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3 years ago

Write the differences between iron-sulphur mixture and iron sulphide compound

Chemistry

1 answer:

Misha Larkins [42]3 years ago

8 0

the mixture can contain more or less iron, but iron sulfide always contains equal amounts of iron and sulphur.

the iron and sulphur atoms are not joined together in the mixture, but they are joined together in iron sulfide.

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A 1.00 liter container holds a mixture of 0.52 mg of He and 2.05 mg of Ne at 25oC. Determine the partial pressures of He and Ne

Ymorist [56]

Answer:

pHe = 3.2 × 10⁻³ atm

pNe = 2.5 × 10⁻³ atm

P = 5.7 × 10⁻³ atm

Explanation:

Given data

Volume = 1.00 L

Temperature = 25°C + 273 = 298 K

mHe = 0.52 mg = 0.52 × 10⁻³ g

mNe = 2.05 mg = 2.05 × 10⁻³ g

The molar mass of He is 4.00 g/mol. The moles of He are:

0.52 × 10⁻³ g × (1 mol / 4.00 g) = 1.3 × 10⁻⁴ mol

We can find the partial pressure of He using the ideal gas equation.

P × V = n × R × T

P × 1.00 L = 1.3 × 10⁻⁴ mol × (0.082 atm.L/mol.K) × 298 K

P = 3.2 × 10⁻³ atm

The molar mass of Ne is 20.18 g/mol. The moles of Ne are:

2.05 × 10⁻³ g × (1 mol / 20.18 g) = 1.02 × 10⁻⁴ mol

We can find the partial pressure of Ne using the ideal gas equation.

P × V = n × R × T

P × 1.00 L = 1.02 × 10⁻⁴ mol × (0.082 atm.L/mol.K) × 298 K

P = 2.5 × 10⁻³ atm

The total pressure is the sum of the partial pressures.

P = 3.2 × 10⁻³ atm + 2.5 × 10⁻³ atm = 5.7 × 10⁻³ atm

6 0

3 years ago

A quantity of CO gas occupies a volume of 0.32 L at 0.90 atm and 323 K . The pressure of the gas is lowered and its temperature

Contact [7]

Mathematical formula of Ideal Gas Law is PV=nRT
where: P-pressure,
V-volume
n-number of moles; m/MW
T-Temperature
m-mass
d-density ; m/V
MW-Molecular Weight
R- Ideal Gas constant. If the units of P,V,n & T are atm, L, mol & K respectively, the value of R is 0.0821 L x atm / K x mol

Substituting the definitions to the original Gas equation becomes:
d= P x MW / (RxT)

Solution : d= .90atm x 28 g/mol (CO) / 0.0821Lxatm / mol x K x 323 K

d = 25.2 g / 26000 mL

d = .0.00096 g/mL is the density of CO under the new conditions

5 0

4 years ago

Water is able to travel up stems to leaves in a plant. this is an example of _____

Anastasy [175]

Your answer should be B :)

5 0

3 years ago

Determine the theoretical yield of HCl if 60.0 g of BC13 and 37.5 g of H20 are reacted according to the following balanced react

Pie

Answer : The theoretical yield of HCl is, 56.1735 grams

Explanation : Given,

Mass of $BCl_3$ = 60 g

Mass of $H_2O$ = 37.5 g

Molar mass of $BCl_3$ = 117 g/mole

Molar mass of $H_2O$ = 18 g/mole

Molar mass of $HCl$ = 36.5 g/mole

First we have to calculate the moles of $BCl_3$ and $H_2O$ .

$\text{Moles of }BCl_3=\frac{\text{Mass of }BCl_3}{\text{Molar mass of }BCl_3}=\frac{60g}{117g/mole}=0.513moles$

$\text{Moles of }H_2O=\frac{\text{Mass of }H_2O}{\text{Molar mass of }H_2O}=\frac{37.5g}{18g/mole}=2.083moles$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$BCl_3(g)+3H_2O(l)\rightarrow H_3BO_3(s)+3HCl(g)$

From the balanced reaction we conclude that

As, 1 mole of $BCl_3$ react with 3 mole of $H_2O$

So, 0.513 moles of $BCl_3$ react with $3\times 0.513=1.539$ moles of $H_2O$

From this we conclude that, $H_2O$ is an excess reagent because the given moles are greater than the required moles and $BCl_3$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $HCl$ .

As, 1 mole of $BCl_3$ react to give 3 moles of $HCl$

So, 0.513 moles of $BCl_3$ react to give $3\times 0.513=1.539$ moles of $HCl$

Now we have to calculate the mass of $HCl$ .

$\text{Mass of }HCl=\text{Moles of }HCl\times \text{Molar mass of }HCl$

$\text{Mass of }HCl=(1.539mole)\times (36.5g/mole)=56.1735g$

Therefore, the theoretical yield of HCl is, 56.1735 grams

7 0

3 years ago

Calculate the pH of a buffer prepared by mixing 20.0 mL of 0.10 M acetic acid and 55.0 mL of 0.10 M sodium acetate

MatroZZZ [7]

Answer:

Calculate the pH of a buffer prepared by mixing 30.0 mL of 0.10 M acetic acid and 40.0 mL of 0.10 M sodium acetate.

3 0

4 years ago