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Lemur [1.5K]
2 years ago
14

) Starting with 5.00 g barium chloride n hydrate yields 4.26 g of anhydrous barium chloride after heating. Determine the integer

n. Round your answer to the nearest integer.
Chemistry
1 answer:
RSB [31]2 years ago
4 0

Answer:

2

Explanation:

Mass of water molecule = mass of hydrated salt - mass of anhydrous salt

Mass of water molecule = 5.00 - 4.26 = 0.74g of water molecule.

Number of moles = mass / molarmass

Molar mass of water = 18.015g/mol

No. of moles of water = 0.74 / 18.015 = 0.0411 moles.

Mass of BaCl2 present =?

1 mole of BaCl2 = 208.23 g

X mole of BaCl2 = 4.26 g

X = (4.26 * 1) / 208.23

X = 0.020

0.020 moles is present in 4.26g of BaCl2

Mole ratio between water and BaCl2 =

0.0411 / 0.020 = 2

Therefore 2 molecules of water is present the hydrated salt.

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Tpy6a [65]

Answer:

Explanation:

In spontaneous reaction , there is decrease in Gibb's free energy .( Δ G is negative ). Out of given reaction , following reactions have negative Δ G so they are spontaneous.

C ₂ H ₄ + H ₂ Rh ( I ) −−−→ C ₂ H ₆ ,  Δ G = − 150.97 kJ / mol

C ₆ H₁₃O₉ P + ATP ⟶ C ₆ H₁₄ O₁₂ P₂ + ADP ,  Δ G = − 14.2 kJ / mol

7 0
1 year ago
The Mond process produces pure nickel metal via the thermal decomposition of nickel tetracarbonyl: Ni(CO)4 (l) → Ni (s) + 4CO (g
Yuki888 [10]

<u>Answer:</u> The volume of CO formed is 254.43 L.

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}  

Given mass of Ni(CO)_4 = 444 g

Molar mass of Ni(CO)_4 = 170.73 g/mol

Putting values in above equation, we get:

\text{Moles of }Ni(CO)_4=\frac{444g}{170.73g/mol}=2.60mol

For the given chemical reaction:

Ni(CO)_4(l)\rightarrow Ni(s)+4CO(g)

By stoichiometry of the reaction:

1 mole of nickel tetracarbonyl produces 4 moles of carbon monoxide.

So, 2.60 moles of nickel tetracarbonyl will produce = \frac{4}{1}\times 2.60=10.4mol of carbon monoxide.

Now, to calculate the volume of the gas, we use ideal gas equation, which is:

PV = nRT

where,

P = Pressure of the gas = 752 torr

V = Volume of the gas = ? L

n = Number of moles of gas = 10.4 mol

R = Gas constant = 62.364\text{ L Torr }mol^{-1}K^{-1}

T = Temperature of the gas = 22^oC=(273+22)K=295K

Putting values in above equation, we get:

752torr\times V=10.4mol\times 62.364\text{ L Torr }mol^{-1}K^{-1}\times 295K\\\\V=254.43L

Hence, the volume of CO formed is 254.43 L.

5 0
2 years ago
A small portion of a crystal lattice is sketched below. What is the name of the unit cell of this lattice? Your answer must be a
Nikolay [14]

This is an incomplete question, the given sketch is shown below.

Answer : The name of given unit cell is, FCC (face-centered cubic unit cell)

Explanation :

Unit cell : It is defined as the smallest 3-dimensional portion of a complete space lattice which when repeated over the and again in different directions produces the complete space lattice.

There are three types of unit cell.

  • SCC (simple-centered cubic unit cell)
  • BCC (body-centered cubic unit cell)
  • FCC (face-centered cubic unit cell)

In SCC, the atoms are arranged at the corners.

Z=\frac{1}{8}\times 8=1

The number of atoms of unit cell = Z = 1

In BCC, the atoms are arranged at the corners and the body center.

Z=\frac{1}{8}\times 8+1=2

The number of atoms of unit cell = Z = 2

The given unit cell is, FCC because the atoms are arranged at the corners and the center of the 6 faces.

Z=\frac{1}{8}\times 8+\frac{1}{2}\times 6=4

The number of atoms of unit cell = Z = 4

Thus, the name of given unit cell is, FCC (face-centered cubic unit cell)

7 0
2 years ago
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frutty [35]

Answer:

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Explanation:

Balanced Chemical reactions

1.-                 N₂(g)  +  3H₂ (g)   ⇒   2NH₃ (g)

2.-                4NH₃ (g) + 5O₂(g)  ⇒  4NO (g)  +  6H₂O (l)

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Process

1.- Calculate the moles of NH₃

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                     x = (20 x 2) / 1

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                    x = (40 x 5) / 4

                    x = 200 / 4

                    x = 50 moles of O₂

3 0
2 years ago
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The volumeof this sphere must be less than 0.7228

5 0
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