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Lady_Fox [76]
2 years ago
12

Consider the following oxides: SO2, Y2O3, MgO, Cl2O, and N2O5. How many are expected to form acidic solutions in water? Consider

the following oxides: , , , , and . How many are expected to form acidic solutions in water?
Chemistry
1 answer:
Snowcat [4.5K]2 years ago
5 0

Answer:

Three of the five oxides are expected to form acidic solutions in water

Explanation:

We have different types of oxides : Acidic oxides, Basic oxides, Amphoteric oxides, Peroxides and Higher oxides.

Only acidic oxides will dissolve in water to give an acidic solution.

Considering the given oxides carefully,

  • SO2 will dissolve in water to produce H2SO3 which is acidic.

  • Y2O3 will dissolve in water to produce Yttrium(III) hydroxide which is basic.

  • MgO will dissolve in water only to produce Mg(OH)2 which is also basic.

  • Cl2O dichlorine mono oxide will dissolve in water to produce HClO which is acidic.

  • N2O5 will dissolve in water to produce HNO3 which is also acidic.

You might be interested in
In run 1, you mix 7.9 mL of the 43 g/L MO solution (MO molar mass is 327.33 g/mol), 3.13 mL of the 0.040 M SnCl2 in 2.0 M HCl so
blsea [12.9K]

Answer:

Concentration of H3O⁺  [H3O⁺] = 0.864 M

Explanation:

Given that:

The mass concentration of MO = 43 g/L

The volume of MO = 7.9 mL = 7.9 × 10⁻³ L

Recall that

The mass number of MO = Mass concentration of MO × Volume of MO

The mass number of MO = (43 g/L) * (7.9 × 10⁻³ L)

The mass number of MO =  0.3397 g

number of  moles of MO = (mass number of MO) / (molar mass of MO)

number of  moles of MO = (0.3397 g) / (327.33 g/mol)

moles of MO = 0.00104 mol

The total volume = 7.9 mL + 3.13 mL + 5.49 mL + 3.43 mL

The total volume = 19.95 mL = 19.95 × 10⁻³ L

Concentration of MO [MO} =(number of moles of MO) / (total volume)

[MO] = 0.00104 mol  /  19.95 × 10⁻³ L

[MO] = 5.2130 × 10⁻⁸ M

the number of moles of H3O⁺ = molarity of HCl in the solution × the volume of HCl in solution

the number of moles of H3O⁺ = [(2.0 M) * (3.13 mL)] + [(2.0 M) * (5.49 mL)]

the number of moles of H3O⁺ = 17.24 mmol

Concentration of H3O⁺  [H3O⁺] = (the number of moles of H3O⁺) / (total volume)

Concentration of H3O⁺  [H3O⁺] = (17.24 mmol) / (19.95 mL)

Concentration of H3O⁺  [H3O⁺] = 0.864 M

5 0
2 years ago
What mass of carbon dioxide (co2) can be produced from 86.17 grams of c6h14 and excess oxygen?
charle [14.2K]
2C6H14 + 13O2 ---> 6CO2 +14H2O

M(C6H14)=12.011*6 +1.008*14 ≈ 86.17 g/mol

86.17 g C6H14 is 1 mole.

                             2C6H14 + 13O2 ---> 6CO2 +14H2O
from reaction        2 mol                         6 mol
from the problem  1 mol                         3 mol

M(CO2)= 12.011 + 2*15.999= 44.009 g/mol
3 mol CO2*44.009 g/1 mol CO2 ≈ 132.0 g CO2
Answer : 132.0 g CO2


3 0
2 years ago
How many moles of lithium chloride will be formed by the reaction of chlorine with 0.046 mol of lithium bromide in the following
Feliz [49]

Answer:

The answer is 0.046 mol.

Explanation:

By looking at the balanced equation, you can form a ratio of lithium chloride and lithium bromide using the coefficient value :

ratio of lithium bromide <u>(</u>2LiBr)

= 2

ratio of lithium chloride (2LiCl)

= 2

So the ratio is 2 : 2 then simplify into 1 : 1 . Which means that 1 mol of lithium bromide is equal to 1 mole of lithium chloride.

In this case, 0.046 mol of lithium bromide will form <u>0</u><u>.</u><u>0</u><u>4</u><u>6</u><u> </u><u>m</u><u>o</u><u>l</u> of lithium chloride.

8 0
2 years ago
Which will not appear in the equilibrium constant expression for the reaction below?
n200080 [17]

Answer:

[C] carbon solid

Explanation:

Pure solids and liquids are never included in the equilibrium constant expression because they do not affect the reactant amount at equilibrium in the reaction, thus since your equation has [C] as solid it will not be part of the equlibrium equation.

5 0
2 years ago
On Earth a package weighs 19.6 newtons. What is the mass of this package on Earth?
kondor19780726 [428]

g = 19.6 N/2.2 kg. g = 8.9 m/s2. 7.

4 0
2 years ago
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