Pyrite is called "fool's gold" because it looks a lot like gold. Which properties can be used to tell gold and pyrite apart?

Potassium nitrate has a lattice energy of -163.8 kcal/mol and a heat of hydration of -155.5 kcal/mol.

Brut [27]

Answer:

293.99 g

OR

0.293 Kg

Explanation:

Given data:

Lattice energy of Potassium nitrate (KNO3) = -163.8 kcal/mol

Heat of hydration of KNO3 = -155.5 kcal/mol

Heat to absorb by KNO3 = 101kJ

To find:

Mass of KNO3 to dissolve in water = ?

Solution:

Heat of solution = Hydration energy - Lattice energy

= -155.5 -(-163.8)

= 8.3 kcal/mol

We already know,

1 kcal/mol = 4.184 kJ/mole

Therefore,

= 4.184 kJ/mol x 8.3 kcal/mol

= 34.73 kJ/mol

Now, 34.73 kJ of heat is absorbed when 1 mole of KNO3 is dissolved in water.

For 101 kJ of heat would be

= 101/34.73

= 2.908 moles of KNO3

Molar mass of KNO3 = 101.1 g/mole

Mass of KNO3 = Molar mass x moles

= 101.1 g/mole x 2.908

= 293.99 g

= 0.293 kg

293.99 g potassium nitrate has to dissolve in water to absorb 101 kJ of heat.

4 0

3 years ago

What is the mass of water that results from combining 2.0 g of hydrogen with 16.0 g of oxygen?

astraxan [27]

The formula for water is H2O so there would have to be two Hyrdogens and one oxygen. Therefore it would be 4g of Hydrogen and 16g of Oxygen leaving you with 20g.
The answer is D.

Hope this helps :) ~

8 0

3 years ago

Convert 9.90km to mm

sukhopar [10]

Answer:

9900000

add 20 LeTtErS or mOrE

3 0

3 years ago

The density of lead is 11.4g/cm3. What volume, in ft3, would be occupied by 10.0 g of lead?

yawa3891 [41]

Answer:

3.10×10¯⁵ ft³.

Explanation:

The following data were obtained from the question:

Density (D) of lead = 11.4 g/cm³

Mass (m) of lead = 10 g

Volume (V) of lead =.?

Density (D) = mass (m) / Volume (V)

D = m/V

11.4 = 10 / V

Cross multiply

11.4 × V = 10

Divide both side by 11.4

V = 10 / 11.4

V = 0.877 cm³

Finally, we shall convert 0.877 cm³ to ft³. This can be obtained as follow:

1 cm³ = 3.531×10¯⁵ ft³

Therefore,

0.877 cm³ = 0.877 cm³ × 3.531×10¯⁵ ft³ /1 cm³

0.877 cm³ = 3.10×10¯⁵ ft³

Thus, 0.877 cm³ is equivalent to 3.10×10¯⁵ ft³.

Therefore, the volume of the lead in ft³ is 3.10×10¯⁵ ft³.

7 0

3 years ago

What is the gravitational force between two masses of 1 kg each kept at the distance of 1m?

Eddi Din [679]

Answer:

Explanation:

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6 0

3 years ago