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user100 [1]
3 years ago
12

Classify each ion according to whether it's produced by an acid or by a base. Drag each ion to the correct location on the

Chemistry
1 answer:
Yanka [14]3 years ago
6 0
If u suck sun I’ll show u 12.43 jp
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Given the standard heats of reaction
ANTONII [103]

Answer:

Explanation:

M(s) → M (g ) + 20.1 kJ --- ( 1 )

X₂ ( g ) → 2X (g ) + 327.3 kJ ---- ( 2 )

M( s) + 2 X₂(g) → M X₄ (g ) - 98.7 kJ ----- ( 3 )

( 3 ) - 2 x ( 2 ) - ( 1 )

M( s) + 2 X₂(g) - 2 X₂ ( g ) - M(s)  → M X₄ (g ) - 98.7 kJ -  2 [ 2X (g ) + 327.3 kJ ] - M (g ) - 20.1 kJ

0 = M X₄ (g ) - 4 X (g ) - M (g ) - 773.4 kJ

4 X (g ) +  M (g ) =  M X₄ (g ) - 773.4kJ

heat of formation of M X₄ (g ) is - 773.4 kJ

Bond energy of one M - X bond =  773.4 / 4 =  193.4 kJ / mole

6 0
3 years ago
Explain why compressional waves are unable to move through vacuum
AnnyKZ [126]
Can travel through<span> most media including a </span>vacuum. The electromagnetic ...Transverse waves<span> are </span>unable<span> to </span>pass through<span> liquids or gases.</span>
4 0
4 years ago
Read 2 more answers
How much does 0.0273 moles of Na2O weigh
devlian [24]

Answer:

1.69 g Na₂O

General Formulas and Concepts:

<u>Chemistry - Atomic Structure</u>

  • Reading a Periodic Table
  • Using Dimensional Analysis

Explanation:

<u>Step 1: Define</u>

0.0273 mol Na₂O

<u>Step 2: Identify Conversions</u>

Molar Mass of Na - 22.99 g/mol

Molar Mass of O - 16.00 g/mol

Molar Mass of Na₂O - 2(22.99) + 16.00 = 61.98 g/mol

<u>Step 3: Convert</u>

<u />0.0273 \ mol \ Na_2O(\frac{61.98 \ g \ Na_2O}{1 \ mol \ Na_2O} ) = 1.69205 g Na₂O

<u>Step 4: Check</u>

<em>We are given 3 sig figs. Follow sig fig rules and round.</em>

1.69205 g Na₂O ≈ 1.69 g Na₂O

6 0
3 years ago
Instead of 6 M NaOH being added to the solution, 6 M HCl is added. How will this affect the test for the presence of ammonium io
Lera25 [3.4K]

Answer:

Ammonia gas(an alkaline gas with characteristics of choking or irritating smell) is not liberated when 6mole of HCl is added to the solution instead of 6mole of NaOH, to test for the presence of ammonium ion in the solution

Explanation:

As expected, when testing for ammonium ion in a solution (precisely ammonium salt solution), Sodium Hydroxide (NaOH) is required as the test reagent.

When NaOH is added to the solution, A gas with characteristics of choking or irritating smell is liberated.

This gas turn red litmus paper blue.

This liberated gas is an alkaline gas, which is confirmed as an ammonia gas(NH3).

If HCl is added instead of NaOH, the ammonia gas will not be liberated, which indicates that the test reagent used is wrong.

3 0
3 years ago
Science need ko po ng help di mo ako kase matalino salamat po
Lorico [155]

Answer:

1.An initial observation is the measurement that you take before you start any process that might cause a change. When you compare your subsequent observations with the initial one, you will see whether any change has taken place, and you will be able to measure the change.

Explanation:

D ko po alam yung iba

6 0
3 years ago
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