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3 years ago

How many total atoms are in 0.280 g of P2O5

2 answers:

5 0

First, divide 0.280 by the molar mass of P2O5; that gives you the number of moles. Then multiply the number of moles by 6.023 * 10^23 (Avogadro's number) to get the number of molecules. THEN you have to multiply that value by 7 (since there are 7 atoms per molecule). *Then* you're done.

liq [111]3 years ago

4 0

Answer:

There are 8.43x10²¹ total atoms in 0.280 g of P₂O₅.

Explanation:

Let's follow some steps to calculate the total atoms.

1st) Calculate the molar mass of P₂O₅

Look for the <u>atomic weight of each atom</u> in the Periodic table:

- Atomic weight of Phosphorus = 31 g/mol

- Atomic weight of Oxygen = 16 g/mol

Then, multiply each atomic weight by its coefficient to calculate the molar mass of P₂O₅:

(Atomic weight of P .2) + (Atomic weight of O .5)= Molar mass of P₂O₅

(31 g/mol.2) + (16 g/mol.5) = 62 g/mol + 80 g/mol = 142 g/mol

2nd) Calculate the moles of P₂O₅ that are contained in 0.280 g

This step is easy using a Rule of three thinking that if 142 g of P₂O₅ represents a mol of P₂O₅ the 0.280 g will be "x" moles:

142 g -------- 1 mol of P₂O₅

0.280 g -------- x = (0.280 g.1 mol)/142 g =0.002 mol of P₂O₅

This means that 0.002 moles of P₂O₅ weights 0.280g.

3rd) Calculate the moles of P and O

To see clear how many moles of each atoms are in the molecule of P₂O₅ we disassociate it:

P₂O₅ → 2P + 5O

From the reaction we know that 1 mol of P₂O₅ produces 2 moles of phosphorus and 5 moles of oxygen. Now we can make a relation and thinking that if 1 mol of P₂O₅ produces 2 moles of P the 0.002 moles of P₂O₅ that we have will produce "x" moles of P:

1 mol of P₂O₅ ----- 2 moles of P

0.002 mol of P₂O₅ ----- x = (0.002 mol.2moles)/1mol = 0.004 moles of P

We use the same reasoning for oxygen:

1 mol of P₂O₅ ------- 5 moles of O

0.002 mol of P₂O₅ ------- x = (0.002 mol.5moles)/1mol = 0.01 moles of O

Up to here we have 0.004 moles of atoms of phosphorus and 0.01 moles of atoms of oxygen.

4th) Calculate the total atoms

To this step it is important to remember that 1 mol of something representa a quantity of 6.022x10²³ (that is called Avogadro's number) So, 1 mol of atoms represents 6.022x10²³ atoms.

Now, if we know that in 1 mol of phosphorus atoms is equal to 6.022x10²³ atoms of phosphorus the 0.004 moles that we have will be equal to "x" atoms:

1 mol of P ------ 6.022x10²³ atoms of P

0.004 mol of P ------ x = (0.004 . 6.022x10²³)/ 1 = 2.41x10²¹ atoms of P

Use the same reasoning for oxygen:

1 mol of O ------ 6.022x10²³ atoms of O

0.01 mol of O ------ x = (0.01 . 6.022x10²³)/ 1 = 6.022x10²¹ atoms of O

Now that we have the number of atoms of phosphorus and oxygen let's sum them to find the total atoms:

2.41x10²¹ atoms of P + 6.022x10²¹ atoms of O = 8.43x10²¹ total atoms

Finally, there are 8.43x10²¹ total atoms (of phosphorus and oxygen) in 0.280 g of P₂O₅.

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No, iron is not

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Answer:

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Explanation:

Given parameters:

Number of moles of magnesium hydroxide = 2.5mol

Unknown:

Mass of Mg(OH)₂ = ?

Solution:

To solve this problem we use the expression below;

Mass of Mg(OH)₂ = number of moles x molar mass

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3 years ago

7. How many formula units are equal to a 0.25 g sample of Chromium (III) sulfate,

ryzh [129]

Answer:

38.541 × 10¹⁹ formula units

Explanation:

Given data:

Mass of chromium sulfate = 0.25 g

Formula units in 0.25 g = ?

