All categories

2 years ago

Zn2+ and Al3+ are amphoteric. Demonstrate

Chemistry

1 answer:

denis23 [38]2 years ago

8 0

Explanation:

1) Amphoteric substances can react both like an acid or a base depending if they are in presence of a base or an acid.

Zinc hydroxide (Zn(OH)2):

In acid: $Zn(OH)_2 + 2 HCl \longrightarrow ZnCl_2 + 2 H_2O
$

In base: $Zn(OH)_2 + 2 NaOH \longrightarrow Na_2[Zn(OH)_4]$

Aluminium hydroxide (Al(OH)3)

In acid: $Al(OH)_3 + 3 HCl \longrightarrow AlCl_3 + 3 H_2O
$

In base: $Al(OH)_3 + NaOH \longrightarrow Na[Al(OH)_4]$

2) Reactions

a) $2 NH_3 + Cu^{2+} + 2 H_2O \longrightarrow 2 NH_4^+ + Cu(OH)_2$

$3 NH_3 + Fe^{3+} + 3 H_2O \longrightarrow 3 NH_4^+ + Fe(OH)_3$

b) $2 NH_3 + Zn^{2+} + 2 H_2O \longrightarrow 2 NH_4^+ + Zn(OH)_2$

$3 NH_3 + Al^{3+} + 3 H_2O \longrightarrow 3 NH_4^+ + Al(OH)_3$

You might be interested in

What is the term?

Alika [10]

Answer:

A DNA molecule

Explanation:

5 0

3 years ago

A piece of wood has a labeled length value of 63.2 cm. You measure its length three times and record the following data: 63.1 cm

Ulleksa [173]

Percent error (%)= $\frac{\left | Accepted value - Measured value \right |}{Accepted value}\times 100$

Accepted value is true value.

Measured values is calculated value.

In the question given Accepted value (true value) = 63.2 cm

Given Measured(calculated values) = 63.1 cm , 63.0 cm , 63.7 cm

1) Percent error (%) for first measurement.

Accepted value (true value) = 63.2 cm, Measured(calculated values) = 63.1 cm

Percent error (%)= $\frac{\left | Accepted value - Measured value \right |}{Accepted value}\times 100$

$Percent error = \frac{\left | 63.2 - 63.1 \right |}{63.2}\times 100$

$Percent error = \frac{0.1}{63.2}\times 100$

$Percent error = 0.00158\times 100$

Percent error = 0.158 %

2) Percent error (%) for second measurement.

Accepted value (true value) = 63.2 cm, Measured(calculated values) = 63.0 cm

Percent error (%)= $\frac{\left | Accepted value - Measured value \right |}{Accepted value}\times 100$

$Percent error = \frac{\left | 63.2 - 63.0 \right |}{63.2}\times 100$

$Percent error = \frac{0.2}{63.2}\times 100$

$Percent error = 0.00316\times 100$

Percent error = 0.316 %

3) Percent error (%) for third measurement.

Accepted value (true value) = 63.2 cm, Measured(calculated values) = 63.7 cm

Percent error (%)= $\frac{\left | Accepted value - Measured value \right |}{Accepted value}\times 100$

$Percent error = \frac{\left | 63.2 - 63.7 \right |}{63.2}\times 100$

$Percent error = \frac{\left | -0.5 \right |}{63.2}\times 100$

$Percent error = \frac{(0.5)}{63.2}\times 100$

$Percent error = 0.00791\times 100$

Percent error = 0.791 %

Percent error for each measurement is :

63.1 cm = 0.158%

63.0 cm = 0.316%

63.7 cm = 0.791%

7 0

3 years ago

Mention the reason why aquatic plants have air filled tissues.

Pepsi [2]

Answer:

Aquatic plants have aur filled tissues to make them easy to float.....

Explanation:

It provides buoyancy

7 0

2 years ago

A 256 mL sample of HCl gas is in a flask where it exerts a force (pressure) of 67.5 mmHg. What is the pressure of the gas if it

Effectus [21]

Answer:

The pressure in the new flask would be $128\; \rm mmHg$ if the $\rm HCl$ here acts like an ideal gas.

Explanation:

Assume that the $\rm HCl$ sample here acts like an ideal gas. By Boyle's Law, the pressure $P$ of the gas should be inversely proportional to its volume $V$ .

For example, let the initial volume and pressure of the sample be $V_1$ and $P_1$ . The new volume $V_2$ and pressure $P_2$ of this sample shall satisfy the equation: $P_1 \cdot V_1 =P_2 \cdot V_2$ .

In this question,

The initial volume of the gas is $V_1= 256\; \rm mL$ .
The initial pressure of the gas is $P_1 = 67.5\; \rm mmHg$ .
The new volume of the gas is $V_2 = 135\; \rm mL$ .

The goal is to find the new pressure of this gas, $P_2$ .

Assume that this sample is indeed an ideal gas. Then the equation $P_1 \cdot V_1 =P_2 \cdot V_2$ should still hold. Rearrange the equation to separate the unknown, $P_2$ . Note: make sure that the units for $V_1$ and $V_2$ are the same before evaluating. That way, the unit of

$\begin{aligned} & P_2\\ &= \frac{P_1 \cdot V_1}{V_2} \\ &= \frac{256\; \rm mL \times 67.5\; \rm mmHg}{135\; \rm mL} \\ & \approx 128\; \rm mmHg\end{aligned}$ .

3 0

3 years ago

What are the properties of metals nonmetals and metalloids?

harkovskaia [24]

Metals:
Distinguishing luster (shine)
Malleable and ductile (flexible) as solids
Conduct heat and electricity
Metallic oxides are basic, ionic
Cations in aqueous solution

Nonmetals:
Non-lustrous, various colors
Brittle, hard or soft
Poor conductors
Nonmetallic oxides are acidic, compounds
Anions, oxyanions in aqueous solution

5 0

3 years ago