of hydrogen peroxide (H2O2)? (5 points)

ratelena [41]

answer: option d (2×molar mass of H +2×molar mass of O

 

7 0

2 years ago

Assume that you have a cylinder with a movable piston. What would happen to the gas pressure inside the cylinder if you do the f

pishuonlain [190]

Answer:

A. The pressure will increase 4 times. P₂ = 4 P₁

B. The pressure will decrease to half its value. P₂ = 0.5 P₁

C. The pressure will decrease to half its value. P₂ = 0.5 P₁

Explanation:

Initially, we have n₁ moles of a gas that occupy a volume V₁ at temperature T₁ and pressure P₁.

What would happen to the gas pressure inside the cylinder if you do the following?

Part A: Decrease the volume to one-fourth the original volume while holding the temperature constant. Express your answer in terms of the variable P initial.

V₂ = 0.25 V₁. According to Boyle's law,

P₁ . V₁ = P₂ . V₂

P₁ . V₁ = P₂ . 0.25 V₁

P₁ = P₂ . 0.25

P₂ = 4 P₁

Part B: Reduce the Kelvin temperature to half its original value while holding the volume constant. Express your answer in terms of the variable P initial.

T₂ = 0.5 T₁. According to Gay-Lussac's law,

$\frac{P_{1}}{T_{1}} =\frac{P_{2}}{T_{2}}\\\frac{P_{1}}{T_{1}} =\frac{P_{2}}{0.5T_{1}}\\\\P_{2}=0.5P_{1}$

Part C: Reduce the amount of gas to half while keeping the volume and temperature constant. Express your answer in terms of the variable P initial.

n₂ = 0.5 n₁.

P₁ in terms of the ideal gas equation is:

$P_{1}=\frac{n_{1}.R.T_{1}}{V_{1}}$

P₂ in terms of the ideal gas equation is:

$P_{2}=\frac{n_{2}.R.T_{1}}{V_{1}}=\frac{0.5n_{1}.R.T_{1}}{V_{1}}=0.5P_{1}$

3 0

2 years ago

In a study of the formation of NOx air pollution, a chamber heated to 2200°C was filled with air (0.790 atm N₂, 0.210 atm O₂). W

Irina-Kira [14]

Answer:

N₂ = 0.7515atm

O₂ = 0.1715atm

NO = 0.0770atm

Explanation:

For the reaction:

N₂(g) + O₂(g) ⇄ 2NO(g)

Where Kp is defined as:

$Kp = \frac{P_{NO}^2}{P_{N_2}P_{O_2}}}$

Pressures in equilibrium are:

N₂ = 0.790atm - X

O₂ = 0.210atm - X

NO = 2X

Replacing in Kp:

0.0460 = [2X]² / [0.790atm - X] [0.210atm - X]

0.0460 = 4X² / 0.1659 - X + X²

0.0460X² - 0.0460X + 7.6314x10⁻³ = 4X²

-3.954X² - 0.0460X + 7.6314x10⁻³ = 0

Solving for X:

X = - 0.050 → False answer. There is no negative concentrations.

X = 0.0385 atm → Right answer.

Replacing for pressures in equilibrium:

N₂ = 0.790atm - X = 0.7515atm

O₂ = 0.210atm - X = 0.1715atm

NO = 2X = 0.0770atm

3 0

3 years ago

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