Solution:

The given problem will solve by using Avogadro number.

It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance.

The number 6.022 × 10²³ is called Avogadro number.

For example,

18 g of water = 1 mole = 6.022 × 10²³ formula units of water

Number of moles of chromium sulfate = Mass / molar mass

Number of moles of chromium sulfate = 0.25 g/ 392.16 g/ mol

Number of moles of chromium sulfate = 6.4 × 10⁻⁴ moles

Number of formula units:

1 mole = 6.022 × 10²³ formula units

6.4 × 10⁻⁴ moles × 6.022 × 10²³ formula units / 1 mol

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7 0

3 years ago

Natural gas is stored in a spherical tank at a temperature of 13°C. At a given initial time, the pressure in the tank is 117 kPa

drek231 [11]

Answer:

1. the absolute pressure in the tank before filling = 217 kPa

2. the absolute pressure in the tank after filling = 312 kPa

3. the ratio of the mass after filling M2 to that before filling M1 = 1.44

The correct relation is option c ( $\frac{M_{2} }{M_{1} } = \frac{P_{2} T_{1} }{P_{1} T_{2} }$ )

Explanation:

To find -

1. What is the absolute pressure in the tank before filling?

2. What is the absolute pressure in the tank after filling?

3. What is the ratio of the mass after filling M2 to that before filling M1 for this situation?

As we know that ,

Absolute pressure = Atmospheric pressure + Gage pressure

So,

Before filling the tank :

Given - Atmospheric pressure = 100 kPa , Gage pressure = 117 kPa

⇒Absolute pressure ( p1 ) = 100 + 117 = 217 kPa

Now,

After filling the tank :

Given - Atmospheric pressure = 100 kPa , Gage pressure = 212 kPa

⇒Absolute pressure (p2) = 100 + 212= 312 kPa

Now,

As given, volume is the same before and after filling,

i.e. $V_{1}$ = $V_{2}$

As we know that, P ∝ M

⇒ $\frac{p_{1} }{p_{2} } = \frac{m_{1} }{m_{2} }$

⇒ $\frac{m_{2} }{m_{1} } = \frac{p_{2} }{p_{1} }$

⇒ $\frac{m_{2} }{m_{1} } = \frac{312 }{217 } = 1.4378$ ≈ 1.44

Now, as we know that PV = nRT

As V is constant

⇒ P ∝ MT

⇒ $\frac{P}{T}$ ∝ M

⇒ $\frac{M_{2} }{M_{1} } = \frac{P_{2} T_{1} }{P_{1} T_{2} }$

So, The correct relation is c option.

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3 years ago

What orbitals are used to form the 10 sigma bonds in propane (ch3ch2ch3)? Label each atom with the appropriate hybridization. Dr

SVETLANKA909090 [29]

Mixing of pure orbitals having nearly equal energy to form equal number of completely new orbitals is said to be hybridization.

For the compound, $CH_3CH_2CH_3$ the electronic configuration of the atoms, carbon and hydrogen are:

Carbon (atomic number=6): In ground state= $1s^{2}2s^{2}2p^{2}$

In excited state: $1s^{2}2s^{1}2p^{3}$

Hydrogen (atomic number=1): $1s^{1}$

All the bonds in the compound is single bond( $\sigma$ -bond) that is they are formed by head on collision of the orbitals.

The structure of the compound is shown in the image.

The Carbon-Hydrogen bond is formed by overlapping of s-orbital of hydrogen to p-orbital of carbon.

In order to complete the octet the required number of electrons for carbon is 4 and for hydrogen is 1. So, the electron in $1s^{1}$ of hydrogen will overlap to the 2p^{3}-orbital of carbon.

Thus, the hybridization of Hydrogen is $s$ -hybridization and the hybridization of Carbon is $sp^{3}$ -hybridization.

The hybridization of each atom is shown in the image.

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3 years ago