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Otrada [13]
3 years ago
10

Summarize the acid-base behavior of the main-group metal and nonmetal oxides in water. How does oxide acidity in water change do

wn a group and across a period?
Chemistry
1 answer:
MaRussiya [10]3 years ago
4 0

Explanation:

Metals react with oxygen to form basic oxides while they react with water to form alkaline solutions. Also, acidic oxides are oxides of nonmetals and they react with water to form acidic solutions.

Trends on the period table shows the variation of metallic character as you move across and down the periodic table. Metallic character of a element decreases across the period on the periodic table from left to right because atoms readily accept electrons in their outermost shell to form stable configurations. Metallic character increases as you move down the group in the periodic table and this is because electrons become easier to lose as the atomic radius increases (more outer shells are added), where there is decreasing attraction between the nucleus and the valence electrons.

So down the group, the acidity of oxide reaction with water decreases because the oxides are more basic down the group while across the period, the acidity of oxide increases because acidic oxides are formed as we move across the period.

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Answer:

a changes more because it going higher

Explanation:

6 0
2 years ago
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What is the molarity of a NaCl stock used to make 750 ml of a dilute 10 mM solution if 5 ml of the concentrated solution was use
ch4aika [34]

Explanation:

The given data is as follows.

          M_{1} = 10 mM = 10 \times 10^{-3} M

          V_{1} = 750 ml,           V_{2} = 5 ml

          M_{2} = ?

Therefore, calculate the molarity of given NaCl stock as follows.

                  M_{1} \times V_{1} = M_{2} \times V_{2}

                  10 \times 10^{-3} \times 750 ml = M_{2} \times 5 ml

                   M_{2} = 1.5 M

Thus, we can conclude that molarity of given NaCl stock is 1.5 M.        

7 0
2 years ago
7. There are 7. 0 ml of 0.175 M H2C2O4 , 1 ml of water , 4 ml of 3.5M KMnO4 what is the molar concentration ofH2C2O4 ?
Illusion [34]

Answer:

7. 0.1021 M

8. 1.167 M

10. Increase in volume of water would lower the rate of reaction

Explanation:

7. What is the molar concentration of H₂C₂O₄ ?

Since we have 7.0 ml of 0.175 M H₂C₂O₄, the number of moles of H₂C₂O₄ present n = molarity of H₂C₂O₄ × volume of H₂C₂O₄ = 0.175 mol/L × 7.0 ml = 0.175 mol/L × 7 × 10⁻³ L = 1.225 × 10⁻³ mol.

Also, the total volume present V = volume of H2C2O4 + volume of water + volume of KMnO4 = 7.0 ml + 1 ml + 4 ml = 12 ml = 12 × 10⁻³ L

So, the molar concentration of H₂C₂O₄, M = number of moles of H₂C₂O₄/volume = n/V

= 1.225 × 10⁻³ mol/12 × 10⁻³ L

= 0.1021 mol/L

= 0.1021 M

8. Using the data from question 7 what is the molar concentration of KMnO₄ ?

Since we have 4.0 ml of 3.5 M KMnO₄, the number of moles of KMnO4 present n' = molarity of KMnO₄ × volume of KMnO₄ = 3.5 mol/L × 4.0 ml = 3.5 mol/L × 4 × 10⁻³ L = 14 × 10⁻³ mol.

Also, the total volume present V = volume of KMnO₄ + volume of water + volume of KMnO₄ = 7.0 ml + 1 ml + 4 ml = 12 ml = 12 × 10⁻³ L

So, the molar concentration of KMnO₄, M' = number of moles of KMnO₄/volume = n'/V

= 14 × 10⁻³ mol/12 × 10⁻³ L

= 1.167 mol/L

= 1.167 M

10. From question number 7, what effect increasing the volume of water has on the reaction rate?

Increase in volume of water would lower the rate of reaction because, the particles of both substances would have to travel farther distances to collide with each other, since there are less particles present in the solution and thus, the concentration of the particles would decrease thereby decreasing the rate of reaction.

3 0
2 years ago
What will be the mass of carbon dioxide
Elza [17]
44.01 g/mol is the mass
4 0
2 years ago
Fe2O3 + 2Al -> Al2O3 + 2Fe
pashok25 [27]
Fe2O3 + 2Al ---> Al2O3 + 2Fe 
Mole ratio Fe2O3 : Al = 1:2 
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No. of moles of Al = 150/27 = 5.555555555 moles. 
Mole ratio 1 : 2. 1.56641604 * 2 = 3.13283208 moles of Al, but you have 5.555555555 moles of Al. So Al is in excess. All of it won't react. 

So take the Fe2O3 and Fe ratio to calculate the mass of iron metal that can be prepared. 
RMM of Fe2O3 / Mass of Fe2O3 = RMM of 2Fe / Mass of Fe 159.6 / 250 = 111.6 / x x = 174.8 g of Fe 
7 0
3 years ago
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