In this part of the periodic table, what type of elements are in the group that includes elements cu, ag, and au?

What do you understand by the corpuscular nature of chemistry

Tom [10]

Answer:

The corpuscular model of matter is a model of the microscopic structure of matter, which seeks to explain the properties and behavior in each state of aggregation

Explanation:

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8 0

2 years ago

Which reaction will occur?

mart [117]

<span>You have to use an activity series and then predict which metal by itself will replace the metal inside of the compound. These all appear to be single replacement reactions.</span>

7 0

3 years ago

Read 2 more answers

Consider four different samples: aqueous LiBr , molten LiBr , aqueous AgBr , and molten AgBr . Current run through each sample p

Charra [1.4K]

Answer:

a) Aqueous LiBr = Hydrogen Gas

b) Aqueous AgBr = solid Ag

c) Molten LiBr = solid Li

c) Molten AgBr = Solid Ag

Explanation:

a) Aqueous LiBr

This sample produces Hydrogen gas, because the H+ (conteined in the water) has a reduction potential higher than the Li+ from the salt. Therefore the hydrogen cation will reduce instead of the lithium one and form the gas.

b) Aqueous AgBr

This sample produces Solid Ag, because the Ag+ has a reduction potential higher than the H+ from the water. Therefore the silver cation will reduce instead of the hydrogen one and form the solid.

c) Molten LiBr

In a molten binary salt like LiBr there is only one cation present in the cathod. In this case the Li+, so it will reduce and form solid Li.

c) Molten AgBr

The same as the item above: there is only one cation present in the cathod. In this case the Ag+, so it will reduce and form solid Ag.

6 0

4 years ago

A chemist prepares a solution of aluminum sulfate Al2SO43 by weighing out 101.g of aluminum sulfate into a 200.mL volumetric fla

qaws [65]

Answer:

50.5 g/dL

Explanation:

From the question given above, the following data were obtained:

Mass of Al₂(SO₄)₃ = 101 g

Volume (V) = 200 mL

Concentration of solution (g/dL) =?

Next, we shall convert 200 mL to decilitre (dL).

This is illustrated below:

1 mL = 0.01 dL

Therefore,

200 mL = 200 mL / 1 mL × 0.01 dL

200 mL = 2 dL

Therefore, 200 mL is equivalent to 2 dL.

Finally, we shall determine the concentration of the solution in g/dL as follow:

Mass of Al₂(SO₄)₃ = 101 g

Volume (V) = 2 dL

Concentration of solution (g/dL) =?

Concentration of solution (g/dL) = mass /volume

Concentration of solution (g/dL) = 101/2

= 50.5 g/dL

Therefore, the concentration of the Al₂(SO₄)₃ solution is 50.5 g/dL

8 0

4 years ago

g Calculate the time (in min.) required to collect 0.0760 L of oxygen gas at 298 K and 1.00 atm if 2.60 A of current flows throu

lozanna [386]

Answer:

7.67 mins.

Explanation:

Data obtained from the question include the following:

Volume (V) = 0.0760 L

Temperature (T) = 298 K

Pressure (P) = 1 atm

Current (I) = 2.60 A

Time (t) =?

Next, we shall determine the number of mole (n) of O2 contained in 0.0760 L.

This can be obtained by using the ideal gas equation as follow:

Note:

Gas constant (R) = 0.0821 atm.L/Kmol

PV = nRT

1 x 0.0760 = n x 0.0821 x 298

Divide both side by 0.0821 x 298

n = 0.0760 / (0.0821 x 298)

n = 0.0031 mole

Next, we shall determine the quantity of electricity needed to liberate 0.0031 mole of O2.

This is illustrated below:

2O²¯ + 4e —> O2

Recall:

1 faraday = 1e = 96500 C

4e = 4 x 96500 C

4e = 386000 C

From the balanced equation above,

386000 C of electricity liberated 1 mole of O2.

Therefore, X C of electricity will liberate 0.0031 mole of O2 i.e

X C = 386000 X 0.0031

X C = 1196.6 C

Therefore, 1196.6 C of electricity is needed to liberate 0.0031 mole of O2

Next, we shall determine the time taken for the process. This can be obtained as follow:

Current (I) = 2.60 A

Quantity of electricity (Q) = 1196.6 C

Time (t) =?

Q = It

1196.6 = 2.6 x t

Divide both side by 2.6

t = 1196.6/2.6

t = 460.23 secs.

Finally, we shall convert 460.23 secs to minute. This can be achieved by doing the following:

60 secs = 1 min

Therefore,

460.23 secs = 460.23/60 = 7.67 mins

Therefore, the process took 7.67 mins.

7 0

4 years ago