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AlexFokin [52]
3 years ago
11

What are the the examples of amphoteric oxide​

Chemistry
2 answers:
Andrew [12]3 years ago
7 0

Answer:

Many metals (such as zinc, tin, lead, aluminium, and beryllium) form amphoteric oxides or hydroxides. Amphoterism depends on the oxidation states of the oxide. Al2O3 is an example of an amphoteric oxide.

WINSTONCH [101]3 years ago
6 0

Answer:

Lead (II) oxide and zinc (II) oxide

You might be interested in
An unsaturated solution ________.
Greeley [361]

Answer:

I think the answer is D.

Explanation:

Because if it is unsaturated then it can dissolve more solutes.

8 0
3 years ago
1. How many joules must be added to 10.0 g of water to raise its temperature from 10°C to<br> 15°C?
weqwewe [10]

Answer:

209.3 Joules require to raise the temperature from 10 °C to 15 °C.

Explanation:

Specific heat capacity:

It is the amount of heat required to raise the temperature of one gram of substance by one degree.

Formula:

Q = m × c × ΔT

Given data:

mass of water = 10 g

initial temperature T1= 10 °C

final temperature T2=  15 °C

temperature change =ΔT= T2-T1 = 15°C - 10°C = 5 °C

Energy or joules added to increase the temperature Q = ?

Solution:

We know that specific heat of water is 4.186 J/g .°C

Q = m × c × ΔT

Q = 10 g × 4.186 J/g .°C × 5 °C

Q = 209.3 J

6 0
3 years ago
If 110g of copper sulphate is present in 550g of solution calculate the concentration of solution
Bingel [31]

Answer:

20%

Explanation:

mass by mass percentage of a solution =(mass of solute)/(mass of solution)

mass of solute=550g

therefore 110×100/550=20%

hope u will understand .:") credit to the owner

7 0
4 years ago
CaCO3(s) ∆→CaO(s) + CO2(g).
sveta [45]

Answer:

74.9%.

Explanation:

Relative atomic mass data from a modern periodic table:

  • Ca: 40.078;
  • C: 12.011;
  • O: 15.999.

What's the <em>theoretical</em> yield of this reaction?

In other words, what's the mass of the CO₂ that should come out of heating 40.1 grams of CaCO₃?

Molar mass of CaCO₃:

M(\text{CaCO}_3) = 40.078 + 12.011 + 3 \times 15.999 = 100.086\;\text{g}\cdot\text{mol}^{-1}.

Number of moles of CaCO₃ available:

\displaystyle n(\text{CaCO}_3) =\frac{m}{M} = \frac{40.1}{100.086} = 0.400655\;\text{mol}.

Look at the chemical equation. The coefficient in front of both CaCO₃ and CO₂ is one. Decomposing every mole of CaCO₃ should produce one mole of CO₂.

n(\text{CO}2) = n(\text{CaCO}_3)= 0.400655\;\text{mol}.

Molar mass of CO₂:

M(\text{CO}_2) = 12.011 + 2\times 15.999 = 44.009\;\text{g}\cdot\text{mol}^{-1}.

Mass of the 0.400655 moles of \text{CO}_2 expected for the 40.1 grams of CaCO₃:

m(\text{CO}_2) = n\cdot M = 0.400655 \times 44.009 = 17.632\;\text{g}.

What's the <em>percentage</em> yield of this reaction?

\displaystyle \textbf{Percentage}\text{ Yield} = \frac{\textbf{Actual}\text{ Yield}}{\textbf{Theoretical}\text{ Yield}}\times 100\%\\\phantom{\textbf{Percentage}\text{ Yield}} = \frac{13.2}{17.632}\times 100\%\\\phantom{\textbf{Percentage}\text{ Yield}} =74.9\%.

7 0
3 years ago
H2(g) + F2(g) → 2 HF(g) ΔH=-546 kJ
lilavasa [31]

Answer: help please !!

Explanation:

7 0
3 years ago
Read 2 more answers
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