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dangina [55]
3 years ago
8

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

Chemistry
1 answer:
Alexxx [7]3 years ago
3 0

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

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kupik [55]

Answer:

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Hope This Helps!

6 0
3 years ago
For the reaction 2 H 2 O ( g ) − ⇀ ↽ − 2 H 2 ( g ) + O 2 ( g ) the equilibrium concentrations were found to be [ H 2 O ] = 0.250
Bess [88]

<u>Answer:</u> The equilibrium constant for the given reaction is 4.224

<u>Explanation:</u>

We are given:

Equilibrium concentration of water = 0.250 M

Equilibrium concentration of hydrogen gas = 0.330 M

Equilibrium concentration of oxygen gas = 0.800 M

For the given chemical reaction:

2H_2O(g)\rightleftharpoons H_2(g)+O_2(g)

The expression of K_{eq} for above reaction follows:

K_{eq}=\frac{[H_2][O_2]}{[H_2O]^2}

Putting values in above expression, we get:

K_{eq}=\frac{0.330\times 0.800}{(0.250)^2}\\\\K_{eq}=4.224

Hence, the equilibrium constant for the given reaction is 4.224

7 0
3 years ago
One of the main contaminants of a nuclear accident, such as that at Chernobyl, is strontium-90, which decays exponentially at an
igor_vitrenko [27]

Answer: a) %(C/Co) = (e^(-0.027t)) × 100

b) t1/2 = 25.67years

c) 5.872%

Explanation:

a) Radioactive reactions always follow a first order reaction dynamic

Let the initial concentration of Strontium-90 be Co and the concentration at any time be C

The rate of decay will be given as:

(dC/dt) = -KC (Minus sign because it's a rate of reduction)

The question provides K = 2.7% per year = 0.027/year

(dC/dt) = -0.027C

(dC/C) = -0.027dt

 ∫ (dC/C) = -0.027 ∫ dt 

Solving the two sides as definite integrals by integrating the left hand side from Co to C and the Right hand side from 0 to t.

We obtain

In (C/Co) = -0.027t

(C/Co) = (e^(-0.027t))

In percentage, %(C/Co) = (e^(-0.027t)) × 100

(Solved)

b) Half life of a first order reaction (t1/2) = (In 2)/K

K = 0.027/year

t1/2 = (In 2)/0.027 = 25.67 years

c) percentage that remains after 105years,

%(C/Co) = (e^(-0.027t)) × 100

t = 105

%(C/Co) = (e^(-0.027 × 105)) × 100 = 5.87%

8 0
3 years ago
Calculate the molar solubility of AgCl in 0.05M NaCl. (Ksp = 1.77 x 10-10)
Aneli [31]

Answer: 3.54 x 10^{-9}

Explanation:

For this Case we have keep in mind the effect of the Common Ion, which is Cl.

You can find the amount of silver ions at equilibrium from Ksp since You already know the chloride ion concentration. Keep in mind that only a few amount of silver chloride will be dissolved. This means the chloride concentration will stay in 0.05 M, Then I can use Ksp to solve for the silver ion concentration. So you can use the following equation.

AgCl Ksp = Ksp/[Cl^-]= (1.77 x 10^{-10})/(0.05) = 3.54 x 10^{-9}

6 0
3 years ago
Convert 145.0 mm to meters and express the answer to the correct number of significant digits
jarptica [38.1K]
You need to use a conversion factor of 1000mm to 1 m. So divide by 1000.  Then you should have 4 digits total by the significant figure rules since the number in the problem is 4 digits.
7 0
4 years ago
Read 2 more answers
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