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Alik [6]
3 years ago
15

Enter the chemical formula of a binary molecular compound of hydrogen and a Group 4A element that can reasonably be expected to

be more acidic in aqueous solution than , e.g. have a larger .
Chemistry
1 answer:
irakobra [83]3 years ago
6 0

Answer: GeH4 (Germanium(IV) Hydride)

Explanation:

A Binary molecular compound Hydrogen and a Group 4A element which is more more acidic than SiH4 in aqueous solution is GeH4.

The pKa of GeH4;

= 25

Whilst that of SiH4

= 35

The lesser the pKa the higher the Ka which means more acidic.

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Propane can be turned into hydrogen by the two-step reforming process. In the first step, propane and water react to form carbon
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Answer:

C₃H₈(g) + 6 H₂O(g) ⇒ + 10 H₂(g) + 3 CO₂(g)

Explanation:

Propane can be turned into hydrogen by the two-step reforming process.

In the first step, propane and water react to form carbon monoxide and hydrogen. The balanced chemical equation is:

C₃H₈(g) + 3 H₂O(g) ⇒ 3 CO(g) + 7 H₂(g)

In the second step, carbon monoxide and water react to form hydrogen and carbon dioxide. The balanced chemical equation is:

CO(g) + H₂O(g) ⇒ H₂(g) + CO₂(g)

In order to get the net chemical equation for the overall process, we have to multiply the second step by 3 and add it to the first step. Then, we cancel what is repeated.

C₃H₈(g) + 3 H₂O(g) + 3 CO(g) + 3 H₂O(g) ⇒ 3 CO(g) + 7 H₂(g) + 3 H₂(g) + 3 CO₂(g)

C₃H₈(g) + 6 H₂O(g) ⇒ + 10 H₂(g) + 3 CO₂(g)

4 0
4 years ago
Who is like, crazy good with Chemistry because I have 2 questions that I am so freaking stuck on! PLEASE HELP!
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6 0
3 years ago
How many moles of HCl are there in 75.0 mL of 0.325 M HCI?​
dangina [55]

Answer: The moles of HCl are 0.0244

Explanation:

Molarity of a solution is defined as the number of moles of solute dissolved per liter of the solution.

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where,

n = moles of solute = ?

V_s = volume of solution in ml

Now put all the given values in the formula of molality, we get

0.325=\frac{moles\times 1000}{75.0}

moles=0.0244

Therefore, the moles of HCl are 0.0244

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3 years ago
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I think the answer is C. 02
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