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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.

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DaniilM [7]

Answer:

The new volume is 5.913*10^4 L

Explanation:

Step 1: Write out the formula to be used:

Using general gas equation;

P1V1 / T1 =P2V2 /T2

V2 = P1V1T2 / P2T1

Step 2: write out the values given and convert to standard unit's where necessary

P1 = 0.995atm

P2 0.720atm

V1 = 5*10^4 L

T1 = 32°C = 32+ 273 = 305K

T2 = -12°C = -12 + 273 = 261K

Step 3: Equate your values and do the calculation:

V2 = 0.995 * 5*10^4 * 261 / 0.720 * 305

V2 = 1298.475 * 10^4 / 219.6

V2 = 5.913 * 10^4 L

So the new volume of the balloon is 5.913*10^4 L

3 0
3 years ago
Use the Periodic Table of the Elements to answer
Juliette [100K]

Answer:

b. lithium

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Li the least likely, to lose an electron.

6 0
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