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What would happen if hydrogen chloride gas is bubbled through a solution of lead (II) nitrate? Explain

Lena [83]

A white precipitate of silver chloride will be formed when Hydrogen chloride gas is passed through silver nitrate.

A white precipitate of silver chloride will be formed when Hydrogen chloride gas is passed through silver nitrate.2HCl+2AgNO³

→2AgCl+2HNO³

The reaction between AgCl and NH4OH will result in the products of Ag(NH3) 2+and Cl - ions .

AgCl+2NH 4OH→Ag(NH3) ²Cl+HH²O

3 0

3 years ago

What volume of a 0.240 m solution of barium nitrate is needed to prepare 0.500 l a solution that is 0.0800 m in nitrate ion?

Doss [256]

Formula for Barium Nitrate = Ba(NO3)2

Thus based on stoichiometry:

1 mole of Ba(NO3)2 contains 2 moles of NO3-

Therefore, concentration of nitrate ion NO3- would be = 2*0.240 = 0.480 M

Use the relation:

V1M1 = V2M2

V1 = V2M2/M1 = 0.500 L * 0.0800/0.480 = 0.0833 L

Thus, 0.0833 L or 83.3 ml solution of Ba(NO3)2 would be required.

4 0

3 years ago

A student dissolves 15.0 g of ammonium chloride(NH4Cl) in 250. 0 g of water in a well-insulated open cup. She then observes the

iren2701 [21]

Answer:

1) Endothermic.

2) $Q_{rxn}=4435.04J$

3) $\Delta _rH=15.8kJ/mol$

Explanation:

Hello there!

1) In this case, for these calorimetry problems, we can realize that since the temperature decreases the reaction is endothermic because it is absorbing heat from the solution, that is why the temperature goes from 22.00 °C to 16.0°C.

2) Now, for the total heat released by the reaction, we first need to assume that all of it is released by the solution since it is possible to assume that the calorimeter is perfectly isolated. In such a way, it is also valid to assume that the specific heat of the solution is 4.184 J/(g°C) as it is mostly water, therefore, the heat released by the reaction is:

$Q_{rxn}=-(15.0g+250.0g)*4.184\frac{J}{g\°C}(16.0-20.0)\°C\\\\ Q_{rxn}=4435.04J$

3) Finally, since the enthalpy of reaction is calculated by dividing the heat released by the reaction over the moles of the solute, in this case NH4Cl, we proceed as follows:

$\Delta _rH=\frac{ Q_{rxn}}{n}\\\\\Delta _rH= \frac{ 4435.04J}{15.0g*\frac{1mol}{53.49g} } *\frac{1kJ}{1000J} \\\\\Delta _rH=15.8kJ/mol$

Best regards!

4 0

3 years ago

For each element write the Noble gas notation

Rom4ik [11]

Answer:

neon is the Noble gas but I have not studied this much

7 0

3 years ago

Which is the largest in size? OA. A. a star O B. a planet O c. an asteroid D. a galaxy E. a supernova E.

Genrish500 [490]

Answer:

galaxy

Explanation:

galaxy is the largest in size

8 0

4 years ago